AP Inter 1st Year Maths 1B Question Paper May 2017

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AP Inter 1st Year Maths 1B Question Paper May 2017

Time : 3 Hours
Max. Marks : 75

Note : This question paper consists of THREE sections A, B and C.

Section – A
(10 × 2 = 20 Marks)

I. Very short answer type questions :

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

Question 1.
Find the equation of the straight line passing through (-2, 4) and making non-zero intercepts whose sum is zero.
Solution:
Let x – intercept = a
y – intercept = b
Given a + b = 0
⇒ b = -a
Intercept form \(\frac{x}{a}+\frac{y}{b}\) = 1
⇒ \(\frac{x}{a}+\frac{y}{-a}\) = 1
⇒ x – y = a
If this line passes through (-2, 4) then
-2 – 4 = a ⇒ a = -6
∴ Required straight line equation is x – y = – 6
⇒ x – y + 6 = 0

Question 2.
Find the value of k, if the straight lines 6x – 10y + 3 = 0 and kx – 5y + 8 = 0 are parallel.
Solution:
Given straight line equations are
6x – 10y + 3 = 0 ………. (1)
kx – 5y + 8 = 0 ………… (2)
If (1), (2) are parallel then
\(\frac{6}{k}\) = \(\frac{-10}{-5}\) ⇒ \(\frac{6}{k}\) = 2
⇒ 2k = 6
⇒ k = 3

Question 3.
Show that the points (5, 4, 2), (6, 2, -1) and (8, -2, -7) are collinear. Find the equation of the plane passing through (1, 1, 1) and parallel to the plane x + 2y + 3z – 7 = 0.
Solution:
Let A = (5, 4, 2)
B = (6, 2, -1)
C = (8, -2, -7)

Question 4.
Find the equation of the plane passing through (1, 1, 1) and parallel to the plane x + 2y + 3z – 7 = 0
Solution:
The equation of the plane passing through (1, 1, 1) and parallel to the plane x + 2y + 3z – 7 = 0 is 1(x – 1) + 2(y – 1) + 3(z – 1) = 0
⇒ x – 1 + 2y – 2 + 3z – 3 = 0
⇒ x + 2y + 3z – 6 = 0

Question 5.
Find : \(\lim _{x \rightarrow 0}\left(\frac{\sqrt{1+x}-1}{x}\right)\)
Solution:

Question 6.
Check the continuity of the function :

Solution:

Question 7.
If f(x) = log(sec x + tan x), then find f(x).
Solution:
Given f(x) = log(sec x + tan x) differentiate w.r. to ‘x’ on bothsides,
we have f'(x) = \(\frac{1}{\sec x+\tan x}\) (sec x tan x + sec²x)
= \(\frac{\sec x(\tan x+\sec x)}{(\sec x+\tan x)}\)
= sec x
∴ f'(x) = sec x

Question 8.
If y = e^t + cos^t t, x = log t + sin t, find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).
Solution:
Given y = e^t + cos t
\(\frac{d y}{d t}\) = e^t – sin t
Given x = log t + sin t

Question 9.
If y = f(x) = x² + x, then find dy and ∆y when x = 10 and ∆x = 0.1.
Solution:
Given y = f(x) = x² + x. x = 10 and ∆x = 0.1
dy = f(x) ∆x
= (2x + 1) ∆x
= [2(10)+1](0.1)]
= (21) (0.1)
= 2.1
∆y = f (x + ∆x) – f (x)
= (x + ∆x)² + (x + ∆x) – (x² + x)
= x² + 2x ∆x + (∆x)² + x + ∆x – x² – x
= 2x ∆x + (∆x)² + ∆x
= 2(10) (0.1) + (0.1)² + 0.1
= 2 + 0.01 + 0.1
= 2.11
∴ dy = 2.1 and ∆y = 2.11

Question 10.
Define Rolle’s theorem.
Solution:
Rolle’s theorem : If f : [a, b] → R is a function such that

1. f is continuous on [a, b] c,
2. f is derivable on (a, b) and
3. f(a) = f(b) then there exists C ∈ (a, b) such that f'(c) = 0

Section – B

II. Short answer type questions : (5 × 4 = 20)

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 the 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 triangle PAB is 9 sq. units.

