Inter 1st Year

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Applications of Derivatives Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B Applications of Derivatives Important Questions

Question 1.
Find dy and ∆y of y = f(x) = x² + x at x = 10 when ∆x = 0.1.
Solution:
As change in y – f(x) is given by ∆y = f(x + ∆x) – f(x), this change at x = 10 with ∆x = 0.1 is
∆y = f(10.1) – f(10)
= {(10.1)² + 10.1} – {10² + 10}
= 2.11.
Since dy = f'(x) ∆x, dy at x = 10 with ∆x = 0.1 is dy = {(2)(10) + 1} 0.1 = 2.1
(since \(\frac{\mathrm{d} y}{\mathrm{dx}}\) = 2x + 1).

Question 2.
Find ∆y and dy for the function y = cos (x) at x = 60° with ∆x = 1° .
Solution:
For the given problem ∆y and dy at x = 60° with ∆x = 1° are
∆y = cos (60° + 1°) – cos (60°) …………. (1)
and dy = -sin(60°) (1°) ………….. (2)
Cos (60°)= 0.5,
Cos (61°)= 0.4848,
Sin (60°) = 0.8660,
1° = 0.0174 radians
∴ ∆y = -0.0152 and
dy = -0.0150.

Question 3.
The side of a square is increased from 3 cm to 3.01 cm. Find the approximate increase in the area of the square.
Solution:
Let x be the side of a square and A be its area.
Then A = x². …… (1)
Clearly A is a function of x. As the side is increased from 3 cm to 3.01 cm we can take x – 3 and ∆x = 0.01 to compute the approximate increase in the area of square. The approximate value of change in area is
∆A ≈ \(\frac{\mathrm{dA}}{\mathrm{dx}}\) ∆x
In view of equation (1), the equation (2) becomes
∆A ≈ 2x∆x
Hence the approximate increase in the area when the side is increased from 3 to 3.01 is
∆A ≈ 2(3)(0.01) = 0.06

Question 4.
If the radius of a sphere is increased from 7 cm to 7.02 cm then find the approximate increase in the volume of the sphere.
Solution:
Let r be the radius of a sphere and V be its volume. Then
V = \(\frac{4 \pi \pi^{2}}{3}\) …………….. (1)
Here V is a function of r. As the radius is increased from 7 cm to 7.02, we can take r = 7 cm and ∆r = 0.02 cm. Now we have to find the approximate increase in the volume of the sphere.
∴ ∆V ≈ \(\frac{\mathrm{dV}}{\mathrm{dr}}\) ∆r = 4πr² ∆r.
Thus, the approximate increase in the volume of the sphere is \(\frac{4(22)(7)(7)(0.02)}{7}\) = 12.32 cm³.

Question 5.
If y = f(x) = k xⁿ then show that the approximate relative error (or increase) in y is n times the relative error (or increase) in x where n and k are constants.
Solution:
The approximate relative error (or increase) in y by the equation (2) of if a number A is very close to a number B but it is not equal to B then A is called an approximate value of B is (\(\frac{f^{\prime}(x)}{f(x)}\)) ∆x = \(\frac{k n x^{n-1}}{k x^{n}}\) ∆x = n(\(\frac{\Delta x}{x}\) = n)
Hence the approximate relative error in y = kxⁿ is n times the relative error in x.

Question 6.
If the increase in the side of a square is 2% then find the approximate percentage of increase in its area.
Solution:
Let x be the side of a square and A be its area.
Then A = x².
Approximate percentage error in area A
= (\(\frac{\frac{\mathrm{dA}}{\mathrm{dx}}}{\mathrm{A}}\)) × 100 × ∆x(by (3) of if a number
A is very close to a number B but it is not equal to B then A is called an approximate value of B with f = A)
= \(\frac{100(2 x) \Delta x}{x^{2}}\) = \(\frac{200 \Delta x}{x}\) = 2(2) = 4
(∵ \(\frac{\Delta x}{x}\) × 100 = 2

Question 7.
If an error of 0.01 cm is made in measuring the perimeter of a circle and the perimeter is measured as 44 cm then find the approximate error and relative error in its area.
Solution:
Let r, p and A be the radius, perimeter and area of the circle respectively. Given that p = 44 cm and ∆p = 0.01. We have to find approximation of ∆A and \(\frac{\Delta \mathrm{A}}{\mathrm{A}}\). Note that A = πr² which is a function of r. As p and ∆p are given we have to transform A = πr² into the form A = f(p). This can be achieved by using the relation, perimeter 2πr = p.
∴ A = π(\(\frac{p}{2 \pi}\))² = \(\frac{p^{2}}{4 \pi}\)
Hence the approximate error in
A = \(\frac{d A}{d p}\)∆p = \(\frac{2 p}{4 \pi}\)∆p = \(\frac{P}{2 \pi}\)∆p
The approximate error in A when p = 44 and ∆p = 0.01 = \(\) (0.01) = 0.07
The approximate relative error

= 0.0004545.

Question 8.
Find the approximate value of \(\sqrt[3]{999}\)
Solution:
This problem can be answered by
f(x + ∆x) ≈ f(x) + f'(x) ∆x …………. (1)
with x = 1000 and ∆x = -1. The reason for taking x = 1000 is to make the calculation of f(x) simpler when f(x) = \(\sqrt[3]{x}\), Suppose
y = f(x) = \(\sqrt[3]{x}\)
The equation (1) becomes
f(x + ∆x) ≈ f(x) = \(\frac{1}{3 x^{\frac{2}{3}}}\) ∆x
Hence f(1000 – 1)
≈ f(1000) + \(\frac{1}{3(1000)^{2 / 3}}\) (-1) = 9.9967.

Question 9.
Find the slope of the tangent to the following curves at the points as indicated.
i) y = 5x² at (-1, 5)
ii) y = \(\frac{1}{x – 1}\) (x ≠ 1) at [3, \(\frac{1}{2}\)]
iii) x = a secθ, y = a tanθ at θ = \(\frac{\pi}{6}\)
iv) (\(\frac{x}{a}\))ⁿ + (\(\frac{x}{b}\))ⁿ = 2 at (a, b)
Solution:
i) y = 5x², then \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 10x
Slope of the tangent at the given points
(\(\frac{\mathrm{dy}}{\mathrm{dx}}\))_(-1.5) -10(-1) = -10

ii) y = \(\frac{1}{x – 1}\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{-1}{(x-1)^{2}}\)
Slope of the tangent at the given point is

Question 10.
Find the equations of the tangent and the normal to the curve y = 5x⁴ at the point (1, 5)
Solution:
y = 5x⁴ implies that \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 20x³
Slope of the tangent to the curve at (1, 5) is
(\(\frac{\mathrm{dy}}{\mathrm{dx}}\))_{(1, 5)} = 20(1)⁵ = 20
Equation of the normal of (1, 5) is
y – 5 = 20 (x – 1) = 20x – 20
y = 20x – 15
Slope of the normal to the curve at (1, 5) is
–\(\frac{1}{m}\) = –\(\frac{1}{20}\)
Equation of the normal of (1, 5) is
y – 5 = –\(\frac{1}{20}\) (x – 1)
20 y – 100 = -x + 1
x + 20y = 101
(or) 20y = 101 – x

Question 11.
Find the equations of the tangent and the normal to the curve y⁴ = ax³. at (a, a).
Solution:
Given curve is y⁴ = ax³
Differentiating w.r.to. x
4y³ . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 3ax²
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{3 a x^{2}}{4 y^{3}}\)
Slope of the tangent at (a, a) = \(\frac{3 a \cdot a^{2}}{4 a^{3}}\) = \(\frac{3}{4}\)
Slope of the normal at (a, a) = –\(\frac{1}{m}\) = –\(\frac{4}{3}\)
Equation of the tangent at (a, a) is
y – a = \(\frac{3}{4}\)(x – a)
4y – 4a = 3x – 3a
4y = 3x + a
Equation of the normal at (a, a) is 4
y – a = – \(\frac{4}{3}\)(x – a)
3y – 3a = -4x + 4a
3y + 4x = 7a

Question 12.
Find the equations of the tangent to the curve y = 3x² – x³, where it meets the X-axis.
Solution:
Equation of the curve is y = 3x² – x³
Equation of X – axis is y = 0
For points is intersection of the curve and X-axis
3x² – x³ = 0 ⇒ x² (3 – x) = 0
x = 0, x = 3
The curve crosses X-axis 0(0, 0) and A(3, 0)
y = -3x² – x³
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 6x – 3x²
At O(0, 0), slope of the tangent = (\(\frac{\mathrm{dy}}{\mathrm{dx}}\))_{(0, 0)} = 0
Equation of the tangent at (0, 0) is y – 0, 0(x – 0) i.e., y = 0
i.e., x-axis is the tangent to the curve at O(0, 0)
At A(3, 0), slope of the tangent = (\(\frac{\mathrm{dy}}{\mathrm{dx}}\))_{(3, 0)}
= 6.3 – 3.3²
= 18 – 27
= -9
Equation of the tangent at A(3, 0) is
y – 0 = -9(x – 3) = -9x + 27
(or) 9x + y = 27

Question 13.
Find the points at which the curve t y = sin x has horizontal tangents.
Solution:
y = sin x

A tangent is horizontal if and anal its slope is
cos x = 0 ⇒ x = (2n + 1)\(\frac{\pi}{2}\), n ∈ Z
Hence the given curve has horizontal tangents at points (x₀, y₀)
⇔ x₀ = (2n + 1) . \(\frac{\pi}{2}\) and
y₀ = (-1)ⁿ for same n ∈ Z

Question 14.
Verify whether the curve y = f(x) = x^1/3 has a vertical tangent at the point with x = 0.
Solution:

The function has a verified tangent at the point whose x co-ordinate is 0.

Question 15.
Find whether the curve y = f(x)= x^2/3 has a vertical tangent at x = 0.
Solution:

Thus left handed be normal \(\frac{1}{h^{1 / 3}}\) as h → 0 is -∞
While the right handed limit is ∞.
Hence does not exist. The vertical tangent does not exist.
At the point x = 0.

Question 16.
Show that the tangent at any point 0 on the curve x = c sec θ, y = c tan θ is y sin θ = x – c cos θ.
Solution:
x = c sec θ, y = c tan θ

Question 17.
Show that the area of the triable formed by the tangent at any point on the curve xy = c (c ≠ 0) with the coordinate axis is constant.
Solution:
Observe that c ≠ 0
If c = 0 the equation xy = 0 represent the coordinate circle which is against the definite.
Let P (x₁, y₁) be a point on the curve xy = c
y = \(\frac{c}{x}\) = 1, \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = – \(\frac{c}{x^{2}}\)
Equation of the tangent at (x₁, y₁) is
y – y₁ = – \(\frac{c}{x^{2}}\) (x – x₁)
x²y – x₁² = -cx + cx₁
cx + x₁² . y = x₁² + cx₁
= cx₁ + cx₁ (x₁y₁ = c)
= 2cx₁
\(\frac{c x}{2 c x_{1}}\) + \(\frac{x_{1}^{2} y}{2 c x_{1}}\) = 1
\(\frac{x}{2 x_{1}}\) + \(\frac{y}{\left(\frac{2 c}{x_{1}}\right)}\) = 1
Area of the triangle formed with co-ordinate axes
= \(\frac{1}{2}\) |OA . OB|
= \(\frac{1}{2}\) (2x₁) (\(\frac{2 c}{x_{1}}\)) = 2c = constant

Question 18.
Show that the equation of the tangent to the curve (\(\frac{x}{a}\))ⁿ + (\(\frac{y}{b}\))ⁿ = 2 (a ≠ 0, b ≠ 0) at the point (a, b) is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 2
Solution:

Equation of the tangent to the curve at the point (a, b) is
y – b = \(\frac{-b}{a}\) (x – a)
\(\frac{y}{b}\) – 1 = – \(\frac{x}{a}\) + 1
\(\frac{x}{a}\) + \(\frac{y}{b}\) = 2

Question 19.
Show that the length of the sub normal at point on the curve y² = 4ax is a constant.
Solution:
Equation of the curve is y² = 4ax
Differentiating w.r.to x
2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 4a
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{4a}{2y}\) = \(\frac{2a}{y}\)
Length of the sub-normal \(\left|\frac{y d y}{d x}\right|=\left|y \cdot \frac{2 a}{y}\right|\)
= 2a = constant

Question 20.
Show that the length of the Sub tangent at any point on the curve y = a^x (a > 0) is a constant.
Solution:
Equation of the curve is y = a^x
Differentiating w.r.to x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = a^x log a = y. log a
Length of the sub-tangent
= \(\left|\frac{y}{\left(\frac{d y}{d x}\right)}\right|\) = \(\left|\frac{y}{\left(\frac{d y}{d x}\right)}\right|\) = \(\frac{1}{\log a}\) constant

Question 21.
Show that the square of the length of subtangent at any point on the curve by² = (x + a)³ (b ≠ 0) varies with the length of the subnormal at that point.
Solution:
Differentiating by² = (x + a)³
w.r.t x, we get
2by y’ = 3(x + a)²
∴ The length of the subnormal at any point (x, y) on the curve
= |y y’| = |\(\frac{3}{2 b}\)(x + a)²| ………………… (1)
The square of the length of subtangent

∴ (length of the subnormal)² ∝ (length of subnormal).

Question 22.
Find the value of k, so that the length of the subnormal at any point on the curve y = a^{1 – k}x^k is a constant.
Solution:
Differentiating y = a^{1 – k}x^k with respect to x,
we get y’ = ka^{1 – k}x^{k – 1}
Length of subnormal at any point P(x, y) on the curve
= |y y’| = |yka^{1 – k}x^{k – 1}|
= |k a^{1 – k}x^ka^{1 – k}x^{k – 1}|
= |ka^2-2kx^2k-1|
In order to make these values a constant, we should have 2k – 1 = 0 i.e., k = \(\frac{1}{2}\).

Question 23.
Find the angle between the curves xy = 2 and x² + 4y = 0.
Solution:
First we find the points of intersection of xy = 2 and x² + 4y = 0
y = \(\frac{-x^{2}}{4}\)
But xy = 2 ⇒ x(\(\frac{-x^{2}}{4}\)) = 2 ⇒ x³ = -8
x = -2
y = \(\frac{-x^{2}}{4}\) = – \(\frac{4}{4}\) = -1
Point of intersection is P(-2, -1)

= \(\left|\frac{-\frac{3}{2}}{\frac{1}{2}}\right|\) = 3
Φ = tan^-1(3)

Question 24.
Find the angle between the curve 2y = e^{\(\frac{-x}{2}\)} and Y-axis.
Solution:
Equation of Y-axis is x = 0.
The point of intersection of the curve
2y = e^{\(\frac{-x}{2}\)} and x = 0 is P(0, \(\frac{1}{2}\))
Let ψ be the angle between the given curves
2y = e^{\(\frac{-x}{2}\)} at P with X – axis is given by
tan ψ = \(\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\left(0, \frac{1}{2}\right)}=\left.\frac{-1}{4} \mathrm{e}^{\frac{-x}{2}}\right|_{\left(0, \frac{1}{2}\right)}=\frac{-1}{4}\)
Further, if Φ is the angle between the Y – axis and 2y = e\(\frac{-x}{2}\), then we have
tan Φ = |tan (\(\frac{\pi}{2}\) – ψ)| = |cot ψ| = 4
∴ The angle between the curve and the Y-axis is tan^-1 4.

Question 25.
Show that the condition of the orthogonality of the curves ax² + by² = 1 and a₁x² + b₁y² = 1 is \(\frac{1}{a}\) – \(\frac{1}{b}\) = \(\frac{1}{a_{1}}\) – \(\frac{1}{b_{1}}\).
Solution:
Let the curves ax² + by² = 1 and a₁x² + b₁y² = 1 intersect at p(x₁, y₁) so that
ax₁² + by₁² = 1 and a₁x₁² + b₁y₁² = 1, from which we get,
\(\frac{x_{1}^{2}}{b_{1}-b}\) = \(\frac{y_{1}^{2}}{a_{1}-a}\) = \(\frac{1}{a b_{1}-a_{1} b}\) …………. (1)
Differentiating ax² + by² = 1 with respect to x,
we get \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{-a x}{b y}\)
Hence, if mt is the slope of the tangent at P(x₁, y₁) to the curve
ax² + by² = 1, m₁ = \(\frac{-a x_{1}}{b y_{1}}\)
Similarly, the slope (m₂) of the tangent at P to
a₁x² + b₁y² = 1 is given by m₂ = \(\frac{-a_{1} x_{1}}{b_{1} y_{1}}\)
Since the curves cut orthogonally we have m₁m₂ = -1,
i.e., \(\frac{\mathrm{a} a_{1} x_{1}^{2}}{\mathrm{~b} \mathrm{~b}_{1} y_{1}^{2}}\) = -1 or \(\frac{x_{1}^{2}}{y_{1}^{2}}=\frac{-b_{1}}{a a_{1}}\) ………………….. (2)
Now from (1) and (2), the condition for the orthogonality of the given curves is
\(\frac{b_{1}-b}{a-a_{1}}\) = \(\frac{b b_{1}}{a a_{1}}\)
or (b – a) a₁b₁ = (b₁ – a₁) ab
or \(\frac{1}{a}\) – \(\frac{1}{b}\) = \(\frac{1}{a_{1}}\) – \(\frac{1}{b_{1}}\)

Question 26.
Show that the curves y² = 4(x + 1)and y² = 36 (9 – x) intersect orthogonally. [Mar 11, May 06, 05]
Solution:
Solving y² = 4(x + 1) and y² = 36 (9 – x) for the points of intersection, we get
4(x + 1) = 36 (9 – x) 10x = 80 or x = 8
i.e., y² = 4(x + 1) ⇒ y² = 4(9) = 36 ⇒ y = ± 6
The points of intersection of the two curves are P(8, 6), Q(8, -6)
y² = 4(x + 1) ⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{2}{y}\)
y² = 36 (9 – x) ⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{-18}{y}\)
Slope of the tangent to the curve
y² = 4(x + 1) at P is
m₁ = \(\frac{2}{6}\) = \(\frac{1}{3}\)
Slope of the tangent to the curve
y² = 36 (9 – x) at P is
m₂ = \(\frac{-18}{6}\) = -3
m₁m₂ = \(\frac{1}{3}\) × -3 = -1
⇒ the curves intersect orthogonally at P.
We can prove, similarly,that the curves intersect orthogonally at Q also.

Question 27.
Find the average rate of change of s = f (t) = 2t² + 3 between t = 2 and t = 4.
Solution:
The average rate of change of s between t = 2 and t = 4 is \(\frac{f(4)-f(2)}{4-2}\) = \(\frac{35-11}{4-2}\) = 12.

Question 28.
Find the rate of change of area of a circle w.r.t. radius when r = 5 cm.
Solution:
Let A be the area of the circle with radius r.
Then A = πr². Now, the rate of change of area A w.r.t. r is given by \(\frac{\mathrm{dA}}{\mathrm{dr}}\) = 2πr. When r = 5 cm.
\(\frac{\mathrm{dA}}{\mathrm{dr}}\) = 10 π.
Thus, the area of the circle is changing at the rate of 10 π cm²/cm.

Question 29.
The volume of a cube is increasing at a rate of 9 cubic centimeters per second. How fast is the surface area increasing when the length of the edge is 10 centimeters ?
Solution:
Let x be the length of the edge of the cube, V be its volume and S be its surface area. Then, V = x³ and S = 6x². Given that rate of change of volume is 9 cm³/sec.
Therefore, \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 9 cm³/sec.
Now differentiating V w.r.t. t, we get,

Question 30.
A particle is moving in a straight line so that after t seconds its distance is s (in cms) from a fixed point on the line is given by s = f(t) = 8t + t³. Find (i) the velocity at time t = 2 sec (ii) the initial velocity (iii) acceleration at t = 2 sec.
Solution:
The distance s and time t are connected by the relation.
s = f(t) = 8t + t³ …………. (1)
∴ velocity (ν) = 8 + 3t² ……………. (2)
and the acceleration is given by
a = \(\frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}\) = 6t
i) The velocity at
t = 2 is 8 + 3 (4) = 20cm/sec.
ii) The initial velocity (t = 0) is 8 cm/sec.
iii) The acceleration at t = 2 is 6(2) = 12 cm/sec²

Question 31.
A container in the shape of an inverted cone has height 12 cm and radius 6 cm at the top. If it is filled with water at the rate of 12 cm³/sec., what is the rate of change in the height of water level when the tank is filled 8 cm ?
Solution:
Let OC be height to water level at t sec. The triangles OAB ad OCD are similar triangles.

Let OC = h and CD = r. Given that AB = 6 cm, OA = 12 cm.
\(\frac{r}{6}\) = \(\frac{h}{12}\)

Hence, the rate of change of water level is \(\frac{3}{4 \pi}\) cm/sec when the water level of the tank is 8 cm.

