AP Inter 1st Year Maths 1B Question Paper March 2019

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AP Inter 1st Year Maths 1B Question Paper March 2019

Time : 3 Hours
Max. Marks : 75

Section – A
(10 × 2 = 20)

I. Very Short Answer Type Questions.

1. Answer all questions.
2. Each question carries two marks.

Question 1.
Find the angle which the straight line y = x – 4 makes with the y-axis.
Solution:
Given straight line is y = \(\sqrt{3}\)x – 4
m = \(\sqrt{3}\) ⇒ tan θ = \(\sqrt{3}\)
⇒ θ = 60°
The straight line y = \(\sqrt{3}\)x – 4 makes angle with the Y – axis is 90° – 60° = 30°.

Question 2.
Find the distance between the parallel straight lines 3x + 4y – 3 = 0 and 6x + 8y – 1 =0.
Solution:
Given straight lines are 3x + 4y – 3 = 0
⇒ 6x + 8y – 6 = 0 ……. (1)
6x + 8y – 1 = 0 ……… (2)
∴ The distance between the parallel straight lines.
(1) and (2) is \(\frac{\left|c_2-c_1\right|}{\sqrt{a^2+b^2}}\)
= \(\frac{|-1-(-6)|}{\sqrt{36+64}}\) = \(\frac{|-1+6|}{\sqrt{100}}\)
= \(\frac{5}{100}\) = \(\frac{1}{2}\)

Question 3.
Find ‘x’, if the distance between (5, -1, 7) and (x, 5, 1) is 9 units.
Solution:
Let A = (5, -1, 7)
B = (x, 5, 1)
Given AB = 9
⇒ \(\sqrt{(x-5)^2+(5+1)^2(1-7)^2}\) = 9
⇒ (x – 5)² + 36 + 36 = 81
⇒ (x – 5)² = 9
⇒ x – 5 = ±3
⇒ x – 5 = 3 (or) x – 5 = -3
⇒ x = 8 (or) x = 2.

Question 4.
Write the equation of the plane 4x – 4y + 2z + 5 = 0 in the intercept form.
Solution:
Given plane equation is 4x – 4y + 2z + 5 = 0
⇒ 4x – 4y + 2z = – 5
⇒ \(\frac{4 x}{-5}-\frac{4 y}{-5}+\frac{2 z}{-5}\) = 1
⇒ \(\frac{x}{-5 / 4}+\frac{y}{5 / 4}+\frac{z}{-5 / 4}\)
x- intercept = \(\frac{-5}{4}\)
y – intercept = \(\frac{5}{4}\)
z – intercept = \(\frac{-5}{2}\)

Question 5.
Compute \(\lim _{x \rightarrow 0} \frac{e^{3+x}-e^3}{x}\)
Solution:

Question 6.
Compute \(\lim _{x \rightarrow 3} \frac{x^2+3 x+2}{x^2-6 x+9}\)
Solution:

Question 7.
Find the derivative of the function tan^-1 (log x).
Solution:
Let y = tan^-1 (logx)
differentiating with respect to ‘x’ on both sides, we have

Question 8.
If y = \(\frac{2 x+3}{4 x+5}\) then find y”.
Solution:
Given y = \(\frac{2 x+3}{4 x+5}\)
differentiating with respect to x on both sides, we have

Question 9.
Define relative error and percentage error of the variable y.
Solution:
Relative error in y = \(\frac{\Delta y}{y}\)
Percentage error in y = \(\frac{\Delta y}{y}\) × 100

Question 10.
Find the absolute extremum of f(x) = x² defined on [-2, 2].
Solution:
Given f(x) = x² defined on [-2, 2]
Clearly f 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 maximum of f is the largest value of f(-2), f(0) and f(2)
f(-2) = (-2)² = 4
f(0) = 0² = 0
f(2) = 2² = 4.
∴ The absolute maximum value is 4.
similarly the absolute minimum is the least value of 4, 0, 4.
∴ The absolute minimum value is 0.