Question 12.
When the origin is shifted to the point (2, 3), the transformed equation of a curve is :
x² + 3xy – 2y² + 17x – 7y – 11 = 0.
Find the original equation of the curve.
Solution:
Equations of transformation are x = x’ + h, y = y’ + k
x’ – x – h = x – 2, y’ = y – 3
Transformed equation is x² + 3xy – if + 17x – 7y – 11 = 0 original equation is (x – 2)² + 3(x – 2) (y – 3) – 2 (y – 3)² + 17(x – 2) – 7(y – 3) – 11 = 0
x² – 4x + 4 + 3xy – 9x – 6y + 18 – 2y² + 12y – 18 + 17x – 34 – 7y + 21 – 11 = 0
x² + 3xy – 2y² + 4x – y – 20 = 0
This is the required original equation.

Question 13.
Find the value of p if the lines
3x + 4y = 5,
2x + 3y = 4 and
px + 4y = 6 are concurrent.
Solution:
Given line equations are
3x + 4y – 5 = 1 ……… (1)
2x + 3y – 4 = 0 ………. (2)
px + 4y – 6 = 0 ………. (3)

∴ The point of intersection of (1) & (2) is (-1, 2)
Since (1), (2) & (3) are concurrent.
∴ The point (-1, 2) lies on (3)
∴ p(-1) + 4(2) – 6 = 0
⇒ -p + 8 – 6 = 0
⇒ -p + 2 = 0
⇒ p = 2

Question 14.
Find : \(\lim _{x \rightarrow 0} \frac{\cos a x-\cos b x}{x^2}\)
Solution:

Question 15.
Find the derivative of tan 2x from the first principle.
Solution:
Let f(x) = tan 2x
By first principle
f'(x) = \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

Question 16.
A stone is dropped into a quiet lake and ripples move in circles at the speed of 5 cm/sec. At the instant when the radius of circular ripple is 8 cm, how fast the enclosed area increases ?
Solution:
Let r, A be the radius, area of the circle respëctivelÿ
Given that t = 5 cm/sec and r = 8
A = πr²
⇒ \(\frac{d A}{d t}\) = \(\pi \cdot 2 r \frac{d r}{d t}\)
⇒ \(\frac{\mathrm{dA}}{\mathrm{dt}}\) = π.2(5).8
= 80 π cm²/sec.

Question 17.
Show that at any point (x, y) on the curve y = b e^x/a, the length of subtangent is a constant and the length of the subnormal \(\frac{y^2}{a}\).
Solution:
Given curve equation is y = be^x/a differentiate w.r.to ‘x’ on both sides, we have
y’ = \(b . e^{x / a} \cdot \frac{1}{a}\)
= \(\frac{b}{a} e^{x / a}\)
∴ The length of the subtangent at any point on the curve
= \(\left|\frac{y}{y^{\prime}}\right|\)
= \(\left|\frac{b e^{x / a}}{\frac{b}{a} e^{x / a}}\right|\) = |a| (constant)
∴ Length of the subtangent is constant.
∴ The length of the subnormal at any point (x, y) on the curve.
= |yy’| = \(\left|y \cdot \frac{b}{a} e^{x / a}\right|\) = \(\left|\mathbf{y} \cdot \frac{\mathbf{y}}{\mathbf{a}}\right|\) = \(\left|\frac{y^2}{a}\right|\)