Question 32.
A particle is moving along a line according to s = f (t) = 4t³ – 3t² + 5t -1 where s is measured in meters and t is measured in seconds. Find the velocity and acceleration at time t. At what time ‘ the acceleration is zero.
Solution:
Since f(t) = 4t³ – 3t² + 5t – 1, the velocity at time t is
v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 12t² – 6t + 5
and the acceleration at time t is a = \(\frac{\mathrm{d}^{2} \mathrm{~s}}{\mathrm{dt}^{2}}\) = 24t – 6.
The acceleration is 0 if 24t – 6 = 0
i.e., t = \(\frac{1}{4}\)
The acceleration of the particle is zero at
t = \(\frac{1}{4}\) sec.

Question 33.
The quantity (in mg) of a drug in the blood at time t (sec) is given by q = 3(0.4)^t. Find the instantaneous rate of change at t = 2 sec.
Solution:
Given that q = 3(0.4)^t
∴ \(\frac{\mathrm{dQ}}{\mathrm{dx}}\) = 3(0.4)^t log_e(0.4) is the instaneous rate of change in q. Hence the instaneous rate of q at time t = 2 sec. is given by
\(\left(\frac{\mathrm{dQ}}{\mathrm{dx}}\right)_{t=2}\) = 3(0.4)² log_e (0.4).

Question 34.
Let a kind of bacteria grow by t³ (t in sec). At what time the rate of growth of the bacteria is 300 bacteria per sec ?
Solution:
Let g be the amount of growth of bacteria at t sec. Then
g(t) = t³ ………… (1)
The growth rate at time t is given by
g'(t) = 3t²
300 = 3t² (given that growth rate is 300)
t = 10 sec.
∴ After t = 10 sec, the growth rate of bacteria should be 300 bacteria/sec.

Question 35.
The total cost C(x) in rupees associated with production of x units of an item is given by C(x) = 0.005 x³ – 0.02x² + 30x + 500. Find the marginal cost when 3 units are produced (marginal cost is the rate of change of total cost).
Solution:
Let M represent the marginal cost. Then
M = \(\frac{\mathrm{dC}}{\mathrm{dx}}\)
Hence,
M = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(0.005x³ – 0.02x² + 30x + 500) dx
= 0.005(3x²) – 0.02(2x) + 30
∴ The Marginal cost at x = 3 is
(M)_{x = 3} = 0.005 (27) – 0.02 (6) + 30 = 30.015.
Hence the required marginal cost is Rs. 30.02 to produce 3 units.

Question 36.
The total revenue in rupees received from the sale of x units of a product is given by R(x) = 3x² + 36x + 5. Find the marginal revenue when x = 5 (marginal revenue is the rate of change of total revenue).
Solution:
Let m denote the marginal revenue. Then
m = \(\frac{\mathrm{dR}}{\mathrm{dx}}\) (since the total revenue is R(x))
Given that R(x) = 3x² + 36x + 5
∴ m = 6x + 36
The marginal revenue at x = 5 is
[m = \(\frac{\mathrm{dR}}{\mathrm{dx}}\)]_{x = 5} = 30 + 36 = 66
Hence the required marginal revenue is Rs. 66.

Question 37.
Verify Rolle’s theorem for the function y = f(x) = x² + 4 in [-3, 3].
Solution:
Here f(x) = x² = 4. f is continuous on [-3, 3] as x² + 4 is a polynomial which is continuous on any closed interval. Further f(3) = f(-3) = 13 and f is differentiable on [-3, 3].
∴ By Rolle’s theorem ∃ c ∈ (-3, 3) such that f'(c) = 0
The point c = 0 ∈ (-3, 3). Thus Rolle’s theorem is verified.

Question 38.
Verify Rolle’s theorem for the function f(x) = x(x + 3)e^-x/2 in [-3, 0].
Solution:
Here f(-3) = 0 and f(0) = 0.
We have
f'(x) = \(\frac{\left(-x^{2}+x+6\right)}{2} e^{\frac{-x}{2}}\)
f'(x) = 0 ⇔ -x² + x + 6 = 0 ⇔ x = -2 or 3. Of these two values -2 is in the open interval (-3, 0) which satisfies the conclusion of Rolle’s theorem.

Question 39.
Let f(x) = (x -1) (x – 2) (x – 3). Prove that there is more than one ‘c’ in (1, 3) such that f'(c) = 0. [Mar 13]
Solution:
Observe that f is continuous on (1, 3) differentiable in (1, 3) and f(1) = f(3) = 0.
f'(x) = (x – 1) (x – 2) + (x – 1) (x – 3) + (x – 2) (x – 3)
= 3x² – 12x + 11
\(\frac{12 \pm \sqrt{144-132}}{6}\) = 2 ± \(\frac{1}{\sqrt{3}}\)
Both these roots lie in the open interval (1, 3) and are such that the derivative vanishes at these points.

Question 40.
On the curve y = x², find a point at which the tangent is parallel to the chord joining (0, 0) and (1, 1).
Solution:
The slope of the chord is \(\frac{1-0}{1-0}\) = 1.
The derivative is \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2x.
We want x such that 2x = 1
i.e., x = \(\frac{1}{2}\)
we not that \(\frac{1}{2}\) is in the open interval (0, 1), as required in the Lagrange’s mean value theorem.
The corresponding point on the curve is (\(\frac{1}{2}\), \(\frac{1}{4}\)).

Question 41.
Show that f(x) = 8x + 2 is a strictly increasing function on R without using the graph of y = f(x).
Solution:
Let x₁, x₂ ∈ R with x₁ < x₂. Then 8x₁ < 8x₂. Adding 2 to both sides of this inequality, we have 8x₁ + 2 < 8x₂ + 2. i.e., f(x₁) < f(x₂).
Thus
x₁ < x₂ ⇒ f(x₁) < f(x₂) ∀ x₁, x₂ ∈ R.
Therefore, the given function f is strictly increasing on R.

Question 42.
Show that f (x) = e^x is strictly increasing on R (without graph).
Solution:
Let x₁, x₂ ∈ R such that x₁ < x₂. we know that if a > b then e^a > e^b
∴ x₁ < x₂ ⇒ e^x₁ < e^x₂
i.e., f(x₁) < f(x₂).
Hence the given function f is a strictly increasing function.

Question 43.
Show that f(x) = – x + 2 is strictly decreasing on R.
Solution:
Let x₁, x₂ ∈ R x₁ < x₂.
Then x₁ < x₂
⇒ -x₁ > -x₂
⇒ -x₁ + 2 > -x₂ + 2
⇒ f(x₁) > f(x₂).
Therefore the given function f is strictly decreasing on R.

Question 44.
Find the intervals on which
f(x) = x² – 3x + 8 is increasing or decreasing ?
Solution:
Given fucntion is f(x) = x² – 3x + 8.
Differentiating it w.r.t. x, we get f'(x) = 2x – 3
f(x) = 0 for x = 3/2.

since f'(x) < 0 in (-∞, 3/2) the function f(x) is strictly decreasing on (-∞, \(\frac{3}{2}\)) Further since f'(x) > 0 in (\(\frac{3}{2}\), -∞), the function f(x) is a strictly increasing function (\(\frac{3}{2}\), -∞).

Question 45.
Show that f(x) = |x| is strictly decreasing on (-∞, 0) and strictly increasing on (0, ∞).
Solution:
The given function is f(x) = |x| i.e.,

Thus f'(c) = 1 if c > 0, f'(c) = -1 if c < 0. Since f'(c) > 0 on (0, ∞), the function f(x) is strictly increasing on (0, ∞). Since f'(c) < 0 on (-∞, 0), the function f(x) is strictly decreasing on (-∞, 0).

Question 46.
Find the intervals on which the function f(x) = x³ + 5x² – 8x + 1 is a strictly increasing function.
Solution:
Given that f(x) = x³ + 5x² – 8x + 1.
∴ f'(x) = 3x² + 10x – 8 = (3x – 2) (x + 4)
= 3(x – \(\frac{2}{3}\)) (x – (-4)).
f'(x) is negative in (-4, \(\frac{2}{3}\)) and positive in (-∞, -4) ∪ (\(\frac{2}{3}\), ∞) .
∴ The function is strictly deceasing in (-4, \(\frac{2}{3}\)) and is strictly decreasing in (-∞, -4)and (\(\frac{2}{3}\), ∞)

Question 47.
Find the Intervals on which f(x) = x^x (x > 0) is increasing and decreasing.
Solution:
Taking logarithms on both sides of f(x) = x^x
we get
log (f(x)) = x log x. Differenetiating it w.r.t. x
we have \(\frac{1}{f(x)}\) f'(x)= 1 + log x
∴ f'(x) = x^x( 1 + log x)
f'(x) = 0 ⇒ x^x(1 + log x) = 0 ……………. (1)
⇒ 1 + log x = 0
⇒ x = 1/e

Suppose x < 1/e log x < log (1/e) (since the base e > 1). i.e., log x < -1
1 + log x < 0 ⇒ x^x (1 + log x) < 0. i.e., f'(x) < 0 Now suppose, x > 1/e. Then log x > log (1/e)
i.e., log x > – 1.
⇒ 1 + log x > 0
⇒ x^x (1 + log x) > 0
⇒ f'(x) > 0
Hence, f is strictly decreasing on (o, 1/e) and it is strictly increasin on (1/e, ∞).

Question 48.
Determine the intervals in which f(x) = \(\frac{2}{(x-1)}\) + 18x ∀ x ∈ R \ {0} is strictly increasing and decreasing.
Solution:
Given that f(x) = \(\frac{2}{(x-1)}\) + 18x. Differenetiating
it w.r.t. x, we get
f'(x) = \(\frac{-1}{(x-1)^{2}}\) . 2 + 18 and f'(x) = 0
⇒ \(\frac{2}{(x-1)^{2}}\) = 18 ⇒ (x – 1)² = 1/9.
∴ f'(x) = 0 if x – 1 = 1/3 or x- 1 = -(1/3).
i.e., x = 4/3 or x – 2/3.
The derivative of f(x) can be expressed as
f'(x) = \(\frac{18}{(x-1)^{2}}\) . (x – /3) (x – 4/3)

∴ The given function f(x) is strictly increasing on (-∞, \(\frac{2}{3}\)) and (\(\frac{4}{3}\), ∞) and it is strictly decreasing on (\(\frac{2}{3}\), \(\frac{4}{3}\)).

Question 49.
Let f(x) = sin x – cos x be defined on [0, 2π]. Determine the intervals in which f(x) is strictly decreasing and strictly increasing.
Solution:
Given that f(x) = sin x – cos x.
∴ f'(x) = cos x + sin x
∴ f'(x) = \(\sqrt{2}\) . sin(x + π/4)
Let 0 < x < 3π/4. Then π/4 < x + π/4 < π. ∴ sin (x + π/4) > 0 i.e., f'(x) > 0.
Similarly it can be shown that f'(x) < 0 in (3π/4 . 7π/4) and f'(x) > 0 in (7π/4, 2π).

Thus the function f(x) strictly increasing in (0, \(\frac{3\pi}{24}\) and (\(\frac{7\pi}{4}\), 2π) it is strictly decreasing in (\(\frac{3\pi}{4}\), \(\frac{7\pi}{4}\)).

Question 50.
If 0 ≤ x ≤ \(\frac{\pi}{2}\) then show that x ≥ sinx.
Solution:
Let f(x) = x – sin x. Then f'(x) = 1 – cos x ≥ 0 ∀ x
∴ f is an increasing function for all x.
Now, f(0) = 0. Hence f(x) ≥ f(0) for all x ∈ (0, \(\frac{\pi}{2}\)). Therefore, x ≥ in x.

Question 51.
Let f : R → R be defined by f(x) = 4x² – 4x + 11. Find the global minimum value and a point of global minimum.
Solution:
We have to look for a value c e R(domain) such that
f(x) ≥ f(c) ∀ x ∈ R
so that f(c) is the global minimum value of f. Consider
f(x) = 4x² – 4x + 11 = (2x – 1)² + 10 ≥ ∀ x ∈ R ……………(1)
Now, f(1/2) = 10
Also f(x) ≥ f(1/2) ∀ x ∈ R
Hence, f(1/2) = 10 is the global minimum value of f(x), and a point of global minimum is x = 1/2.

Question 52.
Let f : [-2, 2] → R be defined by f(x) = |x|. Find the global maximum of f(x) and a point of global minimum.
Solution:
We know that |x| = \(\left\{\begin{array}{ccc}
x & \text { if } & x \geq 0 \\
-x & \text { if } & x<0
\end{array}\right.\)
Therefore, from the graph of the function f on [-2, 2] clearly f(x) ≤ f(2) and f(x) ≤ f(-2) ∀ x ∈ [-2, 2].

∴ f(2) = f(-2) = 2 is the global maximum of f(x), 2 and -2 are the points of global maximum.

Question 53.
Find the global maximum and global minimum of the function f : R → R defined by f (x) = x².
Solution:
We have f(x) ≥ f(0) ∀ x ∈ R.
Hence th global minimum value of f(x) is 0 and a point of global minimum is x = 0.
Suppose f has global maximum at x₀ ∈ R (x₀ > 0). Then as per out assumption we have.
f(x₀) ≥ f(x) ∀ x ∈ R

Choose x₁ = x₀ + 1. Then x₁ ∈ R and x₀ < x₁
∴ x₀² < x₁²
Hence f(x₀) < (fx₁)
Thus we got f(x₁) such that f(x₀) > f(x₀) which is a contradition to Therefore, f(x) has no global maximum on R.

Question 54.
Find the stationary points of f(x) = 3x⁴ – 4x³ + 1, ∀ x ∈ R and state whether the function has local maxima or local minima at those points.
Solution:
Given that f(x) = 3x⁴ – 4x³ +1 and the domain of f is R. Differentiating the function w.r.t. x we have
f'(x) = 12x²(x – 1) ……………………… (1)
The stationary points are the roots of f (x) = 0 i.e., 12x²(x – 1) = 0. Hence x = 0 and x = 1 are the stationary points. Now, we test whether the stationary point x = 1 is a local extreme point or not. For.
f'(0.9) = 12(0.9)² (0.9 – 1) ⇒ f'(0.9) is negative
f'(1.1) = 12(1.1)² (1.1 – 1) ⇒ f'(1.1) is positive
and f(x) is defined in the neighbourhood i.e., (0.8, 1.2) of x = 1 with 8 = 0.2.
By theorem cis a point of local maximum if f(x) changes sign from positive to negative at x = c.
c is a point of local minimum if f'(x) changes sign from negative to positive at x = c.
The given function has local (relative) minimum at x = 1. Hence x = 1 is a local extreme point.
We will now test whether x = 0 is a local extreme point or not.
The function f(x) is defined in the neighbourhood of (-0.2, 0.2).
f'(-0.1) = 12(-0.1)² (-0.1 – 1) ⇒ f'(-0.1) is negative
f'(-0.1) = 12(0.1)² (0.1 -1) ⇒ f'(0.1) is negative
Thus, f(x) has no change in sign at x = 0. Therefore, the function f has no local maximum and no local minimum. Hence, x = 0 is not a local extreme point.

Question 55.
Find the points (if any) of local maxima and local minima of the function f (x) = x³ – 6x² + 12x – 8 ∀ x ∈ R.
Solution:
Given function is f(x) = x³ – 6x² + 12x – 8 and the domain of f is R.
Differentiating the given function w.r.t. x, we get
f'(x) = 3x² – 12x + 12 i.e., f'(x) = 3(x – 2)².
The stationary point of f(x) is x = 2, since 2 is a root of f'(x) = 0.
Choose δ = 0.2 The 0.2- neighbourhood of 2 is (1.8, 2.2). Now
f'(1.9) = 3(1.9 – 2)² ⇒ f'(1.9) is positive
f'(2.1) = 3(2.1- 2)² ⇒ f'(2.1) is positive
Thus f(x) does not change the sign at x = 2. By Theorem c is neither a point of local maximum nor a point of local minimum if f'(x) does not change sign at x = c.
x = 2 is neithere a local maximum nor a local minimum.

Question 56.
Find the points of local minimum and local maximum of the function f(x) = sin 2x ∀ x ∈ [0, 2π]
Solution:
The given function is f(x) = sin 2x and domain is [0, 2π].
f'(x) = 2cos 2x …………… (1)
The critical points are the roots of 2 cos 2x = 0 and lying in the domain [0, 2π].
They are \(\frac{\pi}{4}\) and \(\frac{3\pi}{4}\) .
Now we apply the first derivative test at x = \(\frac{\pi}{4}\)
Clearly (\(\frac{\pi}{4}\) – 0.1 . \(\frac{\pi}{4}\) + 0.1) is a neighbourhood of \(\frac{\pi}{4}\) and the given f is defined on it.
Now
f'(\(\frac{\pi}{4}\) – 0.05) = 2 cos(\(\frac{\pi}{2}\) – 0.1) > 0
f'(\(\frac{\pi}{4}\) + 0.05) = 2 cos(\(\frac{\pi}{2}\) + 0.1) < 0
Thus f'(x) changes sign from positive to negative at x = \(\frac{\pi}{4}\). Therefore f has a local maximum.
Now we apply the first derivative test at x = \(\frac{3\pi}{4}\).
Clearly (\(\frac{3\pi}{4}\) – 0.1 . \(\frac{3\pi}{4}\) + 0.1) is a neighbourhood of \(\frac{3\pi}{4}\) and the given f is defined on it.
Now
f'(\(\frac{3\pi}{4}\) – 0.05) = 2 cos(\(\frac{3\pi}{4}\) – 0.1) < 0 f'(\(\frac{3\pi}{4}\) + 0.05) = 2 cos(\(\frac{3\pi}{4}\) + 0.1) > 0
Thus f'(x) changes sign from positive to negative at x = \(\frac{3\pi}{4}\). Therefore f has a local maximum at x = \(\frac{3\pi}{4}\).

Question 57.
Find the points of local extrema of the function f(x) = x³ – 9x² – 48x + 6 ∀ x ∈ R Also find its local extrema.
Solution:
Given function is
f(x) = x³ – 9x² – 48x + 6 …………… (1)
and the domain of the function is R.
Differentiating (1) w.r.t. x we get
f'(x) = 3x² – 18x – 48 = 3(x – 8) (x + 2) ………….. (2)
Thus the stationary points are – 2 and 8.
Differentiating (2) w.r.t.x we get,
f'(x) = 6(x – 3) ………….. (3)
Let x₁ = -2 and x₂ = 8. Now we have to find f’ at each of these points to know the sign of second derivative.
At x₁ = -2, f'(-2) = – 30. The sign of it is negative.
∴ x₁ = – 2 is a point of local maximum of f and its local maximum value is f(-2) – 58.
Now, at x₂ = 8 f'(8) = 30. Thus the sign of f”(x₂) is positive. Therefore, x = 8 is a point of local minimum of f and its local minimum value is
f(8) = – 442.

Question 58.
Find the points of local extrema of f(x) = x⁶ ∀ x ∈ R. Also find its local extrema.
Solution:
f(x) = x⁶ ………………….. (1)
Differentiating (1) w.r.t. x we get,
f'(x) = 6x⁵ ……………. (2)
and again differentiating (2) w.r.t. x we get
f'(x) = 30x⁴ …………………… (3)
The stationary point of f(x) is x = 0 only (since f'(x) = 0 only at x = 0).
Now f'(0) = 0. At x = 0, we can not conclude anything about the local extrema by the second derivative test. Therefore, we apply the first derivative test. As the domain of f is R, the function f is defined on (-0.2, 0.2) which is a neighbourhood of x = 0. Now
f'(-0.1) = 6(-0.1)5 < 0, f'(0, 1) = 6(0.1) 5 > 0.
Thus f'(x) changes sign form negative of positive at x = 0.
∴ x = 0 is a point of local minimum and its local minimum value is f(0) = 0.

Question 59.
Find the points of local extrema and local extrema for the function f(x) = cos 4x defined on (0, \(\frac{\pi}{2}\))
Solution:
Here f(x) = cos 4x ……………. (1)
and its domain is (0, \(\frac{\pi}{2}\))
∴ f’(x) = -4 sin 4x ………….. (2)
and f”(x) = -16 cos 4x ……………… (3)
The stationary points are the roots of
f'(x) = 0 and lying in the domain (o, \(\frac{\pi}{2}\)).
f'(x) = 0 ⇒ 4 sin 4x = 0
⇒ 4x = 0, π, 2π, 3π, 4π ………………….
⇒ x = 0, π/4, π/2, 3π/4, π …………………..
The point lying in the domain is x = \(\frac{\pi}{4}\) only.
Thus x = \(\frac{\pi}{4}\) is the stationary point of the given function. Now
f”(\(\frac{\pi}{4}\)) = -16 cos(π) = 16 > 0.
The function f has local minimum at x = \(\frac{\pi}{4}\) and its local minimum value is
f(\(\frac{\pi}{4}\)) = -1.