Section – B

II. Short Answer Type Questions.

1. Attempt any five questions.
2. Each question carries four marks.

Question 11.
A (5, 3) and B (3, -2) are two fixed points. Find the equation of locus of P, so that the area of triangle PAB is 9.
Solution:
Given A = (5, 3)
B = (3, – 2)
Let P(x₁, y₁) be any point on the locus.
Given geometric condition is area of ∆PAB is 9.
⇒ \(\frac{1}{2}\left|\begin{array}{cccc}
x_1 & 5 & 3 & x_1 \\
y_1 & 3 & -2 & y_1
\end{array}\right|\) = 9

Question 12.
When the origin is shifted to the point (3, -4) and transformed equation is x + y² = 4. Find the original equation.
Solution:
Here (h, k) = (3, -4)
we know x = x + h ⇒ x = x – h
⇒ x = x – 3
y = y + k ⇒ y = y – k
⇒ y = y – (-4)
⇒ y = y + 4
Given transformed equation is x² + y² = 4
⇒ (x – 3)² + (y + 4)² = 4
⇒ x² – 6x + 9 + y² + 8y + 16 = 4
⇒ x² + y² – 6x + 8y + 21 = 0.
∴ Required original equation is x² + y² – 6x + 8y + 21 = 0.

Question 13.
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 = 3 abc.
Solution:
Given ax + bc + c = 0 ……… (1)
bx + cy + a = 0 ……… (2)
cx + ay + b = 0 ……… (3)
solving (1) and (2)

Question 14.
Compute \(\lim _{x \rightarrow a}\left(\frac{x \sin a-a \sin x}{x-a}\right)\)
Solution:

Question 15.
Find the derivative of the function cot x from the first principle.
Solution:
Let f'(x) = cot x
f (x + h) = cot (x + h)
By first principle

Question 16.
Find the approximate value of \(\sqrt[3]{999}\).
Solution:
Let f (x) = x^1/3, x = 1000, ∆x = – 1

Question 17.
The distance – time formula for the motion of a particle along a straight line is s = t³ – 9t² + 24t – 18. Find when and where the velocity is zero.
Solution:
Given s = t³ – 9t² + 24t – 18.
Velocity v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 3t² – 18t + 24t.
v = 0 ⇒ 3t² – 18t + 24t = 0
⇒ t² – 6t + 8 = 0
⇒ t² – 2t – 4t + 8 = 0
⇒ t (t – 2) – 4 (t – 2) = 0
⇒ (t – 2)(t – 4) = 0
t = 2 (or) t = 4.
When t = 2
s = 2³ – 9 (2)² + 24(2) – 18
= 8 – 36 + 48 – 18
= 56 – 54
= 2 units
When t = 4
s = 4³ – 9 (4)² + 24 (4) – 18
= 64 – 144 + 96 – 18
= 160 – 162 = 2 units.

Section – C

III. Long Answer Type Questions.

1. Attempt any five questions.
2. Each question carries seven marks.

Question 18.
If Q (h, k) is the image of the point P(x₁, y₁) with respect to the straight line ax + by + c = 0 then prove that \(\frac{h-x_1}{a}\) = \(\frac{k-y_1}{b}\) = \(\frac{-2\left(a x_1+b y_1+c\right)}{a^2+b^2}\)
Solution:
Let L ≡ ax + by + c = 0
Given Q (h, k) is the image of p(x₁, y₁)
with respect to the straight line L = 0.

⇒ a(h + x₁) + b(k + y₁) + 2c = 0
⇒ a(x₁ + aλ + x₁) + b(y₁ + b + y₁) + 2c = 0.
⇒ 2ax₁ + a²λ + 2by₁ + b²λ + 2c = 0.
⇒ (a² + b²)λ = -2(ax₁ + by₁ + c)
⇒ λ = \(\frac{-2\left(a x_1+b y_1+c\right)}{a^2+b^2}\)
from (1)
∴ \(\frac{h-x 1}{a}\) = \(\frac{k-y_1}{b}\) = \(\frac{-2\left(a x_1+b y_1+c\right)}{a^2+b^2}\)

Question 19.
If the equation S ≡ ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of parallel straight lines then show that
(i) h² = ab
(ii) af² = bg² and
(iii) the distance between the parallel lines
= \(\sqrt[2]{\frac{g^2-a c}{a(a+b)}}\) = \(\sqrt[2]{\frac{f^2-a c}{b(a+b)}}\)
Solution:
Let the two parallel straight lines represented by S = 0 be
lx + my + n₁ = 0 ………… (1)
and lx + my + n₂ = 0 ……. (2)
Then, S ≡ λ(lx + my + n₁) (lx + my + n₂), for some real λ ≠ 0.
From this we have
l² = \(\frac{a}{\lambda}\), m² = \(\frac{b}{\lambda}\), n₁n₂ = \(\frac{c}{\lambda}\), lm = \(\frac{h}{\lambda}\)
l(n₁ + n₂) = \(\frac{2 \mathrm{~g}}{\lambda}\) and m(n₁ + n₂) = \(\frac{2 f}{\lambda}\).
Now

(i) h² = λ²l²m² = (λl²)(λm²) = ab.