Section – C

III. Long answer type questions :

1. Attempt ANY Five questions,
2. Each question carries SEVEN marks.

Question 18.
Find the circumcenter of the triangle whose vertices are (1, 3), (-3, 5) and (5,-1).
Solution:
Let A = (1, 3).
B = (-3, 5)
C = (5, -1)
Let s(α, β) be the circumcenter of ∆ABC
∴ SA = SB = SC
SA = SB ⇒ SA² = SB²
⇒ (α – 1)² + (β – 3)² = (α + 3)² + (β – 5)²
⇒ α² – 2α + 1 + β² – 6β + 9 = α² + 6α + 9 + β² – 10β + 25
⇒ 8 – 4β + 24 = 0
⇒ 2α – β + 6 = 0 ……….. (1)
SB = SC ⇒ SB² = SC²
⇒ (α – 3)² + (β – 5)² = (α – 5)² + (β + 1)²
⇒ α² – 6α + 9 + β² – 10β + 25 = α² + 10α + 25 + β² – 2β + 1
⇒ 16α – 12β + 8 = 0
⇒ 4α – 3β + 2 = 0 …….. (2)
Solving (1) and (2)

Question 19.
Show that the product of the perpendicular distances from a point (α, β) to the pair of straight lines ax² + 2hxy + by² = 0 is :
\(\frac{\left|a \alpha^2+2 h \alpha \beta+b \beta^2\right|}{\sqrt{(a-b)^2+4 h^2}}\)
Solution:
Combined equation of OA, OB is
(x + 2a)² + [late](\sqrt{3} y)^2[/latex] = 0
(x + 2a)² – \((\sqrt{3} y)^2\) = 0

Equation of OA is x + \(\sqrt{3} y\) + 2a = 0 …….. (1)
Equation of OB is x – \(\sqrt{3} y\) + 2a = 0 ……… (2)
Equation of AB is x – a = 0

Question 20.
Write down the equation of the pair of straight lines joining the origin to the points of intersection of the line 6x – y + 8 = 0 with pair of straight lines :
3x² + 4xy – 4y² – 11 x + 2y + 6 = 0.
Solution:
Given pair of lines is
3x² + 4xy – 4y² – 11x + 2y + 6 = 0 ……….. (1)
Given line is

is eq. of pair of lines joining the origin to the point of intersection in.
The eq. pair of angle bisectors to (3) is
h(x² – y²) – (a – b)xy = 0
0(x² – y²) – (4 – 1 )xy = 0
⇒ xy = 0
x = 0 ory = 0 [Eqs. is of coordinate axes]
∴ The pair of lines are equally inclined to the co-ordinate axes.

Question 21.
Find the angle between the lines whose direction cosines satisfy the equations :
3l + m 5n = 0 and
6mn – 2nl + 5lm = 0
Solution:
Given 3l + m + 5n = 0 ……….. (1)
6mn – 2nl + 5lm = 0
From (1), m = – (3l + 5n)
Substituting in (2)
-6n(3l + 5n) – 2nl – 5l (3l + 5n) = 0
-18ln – 30n² – 2nl – 15l² – 25ln = 0
-15l² – 45ln – 30n² = 0
l² + 3ln + 2n² = 0
l² + 2ln + ln + 2n² = 0
l (l + 2n) + n(l + n) = 0
(l + 2n) (l + n) = 0
l + 2n = 0 or l + n = 0

Suppose ‘θ’ is the angle between the lines l₁ and l₂

Question 22.
Find the derivative of the function :
y = (sin x)^{log x} + x^{sin x}
Solution:
Given y = (sin x)^{log x} + x^{sin x}
Let u = (sin x)^{log x}
Taking logarithms on bothsides, we have
log u = log x. log (sin x)
Differentiating w.r.to ‘x’ on bothsides, we have

Let v = x^{sin x}
Taking logarithems on bothsides, we have
log v = sin x. log x
Differentiating w.r.to x on bothsides, we have

Question 23.
If the tangent at any point on the curve
\(x^{\frac{2}{3}}+y^{\frac{2}{3}}\) = \(a^{\frac{2}{3}}\) intersects the co-ordinate axes at A and B, then show that the length AB is a constant.
Solution:
Equation of the curve is x^2/3 + y^2/3 = a^2/3
Differentiating w.r.to x

Question 24.
A window is in the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 20 ft, find the maximum area.
Solution:

Suppose ‘r’ is the radius and h be the height of the cylinder.
From ∆ OAB, OA² + AB² = OB²