Question 60.
Find two positive number whose sum is 15 so that the sum of their squares is minimum.
Solution:
Suppose one numbers is x and the other number 15 – x. Let S be the sum of squares of these numbers. Then S = x² + (15 – x)² ………………… (1)
Note that the quantity S, to be minimized, is a function of x.
Differentiating (1) w.r.t. x, we get
\(\frac{\mathrm{dS}}{\mathrm{dx}}\) = 2x + 2(15 – x) (-1)
= 4x – 30 ………………. (2)
and again differentiating (2) w.r.t.x, we get
\(\frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dx}^{2}}\) = 4 ……………….. (3)
The stationary point can be obtained by solving \(\frac{\mathrm{dS}}{\mathrm{dx}}\) = 0 i.e., 4x – 30 = 0.
∴ x = 15/2 is the stationary point of (1).
Since \(\frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dx}^{2}}\) = 4 > 0, S is minimum at x = \(\frac{15}{2}\)
∴ The two numbers are \(\frac{15}{2}\), 15 – \(\frac{15}{2}\) i.e., \(\frac{15}{2}\) and \(\frac{15}{2}\).

Question 61.
Find the maximum area of the rectangle that can be formed with fixed perimeter 20.
Solution:
Let x and y denote the length and the breadth of a rectangle respectively. Given that the perimeter of the rectangle is 20.
i.e., 2(x + y) = 20
i.e., x + y = 10 ……………. (1)
Let A denote the area of rectangle.
Then A = xy ………….. (2)
Which is to be minimized. Equation (1) can be expressed as
y = 10 – x …………… (3)
From (3) and (2), we have
A = x (10 – x)
A = 10x – x² ……………… (4)
Differentiating (4) w.r.t. x we get
\(\frac{\mathrm{dA}}{\mathrm{dx}}\) = 10 – 2x ……………….. (5)
The stationary point is a root of 10 – 2x = 0
∴ x = 5 is the stationary point.
Differentiating (5) w.r.t. x, we get
\(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\) = -2
which is negative. Therefore by second derivative test the area A is maximized at x = 5 and hence y = 10 – 5 = 5, and the maximum area is A = 5(5) = 25.

Question 62.
Find the point on the graph y² = x which is the nearest to the point (4, 0).
Solution:

Let P(x, y) be any point on y² = x and A(4, 0). We have to find P such that PA is minimum
Suppose PA = D. The quantity to be minimized is D.
D = (\(\sqrt{(x-4)^{2}+(y-0)^{2}}\)) ………………. (1)
P(x, y) lies on the curve, therefore
y² = x …………………… (2)
From (1) and (2), we have
D = \(\sqrt{\left((x-4)^{2}+x\right)}\)
D = \(\sqrt{\left(x^{2}-7 x+16\right)}\) ……………………. (3)
Differentiating (3) w.r.t. x, we get
\(\frac{\mathrm{dD}}{\mathrm{dx}}\) = \(=\frac{2 x-7}{2} \cdot \frac{1}{\sqrt{x^{2}-7 x+16}}\)
Now \(\frac{\mathrm{dD}}{\mathrm{dx}}\) = 0
gives x = \(\frac{7}{2}\). Thus \(\frac{7}{2}\) is a stationary point of the function D. We apply the first derivative test to verify whether D is minimum at x = \(\frac{1}{2}\)

Question 63.
Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.
Solution:
Let O be the centre of the circular base of the cone and its height be h. Let r be the radius of the circular base of the cone.
Then AO = h, OC = r.
Let a cylinder with radius x(OE) be inscribed in the given cone. Let its height be u.

i.e., RO = QE = PD = u
Now the triangles AOC and QEC are similar.
Therefore,
\(\frac{\mathrm{QE}}{\mathrm{OA}}\) = \(\frac{\mathrm{EC}}{\mathrm{OC}}\)
i.e., \(\frac{u}{h}\) = \(\frac{r-x}{r}\)
∴ u = \(\frac{h(r-x)}{r}\) ………………… (1)
Let S denote the curved surface area of the chosen cylinder.Then
S = 2 π xu.
As the cone is fixed one, the values of r and h are constants. Thus S is function of x only. Now,
\(\frac{\mathrm{dS}}{\mathrm{dx}}\) = 2 πh (r – 2x)/r and \(\frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dx}^{2}}\) = -4πh/r.
The stationary point of S is a root of
\(\frac{\mathrm{dS}}{\mathrm{dx}}\) = 0
i.e., π(r – 2x)/r = 0
i.e., x = r/2
\(\frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dx}^{2}}\) < 0 for all x, Therefore (\(\frac{\mathrm{d}^{2} \mathrm{~S}}{\mathrm{dx}^{2}}\))_x=r/2 < 0.
Hence, the radius of the cylinder of greatest curved surface area which can be inscribed in a given cone is r/2.

Question 64.
The profit function P(x) of a company, selling x items per day is given by P(x) = (150 – x)x – 1600. Find the number of items that the company should sell to get maximum profit. Also find the maximum profit.
Solution:
Given that the profit function is
P(x) = ( 150 – x)x – 1600 ………….. (1)
For maxima or minima \(\frac{\mathrm{dP}(\mathrm{x})}{\mathrm{dx}}\) = 0
∴ (150 – x) (1) + x (-1) = 0
i.e., x = 75
Now \(\frac{d^{2} P(x)}{d x^{2}}\) = -2 and \(\left[\frac{d^{2} P(x)}{d x^{2}}\right]_{x=75}\) < 0.
∴ The profit P(x) is maximum for x = 75.
∴ The company should sell 75 items a day to make maximum profit.
The maximum profit will be P(75) = 4025.

Question 65.
A manufacturer can sell x items at a price of rupees (5 – x/100) each. The cost price of x items is Rs. (x/5 + 500). Find the number of items that the manufacturer should sell to earn maximum profits.
Solution:
Let S(x) be the selling price of x items and C(x) be the cost price of x items. Then, we have
S(x) = {cost of each item}, x .
∴ S(x) = (5 – x/100) x = 5x – x²/100 and C(x) = x/5 + 500
Let P(x) denote the profit function. Then
P(x) = S(x) – C(x)
P(x) = (5x – x²/100) – (x/5 + 500)
– (24x/5) – (x²/100) – 500 …………….. (1)
For maxima or minima
\(\frac{\mathrm{dP}(\mathrm{x})}{\mathrm{d} x}\) = 0
i.e., 24/5 – x/50 = 0
The stationary point of P(x) is x = 240 and
\(\left[\frac{d^{2} P(x)}{d x^{2}}\right]\) = –\(\frac{1}{50}\) for all x.
Hence the manufacturer can earn maximum profit if he sells 240 items.

Question 66.
Find the absolute extrement of f(x) = x² defined on [-2, 2].
Solution:
The given function f(x) = x² is continuous on [-2, 2]. It can be shown that it has only local minimum and the point of local minimum is 0. The absolute(global) maximum of f is the largest value of f(-2), f(0) and f(2) i.e., 4, 0, 4.
Hence, the absolute maximum value is 4. Similarly the absolute minimum is the least value of 4, 0, 4. Hence 0 is the absolute minimum value.

Question 67.
Find the absolute maximum of x⁴⁰ – x²⁰ on the interval [0, 1]. Find also its absolute maximum value.
Solution:
Let f(x) = x⁴⁰ – x²⁰ ∀ x ∈ [0,1] ………… (1)
The function f is continuous on [0,1 j and the interval [0,1] is closed.
From (1) we have
f'(x) = 40 x³⁹ – 20 x¹⁹ = 20x¹⁹ (2x²⁰ – 1).
Thus f'(x) = 0 at x = 0 or
x = \(\left(\frac{1}{2}\right)^{\frac{1}{20}}\)
Therefore, the critical points of f are and \(\left(\frac{1}{2}\right)^{\frac{1}{20}}\) and 0 is one of the end points of the domain. Therefore no local maximum exists at x = 0. Now
f'(x) = 40(39) x³⁸ – 20(19) x¹⁸
= 20x¹⁸ (78 x²⁰ – 19)
[f”(x)]_{x = \(\left(\frac{1}{2}\right)^{\frac{1}{20}}\)} = 20 (1/2)^(18/20)[39 – 19] > 0.
Therefore f has local minimum at
x = (1/2)(1/20)
and its value is \(f\left(\left(\frac{1}{2}\right)^{\frac{1}{20}}\right)\) = –\(\frac{1}{4}\)
Therefore the absolute maximum value of the function f is the largest value of f(0), f(1) and \(f\left(\left(\frac{1}{2}\right)^{\frac{1}{20}}\right)\) i.e., the largest value of {0, 0, –\(\frac{1}{4}\)}
Hence, the absolute maximum of f is 0 and the points of absolute maximum are 0 and 1. Further the absolute minimum is the least of 0, 0, -latex]\frac{1}{4}[/latex].
Hence the absolute minimum is -latex]\frac{1}{4}[/latex] and the point of absolute minimum is x = \(\left(\frac{1}{2}\right)^{\frac{1}{20}}\)

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B The Straight Line Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B The Straight Line Important Questions

Question 1.
Find the equation of the straight line pass-ing through the point (2, 3) and making non-zero intercepts on the axes of co-ordinates whose sum is zero. [Mar 12]
Sol. Equation of the line in the intercept form is
\(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
Given b = -a
Equation of the line is \(\frac{x}{a}\) – \(\frac{y}{a}\) = 1
⇒ x – y = a
This line passes through (2, 3)
2 – 3 = a ⇒ a = -1
Equation of the line is
x – y = -1 or x – y + 1 = 0

Question 2.
Find the equation of the straight line passing through the points (at₁², 2at₁ ) and (at₂², 2at₁ ).
Solution:
Equation of the given points.
(x – x₁) (y – y₂) = (y – y₁) (x₁ – x₂)
( x -at₁²) (2at₁ – 2at₂)
= (y – 2at₁) (at, — at2)
(x – at₁²) .2a(t₁ – t₂) = (y – 2at₁) a. (t₁² – t₂²)
2x – 2at₁² = y(t₁ + t₂) – 2at₁² + 2at₁t₂ = 0
2x – (t₁ + t₂)y + 2at₁t₂ = 0

Question 3.
Find the equation of the straight line passing through A (-1, 3) and i) parallel ii) perpendicular to the straight line passing through B(2, – 5) and C(4, 6). [Mar 11]
Solution:
Slope of BC = \(\frac{-5-6}{2-4}\) = \(\frac{-11}{-2}\) = \(\frac{11}{2}\)
i) The required line is parallel to BC and passes through A(- 1, 3)
Equation of the parallel line is
y – 3 = \(\frac{11}{2}\)(x + 1)
2y – 6 = 11x + 11
11x – 2y + 17 = 0

ii) The required line is perpendicular to BC.
Slope of the required line = –\(\frac{1}{m}\) = –\(\frac{2}{11}\)
This line passes through A (-1, 3)
Equation of the required line is
y – 3 = –\(\frac{2}{11}\)(x + 1)
11y – 33 = -2x – 2
2x + 11y – 31 = 0

Question 4.
Prove that the points (1, 11), (2, 15) and (-3, -5) are collinear and find the equation of the straight line containing them.
Solution:
A(1, 11), B(2, 15) and C (-3, -5) are the given points.
Equation of AB is
(y – y₁) (x₁ – x₂) = (x – x₁) (y₁ – y₂)
(y – 11) (1 -2) = (x – 1) (11 – 15)
-(y – 11) = -4 (x – 1)
-y + 11 = – 4x + 4
4x – y + 7 = 0
C (-3, -5) .
4x – y + 7 = 4(-3) + 5 + 7
= -12 + 12 = 0
C lines on AB ⇒ A, B, C are collinear.
Equation of the line containing them is 4x – y + 7 = 0

Question 5.
A straight line passing through A(1, -2) makes an angle tan^-1 \(\frac{4}{3}\) with the positive direction of the X – axis in the anticlock-wise sense. Find the points on the straight line whose distance from A is 5.
Solution:

Given α = tan^-1 \(\frac{4}{3}\) ⇒ tan α = \(\frac{4}{3}\)

cos α = \(\frac{3}{5}\), sin α = \(\frac{4}{5}\)
(x₁, y₁) = (1, -2) = x₁ = 1, y₁ = -2
Case i) .
r = 5
x = x₁ + r cos α = 1 + 5 . \(\frac{4}{5}\) = 1 + 4 = 5
y = y₁ + r sin α= – 2 + 5 . \(\frac{3}{5}\) = -2 + 3 = 1
Co-ordinates of B are (5, 1)

Case ii) r = -5
x = x₁ + r cos α = 1 – 5 . \(\frac{4}{5}\) = 1 – 4 = -3
y = y₁ + r sin α = -2 – 5 . \(\frac{3}{5}\) = -2 – 3 = -5
Co-ordinates of C are (- 3, – 5)

Question 6.
A straight line parallel to the line y = \(\sqrt{3} x\) passes through Q(2, 3) and cuts the line 2x + 4y – 27 = 0 at P. Find the length of PQ.
Solution:
PQ is parallel to the straight line y = \(\sqrt{3} x\)
tan α = \(\sqrt{3}\) = tan 60°
α = 60°
Q(2, 3) is a given point

Co-ordinates of any point P are
(x₁ + r cos α, y₁ + r sin α)
(2 + r cos 60°, 3 + r sin 60°)

Question 7.
Transform the equation
3x + 4y+ 12 = 0 into
i) slope—intercept form
ii) Intercept form and
in) normal form
Solution:
The given equation is 3x + 4y + 12=0
i) Slope-intercept form
4y = -3x – 12
y = (-\(\frac{3}{4}\))x + (-3)
Slope = –\(\frac{3}{4}\) , y – intercept = -3.

ii) Intercept form
-3x – 4y = 12
–\(\frac{3 x}{12}\) – \(\frac{4 y}{12}\) = 1
\(\frac{x}{(-4)}\) + \(\frac{y}{(-3)}\) = 1
x – intercept = -4, y – intercept = -3

iii) Normal form
-3x – 4y = 12
Dividing with \(\sqrt{9+16}\) = 5
(-\(\frac{3}{5}\))x + (-\(\frac{4}{5}\))y = \(\frac{12}{5}\)
Let cos α = \(\frac{-3}{5}\) and sin α = –\(\frac{4}{5}\)
p = \(\frac{12}{5}\) so that
x cos α + y sin α = p
α lies in third quadrant so that
α = π + tan^-1(\(\frac{4}{3}\))

Question 8.
If the area of the triangle formed by the straight line x = 0, y = 0 and 3x + 4y = a (a > 0) is 6, find the value of a. [May 11]
Solution:
Equation of the line is 3x + 4y = a

Given \(\frac{a^{2}}{2 a}\) = 6 ⇒ a² = 144
a = ±12
But a > 0
a = 12

Question 9.
Find the value of k, if the lines 2x – 3y + k = 0, 3x – 4y -13 = 0 and 8x – 11y – 33 = 0 are concurrent.
Solution:
Let L₁, L₂, L₃ be the straight lines whose equations are respectively
2x – 3y + k = 0 ………………. (1)
3x – 4y – 13 = 0 ……………….. (2)
8x – 11y – 33 = 0 ………………. (3)
Solving (2) and (3) for x and y

Point of the lines (2) and (3) is (11, 5)
The given lines L₁, L₂, L₃ are concurrent
∴ L₁ contain (11, 5)
∴ 2(11) – 3(5) + k = 0
k = -7

Question 10.
If the straight lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 are concurrent, then prove that a³ + b³ + c³ = 3abc.
Solution:
The equation of the given lines are
ax + by +c = 0 ………………… (1)
bx + cy + a = 0 ……………….. (2)
cx + ay + b = 0 ……………….. (3)
Solving (1) and (2) point of intersection is got by

c(ab – c²) +a (bc – a²) + b(ca – b²) = 0
abc – c³ + abc – a³ + abc – b³ = 0
∴ a³ + b³ + c³ = 3abc.

Question 11.
A variable straight line drawn through the point of intersection of the straight lines \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 and \(\frac{x}{b}\) + \(\frac{y}{a}\) = 1 meets the co-ordinate axes at A and B. Show that the locus of the mid point of \(\overline{\mathrm{AB}}\) is 2(a + b) xy = ab(x + y)
Solution:
Equations of the given lines are \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
and \(\frac{x}{b}\) + \(\frac{y}{a}\) = 1
Solving the point of intersection P(\(\frac{a b}{a+b}\), \(\frac{a b}{a+b}\))
Q (x₀, y₀) is any point on the locus
⇔ The line with x – intercept 2x₀, y – intercept 2y₀, passes through P
⇔ P lies on the straight line \(\frac{x}{2 x_{0}}\) + \(\frac{y}{2 y_{0}}\) = 1
i.e., \(\frac{a b}{a+b}\)(\(\frac{1}{2 x_{0}}\) + \(\frac{1}{2 y_{0}}\)) = 1
⇒ \(\frac{a b}{a+b}\) . \(\frac{x_{0}+y_{0}}{2 x_{0} y_{0}}\) = 0
ab(x₀ + y₀) = 2(a + b) x₀y₀
Q(x₀, y₀) lies on the curve
2(a + b)xy = ab(x + y)
Locks the mid of point of
AB is 2(a + b)xy = ab(x +y)

Question 12.
If a, b, c are in arithmetic progression, then show that the equation ax + by + c = 0 represents a family of concurrent lines and find the point of concurrency.
Solution:
a, b, c are in A.P.
2b = a + c
a – 2b + c = 0
a.1 + b(-2) + c = 0
Each number of family of straight lines
ax + by + c = 0
passes through the fixed point (1,-2)
∴ Set of lines ax + by + c = 0 for parametric values of a, b and c is a family of concurrent lines.
∴ Point of concurrency is (1, -2).

Question 13.
Find the value of k, if the angle between the straight lines 4x – y + 7 = 0 and kx – 5y – 9 = 0 is 45°.
Solution:

Squaring and cross multiplying
2(4k + 5)² = 17(k² + 25)
2(16k² + 40k + 25) = 17k² + 425
32k² + 80k + 50 = 17k² + 425
15k² + 80k – 375 = 0
3k² + 16k – 75 = 0
(k – 3) (3k + 25) = 0
k = 3 or -25/3

Question 14.
Find the equation of the straight line passing through (x₀, y₀) and (i) parallel (ii) perpendicular to the straight line
ax + by + c = 0.
Solution:
Equation of the given line is ax + by + c = 0
i) Equation of the parallel line is ax + by = k ………………… (1)
This line passes through P(x₀, y₀) ⇒ ax₀ + by₀ = k ……………… (2)
Subtracting (2) from (1) equation of the required line is a(x – x₀) + b(y – y₀) = 0

ii) Equation of the perpendicular line is bx – ay = k.
This line passes through P(x₀, y₀) ⇒ bx₀ – ay₀ = k
Subtracting, equation of the required line is b(x – x₀) – a(y – y₀) = 0

Question 15.
Find the equation of the straight line perpendicular to the line 5x – 2y = 7 and passing through the point of intersection of the lines 2x + 3y = 1 and 3x + 4y = 6.
Solution:
Given lines are L₁ = 2x + 3y – 1 = 0
L₂ = 3x + 4y – 6 = 0
Equation of the line passing through the intersection of L₁ = 0, L₂ = 0 is
L₁ + kL₂ = 0
(2x + 3y – 1) + k(3x + 4y – 6) = 0
(2 + 3k)x + (3 + 4k)y – (1 + 6k) = 0 …………………. (1)
This line is perpendicular to 5x – 2y = 7 ………………… (2)
a₁a₂ + b₁b₂ = 0
5(2 + 3k) – 2(3 + 4k) = 0
10 + 15k – 6 – 8k = 0
7k = -4 ⇒ k = – 4/7
Substituting in (1) equation of the required lines
(2 – \(\frac{12}{7}\))x + (3 – \(\frac{16}{7}\))x – (1 – \(\frac{24}{7}\)) = 0
\(\frac{2}{7}\)x + \(\frac{5}{7}\)y + \(\frac{17}{7}\) = 0 ⇒ 2x + 5y + 17 = 0

Question 16.
If 2x – 3y – 5 = 0 is the perpendicular bi-sector of the line segment joining (3, – 4) and (α, β) find α + β.
Solution:
(α, β) is the reflection of (3, -4) is the line
2x – 3y – 5 = 0
\(\frac{\alpha-3}{2}\) = \(\frac{\beta+4}{-3}\) = \(\frac{-2(6+12-5)}{4+9}\) = -2
α – 3 = -4 ⇒ α = -1
β + 4 = 6 ⇒ β = 2
α + β = -1 + 2 = 1

Question 17.
If the four straight lines ax + by + p = 0, ax + by + q = 0, cx + dy + r = 0 and cx + dy + s = 0 form a parallelogram, show that the area of the parallelogram so formed is \(\left|\frac{(p-a)(r-s)}{b c-a d}\right|\).
Solution:
Let L₁, L₂, L₃, L₄ be the lines given by
L₁ = ax + by + p = 0
L₂ = ax + by + q = 0
L₃ = cx + dy + r = 0
L₄ = cx + dy + s = 0
L₁ and L₂ are parallel: L₃ and L₄ are parallel
Area of the parallelogram = \(\frac{\mathrm{d}_{1} \mathrm{~d}_{2}}{\sin \theta}\)
d₁ = distance between L₁ and L₂ = \(\frac{|p-a|}{\sqrt{a^{2}+b^{2}}}\)
d₂ = distance between L₃ and L₄ = \(\frac{|r-s|}{\sqrt{c^{2}+d^{2}}}\)

Question 18.
The hypotenuse of a right angled isosceles triangle has its ends at the points (1, 3) and (-4, 1). Find the equations of the legs of the triangle.
Sol:
LetA =(1, 3)and B = (-4, 1) and ABC b a right isosceles triangle with \(\stackrel{\leftrightarrow}{A B}\) as hypotenuse.
We require, therefore, the equations of
\(\stackrel{\leftrightarrow}{A C}\) and \(\stackrel{\leftrightarrow}{B C}\)
Slope of \(\stackrel{\leftrightarrow}{A B}\) is \(\frac{1-3}{-4-1}\) = \(\frac{2}{5}\)
Since the slope of \(\stackrel{\leftrightarrow}{A B}\) is \(\), neither \(\frac{2}{5}\) nor \(\stackrel{\leftrightarrow}{A C}\) is vertical.