(ii) 4af² = (λ²l²) (λ²m²) (n₁ + n₂)²
= (λm²) (λ²l²) (n₁ + n₂)² = b(4g²)
so that af² = bg².

(iii) Distance between the parallel lines

Question 20.
Find the value of k, if the lines joining the origin to the points of intersection of the curves 2x² – 2xy + 3y² + 2x – y – 1 =0 and the line x + 2y = k are mutually perpendicular.
Solution:

Equation of the circle is x² + y² = a² …… (1)
Equation of AB is lx + my = 1 ……….. (2)
Homogenising (1) with the help of (2)
Combined equation of OA, OB is
x² + y² = a².1²
x² + y² = a² (lx + my)²
= a²(l²x² + m²y² + 2lmxy)
= a²l²x² + a²m²y² + 2a²lmxy
i.e., a²l²x² + 2a² lmxy + a² m²y² – x² – y² = 0
(a²l² – 1) x² + 2a² lmxy + (a²m² – 1) y² = 0
Since OA, OB are perpendicular
Co-efficient of x² + co-efficient of y² = 0
a²l² – 1 + a²m² – 1 = 0
a²(l² + m²) = 2
This is the required condition.

Question 21.
Find the angle between the lines whose direction cosines satisfy the equations l + m + n = 0, l² + m² – n² = 0.
Solution:
Given l + m + n = 0 …… (1)
l² + m² – n² = 0 ……. (2)
From (1) l = – m – n …… (3)
Substituting in (2)
(- m – n)² + (m² – n²) = 0.
⇒ (m + n)² + (m + n) (m – n) = 0.
⇒ (m + n) [m + n + m – n] = 0
⇒ 2m (m + n) = 0
⇒ m = 0, m + n = 0
When m = 0
From (3), l = -n
∴ \(\frac{l}{l}\) = \(\frac{m}{0}\) = \(\frac{n}{-l}\)
When m = – n
From (3) l = -(-n) – n
= n – n
= 0.
∴ \(\frac{l}{0}\) = \(\frac{m}{l}\) = \(\frac{n}{-l}\)
∴ The d.rs of the two lines are (1, 0, -1) and (0, 1, -1) let ‘θ’ be the angle between the lines.

Question 22.
If y = x \(\sqrt{a^2+x^2}\) + a² log (x + \(\sqrt{a^2+x^2}\)) then prove that \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(2 \sqrt{a^2+x^2}\)
Solution:

Question 23.
Show that the condition for 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 P(x₁, y₁) be the point of intersection of the curves
Given curve equation is ax² + by² = 1
Differentiating (1) w.r. to y on both sides, we have

Since ‘P’ lies on ax² + by² = 1 and a₁x² + b₁y² = 1
Solving these two equations, we get

Question 24.
Find the points of local extrema and local extrema for the function f(x) = cos 4x defined on \(\left(0, \frac{\pi}{2}\right)\)
Solution:
The function f(n) = cos 4x defined on (0, \(\frac{\pi}{2}\)) …….. (1)
f'(x) = – 4 sin 4x ……… (2)
f”(x) = -16 cos 4x …….. (3)
f'(x) = 0 ⇒ – 4 sin 4x = 0
⇒ sin 4x = 0.
⇒ 4x = 0, π, 2π, 3π, 4π, ….
⇒ x = 0, \(\frac{\pi}{4}\), \(\frac{\pi}{2}\), \(\frac{3 \pi}{4}\), π
Since x ∈ (0, \(\frac{\pi}{2}\))
∴ x = \(\frac{\pi}{4}\)
f”(\(\frac{\pi}{4}\)) = -16 cos 4 (\(\frac{\pi}{4}\))
= -16 cos π
= – 16 (-1)
= 16 > 0.
∴ The function f has local minimum at x = \(\frac{\pi}{4}\),
∴ Local minimum value = f(\(\frac{\pi}{4}\))
= cos 4(\(\frac{\pi}{4}\))
= cos π
= – 1.