Taking the slope of \(\stackrel{\leftrightarrow}{A C}\) as \(\frac{7}{3}\), the slope of \(\stackrel{\leftrightarrow}{B C}\) would be –\(\frac{3}{7}\). Therefore, the equations of \(\stackrel{\leftrightarrow}{A C}\) and \(\stackrel{\leftrightarrow}{B C}\) are respectively.
y – 3 = \(\frac{7}{3}\) (x – 1) and y – 1 = – \(\frac{3}{7}\) (x + 4),
which become 7x – 3y + 2 = 0 and 3x + 7y + 5 = 0
If the lines drawn through A and B respectively parallel to \(\stackrel{\leftrightarrow}{B C}\) and \(\stackrel{\leftrightarrow}{A C}\) meet at D, then ∆ABD is also right isosceles, having \(\stackrel{\leftrightarrow}{A B}\) as its hypotenuse.
Therefore, the equations of \(\stackrel{\leftrightarrow}{A D}\) and \(\stackrel{\leftrightarrow}{B D}\) are respectively,
3(x – 1) + 7(y – 3) = 0 and 7(x + 4) – 3(y – 1) = 0
⇒ 3x + 7y – 24 = 0 and 7x – 3y + 31 = 0.
Therefore, the two pairs of legs required are
7x – 3y + 2 = 0, 3x + 7y + 5 = 0 and
3x + 7y – 24 = 0, 7x – 3y + 31 = 0.
Note : ADBC is square.

Question 19.
A line is such that its segment between the lines 5x – y + 4 = 0 and 3x + 4y – 4 = 0 is . bisected at the point (1, 5). Obtain its equation.
Solution:
Let the required line meet 3x + 4y – 4 = 0 at A and 5x – y + 4 = 0 at B, so that AB is the segment between the given lines, with its mid-point at C = (1, 5).
The equation 5x – y + 4 = 0 can be written as
y = 5x + 4 so that any point on \(\stackrel{\leftrightarrow}{B X}\) is (t, 5t + 4) for all real t.
∴ B = (t, 5t + 4) for some t. Since (1, 5) is the mid-point of \(\stackrel{\leftrightarrow}{A B}\)
A = [2 – t, 10 – (5t + 4)]
= [2 – t, 6 – 5t]
Since A lies on 3x + 4y – 4 = 0,
3(2 – 1) + 4(6 – 5t) – 4 = 0
⇒ -23t + 26 = 0
⇒ t = \(\frac{26}{23}\)

equation of \(\stackrel{\leftrightarrow}{A B}\) is y – 5 = \(\frac{107}{3}\) (x – 1)
⇒ 3y – 15 = 107x – 107
⇒ 107x – 3y – 92 = 0.

Question 20.
An equilateral triangle has its incentre at the origin and one side as x + y – 2 = 0. Find the vertex opposite to x + y – 2 = 0.
Solution:
Let ABC be the equilateral triangle and
x + y – 2 = 0 represent side \(\stackrel{\leftrightarrow}{B C}\).
Since O is the incentre of the triangle, \(\stackrel{\leftrightarrow}{A D}\) is the bisector of ∠BAC .
Since the triangle is equilateral, \(\stackrel{\leftrightarrow}{A D}\) is the perpendicular bisector of \(\stackrel{\leftrightarrow}{B C}\).
Since O is also the centroid, AO : OD = 2 : 1. [The centroid, circumcentre incentre and orthocentre coincide]
Let D = (h, k).
Since D is the foot of the perpendicular from O onto \(\stackrel{\leftrightarrow}{B C}\), D is given by

∴ x₁ = -2, y₁ = -2.
∴ A = (-2, -2), the required vertex.

Question 21.
Find the orthocentre of the triangle whose vertices are (-5,-7) (13, 2) and (-5, 6)
Solution:
Let A(-5, -7), B(13,2) and C (-5, 6) be the vertices of a triangle. Let \(\stackrel{\leftrightarrow}{A D}\) be the perpendicular drawn from A to \(\stackrel{\leftrightarrow}{B C}\) and \(\stackrel{\leftrightarrow}{B E}\) be the perpendicular drawn from B to \(\stackrel{\leftrightarrow}{A C}\).

Now slope of \(\stackrel{\leftrightarrow}{B C}\) = \(\frac{6-2}{-5-13}\) = \(\frac{-2}{9}\)
Since \(\stackrel{\leftrightarrow}{A D}\) ⊥ \(\stackrel{\leftrightarrow}{B C}\) , slope of \(\stackrel{\leftrightarrow}{A D}\) = \(\frac{9}{2}\) and so, the equation of \(\stackrel{\leftrightarrow}{A D}\) is
9x – 2y = – 45 + 14 = -31 ………………… (1)
Equation of \(\stackrel{\leftrightarrow}{A C}\) is x = -5 which is a vertical line and therefore equation of \(\stackrel{\leftrightarrow}{B E}\) is y = 2. ……………… (2)
Point of intersection of the lines (1) and (2) is (-3, 2) which is the orthocentre of ∆ ABC.

Question 22.
If the equations of the sides of a triangle are 7x + y – 10 = 0, x – 2y + 5 = 0 and x + y + 2 = 0, find the orthocentre of the triangle.
Solution:
Let ∆ ABC be the given triangle
Let the equations x – 2y + 5 = 0 ……………… (1)
7x + y – 10 = 0 ……………… (2)
and x + y + 2 = 0 ……………….. (3)
represent the sides \(\overleftrightarrow{\mathrm{AB}}\), \(\overleftrightarrow{\mathrm{BC}}\), and \(\overleftrightarrow{\mathrm{CA}}\)

Let \(\overleftrightarrow{\mathrm{AD}}\) and \(\overleftrightarrow{\mathrm{BE}}\) be the altitudes drawn from A and B respectively to the sides \(\overleftrightarrow{\mathrm{BC}}\) & \(\overleftrightarrow{\mathrm{CA}}\)
Solving the equations (1) and (3), we obtain A =(-3,1).
Since \(\overleftrightarrow{\mathrm{AD}}\) ⊥ \(\overleftrightarrow{\mathrm{BC}}\), the equations of \(\overleftrightarrow{\mathrm{AD}}\) is
x – 7y = -3 – 7 = -10 …………………… (4)
Solving the equation (1) and (2), we obtain B = (1, 3).
Since \(\overleftrightarrow{\mathrm{BE}}\) ⊥ \(\overleftrightarrow{\mathrm{AC}}\) , the equation of BE is
x – y = 1 – 3 = -2 …………… (5)
Point of intersection of the lines (4) and (5) is H(\(\frac{-2}{3}\), \(\frac{4}{3}\)) which is the orthocentre of ∆ ABC.

Question 23.
Find the circumcentre of the triangle whose vertices are (1, 3), (-3, 5) and (5, -1).
Solution:
Let the vertices of the triangle be
A(1, 3), B(-3, 5) and (5,-1).
The midpoints of the sides \(\overline{\mathrm{BC}}\), \(\overline{\mathrm{CA}}\) are respectively D(1, 2) and E(3, 1).
Let S be the point of intersection of the perpendicular bisectors of the sides \overline{\mathrm{BC}} and \overline{\mathrm{CA}}

Since the slope of \(\overleftrightarrow{\mathrm{BC}}\) = \(\frac{5+1}{-3-5}\) = \(\frac{-3}{4}\), the slope \(\overleftrightarrow{\mathrm{SD}}\) is \(\frac{4}{3}\) and therefore its equation is 4x – 3y = 4 – 6 = -2 …………….. (1)
Slope of \(\overleftrightarrow{\mathrm{AC}}\) = \(\frac{3+1}{1-5}\) = -1 ⇒ Slope of –\(\sqrt{3x}\) = 1
∴ Equation of \(\overleftrightarrow{\mathrm{SE}}\) is x – y = 3 – 1 = 2 ………………….. (2),
Solving the equations (1) and (2),we obtain S = (-8, -10) which is the circumcentre of ∆ ABC.

Question 24.
Find the circumcentre of the triangle whose sides are 3x – y – 5 = 0, x + 2y – 4 = 0 and 5x + 3y + 1 = 0. [May 11, 05; Mar. 06]
Solution:
Let the given equations 3x – y – 5 = 0, x + 2y – 4 = 0 and 5x + 3y + 1 = 0 represent the sides \(\overleftrightarrow{\mathrm{BC}}\), \(\overleftrightarrow{\mathrm{CA}}\) and \(\overleftrightarrow{\mathrm{AB}}\) respectively of ∆ ABC.
Solving the above equations two by two, we obtain the vertices A(-2, 3), B(1, -2) and (2, 1) of the given triangle.
The midpoints of the sides \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{CA}}\) are respectively D = (\(\frac{3}{2}\), \(\frac{-1}{2}\)) and E = (0, 2).

Equation of \(\overleftrightarrow{\mathrm{SD}}\) the perpendicular bisector of \(\overline{\mathrm{BC}}\) is x + 3y = 0 and the equation of \(\overleftrightarrow{\mathrm{SE}}\) the perpendicular bisector of \(\overline{\mathrm{AC}}\) is 2x – y + 2 = 0. Solving these two equations, we obtain the point of intersection of the lines \(\overleftrightarrow{\mathrm{SD}}\) and \(\overleftrightarrow{\mathrm{SE}}\) which is S(\(\frac{-6}{7}\), \(\frac{2}{7}\)) the circumcentre of ∆ ABC.

Question 25.
Find the incentre of the triangle formed by the straight lines y = \(\sqrt{3}\)x, y = –\(\sqrt{3}\)x and y = 3.
Solution:

The straight lines y = \(\sqrt{3x}\) and y = –\(\sqrt{3x}\) respectively make angles 60° and 120° with the positive direction of X – axis.
Since y = 3 is a horizontal line, the triangle formed by the three given lines is equilateral. So in-centre is same and centroid.
Vertices of the triangle and O(0, 0), A(\(\sqrt{3}\), 3) and D(\(\sqrt{3}\), 3).
∴ Incentre is (\(\frac{0+\sqrt{3}-\sqrt{3}}{3}\), \(\frac{0+3+3}{3}\)) = (0, 2)

Question 26.
Find the equation of the straight line whose distance from the origin is 4, If the normal ray from the origin to the straight line makes an angle of 135° with the positive direction of the x-axis.
Solution:
The equation of the given line is
x cos α + y sin α = p where p = 4 and α = 135°
∴ x(\(\frac{-1}{\sqrt{2}}\)) + y(\(\frac{1}{\sqrt{2}}\)) = 4 ⇒ or x – y + 4\(\sqrt{2}\) = 0

(i). Transform the equation x + y + 1 = 0 in to normal from.
Solution:
x + y + 1 = 0
⇔ (\(\frac{-1}{\sqrt{2}}\))n + (\(\frac{-1}{\sqrt{2}}\)) y = \(\frac{1}{\sqrt{2}}\)
⇔ x cos \(\frac{5\pi}{42}\) + y sin \(\frac{5\pi}{4}\) = \(\frac{1}{\sqrt{2}}\)
Hence the normal form of the equation of the given straight line is x cos \(\frac{5\pi}{4}\) + y sin \(\frac{5\pi}{4}\) = \(\frac{1}{\sqrt{2}}\) and the distance of the line from the origin is = \(\frac{1}{\sqrt{2}}\)

ii) A straight line passing through A(1, -2) makes and angle Tan^-1\(\frac{4}{3}\) with the positive direction of the x – axis in the anti-clock wise sense. Find the points of the straight line whose distance from A is 5.
Solution:
The paramatic equation of the lines A (1, -2)
Slope is \(\frac{4}{3}\) (∵ tan θ = \(\frac{4}{3}\))
are x = 1 + r cos θ = 1 + r(\(\frac{3}{5}\)) and y
= -2 + r sinθ
= -2 + r (\(\frac{4}{5}\))
The points on the above line a + a distance of (r) = 5 correspond to r = ±5 in the above equation and are
∴ (4, 2) and (-2, -6)

Question 27.
Trans form the equation 3x + 4y + 12 = 0 in to (i) slope – intercept from (ii) intercept form (iii) normal form.
Solution:
(i) slop – intercept form
3x + 4y + 12 = 0
⇔ 4y = -3x – 12
⇔ y = (\(\frac{-3}{4}\)) x + (-3)
∴ Slope = (\(\frac{-3}{4}\)) and y – intercept = -3,

ii) Intercept form :
3x = 4y + 12 = 0
⇔ \(\frac{-3y}{12}\) – \(\frac{-4y}{12}\) = 1
⇔ \(\frac{x}{(-4)}\) + \(\frac{y}{(-3)}\) = 1
∴ x-intercept of the line is -4, and tan y-intercept is -3.

iii) Normal form
3x = 4y + 12 = 0
⇔ -3x + -4y – 12 = 0
⇔ (\(\frac{-3}{5}\))x + (\(\frac{-4}{5}\))y = \(\frac{12}{5}\)
⇔ x cos α + y sin α = p ⇒ p\(\frac{12}{5}\) and cos α = \(\frac{-3}{5}\), sin α = \(\frac{-4}{5}\) determine the angle α in (0, 2π).

Question 28.
Find the angle between the lines 2x + y + 4 = 0 and y – 3x = 7.
Solution:
The angle between the given lines
= cos^-1 \(\frac{-6+1}{\sqrt{5 \times 10}}\)
= cos^-1 [latex]\frac{5}{\sqrt{2}}[/latex] = cos^-1(\(\frac{1}{\sqrt{2}}\)) = \(\frac{\pi}{4}\)

Question 29.
Find Q(h,k) in the foot of the perpendicular from p(x₁, y₁) on the straight lines ax + by + c = 0 then (h – x₁) ; a = (k – y₁); b = -(ax₁ + by₁ + c); (a² + b²)
Solution:
Equation of \(\stackrel{\leftrightarrow}{P Q}\) which is normal to the given straight line
L : ax + by + c = 0
bx – ay = bx₁ – ay₁

∴ Q ∈ \(\stackrel{\leftrightarrow}{P Q}\) we have
bh – ak = bx₁ – ay₁
∴ b(h – x₁) = a(k – y₁)
or (h – x₁) a = (k – y₁); b.
But, this implus that h = aλ + x₁, and k = bλ + y₁
for some λ ∈ R. sine Q(h, k) in point on L.
a(aλ + x₁) + b(bλ + y₁) + c = 0
i.e., λ = \(\frac{\left(a x_{1}+b y_{1}+c\right.}{\left(a^{2}+b^{2}\right)}\)
∴ (h – x₁) ; a = (k – y₁); b
= – (ax₁ + by₁ + c); (a² + b²)

Question 30.
Find the distance between the parallel straight lines 3x + 4y – 3 = 0 and 6x + 8y -1 = 0.
Solution:
The equation of the given straight lines is 3x + 4y – 3 = 0 and 6x + 8y – 1 = 0.
formula : \(\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\) = \(\frac{-6+1}{\sqrt{6^{2}+8^{2}}}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\)

Question 31.
Find the condition for the points (a, o) (h, k) and (o, b) when ab ≠ 0 to be collinear. [Mar 10]
Solution:
The equation of the line passing through (a, o) and (o, b) is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
The given points are collinear ⇒ (h, k) lies on the above line ⇒ \(\frac{h}{a}\) + \(\frac{k}{b}\) = 1
⇒ hb + ka = ab

Question 32.
Find the area of the triangle formed by the straight lines x cos a + y sin a = p and the axes of co-ordinates. [Mar 10]
Solution:
The area of the triangle formed by the line ax + by + c = 0
and the co-ordinate axes is \(\frac{c^{2}}{2|a b|}\)
∴ Area of the triangle = \(\frac{p^{2}}{2|\cos \alpha \cdot \sin \alpha|}\)
= \(\frac{p^{2}}{|\sin 2 \alpha|}\)

Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1A Trigonometric Ratios up to Transformations Important Questions

Question 1.
Find the values of
(i) sin \(\frac{5 \pi}{3}\)
(ii) tan (855°)
(iii) sec \(\left(\frac{13 \pi}{3}\right)\)
Solution:
i) sin \(\frac{5 \pi}{3}\) = sin \(\left(2 \pi-\frac{\pi}{3}\right)\) = -sin \(\frac{\pi}{3}\) = \(-\frac{\sqrt{3}}{2}\)

ii) tan (855°) = tan (2 × 360° + 135°)
= tan (135°)
= tan (180° – 45°)
= -tan 45° = -1

iii) sec \(\left(13 \frac{\pi}{3}\right)\) = sec \(\left(4 \pi+\frac{\pi}{3}\right)\)
= sec \(\frac{\pi}{3}\) = 2

Question 2.
Simplify.
i) Cot (θ – \(\frac{13 \pi}{2}\))
ii) tan \(\left(-23 \frac{\pi}{3}\right)\)
Solution:

Question 3.
Find the value of sin² \(\frac{\pi}{10}\) + sin² \(\frac{4 \pi}{10}\) + sin² \(\frac{6 \pi}{10}\) + sin² \(\frac{9 \pi}{10}\)
Solution:

Question 4.
If sin θ = \(\frac{4}{5}\) and θ is not in the first qua-drant, find the value of cos θ.
Solution:
∵ sin θ = \(\frac{4}{5}\) and θ is not in the first quadrant
⇒ θ lies in 2^nd quadrant, ∵ sin θ is +ve

Question 5.
If sec θ + tan θ = \(\frac{2}{3}\), find the value of sin θ and determine the quadrant in which θ lies.
Solution:
∵ sec θ + tan θ = \(\frac{2}{3}\)

∵ tan θ is negative, sec θ is positive
⇒ θ lies in fourth quadrant.

Question 6.
Prove that
cot\(\frac{\pi}{16}\).cot\(\frac{2 \pi}{16}\).cot\(\frac{3 \pi}{16}\)…..cot\(\frac{7 \pi}{16}\) = 1
Solution:

Question 7.
If 3 sin θ + 4 cos θ = 5, then find the value of 4 sin θ – 3 cos θ.
Solution:
Given 3 sin θ + 4 cos θ = 5 and let 4 sin θ – 3 cos θ = x
By squaring and adding, we get
(3 sin θ + 4 cos θ)² + (4 sin θ – 3 cos θ)² = 5² + x²
⇒ 9 sin² θ + 16 cos² θ + 24 sin θ cos θ + 16 sin² θ + 9 cos² θ – 24 sin θ cos θ = 25 + x²
⇒ 9 + 16 = 25 + x²
⇒ x² = 0 ⇒ x = 0
∴ 4 sin θ – 3 cos θ = 0.

Question 8.
If cos θ + sin θ = \(\sqrt{2}\) cos θ, prove that cos θ – sin θ = \(\sqrt{2}\) sin θ. (May ’11)
Solution:
Given cos θ + sin θ = \(\sqrt{2}\) cos θ
⇒ sin θ = (\(\sqrt{2}\) – 1) cos θ
Now multiplying both sides by (\(\sqrt{2}\) + 1)

Question 9.
Find the value of 2(sin² θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ)
Solution:
2 (sin⁶ θ + cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ)
= 2[(sin² θ)³ + (cos² θ)³] – 3[(sin² θ)² + (cos² θ)²]
= 2[(sin² θ + cos² θ)³ – 3 sin² θ cos² θ (sin² θ + cos² θ)] – 3[(sin² θ + cos² θ)² – 2 sin² θ cos² θ]
= 2[1 – 3 sin² θ cos² θ] – 3[1 – 2 sin² θ cos² θ]
= 2 – 6 sin² θ cos² θ – 3 + 6 sin² θ cos² θ
= -1

Question 10.
Prove that (tan θ + cot θ)² = sec² θ + cosec² θ = sec² θ cosec² θ.
Solution:
(tan θ + cot θ)²
=tan² θ + cot ² θ + 2 tan θ cot θ
= tan² θ + cot² θ + 2
= (1 + tan² θ) + (1 + cot² θ)
= sec² θ + cosec² θ

Question 11.
If cos θ > 0, tan θ + sin θ = m and tan θ – sin θ = n, then show that m² – n² = n, then show that m² – n² = 4\(\sqrt{m n}\)
Solution:
Given that m = tan θ + sin θ
n = tan θ – sin θ
⇒ m + n = 2 tan θ and m – n = 2 sin θ
Now(m + n)(m – n) = 4 tan θ sin θ

Question 12.
If tan 20° = λ, then show that \(\frac{\tan 160^{\circ}-\tan 110^{\circ}}{1+\tan 160^{\circ} \cdot \tan 110^{\circ}}\) = \(\frac{1-\lambda^{2}}{2 \lambda}\) (A.P.) Mar. ’16
Solution:
Given tan 20° = λ

Question 13.
Find the values of sin 75°, cos 75°, tan 75° and cot 75°.
Solution:
i) sin 75° = sin (45° + 30°)
= sin 45°. cos 30° + cos 45°. sin 30°

ii) cos (75°) = cos (45° + 30°)
= cos 45° cos 30° – sin 45° sin 30°

Question 14.
If 0 < A, B < 90°.cos A = \(\frac{5}{13}\) and sin B = \(\frac{4}{5}\)-then find sin(A + B).
Solution:
0 < A < 90°and cos A = \(\frac{5}{13}\) ⇒ sin A = \(\frac{12}{13}\)
0 < B < 90°and sin B = \(\frac{4}{5}\) ⇒ cos B = \(\frac{3}{5}\)
∴ sin (A + B) = sin A cos B + cos A sin B
= \(\frac{12}{13}\) • \(\frac{3}{5}\) + \(\frac{5}{13}\) • \(\frac{4}{5}\) = \(\frac{56}{65}\)

Question 15.
Prove that
sin² \(\left(52 \frac{1}{2}\right)^{\circ}\) – sin² \(\left(22 \frac{1}{2}\right)^{\circ}\) = \(\frac{\sqrt{3}+1}{4 \sqrt{2}}\)
Solution:

Question 16.
Prove that tan 70° – tan 20° = 2 tan 50°.
Solution:
tan 50° = tan (70° – 20°)
= \(\frac{\tan 70^{\circ}-\tan 20^{\circ}}{1+\tan 20^{\circ} \cdot \tan 70^{\circ}}\)
⇒ tan 70° – tan 20°
= tan 50° [1 + tan 20° . tan (90° – 20°)]
= tan 70° – tan 20° = tan 50°[1 + tan 20° cot 20°]
⇒ tan 70° – tan 20° = 2 tan 50°.

Question 17.
If A + B = π/4, then prove that
i)(1 + tan A)(1 + tan B) = 2. (May ’11; Mar. ’07)
ii) (cot A – 1)(cot B – 1) = 2
Solution:
i) Given that A + B = π/4 (T.S) (Mar. ’16)
⇒ tan (A + B) = tan (π/4)
⇒ \(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = 1
⇒ tan A + tan B = 1 – tan A tan B
⇒ tan A + tan B + tan A tan B = 1
Add 1 on both sides
⇒ 1 + tan A + tan B + tan A tan B = 1 + 1
⇒ (1 + tan A)(1 + tan B) = 2

ii) Given A + B = π/4
⇒ cot (A + B) = cot π/4
⇒ \(\frac{\cot A \cot B-1}{\cot B+\cot A}\) = 1
⇒ cot A cot B – 1 = çot B + cot A
⇒ cot A cot B – cot A – cot B = 1
Again add 1 on both sides
cot A cot B – cot A – cot B + 1 = 1 + 1
∴ (cot A – 1) (cot B – 1) = 2.

Question 18.
If sin α = \(\frac{1}{\sqrt{10}}\), sin β = \(\frac{1}{\sqrt{5}}\) and α, β are acute, show that α + β = π/4.
Solution:
Given α is acute and sin α = \(\frac{1}{\sqrt{10}}\)
⇒ tan α = \(\frac{1}{3}\)
β is acute and sin β = \(\frac{1}{\sqrt{5}}\) ⇒ tan β = \(\frac{1}{2} .\)

Question 19.
If sin A = \(\frac{12}{13}\), cos B = \(\frac{3}{5}\) and neither A nor B is in the first quadrant, then find the quadrant in which A + B lies.
Solution:
sin A = \(\frac{12}{13}\) and A is not in first quadrant
⇒ A lies in second quadrant, ∵ sin A is +ve
cos B = \(\frac{3}{5}\) and B is not in first quadrant
⇒ B lies in fourth quadrant, cos B is +ve
∵ sin A = \(\frac{12}{13}\) ⇒ cos A = –\(\frac{5}{13}\)
cos B = \(\frac{3}{5}\) ⇒ sin B = –\(\frac{4}{5}\)
sin (A + B) = sin A cos B + cos A sin B

cos (A + B) = cos A cos B – sin A sin B

∵ sin (A + B) and cos (A + B) are positive
⇒ (A + B) lies in first quadrant.

Question 20.
Find (i) tan \(\left(\frac{\pi}{4}+\mathbf{A}\right)\) interms of tan A
and (ii) cot \(\left(\frac{\pi}{4}+\mathbf{A}\right)\) interms of cot A.
Solution:

Question 21.
Prove that \(\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}\) = cot 36°. (A.P) (Mar ’15, Mar. ’11)
Solution:

Question 22.
Show that
cos 42° + cos 78° + cos 162° = 0 (May. ’11)
Solution:
L.H.S. = cos 42° + cos 78° + cos 162°
= 2 cos \(\left(\frac{42^{\circ}+78^{\circ}}{2}\right)\). cos \(\left(\frac{42^{\circ}-78^{\circ}}{2}\right)\) + cos (180° – 18°)
= 2 cos 60° . cos (-18°) + cos (180° – 18°)
= 2\(\left(\frac{1}{2}\right)\) cos 18° – cos 18° = 0
∴ cos 42° + cos 78° + cos 162° = 0

Question 23.
Express \(\sqrt{3}\) sin θ + cos θ as a sine of an angle.
Solution:
\(\sqrt{3}\) sin θ + cos θ = 2(\(\frac{\sqrt{3}}{2}\) sin θ + cos θ)
= 2(cos \(\frac{\pi}{6}\) sin θ + sin \(\frac{\pi}{6}\) cos θ)
= 2. sin[θ + \(\frac{\pi}{6}\)]

Question 24.
Prove that sin² θ + sin² [θ – \(\frac{\pi}{3}\)] + sin² [θ – \(\frac{\pi}{3}\)] = \(\frac{3}{2} .\)
Solution:

Question 25.
If A, B, C are angles of a triangle and if none of them is equal to \(\frac{\pi}{2}\), then prove that
i) tan A + tan B + tan C = tan A tan B tan C
ii) cot A cot B + cot B cot C + cot C cot A = 1
Solution:
i) ∵ A, B, C are angles of a triangle
⇒ A + B + C = 180°
⇒ A + B = 180° – C
⇒ tan (A + B) = tan (180° – C)
⇒ \(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = -tan C
⇒ tan A + tan B = -tan C + tan A tan Btan C
⇒ tan A + tan B + tan C = tan A tan B tan C

ii) ∵ A + B + C = 180°
⇒ A + B = 180° – C
⇒ cot (A + B) = cot (180° – C)
⇒ \(\frac{\cot A \cot B-1}{\cot B+\cot A}\) = -cot C
⇒ cot A cot B – 1 = -cot B cot C – cot C. cot A
⇒ cot A cot B + cot B cot C + cot C cot A = 1

Question 26.
Let ABC be a triangle such that cot A + cot B + cot C = \(\sqrt{3}\). Then prove that ABC is an equilateral triangle.
Solution:
Given that A + B + C = 180°
Then we know that
cot A cot B + cot B cot C + cot C cot A = 1
i.e., Σ(cot A cot B) = 1 —– (1)
Now Σ (cot A – cot B)²
= Σ (cot² A + cot² B – 2 cot A cot B)
= 2 cot² A + 2 cot² B + 2 cot² C – 2 cot A cot B – 2 cot B cot C – 2 cot C cot A
= 2 [cot A + cot B + cot C]² – 6 (cot A cot B + cot B cot C + cot C cot A)
= 2
= 6 – 6 = 0
∴ Σ (cot A – cot B)² = 0 .
⇒ cot A = cot B = cot C
⇒ cot A = cot B = cot C = \(\frac{\sqrt{3}}{3}\) = \(\frac{1}{\sqrt{3}}\)
(∵ cot A + cot B + cot C = \(\sqrt{3})\))
⇒ A = B = C = 60°
∴ ΔABC is an equilateral triangle.

Question 27.
Suppose x = tan A, y = tan B, z = tan C. Suppose none of A, B, C, A – B, B – C, C – A is an odd multiple of \(\frac{\pi}{2}\).Then prove that \(\sum\left(\frac{x-y}{1+x y}\right)\) = \(\pi\left(\frac{x-y}{1+x y}\right)\)
Solution:
∵ x = tan A, y = tan B, z = tan C

Write P = A – B, Q = B – C, R = C – A.
Then P + Q + R = O
⇒ tan (P + Q) = tan (-R)
⇒ \(\frac{\tan P+\tan Q}{1-\tan P \tan Q}\) = -tan R
⇒ tan P + tan Q = -tan R + tan P tan Q tan R
⇒ tan P + tan Q + tan R = tan P tan Q tan R
⇒ Σ(tan P) = π (tan P)
⇒ Σtan (A – B) = π tan (A – B)
∴ Σ\(\left(\frac{x-y}{1+x y}\right)\) = π\(\left(\frac{x-y}{1+x y}\right)\)

Question 28.
Find the values of
i) sin 22\(\frac{1}{2}\)°
ii) cos 22\(\frac{1}{2}\)°
iii) tan 22\(\frac{1}{2}\)°
iv) cot 22\(\frac{1}{2}\)°
Solution:

Question 29.
Find the values of
i) sin 67\(\frac{1}{2}\)°
ii) cos 67\(\frac{1}{2}\)°
iii) tan 67\(\frac{1}{2}\)°
iv) cot 67\(\frac{1}{2}\)°
Solution:
Let A = 67\(\frac{1}{2}\)° ⇒ = 2A = 135°

Question 30.
Simplify: \(\frac{1-\cos 2 \theta}{\sin 2 \theta}\)
Solution:
\(\frac{1-\cos 2 \theta}{\sin 2 \theta}\) = \(\frac{2 \sin ^{2} \theta}{2 \sin \theta \cos \theta}\) = \(\frac{\sin \theta}{\cos \theta}\) = tan θ

Question 31.
If cos A = \(\sqrt{\frac{\sqrt{2}+1}{2 \sqrt{2}}}\), find the value of cos 2A.
Solution:
cos 2A = 2 cos²A – 1 = 2 \(\left(\frac{\sqrt{2}+1}{2 \sqrt{2}}\right)\) – 1
= \(\frac{\sqrt{2}+1}{\sqrt{2}}\) – 1 = \(\frac{1}{\sqrt{2}}\)

Question 32.
If cos θ = \(\frac{-5}{13}\) and \(\frac{\pi}{2}\) < θ < π, find the value of sin 2θ.
Solution:
\(\frac{\pi}{2}\) < θ < π ⇒ sin θ > 0 and cos θ = –\(\frac{5}{13}\)
⇒ Sin θ = \(\frac{12}{13}\)
∴ sin 2θ = 2 sin θ cos θ
= 2 . \(\frac{12}{13}\)(-\(\frac{5}{13}\)) = –\(\frac{120}{169}\)

Question 33.
For what values of x in the first quadrant \(\frac{2 \tan x}{1-\tan ^{2} x}\) is positive?
Solution:
\(\frac{2 \tan x}{1-\tan ^{2} x}\) > 0 ⇒ tan 2x > 0
⇒ 0 < 2x < \(\frac{\pi}{2}\) (since x is in the first quadrant)
⇒ 0 < x < \(\frac{\pi}{4}\)

Question 34.
If cos θ = \(\frac{-3}{5}\) and π < θ < \(\frac{3 \pi}{2}\), find the value of tan θ/2.
Solution:

Question 35.
If A is not an integral multiple of \(\frac{\pi}{2}\), prove that
i) tan A + cot A = 2 cosec 2A
ii) cot A – tan A = 2 cot 2A
Solution:
i) tan A + cot A = \(\frac{\sin A}{\cos A}\) + \(\frac{\cos A}{\sin A}\)
= \(\frac{\sin ^{2} A+\cos ^{2} A}{\sin A \cos A}\)

Question 36.
If A is not an integral multiple of prove that
i) tan A + cot A = 2 cosec 2A
ii) cot A – tan A = 2 cot 2A
Solution:

Question 37.
If θ is not an integral multiple of \(\frac{\pi}{2}\), prove that tan θ + 2 tan 2θ + 4 tan 4θ +
8 cot 8θ = cot θ.
Solution:
From cot A – tan A = 2 cot 2A above tan A = cot A – 2 cot 2A ….(1)
∴ tan θ + 2 tan 2θ + 4 tan 4θ + 8 cot 8θ = (cot θ – 2 cot 2θ) + 2 (cot 2θ – 2 cot 4θ) + 4 (cot 4θ – 2 cot 8θ) + 8 cot 8θ (by (1) above)
= cot θ

Question 38.
For A ∈ R, prove that
i) sin A.sin(π/3 +A).sin(π/3 – A) = \(\frac{1}{4}\) . sin3A
ii) cosA . cos (π/3 + A). cos(π/3 – A)
= \(\frac{1}{4}\) cos 3A and hence deduce that
iii) sin 20° sin 40° sin 60° sin 80° = \(\frac{3}{16}\)
iv) cos \(\frac{\pi}{9}\), cos \(\frac{2 \pi}{9}\), cos \(\frac{3 \pi}{9}\). cos \(\frac{4 \pi}{9}\) = \(\frac{1}{16}\).
Solution:
i) sin A. sin (π/3 + A). sin (π/3 – A)
= sin A [sin² π/3 – sin² A]


iii) ∵ sin A. sin (60° + A). sin (60° – A)
= \(\frac{1}{4}\) sin 3A
Put A = 20°
⇒ sin 20°. sin (60° + 20°). sin (60° – 20°)
= \(\frac{1}{4}\). sin(3 × 20°) .
⇒ sin 20°. sin 40°. sin 80° = \(\frac{1}{4}\) sin 60°
Multiplying on both sides with sin 60°
We get, sin 20° sin 40° sin 60° sin 80°
= \(\frac{1}{4}\) sin² 60°
= \(\frac{1}{4}\left(\frac{\sqrt{3}}{2}\right)^{2}[latex] = [latex]\frac{3}{16}\)

iv) ∵ cos A. cos (60° + A). cos (60° – A) = \(\frac{1}{4}\)cos 3A
Put A = 20°
⇒ cos 20°. cos (60° + 20°) cos (60° – 20°)
= \(\frac{1}{4}\). cos(3 × 20°)
⇒ cos 20°. cos 40°. cos 80° = cos 60°
On multiplying both sides by cos 60°, we get cos 20°. cos 40°. cos 60°. cos 80°
= \(\frac{1}{4}\). cos² 60°

Question 39.
If 3A is not an odd multiple of \(\frac{\pi}{2}\), prove that tan A. tan (60° + A). tan (60° – A)
= tan 3A and hence find the value of tan 6°. tan 42°. tan 66°. tan 78°.
Solution:

∴ tan A . tan \(\left(\frac{\pi}{3}+\mathrm{A}\right)\) tan \(\left(\frac{\pi}{3}-A\right)\) = tan 3A
i.e., tan A tan (60° – A) tan (60° + A) = tan 3A —— (1)
Put A = 6°
⇒ tan 6° tan 54° tan 66° = tan 18° ——- (2)
Put A = 18° in (1)
tan 18° tan 42° tan 78° = tan 54° ——- (3)
put (2) in (3),
(tan 6° tan 54° tan 66°) tan 42° tan 78°
= tan 54
⇒ tan 6° tan 42° tan 66° tan 78° = 1

Question 40.
For α, β ∈ R, prove that (cos α + cos β)² + (sin α + sin β)² = 4 cos² \(\left(\frac{\alpha-\beta}{2}\right)\).
Solution:
LH.S. = (cos α + cos β)² + (sin α + sin β)²
= (cos² α + cos² β + 2 cos α cos β + (sin² α + sin² β + 2 sin α sin β)
= 1 + 1 + 2 (cos α cos β + sin α sin β)
= 2 + 2 cos (α – β) = 2 (1 + cos (α – β)]
= 2[2cos²\(\left(\frac{\alpha-\beta}{2}\right)\)]
= 4.cos²\(\left(\frac{\alpha-\beta}{2}\right)\); [∵ 1 + cos θ = 2 cos² \(\frac{\theta}{2}\)]

Question 43.
If a, b, c are non zero real numbers and α, β are solutions of the equation a cos θ + b sin θ = c then show that
(i) sin α + sin β = \(\frac{2 b c}{a^{2}+b^{2}}\) and
(ii) sin α. sin β = \(\frac{c^{2}-a^{2}}{a^{2}+b^{2}}\)
Solution:
∵ a cos θ + b sin θ = c
= a cos θ = c – b sin θ
⇒ (a cos θ)² = (c – b sin θ)²
⇒ a² cos² θ = c² + b² sin² θ – 2bc sinθ
⇒ a² (1 – sin² θ) = c² + b² sin² θ – 2bc sin θ
⇒ (a² + b²) sin² θ – 2bc sin θ + (c² – a²) = 0
This is a quadratic equatioñ in sin θ, whose roots are sin α and sin β (given that α, β are two solutions of θ)
∴ Sum of the roots sin α + sin β = \(\frac{2 b c}{a^{2}+b^{2}}\)
Product of the roots sin α sin β = \(\frac{c^{2}-a^{2}}{a^{2}+b^{2}}\)

Question 42.
If θ is not an odd multiple of \(\frac{\pi}{2}\) and cos θ ≠ \(\frac{-1}{2}\), prove that
\(\frac{\sin \theta+\sin 2 \theta}{1+\cos \theta+\cos 2 \theta}\) = tan θ.
Solution:

Question 43.
Prove that sin⁴ \(\frac{\pi}{8}\) + sin⁴ \(\frac{3 \pi}{8}\) + sin⁴ \(\frac{5 \pi}{8}\) + sin⁴ \(\frac{7 \pi}{8}\) = \(\frac{3}{2}\)
Solution:

Question 44.
If none of 2A and 3A is an odd multiple of \(\frac{\pi}{2}\), then prove that tan 3A. tan 2A. tanA = tan 3A – tan 2A – tan A.
Solution:
∵ tan 3A = tan (2A + A)
= \(\frac{\tan 2 A+\tan A}{1-\tan 2 A \tan A}\)
⇒ tan 3A(1 – tan 2A tan A) = tan 2A + tan A
⇒ tan 3A – tan A tan 2A tan 3A
= tan 2A + tan A
∴ tan 3A – tan 2A – tan A
= tan A tan 2A tan 3A

Question 45.
Prove that sin 78° + cos 132° = \(\frac{\sqrt{5}-1}{4}\).
Solution:
L.H.S. = sin 78° + cos 132°
= sin 78° + cos (90° + 42°)
= sin 78° – sin 42°

Question 46.
Prove that sin 21° cos 9° – cos 84° cos 6° = \(\frac{1}{4}\).
Solution:
LH.S. = sin 21° cos 9° – cos 84° cos 6°
= \(\frac{1}{2}\)[2 sin 21° cos 9° – 2 cos 84° cos 6°]
= \(\frac{1}{2}\)[sin (21° + 9°) + sin (21° – 9°) – 2 cos (90° – 6°) cos 6°]
= \(\frac{1}{2}\)[sin 30° + sin 12° – 2 sin 6° cos 6°]
= \(\frac{1}{2}\)[\(\frac{1}{2}\) + sin 12° – sin (2 × 16°)
= \(\frac{1}{4}\) = R.H.S.

Question 47.
Find the value of sin 34° + cos 64° – cos 4°.
Solution:
sin 34° + (cos 64° – cos 4°)

Question 48.
Prove that cos² 76° + cos² 16° – cos 76° cos 16° = \(\frac{3}{4} .\)
Solution:
L.H.S. = cos² 76° + cos² 16° – cos 76° cos 16°
= cos² 76° + (1 – sin² 16°) – \(\frac{1}{2}\)(2 cos 76° cos 16°)
= 1 + (cos² 76° – sin² 16°) – \(\frac{1}{2}\)[cos (76° + 16°) + cos (76° – 16°)]
= 1 + cos (76° + 16°). cos (76° – 16°) – \(\frac{1}{2}\)[cos 92° + cos 60°]
= 1 + cos (92°). cos 60° – \(\frac{1}{2}\)cos 92° – \(\frac{1}{2}\)cos 60°
= 1 + \(\frac{1}{2}\). cos 92° – \(\frac{1}{2}\). cos 92° – \(\frac{1}{2}\) × \(\frac{1}{2}\)
= 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\) = R.H.S.

Question 49.
If a, b, ≠ 0 and sin x + sin y = a and cos x + cos y = b, find two values of
i) tan \(\left(\frac{x+y}{2}\right)\)
ii) sin \(\left(\frac{x-y}{.2}\right)\) interms of a and b.
Solution:

First method:

ii) a² + b² = (sin x + sin y)² + (cos x + cos y)²
= sin² x + sin² y + 2 sin x sin y + cos² x + cos² y + 2 cos x cos y
= 2 + 2(cos x cos y + sin x sin y)
= 2 + 2 cos (x – y)
= 2[1 + cos (x – y)]
a² + b² – 2 = 2 cos (x – y)

Second method:

Question 50.
Prove that cos 12° + cos 84° + cos 132° + cos 156° = –\(\frac{1}{2}\).
Solution:
L.H.S. = cos 12° + cos 84° + cos 132° + cos 156°

Question 51.
Show that for any θ ∈ R
4 sin \(\frac{5 \theta}{2}\) cos \(\frac{3 \theta}{2}\) cos 3θ = sin θ – sin 2θ + sin 4θ + sin 7θ
Solution:
R.H.S. = 4 sin \(\frac{5 \theta}{2}\) cos \(\frac{3 \theta}{2}\) cos 3θ

= 2 [(sin 4θ + sin θ) cos 3θ]
= 2 sin 4θ cos 3θ + 2 sinθ cos 3θ
= sin (4θ + 3θ) + sin (4θ – 3θ) + sin (θ + 3θ) + sin(θ – 3θ)
= sin 7θ + sin θ + sin 4θ – sin 2θ = R.H.S.

Question 52.
If none of A, B, A + B is an integral multiple of π, then prove that

Solution:

Question 53.
For any α ∈ R, provethat cos² (α – π/4) + cos² (α + π/12) – cos² (α – π/12) = \(\frac{1}{2}\).
Solution:

Question 54.
Suppose (α – β) is not an odd multiple of \(\frac{\pi}{2}\), m is a non – zero real number such that m ≠ -1 and \(\frac{\sin (\alpha+\beta)}{\cos (\alpha-\beta)}\) = \(\frac{1-m}{1+m}\), Then prove that tan \(\left(\frac{\pi}{4}-\alpha\right)\) = m. tan \(\left(\frac{\pi}{4}+\beta\right)\)
Solution:

Question 55.
If A, B, C are the angles of a triangle, prove that
i) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
ii) sin 2A + sin 2B – sin 2C = 4 cos A cos B sin C
Solution:
i) ∵ A, B, C are angles of a triangle
⇒ A + B + C = 180° ——- (1)
L.H.S. = sin 2A + sin 2B + sin 2C

= 2 sin (A + B) cos (A – B) + sin 2C
= 2 sin (180° – C) cos (A – B) + sin 2C
By(1)
= 2 sin C. cos (A — B) + 2 sin C. cos C
= 2 sin C [cos (A — B) + cos C]
= 2 sin C [cos (A — B) + cos (180° — \(\overline{A+B}\))]
= 2 sin C [cos (A – B) — cos (A + B)]
= 2 sin C (2 sin A sin B)
= 4 sin A sin B sin C = R.H.S.

ii) LH.S. = sin 2A + sin 28 — sin 2C

= 2 sin(A+ B)cos(A—B)—sin2C
= 2 sin (180° — C) cos (A — B) — sin 2C
= 2 sin C. cos(A – B) – 2 sin C cos C
= 2 sin C [cos (A – B) – cos C]
= 2 sin C [cos (A — B)—cos (180°— \(\overline{A+B}\))]
By(1)
= 2 sin C [cos (A — B) + cos (A + B)]
= 2 sin C [2 cos A cas B]
= 4 cos A cos B sin C = R.H.S.

Question 56.
If A, B, C are angles of a triangle, prove that
i) cos 2A + cos 2B + cos 2C
= -4 cos A cos B cos C – 1

ii) cos 2A + cos 2B – cos 2C
= 1 – 4 sin A sin B cos C
Solution:
i) ∵ A, B, C are angles of a triangle
⇒ A + B + C = 180° ———— (1)
L.H.S. = (cos 2A + cos 2B) + cos 2C
= 2 cos(A + B) cos (A – B) + 2 cos² C – 1
= -2 cos C cos (A – B) – 2 cos C cos (A + B) – 1
[∵ cos (A + B) = -cos C]
= – 1 – 2 cos C[cos (A – B) + cos(A – B)]
= – 1 – 2 cos C[2 cos A cos B]
= -1 – 4 cos A cos B cos C = R.H.S.

ii) L.H.S. = cos 2A + cos 2B – cos 2C

= 2 cos (A + B). cos (A – B) – (2 cos² C – 1)
= 2 cos (180° – C) cos (A – B) – 2 cos² C + 1
= 1 – 2 cos C cos (A – B) – 2 cos² C
= 1 – 2 cos C [cos (A – B) + cos C]
= 1 – 2 cos C[cos(A – B) + cos(180° – \(\overline{\mathrm{A}+\mathrm{B}}\))]
= 1 – 2 cos C [cos (A – B) – cos (A + B)]
= 1 – 2 cos C [2 sin A sin B]
= 1 – 4 sin A sin B cos C = R.H.S.

Question 57.
If A B, C are angles in a triangle, then prove that
i) sin A + sin B + sin C = 4 cos \(\frac{A}{2}\) cos \(\frac{B}{2}\) cos \(\frac{C}{2}\)
ii) cos A + cos B + cos C = 1 + 4 sin \(\frac{A}{2}\) sin \(\frac{B}{2}\) sin \(\frac{C}{2}\)
Solution:

Question 58.
If A + B + C = π/2, then show that
i) sin² A + sin² B + sin² C = 1 – 2 sin A sin B sin C
ii) sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C
Solution:
A + B + C = π/2 ——– (1)
L.H.S. = sin² A + sin² B + sin² C
= \(\frac{1}{2}\)(1 – cos2A + 1 – cos 2B + 1 – cos 2C]
= \(\frac{1}{2}\)[3 – (cos 2A + cos 2B + cos 2C)]
= \(\frac{1}{2}\)[3 – (1 + 4 sin A sin B sin C)
(By Problem No. 57(ii)]
(∵ 2A + 2B + 2C = 180°)
= \(\frac{1}{2}\)[2 – 4 sin A sin B sin C]
= 1 – 2 sin A sin B sin C

ii) A + B + C = 90°
⇒ 2A + 2B + 2C = 180°
2A + 2B + 2C 180°
sin 2A + sin 2B + sin 2C
= 4 cos A cos B cos C
By Problem No. 57(i).

Question 59.
If A + B + C = 3π/2, prove that cos 2A + cos 2B + cos 2C = 1 – 4 sin A sin B sin C.
Solution:
A + B + C = 3π/2 —— (1)
L.H.S. = cos 2A + cos 2B + cos 2C
= 2 cos (A + B). cos(A – B) +1 – 2 sin² C
= 2 cos (270° – C). cos (A – B) + 1 – 2 sin² C
= 1 – 2 sin C cos (A – B) – 2 sin²C
= 1 – 2 sin C [cos(A – B) + sin C]
= 1 – 2 sin C[cos (A – B) + sin (270° – \(\overline{A+B}\))]
= 1 – 2 sin C[cos(A – B) – cos(A + B)]
= 1 – 2 sin C[2 sin A sin B]
= 1 – 4 sin A sin B sin C

Question 60.
If A B, C are angles of a triangle, then prove that
sin² \(\frac{A}{2}\) + sin² \(\frac{B}{2}\) – sin² \(\frac{C}{2}\) = 1 – 2 cos \(\frac{A}{2}\) cos \(\frac{B}{2}\) sin \(\frac{C}{2}\) (Mar. ’16, May ’12, ’11)
Solution:

Question 61
If A. B, C are angles of a triangle, then prove that sin \(\frac{\mathbf{A}}{\mathbf{2}}\) + sin \(\frac{\mathbf{B}}{\mathbf{2}}\) + sin \(\frac{\mathbf{C}}{\mathbf{2}}\) = 1 + 4 sin \(\frac{\pi-\mathbf{A}}{4}\) sin \(\frac{\pi-B}{4}\) sin \(\frac{\mathrm{C}}{2}\)
Solution:
A + B + C = 180° ——- (1)

Question 62.
If A + B + C = 0, then prove that cos² A + cos² B + cos² C = 1 + 2 cos A cos B cos C.
Solution:
A + B + C = 0 —— (1)
L.H.S. = cos² A + cos² B + cos² C
= cos² A + (1 – sin² B) + cos² C
= 1 + (cos² A – sin² B) + cos² C
= 1 + cos (A + B) cos (A – B) + cos² C
= 1 + cos(-C) cos(A – B) + cos² C
By (1)
= 1 + cos C cos (A – B) + cos² C
= 1 + cos C [cos (A – B) + cos c]
= 1 + cos C[cos(A – B) + cos(-B – A)]
By (1)
1 + cos C[cos (A – B) + cos(A + B)]
= 1 + cos C (2 cos A cos B)
= 1 + 2 cos A cos B cos C = R.H.S.

Question 63.
If A + B + C = 25, then prove that cos (S — A) + cos (S — B) + cos (S — C) + cos (S) = 4 cos \(\frac{A}{2}\) cos \(\frac{B}{2}\) cos \(\frac{C}{2}\).
Solution:

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B Differentiation Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1B Differentiation Important Questions

Question 1.
If f(x) = x² (x ∈ R), prove that f is differentiable on R and find its derivative.
Solution:
Given that f(x) = x²
for x, h ∈ R, f(x + h) – f(x)(x + h)² – x²
= x² + h² + 2hx – x²
= 2hx + h² = h(2x + h)

∴ f is differentiable on R and f'(x) = 2x for each x ∈ R

Question 2.
Suppose f(x) = \(\sqrt{x}\) (x > 0). Prove that f is differentiable on (0, ∞) and find f(x).
Solution:
Let x ∈ (0, ∞) h ≠ 0 and |h| < 0

Question 3.
If f(x) = \(\frac{1}{x^{2}+1}\) (x ∈ R), prove that f is differentiable on R and find f'(x).
Solution:
Let x ∈ R and h ≠ 0

= –\(\frac{2 x}{\left(x^{2}+1\right)^{2}}\)
∴ f is differentiable and f'(x) = –\(\frac{2 x}{\left(x^{2}+1\right)^{2}}\) for each x ∈ R .

Question 4.
If f(x) = sin x (x ∈ R), then show that f is differentiable on R and f'(x) = cosx.
Solution:
Let x ∈ R and h ≠ 0

∴ f is differentiable on R and f'(x) = cos x for each x ∈ R.

Question 5.
Show that f(x) = |x| (x ∈ R) is not differentiable at zero and is differentiable at any x ≠ 0.
Solution:
Given f(x) = |x|
∴ f(x) = x if x ≥ 0
if h ≠ 0
\(\frac{f(0+h)-f(0)}{h}\) = \(\frac{f(h)}{h}\) = \(\left\{\begin{array}{r}
1 \text { if } h>0 \\
-1 \text { if } h<0
\end{array}\right.\)
f'(0⁺) = 1, f'(0^–) = -1
∴ is not differentiable at zero it can be easily proved that f is differentiable at any x ≠ 0 and that f'(x) = \(\left\{\begin{array}{l}
1 \quad \text { if } x>0 \\
-1 \text { if } x<0
\end{array}\right.\)

Question 6.
Check whether the following function is differentiable at zero f(x) = \(\left\{\begin{array}{l}
3+x \text { if } x \geq 0 \\
3-x \text { if } x<0
\end{array}\right.\)
Solution:

f has the left hand derivative at zero and f'(0^–) = -1
∴ f'(0⁺) ≠ f'(0^–)
f(x) is not differentiable at zero.

Question 7.
Show that the derivative of a constant function on an interval is zero.
Solution:
let f be a constant function on an interval I.
f(x) = C ∀ x ∈ I for some constant.
Let a ∈ I, for h ≠ 0 \(\frac{f(a+h)-f(a)}{h}\) = \(\frac{c-c}{h}\) = 0
for sufficiently small (h)

∴ f is differentiable 0 and f'(0).

Question 8.
Suppose for all x, y ∈ R f(x + y) = f(x). f(y) and f'(0) exists. Then show that f(x) exists and equals to f(x) f'(0)for all x ∈ R.
Solution:
Let x ∈ R, for h ≠ 0, we have
\(\frac{f(x+h)-f(x)}{h}\) = \(\frac{f(x) f(h)-f(x)}{h}\)
= f(x) \(\frac{[f(h)-1]}{h}\) ………………….. (1)
f(0) = f(0 + 0) = f(0) f(0) ⇒ f(0) (1 —f(0)) = 0
∴ f(0) = 0 or f(0) = 1
Case (1) : Suppose f(0) = 0
f(x) = f(x + 0) = f(x) f(0) = 0 ∀ x ∈ R
∴ f(x) is a constant function = f'(x) = 0 for all x ∈ R
∴ f'(x) = 0 = f(x) . f'(0)

Case (ii): Suppose f(0) = 1

∴ f is differentiable and f'(x) = f'(x) f'(0).

Question 9.
If f(x) = (ax + b)ⁿ, (x > –\(\frac{b}{a}\)), then find f'(x).
Sol:
Let u = ax + b so that y = uⁿ
f'(x) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (uⁿ) \(\frac{\mathrm{du}}{\mathrm{dx}}\)
= n.u^n-1a.
= an(ax+b)^{n – 1}

Question 10.
Find the derivative of f(x) = e^x (x² + 1)
Solution:
Let u = e^x, V = x² + 1
\(\frac{\mathrm{du}}{\mathrm{dx}}\) = e^x, \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = 2x
f'(x) = u(x) v'(x) + u'(x) . v(x)
= e^x2x + (x² + 1) e^x
= e^x(2x + x² + 1)
= e^x(x + 1)²

Question 11.
If y = \(\frac{a-x}{a+x}\) (x ≠ -a), find \(\frac{\mathrm{d} y}{\mathrm{dx}}\).
Solution:
Let u = a – x and v = a + x so that y = \(\frac{\mathrm{u}}{\mathrm{v}}\)

Question 12.
If f(x) = e^2x . log x (x > 0), then find f'(x).
Solution:
Let u = e^2x, v = log x so that
\(\frac{\mathrm{du}}{\mathrm{dx}}\) = 2 . e^2x , \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = 1
f(x) = u.v
f'(x) = u . \(\frac{\mathrm{dv}}{\mathrm{dx}}\) + y . \(\frac{\mathrm{du}}{\mathrm{dx}}\)
= e^2x . \(\frac{1}{x}\) + log x (2e^2x)
= e^2x (\(\frac{1}{x}\) + 2 logx)

Question 13.
If f(x) = \(\sqrt{\frac{1+x^{2}}{1-x^{2}}}\) (|x| < 1), then find f'(x)
Solution:

Question 14.
If f(x) = x², 2^x log x (x > 0), find f'(x).
Solution:
u = x², v = 2^x, w = logx
\(\frac{\mathrm{du}}{\mathrm{dx}}\) = 2x, \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = 2^x . log2, \(\frac{\mathrm{dw}}{\mathrm{dx}}\) = \(\frac{1}{x}\)
f'(x) = uv . \(\frac{\mathrm{dw}}{\mathrm{dx}}\) + vw . \(\frac{\mathrm{du}}{\mathrm{dx}}\) + uw . \(\frac{\mathrm{dv}}{\mathrm{dx}}\)
= x²2^x . \(\frac{1}{x}\) + 2^x . logx(2x) + x² . logx . 2^x log2
= x . 2^x (logx² + xlogx (log 2) + 1)

Question 15.
If y = \(\left|\begin{array}{l}
f(x) g(x) \\
\phi(x) \psi(x)
\end{array}\right|\) then show that \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\left|\begin{array}{l}
f^{\prime}(x) g^{\prime}(x) \\
\phi(x) \psi(x)
\end{array}\right|\) + \(\left|\begin{array}{l}
f(x) g \backslash(x) \\
\phi^{\prime}(x) \Psi(x)
\end{array}\right|\)
Solution:
Given y = \(\left|\begin{array}{l}
f(x) g(x) \\
\phi(x) \psi(x)
\end{array}\right|\)
= f(x) ψ(x) – Φ(x) g(x)
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = f(x) ψ'(x) + ψ(x) f'(x) – [Φ(x). g'(x) + g(x). Φ'(x)]
= [f(x) ψ'(x) – g(x) Φ’x)] + [f'(x) ψ(x) – Φ(x).g'(x)]
= \(\left|\begin{array}{l}
f(x) g(x) \\
\phi^{\prime}(x) \psi^{\prime}(x)
\end{array}\right|\) + \(\left|\begin{array}{cc}
f^{\prime}(x) & g^{\prime}(x) \\
\phi(x) & \psi(x)
\end{array}\right|\)

Question 16.
If f(x) = 7^x2+3x (x > 0), then find f'(x).
Solution:
Let u = x³ + 3x ⇒ \(\frac{\mathrm{du}}{\mathrm{dx}}\) = 3x² + 3 = 3(x² + 1)
f(X) = 7^u
f'(x) = \(\frac{d f}{d u}\) . \(\frac{\mathrm{du}}{\mathrm{dx}}\) (7^u . l0g 7) [3(x² + 1)]
= 3(x² + 1) 7^x2+3x log 7

Question 17.
If f(x) = x e^x sin x, then find f(x).
Solution:
Let u = x ,v = e^x, w = sin x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 1, \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = e^x . \(\frac{\mathrm{dw}}{\mathrm{dx}}\) = cos x
f(x) = u.v.w
f'(x) = uv . \(\frac{\mathrm{dw}}{\mathrm{dx}}\) + uw \(\frac{\mathrm{dv}}{\mathrm{dx}}\) + vw \(\frac{\mathrm{du}}{\mathrm{dx}}\)
= xe^x cos x + x . sinx e^x + e^x sin ^x.

Question 18.
If f(x) = sin (log x), (x > 0), find f'(x).
Solution:
Let u = logx, y = f(x) so that y = sin u
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{d y}{d u}\) . \(\frac{\mathrm{du}}{\mathrm{dx}}\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = cos u, \(\frac{\mathrm{du}}{\mathrm{dx}}\) = \(\frac{1}{x}\)
f'(x) = \(\frac{1}{x}\) . cos u = \(\frac{1}{x}\) cos (log x)

Question 19.
If f(x) =(x³ + 6x² + 12x – 13)¹⁰⁰; find f'(x).
Solution:
u = x³ + 6x² + 12x – 13
⇒ \(\frac{\mathrm{du}}{\mathrm{dx}}\) = 3x² + 12x + 12
= 3(x² + 4x + 4)
= 3(x + 2)²
f(x) = u¹⁰⁰
f'(x) = 100 . u⁹⁹ . \(\frac{\mathrm{du}}{\mathrm{dx}}\)
= 100 (x³ + 6x² + 12x – 13)⁹⁹ . 3(x + 2)²
= 300 (x + 2)² (x³ + 6x² + 12x – 13)⁹⁹

Question 20.
Find the derivative of f(x) = \(\frac{x \cos x}{\sqrt{1+x^{2}}}\)
Solution:
Let u = x cos x, and v = \(\sqrt{1+x^{2}}\) so that

Question 21.
li f(x) = log (secx + tan x), find f'(x). [Mar 14, May 11]
Solution:
Let u = sec x + tan x and y = log u
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{u}\), \(\frac{\mathrm{du}}{\mathrm{dx}}\) = sec x. tan x + sec² x
= sec x (sec x + tan x)
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{dy}}{\mathrm{dx}}\) . \(\frac{\mathrm{du}}{\mathrm{dx}}\)
= \(\frac{1}{\sec x+\tan x}\) . sec x(sec x + tan x) = sec x

Question 22.
If y = sin^-1\(\sqrt{x}\), find \(\frac{\mathrm{dy}}{\mathrm{dx}}\)
Solution:
u = \(\sqrt{x}\), y = sin^-1 x.

Question 23.
If y = sec (\(\sqrt{\tan x}\)), find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
u = \(\sqrt{\tan x}\), v = tanx
Then y = sec u, u = \(\sqrt{\mathrm{v}}\), v = tan x

Question 24.
If y = \(\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}\), find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Let u = x sin^-1x, v = \(\sqrt{1-x^{2}}\)

Question 25.
If y = log (cosh 2x), find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Let u = cosh 2x, so that y = log u
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{u}\); \(\frac{\mathrm{du}}{\mathrm{dx}}\) = 2 sin h2x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{d y}{d u}\) . \(\frac{\mathrm{du}}{\mathrm{dx}}\)
= 2 sin h 2x . \(\cosh 2 x\) = 2 tan h 2x

Question 26.
If y = log (sin (log x)), find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Let v = log x, u = sin v so that y = log u.
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{u}\); \(\frac{d u}{d v}\) = cos u; \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = \(\frac{1}{x}\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{d y}{d u}\) . \(\frac{d u}{d v}\) . \(\frac{\mathrm{dv}}{\mathrm{dx}}\)
= \(\frac{1}{\sin (\log x)}\) . cos (logx) \(\frac{1}{x}\) = \(\frac{\cot (\log x)}{x}\)

Question 27.
If y = (cot^-1x³)², find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
u = cot^-1x³, u = x³, y = u²

Question 28.
If y = cosec^-1(e^2x+1), find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
u = e^2x+1, y = cosec^-1u

Question 29.
If y = tan^-1 (cos \(\sqrt{x}\)), find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
v = \(\sqrt{x}\) and u = cos v, y = tan^-1u
\(\frac{\mathrm{dv}}{\mathrm{dx}}\) = \(\frac{1}{2 \sqrt{x}}\), \(\frac{d u}{d v}\) = – sin u; \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{1+u^{2}}\)
= – sin \(\sqrt{x}\) = \(\frac{1}{1+\cos ^{2}(\sqrt{x})}\)

Question 30.
If y = Tan^-1 \(\sqrt{42}\) for 0 < |x| < 1, find \(\frac{\mathrm{dy}}{\mathrm{dx}}\). [May, Mar 12]
Solution:
Put x² = cos 2θ

Question 31.
If y = x²e^xsin x, find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
log y = log x². e^x. sin x
= log x² + log e^x + log sin x
= 2 log x + log e^x + log sin x
Differentiating w.r.to by sin x
\(\frac{1}{y}\) . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{2}{x}\) + 1 + \(\frac{1}{sin x}\) . cos x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = y(\(\frac{2}{x}\) + 1 + cot x)

Question 32.
If y = x^tanx + (sin x)^{cos x}, find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) [Mar. 14, 11]
Solution:
Let u = x^tanx and v = (sin x)^{cos x}
log u logx ^tanx = (tan x) log x

Question 33.
If x = a(cos t + log tan (\(\frac{t}{2}\))), y = a sin t, find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:

Question 34.
If x^y = e^x-y, than show that \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\log x}{(1+\log x)^{2}}\)    [May 07]
Solution:
x^y = e^x-y
log x^y = log e^x-y
y log x = (x – y) (log e = 1)
y(1 + log x) = x

Question 35.
If siny = x sin (a + y), then show that \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\sin ^{2}(a+y)}{\sin a}\) (a is not a multiple of π)
Solution:

Question 36.
If y = x⁴ + tan x, then find y”.
Solution:
y = x⁴ + tan x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 4x³ + sec² x
\(\frac{d^{2} y}{d x^{2}}\) = 12x² + 2 sec x (sec x tan x)
= 12x² + 2 sec²x . tan x

Question 37.
If f(x) = sinx, sin 2x sin 3x, find f”(x).
Solution:
f(x) = \(\frac{1}{2}\) sin 2x(2 sin 3x sin x)
= \(\frac{1}{2}\) (sin 2x) (cos 2x – cos4x)
= \(\frac{1}{4}\) (2 sin 2x cos 2x – 2 sin 2x cos 4x)
= \(\frac{1}{4}\) (sin2x + sin4x – sin6x)
Therefore,
f'(x)= \(\frac{1}{4}\)[2 cos 2x+ 4cos 4x – 6cos 6x]
Hence,
f”(x) = \(\frac{1}{4}\) (-4 sin 2x – 16 sin 4x + 36 sin 6x)
= 9 sin 6x – 4 sin 4x – sin 2x.

Question 38.
Show that y = x + tan x satisfies cos2x \(\frac{d^{2} y}{d x^{2}}\) + 2x = 2y.
Solution:
y = x + tan x implies that y’ = 1 + sec² x
That is, y’ cos²x = 1 + cos²x.
Differentiating both sides of the above equation we get
y” cos²x + y’ . 2 cos x (-sin x) = 2 cos x (- sin x)
∴ y” cos² x = 2(y’ – 1) sin x cos x
= 2 sec²x sin x cos x = 2 tan x = 2(y – x)
This proves the result.

Question 39.
If x = a(t – sin t),y = a(1 + cost), find \(\frac{d^{2} y}{d x^{2}}\).
Solution:

Question 40.
Find the second order derivative of y = tan^-1(\(\frac{2 x}{1-x^{2}}\))
Solution:
Put x = tan θ, Then
y = tan^-1 (\(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\))
= tan^-1 (tan 2θ)
= 2θ = 2 tan^-1x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{2}{1+x^{2}}\) and \(\frac{d^{2} y}{d x^{2}}\) = \(\frac{-4 x}{\left(1+x^{2}\right)^{2}}\).

Question 41.
If y = sin (sin x), show that y” + (tan x) y’ + y cos²x = 0.
Solution:
y = sin (sin x) implies that .
y’ = cos x . cos (sin x) and
y” = -cos²x sin (sin x) – sin x cos (sin x)
= – y cos²x – sin x (\(\frac{y^{\prime}}{\cos x}\))
= -y cos²x – y’ tan x
∴ y” + (tan x)y’ + y cos² x = 0.

Question 42.
If f(x) = e^x(x ∈ R), then show that f(x) = e^x by first principle.
Solution:
From f(x) = e^x we have for h ≠ 0

Question 43.
If f(x) = log x (x > 0), then show that f(x) = \(\frac{1}{x}\) by first principle.
Solution:
Now for h ≠ 0

\(\frac{d}{dx}\) (log x) = \(\frac{1}{x}\)

Question 44.
If 1(x) = a^x (x ∈ R) (a > 0), then show that f'(x) = a^x log a by first principle.
Solution:
For h ≠ 0
\(\frac{f(x+h)-f(x)}{h}\) = \(\frac{a^{x+h}-a^{x}}{h}\) = a^x [latex]\frac{a^{h}-1}{h}[/latex]
We know that \(\frac{a^{h}-1}{h}\) → log a as h → 0
Hence f'(x) = a^x . log a.
\(\frac{d}{d x}\) = (a^x) = a^x log a

Question 45.
If y = Tan^-1 \(\sqrt{\frac{1-x}{1+x}}\) (|x| < 1), we shall find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Substituting x cos u (u ∈ (0, π)) in y, we get


observe that Tan^-1x, \(\sqrt{\frac{1-x}{1+x}}\) and cos u are the functions that stand for f(x), g(x) and h(u) respectively, mentioned in the method.

Question 46.
If y = Tan^-1[latex]\frac{2 x}{1-x^{2}}[/latex] (|x| < 1) then we shall \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Substituting x = tan u

Question 47.
If x = a cos³t, y = a sin³t, find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Here \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 3a cos²t (-sin t) and
\(\frac{d y}{d t}\) = 3a sin²t. cost.
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}\) = -tan t

Question 48.
If y = e^t +cost, x = log t + sin t find \(\frac{\mathrm{dy}}{\mathrm{dx}}\)
Solution:
Here \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^t – sin t and \(\frac{d x}{d t}\) = \(\frac{1}{t}\) + cos t
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{t\left(e^{t}-\sin t\right)}{(1+t \cos t)}\)

Question 49.
To find the derivative of f(x) = x^{\(\sin ^{\frac{1}{x}}\)} with respect to g(x) = sin^-1x, we have to compute \(\frac{d f}{d g}\)
Solution:
Now f(x) = x^{\(\sin ^{\frac{1}{x}}\)} implies that
log f(x) = sin^-1x . log x so that

Question 50.
If x³ + y³ – 3axy = 0, find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Let the given equation define the function.
y = 1(x) that is x³ + (f(x))³ – 3axf(x) = 0
Differentiating both sides of this equation with respect to x, we get
3x² + 3 (f(x))2 f'(x) – [3a. f(x) + 3axf'(x)] = 0
Hence 3x² + 3y² f'(x) – [3ay + 3ax f'(x)] = 0
∴ f'(x) = \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{a y-x^{2}}{y^{2}-a x}\)

Question 51.
If 2x² – 3xy + y² + x + 2y – 8 = 0, find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Treating y as a function of x and then differentiating with respect to x,
we get 4x – 3y – 3xy’ + 2yy’ + 1 + 2y’ = 0
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = y’ = \(\frac{3 y-4 x-1}{2 y-3 x+2}\)

Question 52.
If y = x^x (x > 0), we shall find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Taking logarithms on both the sides of
y = x^x we obtain log y = x log x
Differentiating with respect to x,
We get \(\frac{y^{\prime}}{y}\) = x . \(\frac{1}{x}\) + log x = 1 + log x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = y’ = y(1 + log x) = x^x (1 + log x)

Question 53.
If y = (tan x)^{sin x} [o < x < \(\frac{\pi}{2}\) ] compute \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Taking logarthms on both sides of
y= (tan x)^{sin x}, we get
log y = sin x . log (tan x)
Differentiating with respect to x, we get
\(\frac{y^{\prime}}{y}\) = \(\frac{\sin x}{\tan x}\) . sec²x + cosx . log (tan x)
= sec x + cos x . log (tan x)
Hence \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (tan x)^{sin x} [sec x + cos x log (tan x)]

Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Products of Vectors Important Questions which are most likely to be asked in the exam.

Intermediate 1st Year Maths 1A Products of Vectors Important Questions

Question 1.
If \(\bar{a}\) = 6\(\bar{i}\) + 2\(\bar{j}\) + 3\(\bar{k}\) and \(\bar{b}\) = 2\(\bar{i}\) – 9\(\bar{j}\) + 6\(\bar{k}\), then find \(\bar{a}\) . \(\bar{b}\) and the angle between \(\bar{a}\) and \(\bar{b}\).
Solution:

Question 2.
If a = i + 2j – 3k, b = 3i – j + 2k then show that a + b and a – b are perpendicular to each other. (A.P) (Mar. ’15, May ’11)
Solution:
a + b = (1 + 3)i + (2 – 1)j+ (-3 + 2) k
= 4i + j – k
Similarly a – b = – 2i + 3j – 5k
(a + b). (a – b) = 4(-2) + 1(3) + (-1)(-5) = 0
Hence (a + b) and (a – b) are mutually perpendicular.

Question 3.
Let \(\bar{a}\) and \(\bar{b}\) be non- zero, non-collinear vectors. If |\(\bar{a}\) + \(\bar{b}\) | = | \(\bar{a}\) – \(\bar{b}\) |, then find the angle between \(\bar{a}\) and \(\bar{b}\).
Solution:
| \(\bar{a}\) + \(\bar{b}\) |² = | a – b |².
⇒ |\(\bar{a}\)|² + |\(\bar{b}\)|² + 2 (\(\bar{a}\). \(\bar{b}\))
⇒ |\(\bar{a}\)|² + |\(\bar{b}\)|² – 2 (\(\bar{a}\). \(\bar{b}\))
⇒ 4(\(\bar{a}\) . \(\bar{b}\)) = 0 ⇒ \(\bar{a}\) . \(\bar{b}\) = 0
⇒ Angle between \(\bar{a}\) and \(\bar{b}\) is 90°.

Question 4.
If |\(\bar{a}\)| = 11, |\(\bar{b}\)| = 23 and | \(\bar{a}\) – \(\bar{b}\) | = 30, then find the angle between the vectors \(\bar{a}\), \(\bar{b}\) and also find | \(\bar{a}\) + \(\bar{b}\) |.
Solution:
Given that | \(\bar{a}\) | = 11, | \(\bar{b}\) | = 23 and | \(\bar{a}\) – \(\bar{b}\) | = 30, Let (\(\bar{a}\), \(\bar{b}\)) = θ.

Question 5.
If \(\bar{a}\) = \(\bar{i}\) – \(\bar{j}\) – \(\bar{k}\) and \(\bar{b}\) = 2\(\bar{i}\) – 3\(\bar{j}\) + \(\bar{k}\), then find the projection vector of \(\bar{b}\) on \(\bar{a}\) and its magnitude.
Solution:

Question 6.
If P, Q, R and S are points whose position vectors are \(\bar{i}\) – \(\bar{k}\), –\(\bar{i}\) +2\(\bar{j}\), 2\(\bar{i}\) – 3\(\bar{k}\) and 3\(\bar{i}\) – 2\(\bar{j}\) – \(\bar{k}\) respectively, then find the component of \(\overline{\mathbf{R S}}\) on \(\overline{\mathbf{P Q}}\).
Solution:
Let ‘O’ be the origin.
\(\overline{\mathrm{OP}}\) = \(\overline{\mathbf{i}}\) – \(\overline{\mathbf{k}}\), \(\overline{\mathrm{OQ}}\) = –\(\overline{\mathbf{i}}\) + 2\(\overline{\mathbf{j}}\), \(\overline{\mathrm{OR}}\) = 2\(\overline{\mathbf{i}}\) – 3\(\overline{\mathbf{k}}\) and \(\overline{\mathrm{OR}}\) = 3\(\overline{\mathbf{i}}\) – 2\(\overline{\mathbf{j}}\) – \(\overline{\mathbf{k}}\)

The component of \(\overline{\mathrm{RS}}\) and \(\overline{\mathrm{PQ}}\)

Question 7.
If the vectors λ\(\overline{\mathbf{i}}\) – 3\(\overline{\mathbf{j}}\) + 5\(\overline{\mathbf{k}}\) and 2 λ\(\overline{\mathbf{i}}\) – λ\(\overline{\mathbf{j}}\) + \(\overline{\mathbf{k}}\) are perpendicular to each other, find λ. (T.S) (Mar. ’16)
Solution:
Since the vectors are perpendicular
⇒ (λ\(\overline{\mathbf{i}}\) – 3\(\overline{\mathbf{j}}\) + 5\(\overline{\mathbf{k}}\)) . (2λ\(\overline{\mathbf{i}}\) – λ\(\overline{\mathbf{j}}\) – \(\overline{\mathbf{k}}\)) = 0
⇒ λ(2λ) + (-3)(-λ) + 5(-1) = 0
⇒ 2λ² + 3λ – 5 = 0
⇒ (2λ + 5)(λ – 1) = 0
∴ λ = 1 (or) \(\frac{-5}{2}\)

Question 8.
Let \(\overline{\mathbf{a}}\) = 2\(\overline{\mathbf{i}}\) + 3\(\overline{\mathbf{j}}\) + \(\overline{\mathbf{k}}\), \(\overline{\mathbf{b}}\) = 4\(\overline{\mathbf{i}}\) + \(\overline{\mathbf{j}}\) and \(\overline{\mathbf{c}}\) = \(\overline{\mathbf{i}}\) – 3\(\overline{\mathbf{j}}\) – 7\(\overline{\mathbf{k}}\). Find the vector \(\overline{\mathbf{r}}\) such that \(\overline{\mathbf{r}}\) . \(\overline{\mathbf{a}}\) = 9, \(\overline{\mathbf{r}}\) . \(\overline{\mathbf{b}}\) = 7 and \(\overline{\mathbf{r}}\) . \(\overline{\mathbf{c}}\) = 6.
Solution:
Let \(\overline{\mathbf{r}}\) = x\(\overline{\mathbf{i}}\) + y\(\overline{\mathbf{j}}\) + z\(\overline{\mathbf{k}}\)
∴ \(\overline{\mathbf{r}}\) . \(\overline{\mathbf{a}}\) = 9 ⇒ 2x + 3y + z = 9
\(\overline{\mathbf{r}}\) . \(\overline{\mathbf{b}}\) = 7 ⇒ 4x + y = 7 and \(\overline{\mathbf{r}}\) . \(\overline{\mathbf{c}}\) = 6 ⇒ x – 3y – 7z = 6
By solving these equations, we get x = 1, y = 3 and z = -2
∴ \(\overline{\mathbf{r}}\) = \(\overline{\mathbf{i}}\) + 3\(\overline{\mathbf{j}}\) – 2\(\overline{\mathbf{k}}\)

Question 9.
Show that the points 2\(\overline{\mathbf{i}}\) – \(\overline{\mathbf{j}}\) + \(\overline{\mathbf{k}}\), \(\overline{\mathbf{i}}\) – 3\(\overline{\mathbf{j}}\) – 5\(\overline{\mathbf{k}}\) and 3\(\overline{\mathbf{i}}\) – 4\(\overline{\mathbf{j}}\) – 4\(\overline{\mathbf{k}}\) are the vertices of a right angle triangle. Also, find the other angles.
Solution:
Let ‘O’ be the origin and A, B, C be the given points, then

Question 10.
Prove that the angle θ between any two diagonals of a cube is given by cos θ = \(\frac{1}{3}\). (May ’11)
Solution:
Let us suppose that the cube is a unit cube

Question 11.
Let \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\), \(\overline{\mathbf{c}}\) be non—zero mutually orthogonal vectors, if x \(\overline{\mathbf{a}}\) + y\(\overline{\mathbf{b}}\) + z = 0, then x = y = z = 0.
Solution:
x\(\overline{\mathbf{a}}\) + y\(\overline{\mathbf{b}}\) + z\(\overline{\mathbf{c}}\) = 0
⇒ \(\overline{\mathbf{a}}\) • (x\(\overline{\mathbf{a}}\) + y\(\overline{\mathbf{b}}\) + z\(\overline{\mathbf{c}}\)) = 0
⇒ x(\(\overline{\mathbf{a}}\) • \(\overline{\mathbf{a}}\)) = 0
⇒ x = 0 (∵ \(\overline{\mathbf{a}}\) • \(\overline{\mathbf{a}}\) ≠ 0)
Similarly y = 0, z = 0.

Question 12.
Let \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) be mutually orthogonal vectors of equal magnitudes. Prove that the vector \(\overline{\mathbf{a}}\) + \(\overline{\mathbf{b}}\) + \(\overline{\mathbf{c}}\) is equally inclined to each of \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\), the angle of inclination being cos^-1\(\left(\frac{1}{\sqrt{3}}\right)\).
Solution:

Similarly, it can be proved that \(\overline{\mathbf{a}}\) + \(\overline{\mathbf{b}}\) + \(\overline{\mathbf{c}}\) inclines at an angle of cos^-1\(\left(\frac{1}{\sqrt{3}}\right)\) with \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\).

Question 13.
The vectors \(\overline{\mathbf{A B}}\) = 3\(\bar{i}\) – 2\(\bar{j}\) + 2\(\bar{k}\) and \(\overline{\mathbf{A D}}\) = \(\bar{i}\) – 2\(\bar{k}\) represent the adjacent sides of a parallelogram ABCD. Find the angle between the diagonals.
Solution:

Question 14.
For any two vectors a and b, show that
i) |a • b| (Cauchy – Schwartz inequality)
ii) || a + b | ≤ ||a|| + ||b| (triangle inequality
Solution:
i)
If a = 0 or b = 0, the inequalities hold trivially.

ii)

Question 15.
Find the cartesian equation of the plane passing through the point (-2, 1, 3) and perpendicular to the vector 3\(\bar{i}\) + \(\bar{j}\) + 5\(\bar{k}\).
Solution:
Let A = -2\(\bar{i}\) + \(\bar{j}\) + 3\(\bar{k}\) and \(\bar{r}\) = x\(\bar{i}\) + y\(\bar{j}\) + z\(\bar{k}\) be any point in the plane
∴ \(\overline{\mathrm{AP}}\) = (x + 2)\(\bar{i}\) + (y – 1)\(\bar{j}\) + (z – 3)\(\bar{k}\)
∴ \(\overline{\mathrm{AP}}\) is perpendicular to 3\(\bar{i}\) + \(\bar{j}\) + 5\(\bar{k}\)
⇒ (x + 2)3 + (y – 1)1 + (z – 3)5 = 0
⇒ 3x + y + z – 10 = 0

Question 16.
Find the cartesian equation of the plane through the point A (2, -1, -4) and parallel to the plane 4x – 12y – 3z – 7 = 0.
Solution:
The equation of the plane parallel to
4x – 12y – 3z – 7 = 0 be
4x -12y – 3z = p
∴ If passes through the point A(2, -1, -4)
⇒ 4(2) – 12(-1) – 3(-4) = p
⇒ p = 32
∴ The equation of the required plane is
4x – 12y – 3z = 32

Question 17.
Find the angle between the planes 2x – 3y – 6z = 5 and 6x + 2y – 9z = 4.
Solution:
Equation to the planes is 2x – 3y – 6z = 5
⇒ \(\overline{\mathbf{r}}\) .(2\(\overline{\mathbf{i}}\) – 3\(\overline{\mathbf{j}}\) – 6\(\overline{\mathbf{k}}\)) = 5 ———- (1)
and 6x + 2y – 9z = 4 —— (2)
∴ The angle between the planes \(\overline{\mathbf{r}} \cdot \overline{\mathbf{n}}_{1}\) = p, and \(\overline{\mathbf{r}} \cdot \overline{\mathbf{n}}_{2}\) = q is θ, then

Question 18.
Find unit vector orthogonal to the vector 3\(\bar{i}\) + 2\(\bar{j}\) +6\(\bar{k}\) and coplanar with thevectors 2\(\bar{i}\) + \(\bar{j}\) + \(\bar{k}\) and \(\bar{i}\) – \(\bar{j}\) + \(\bar{k}\).
Solution:

⇒ (2x + y)3 + (x – y)2 + (x + y)6 = 0
⇒ 6x + 3y + 2x – 2y + 6x + 6y = 0
⇒ 14x + 7y = 0
⇒ y = -2x —— (1)
∵ |\(\overline{\mathbf{r}}\)| = 1
⇒ (2x + y)² + (x – y)² + (x + y)² = 1
⇒ 4x² + 4xy + 9 + 2(x² + y²) = 1
⇒ 6x² + 4x (-2x) + 3(-2x)² = 1 (By using (1))
⇒ 10x² = 1

Question 19.
If \(\overline{\mathbf{a}}\) = 2\(\overline{\mathbf{i}}\) – 3\(\overline{\mathbf{j}}\) + 5\(\overline{\mathbf{k}}\), \(\overline{\mathbf{b}}\) = –\(\overline{\mathbf{i}}\) + 4\(\overline{\mathbf{j}}\) + 2\(\overline{\mathbf{k}}\), then find \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\) and unit vector perpendicular to both \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\).
Solution:
\(\overline{\mathbf{a}}\) = 2\(\overline{\mathbf{i}}\) – 3\(\overline{\mathbf{j}}\) + 5\(\overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}\) = –\(\overline{\mathbf{i}}\) + 4\(\overline{\mathbf{j}}\) + 2\(\overline{\mathbf{k}}\)

Question 20.
If \(\bar{a}\) = 2\(\bar{i}\) – 3\(\bar{j}\) + 5\(\bar{k}\), \(\bar{b}\) = –\(\bar{i}\) + 4\(\bar{j}\) + 2\(\bar{k}\), then find (\(\bar{a}\) + \(\bar{b}\)) × (\(\bar{a}\) – \(\bar{b}\)) and unit vector perpendicular to both \(\bar{a}\) + \(\bar{b}\) and \(\bar{a}\) – \(\bar{b}\).
Solution:

Question 21.
Find the area of the parallelogram for which the vectors \(\bar{a}\) = 2\(\bar{i}\) – 3\(\bar{j}\) and \(\bar{b}\) = 3\(\bar{i}\) – \(\bar{k}\) are adjacent sides. (Mar. ; May ‘08)
Solution:
The vector area of the given parallelogram

Question 22.
If \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\), \(\overline{\mathbf{c}}\) and \(\overline{\mathbf{d}}\) are vectors such that \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\) = \(\overline{\mathbf{c}}\) × \(\overline{\mathbf{d}}\) and \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{c}}\) = \(\overline{\mathbf{b}}\) × \(\overline{\mathbf{d}}\) then show that the vectors \(\overline{\mathbf{a}}\) – \(\overline{\mathbf{d}}\) and \(\overline{\mathbf{b}}\) – \(\overline{\mathbf{c}}\) are parallel.
Solution:

Question 23.
If \(\overline{\mathbf{a}}\) = \(\overline{\mathbf{i}}\) + 2\(\overline{\mathbf{j}}\) + 3\(\overline{\mathbf{k}}\) and \(\overline{\mathbf{b}}\) = 3\(\overline{\mathbf{i}}\) + 5\(\overline{\mathbf{j}}\) – \(\overline{\mathbf{k}}\) are two sides of a triangle, then find
its area.
Solution:
Area of the triangle = \(\frac{1}{2}|\bar{a} \times \bar{b}|\)

Question 24.
In Δ ABC, if \(\overline{\mathrm{BC}}\) = \(\bar{a}\), \(\overline{\mathrm{CA}}\) = \(\bar{b}\), \(\overline{\mathrm{AB}}\) = \(\bar{c}\) then show that \(\bar{a}\) × \(\bar{b}\) = \(\bar{b}\) × \(\bar{c}\) = \(\bar{c}\) × \(\bar{a}\).
Solution:

Question 25.
Let \(\bar{a}\) = 2\(\bar{i}\) – \(\bar{j}\) + \(\bar{k}\) and \(\bar{b}\) = 3\(\bar{i}\) + 4\(\bar{j}\) – \(\bar{k}\). If θ is the angle between \(\bar{a}\) and \(\bar{b}\) then find sin θ.
Solution:

Question 26.
Let \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) be such that \(\bar{c}\) ≠ 0,
\(\bar{a}\) × \(\bar{b}\) = \(\bar{c}\), \(\bar{b}\) × \(\bar{c}\) = \(\bar{a}\). Show that \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) are pair wise orthogonal vectors and |\(\bar{b}\)| = 1, |\(\bar{c}\)| = |\(\bar{a}\)|.
Solution:

Question 27.
Let \(\overline{\mathbf{a}}\) = 2\(\overline{\mathbf{i}}\) + \(\overline{\mathbf{j}}\) – 2\(\overline{\mathbf{k}}\) ; \(\overline{\mathbf{b}}\) = \(\overline{\mathbf{i}}\) + \(\overline{\mathbf{j}}\). If \(\overline{\mathbf{c}}\) is a vector such that \(\overline{\mathbf{a}}\) . \(\overline{\mathbf{c}}\) = |\(\overline{\mathbf{c}}\)|, |\(\overline{\mathbf{c}}\) – \(\overline{\mathbf{a}}\)| = \(2 \sqrt{2}\) and the angle between \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\) = \(2 \sqrt{2}\) and the angle between \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) is 30°, then find the value of |(\(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\)) × \(\overline{\mathbf{c}}\)|
Solution:

Question 28.
Let \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\) be two non-collinear unit vectors. If \(\bar{\alpha}\) = \(\overline{\mathbf{a}}\) – (\(\overline{\mathbf{a}}\) • \(\overline{\mathbf{b}}\))\(\overline{\mathbf{b}}\) and \(\bar{\beta}\) =
= \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\), then show that |\(\overline{\boldsymbol{\beta}}\)| = |\(\bar{\alpha}\)|.
Solution:

Question 29.
A non-zero vector \(\bar{a}\) is parallel to the line of intersection of the plane determined by the vectors \(\bar{i}\), \(\bar{i}\) + \(\bar{j}\) and the plane determined by the vectors \(\bar{i}\) – \(\bar{j}\), \(\bar{i}\) + \(\bar{k}\). Find the angle between \(\bar{a}\) and the vector \(\bar{i}\) – 2\(\bar{j}\) + 2\(\bar{k}\).
Solution:
Let l be the line of intersection of the planes determined by the pairs \(\bar{i}\), \(\bar{i}\) + \(\bar{j}\) and \(\bar{i}\) – \(\bar{j}\), \(\bar{i}\) + \(\bar{k}\)

∴ \(\bar{n}_{1}\) is perpendicular to l and \(\bar{n}_{2}\) is also perpendicular to l.
∴ \(\bar{a}\) is parallel to the line l, follows that \(\bar{a}\) is perpendicularto both \(\bar{n}_{1}\), and \(\bar{n}_{2}\)

Question 30.
Let \(\overline{\mathbf{a}}\) = 4\(\overline{\mathbf{i}}\) + 5\(\overline{\mathbf{j}}\) – \(\overline{\mathbf{k}}\), \(\overline{\mathbf{b}}\) = \(\overline{\mathbf{i}}\) – 4\(\overline{\mathbf{j}}\) + 5\(\overline{\mathbf{k}}\) and \(\overline{\mathbf{c}}\) = 3\(\overline{\mathbf{i}}\) + \(\overline{\mathbf{j}}\) – \(\overline{\mathbf{k}}\). Find the vector \(\bar{\alpha}\) which is perpendicular to both \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\) and \(\bar{\alpha} \cdot \mathbf{c}\) = 21.
Solution:

Question 31.
For any vector a, show that |a × i|² + |a × j|² + |a × k|² = 2|a|².
Solution:
Let a = x i + y j + z k.
Then a × i = -yk + zj.
∴ |a × i|² = y² + z²
Similarly |a × j|² = z² + x² and |a × k|² = x²
+ y²
∴ |a × i|² + |a × j|² + |a × k|² = 2(x² + y² + z²) = 2|a|²

Question 32.
If \(\overline{\mathbf{a}}\) is a non-zero vector and \(\overline{\mathbf{b}}\), \(\overline{\mathbf{c}}\) are two vector such that \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\) = \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{c}}\) and \(\overline{\mathbf{a}}\) • \(\overline{\mathbf{b}}\) = \(\overline{\mathbf{a}}\) • \(\overline{\mathbf{c}}\), then prove that \(\overline{\mathbf{b}}\) = \(\overline{\mathbf{c}}\).
Solution:

Question 33.
Prove that the vectors \(\bar{a}\) = 2\(\bar{i}\) – \(\bar{j}\) + \(\bar{k}\), \(\bar{b}\) = \(\bar{i}\) – 3\(\bar{j}\) – 5\(\bar{k}\) and \(\bar{c}\) = 3\(\bar{i}\) – 4\(\bar{j}\) – 4\(\bar{k}\) are coplanar.
Solution:

Question 34.
Find the volume of the parallelopiped whose cotersminus edges are represented by the vectors 2\(\bar{i}\) – 3\(\bar{j}\) + \(\bar{k}\), \(\bar{i}\) – \(\bar{j}\) + 2\(\bar{k}\) and 2\(\bar{i}\) + \(\bar{j}\) – \(\bar{k}\).
Solution:
Let \(\bar{a}\) = 2\(\bar{i}\) – 3\(\bar{j}\) + \(\bar{k}\)

Question 35.
If the vectors \(\bar{a}\) = 2\(\bar{i}\) – \(\bar{j}\) + \(\bar{k}\), \(\bar{b}\) = \(\bar{i}\) + 2\(\bar{j}\) – 3\(\bar{k}\) and \(\bar{c}\) = 3\(\bar{i}\) + p\(\bar{j}\) + 5\(\bar{k}\) are coplanar, then find p.
Solution:

Question 36.
Show that
\(\overline{\mathbf{i}}\) × (\(\overline{\mathbf{a}}\) × \(\overline{\mathbf{i}}\)) + \(\overline{\mathbf{j}}\) × (\(\overline{\mathbf{a}}\) × \(\overline{\mathbf{j}}\)) + k × (\(\overline{\mathbf{a}}\) × \(\overline{\mathbf{k}}\)) = 2\(\overline{\mathbf{a}}\) for any vector \(\overline{\mathbf{a}}\).
Solution:

Question 37.
Prove that for any three vectors \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\), \(\overline{\mathbf{c}}\), [\(\overline{\mathbf{b}}\) + \(\overline{\mathbf{c}}\) \(\overline{\mathbf{c}}\) + \(\overline{\mathbf{a}}\) \(\overline{\mathbf{a}}\) + \(\overline{\mathbf{b}}\)] = 2[\(\overline{\mathbf{a}}\) \(\overline{\mathbf{b}}\) \(\overline{\mathbf{c}}\)]
Solution:

Question 38
For any three vectors \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\), \(\overline{\mathbf{c}}\) prove that [\(\overline{\mathbf{b}}\) × \(\overline{\mathbf{c}}\) \(\overline{\mathbf{c}}\) × \(\overline{\mathbf{a}}\) \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\)] = [\(\overline{\mathbf{a}}\) \(\overline{\mathbf{b}}\) \(\overline{\mathbf{c}}\)]²
Solution:

Question 39.
Let \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) be unit vectors such that \(\overline{\mathbf{b}}\) is not parallel to \(\overline{\mathbf{c}}\) and \(\overline{\mathbf{a}}\) × (\(\overline{\mathbf{b}}\) × \(\overline{\mathbf{c}}\)) = \(\frac{1}{2} \bar{b}\). Find the angles made by \(\overline{\mathbf{a}}\) with each of \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\).
Solution:

Since \(\overline{\mathbf{b}}\) and \(\overline{\mathbf{c}}\) are non—collinear vectors. Equating corresponding coeffs. on both sides

Question 40.
Let \(\overline{\mathbf{a}}\) = \(\overline{\mathbf{i}}\) + \(\overline{\mathbf{j}}\) + \(\overline{\mathbf{k}}\), \(\overline{\mathbf{b}}\) = 2\(\overline{\mathbf{i}}\) – \(\overline{\mathbf{j}}\) + 3\(\overline{\mathbf{k}}\), \(\overline{\mathbf{c}}\) = \(\overline{\mathbf{i}}\) – \(\overline{\mathbf{j}}\) and \(\overline{\mathbf{d}}\) = 6\(\overline{\mathbf{i}}\) + 2\(\overline{\mathbf{j}}\) + 3\(\overline{\mathbf{k}}\). Express \(\overline{\mathbf{d}}\) interms of \(\overline{\mathbf{b}}\) × \(\overline{\mathbf{c}}\), \(\overline{\mathbf{c}}\) × \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\)  (May. ’12)
Solution:

Then from the above problem,

Question 41.
For any four vectors \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\), \(\overline{\mathbf{c}}\) and \(\overline{\mathbf{d}}\) prove that
(\(\overline{\mathbf{b}}\) × \(\overline{\mathbf{c}}\)) . (\(\overline{\mathbf{a}}\) × \(\overline{\mathbf{d}}\)) + (\(\overline{\mathbf{c}}\) × \(\overline{\mathbf{a}}\)) . (\(\overline{\mathbf{b}}\) × \(\overline{\mathbf{d}}\)) + (\(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\)) . (\(\overline{\mathbf{c}}\) × \(\overline{\mathbf{d}}\)) = 0.
Solution:

Question 42.
Find the equation of the plane passing through the points A = (2, 3, -1), B = (4, 5, 2) and C = (3, 6, 5).
Solution:
Let ‘O’ be the origin

Let P be any point on the plane passing through the points A, B, C.

Hence, equation of the required plane is

Question 43.
Find the equation of the plane passing through the point A (3, -2, -1) and parallel to the vector \(\bar{b}\) = \(\bar{i}\) – 2\(\bar{j}\) + 4\(\bar{k}\) and \(\bar{c}\) = 3\(\bar{i}\) + 2\(\bar{j}\) – 5\(\bar{k}\).
Solution:
The required plane passes through A = (3, -2, -1) and is parallel to the vectors
\(\bar{b}\) = \(\bar{i}\) – 2\(\bar{j}\) + 4\(\bar{k}\) and \(\bar{c}\) = 3\(\bar{i}\) + 2\(\bar{j}\) – 5\(\bar{k}\). Hence if P is a point in the plane, then AP, b and C are coplanar and [AP b c] = 0.

i.e., 2x + 17y + 8z + 36 = 0 is the equation of the required plane.

Question 44.
Find the vector equation of the plane passing through the intersection of the planes r. (i + j + k) = 6 and r. (2i + 3j+ 4k) = -5 and the point (1, 1, 1).
Solution:

Substituting this value of λ in equation (1), we get

Question 45.
Find the distance of a point (2, 5, -3) from the plane r. (6i – 3j + 2k) = 4.
Solution:
Here a = 2i + 5j – 3k, N = 6i – 3j + 2k and d = 4.
∴ The distance of the point (2, 5, -3) from the given plane is

Question 46.
Find the angle between the line \(\frac{x+1}{2}\) = \(\frac{y}{3}\) = \(\frac{z-3}{6}\) and the plane 10x + 2y – 11z = 3.
Solution:
Let φ be the angle between the given line and the normal to the plane.
Converting the given equations into vector form, we have
r = (-i + 3k) + λ(2i + 3j + 6k)
and r. (10i + 2j – 11k) = 3
Here b = 2i + 3j + 6k and n = 10i + 2j – 11k

Question 47.
For any four vectors a, b, c and d, (a × b) × (c × d) = [a c d]b – [b c d]a and (a × b) × (c × d) = [a b d]c – [a b c]d.
Solution:
Let m = c × d
∴ (a × b) × (c × d) = (a × b) × m
= (a.m)b – (b.m)a
= (a. (c × d))b – (b. (c × d))a
= [a c d]b – [b c d]a.
Again Let a × b = n.
Then(a × b) × (c × d) = n × (c × d)
= (n .d)c – (n.c)d
= ((a × b). d) c – ((a × b). c)d
= [a b d]c – [a b c]d.

Question 48.
Find the shortest distance between the skew line r = (6i + 2j + 2k) + t(i – 2j + 2k) and r = (-4i – k) + s (3i – 2j – 2k).
Solution:
The first line passes through the point
A(6, 2, 2) and is parallel to the vector b = i – 2j + 2k. Second line passes through the point C(-4, 0, -1) and is parallel to the vector d = 3i – 2j – 2k

Question 49.
The angle ¡n a semi circle is a right angle. (May ’08)
Solution:
Let O be the centre and AOB be the diameter of the given šemi circle. Let P be a point on the semi circle. With reference to O as the origin, if the position vectors of A and P are taken as \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{p}}\),then

Question 50.
The altitude of a triangle are concurrent.
ii) The perpendicular bisectors of the sides of a triangle are concurrent.
Solution:
i) In the given triangle ABC, let the altitudes from A, B drawn to BC. CA respectively intersect them at D, E. Assume that AD and BE intersect at O, join CO and produce it to meet AB at F. With reference to O, let the position vectors of A, B and C be a, b and c respectively.

From Fig. we have

ii) In the given ABC, let the mid-points of BC, CA and AB be D, E and F respectively. Let the perpendicular bisectors drawn to BC and CA at D and E meet at O. Join OF. With respect to O, let the position vectors of A, B and C be \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) respectively.


On adding equations (1) and (2), we obtain

Question 51.
Show that the vector area of the quadrilateral ABCD having diagonals \(\overline{\mathbf{A C}}\), \(\overline{\mathbf{B D}}\) is \(\frac{1}{2}(\overline{A C} \times \overline{B D})\)
Solution:
If the point of intersection of the diagonals AC, BD of the given quadrilateral ABCD is assumed as Q, the vector area of the quadrilateral ABCD.

= Sum of the vector areas of ΔQAB, ΔQBC, ΔQCD and ΔQDA.

Question 52.
For any vectors a, b and c (a × b) × c = (a.c) b – (b.c) a (AP) (Mar. 15 May ‘06; June ’04)
Solution:
Equation (1) is evidently true if a and b are parallel.
Now suppose that a and b are non-parallel.
Let O denote the origin. Choose points A and B such that \(\overline{\mathrm{OA}}\) = \(\overline{\mathbf{a}}\) and \(\overline{\mathrm{OB}}\) = \(\overline{\mathbf{b}}\). Since \(\overline{\mathbf{a}}\) and \(\overline{\mathbf{b}}\) are non-parallel, the points O, A and B are non-collinear. Hence they determine a plane.
Let i denote the unit vector along \(\overline{\mathrm{OA}}\).
Let j be a unit vector in the OAB plane perpendicular to i. Let k = i × j. Then

Question 53.
For any four vectors a, b, c and d(a × b). (c × d) = \(\left|\begin{array}{cc}
a \cdot c & a \cdot d \\
b . c & b . d
\end{array}\right|\) and in particular (a × b)² = a²b² – (a. b)².
Solution:

Question 54.
In the above fou rmula if \(\overline{\mathbf{a}}\), \(\overline{\mathbf{b}}\), \(\overline{\mathbf{c}}\) and \(\overline{\mathbf{d}}\)(\(\overline{\mathbf{a}}\) × \(\overline{\mathbf{b}}\)). (\(\overline{\mathbf{c}}\) × \(\overline{\mathbf{d}}\)) = (\(\overline{\mathbf{a}}\) . \(\overline{\mathbf{c}}\))(\(\overline{\mathbf{b}}\) . \(\overline{\mathbf{d}}\)) – (\(\overline{\mathbf{a}}\) . \(\overline{\mathbf{d}}\))(\(\overline{\mathbf{b}}\) . \(\overline{\mathbf{c}}\)) = \(\left|\begin{array}{ll}
\overline{\mathbf{a}} \cdot \overline{\mathbf{c}} & \overline{\mathbf{a}} \cdot \overline{\mathbf{d}} \\
\overline{\mathbf{b}} \cdot \overline{\mathbf{c}} & \bar{b} \cdot \overline{\mathbf{d}}
\end{array}\right|\).
Solution:

Since this is a scalar, this is also called the scalar product of four vectors.