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

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AP Inter 1st Year Maths 1B Question Paper March 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 slopes of the lines x + y = 0 and x – y = 0.
Solution:
Given line equation is x + y = 0
⇒ y = -x
⇒ y = (-1)x
∴ slope = -1
Given line equation is x – y = 0
⇒ y = x
⇒ y = 1(x)
∴ slope = -1.

Question 2.
Transform the equation x + y + 1 =0 in to normal form.
Solution:
Given line equation is x + y + 1 = 0
⇒ x + y = -1
⇒ x + y = 1
⇒ (-1)x + (-1)y = 1
⇒ \(\frac{-1}{\sqrt{2}}\)x + \(\left(\frac{-1}{\sqrt{2}}\right)\)y = \(\frac{1}{\sqrt{2}}\)
⇒ x cos\(\left(\frac{5 \pi}{4}\right)\) + y sin\(\left(\frac{5 \pi}{4}\right)\) = \(\frac{1}{\sqrt{2}}\)
Which is the required normal form.

Question 3.
If (3, 2, -1), (4, 1, 1) and (6, 2, 5) are three vertices and (4, 2, 2) is the centroid of a tetrahedron, find the fourth vertex.
Solution:
Let A = (3, 2, -1)
B = (4, 1, 1)
C = (x, y, z)
D = (x, y, z)
Given centroid of a tetrahedran ABCD = (4, 2, 2)
⇒ \(\left(\frac{3+4+6+x}{4}, \frac{2+1+2+y}{4}, \frac{-1+1+5+z}{4}\right)\) = (4, 2, 2)

Question 4.
Find the angle between the planes 2x – y + z – 6 and x + y + 2z – 7.
Solution:
Given plane equations are 2x – y + z – 6 = 0 ……. (1)
x + y + 2z – 7 = 0 …… (2)
Let ‘θ’ be the acute angle between the planes (1) and (2)

Question 5.
Compute \(\lim _{x \rightarrow 0} \frac{e^{7 x}-1}{x}\)
Solution:
\(\lim _{x \rightarrow 0} \frac{e^{7 x}-1}{x}\) = 7\(\lim _{7 x \rightarrow 0} \frac{e^{7 x-1}}{7 x}\)
= 7(1)
= 7
∴ \(\lim _{x \rightarrow 0} \frac{e^{7 x}-1}{x}\) = 7

Question 6.
Compute \(\lim _{x \rightarrow \infty} \frac{x^2+5 x+2}{2 x^2-5 x+1}\)
Solution:

Question 7.
Find the derivative of 5 sin x + e^x log x.
Solution:
Let y = 5 sin x + e^x log x
differentiate w.r.to ‘x’ both sides, we have

Question 8.
Find the derivative of sec^-1\(\left(\frac{1}{2 x^2-1}\right)\), (0 < x < \(\frac{1}{\sqrt{2}}\))
Solution:
Let y = sec^-1\(\left(\frac{1}{2 x^2-1}\right)\)
Put x = cos θ ⇒ e = cos^-1x

Question 9.
Find dy and ∆y of y = f(x) = x² + x at x = 10 when ∆x = 0.1.
Solution:
Given y = f (x) = x² + x
dy = f¹(x)∆x
= (2x + 1) ∆x
At x = 10 when ∆x = 0.1
∴ dy = [2(10) + 1](0.1)
= (20 + 1) (0.1)
= (2, 1) (0.1)
= 2.1
∆y= f(x)+ ∆x) – f(x)
= (x + ∆x)² + (x + ∆x) – (x² + x)
= x² + 2x ∆x + x + ∆x – x² – x
= 2x∆x + (∆x)² + ∆x
At x = 10 when ∆x = 0.1
∆y = 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.
Verify Rolle’s theorem for the function y = f (x) = x² + 4 in [-3, 3]
Solution:
Given y = f(x) = x² + 4
Since f is a second degree polynomial
∴ f is continuous on [-3, 3] and f is derivable on (-3, 3)
Also f(-3) = (-3)² + 4 = 9 + 4 = 13
f (3) = 3² + 4 = 9 + 4 = 13
∴ f(-3) = f(3)
∴ f satisfies all the conditions of Roile’s theorem.
∴ there exists ct (-3, 3) such that f¹(c) = 0
f (x) = x² + 4
⇒ f¹(x) = 2x
⇒ f¹(c) = 2c
⇒ 0 = 2c ⇒ c = 0t (-3, 3)
Hence Rolle’s theorem is verified.

Section – B

II. Short answer type questions.

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

Question 11.
A (1, 2), B (2, -3) and C (-2, 3) are three points. A point P moves such that PA² + PB² = 2PC² then find the equation of locus of P.
Solution:
Given A = (1, 2) B = (2, -3) and c = (-2, -3)
Let P(x₁, y₁) be any point on the locus
Given geometric condition is
PA² + PB² = 2PC²

⇒ -6x₁ + 2y₁ + 18 = 8x₁ – 12y₁ + 26
⇒ 14x₁ – 14y₁ + 8 = 0
⇒ 7x₁ – 7y₁ + 4 = 0
∴ Locus of P is 7x – 7y + 4 = 0

Question 12.
When the axes are rotated through an angle \(\frac{\pi}{4}\), find the transformed equation of 3x² + 10 xy + 3y² = 9.
Solution:
Given equation is 3x² + 10xy + 3y² = 9 ………. (1)

⇒ 3 (x² – 2xy + y² ) + 10 (x² – y² ) + 3(x² + 2xy + y² ) = 18
⇒ 3x² – 6xy + 3y² + 10x² – 10y² + 3x² + 6xy + 3y² = 18
⇒ 16x² – 4y² = 18
⇒ 8x² – 2y² = 9

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

∴ Point of intersection of (1) and (2) is (-1, 2) since (1), (2), (3) are concurrent.
∴ (-1, 2) lies on (3)
∴ p(-1) + 4(2) – 6 = 0
⇒ -p + 8 – 6 = 0
⇒ -p + 2 = 0
⇒ P = 2

Question 14.
Check the continuity of the following function at 2.

Solution:

Question 15.
Find the derivative of cot x from the first principle.
Solution:

Question 16.
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 given by S = f(t) = 8t + t³. Find
(i) the velocity at time t = 2 sec,
(ii) the initial velocity and
(iii) acceleration at t = 2 sec.
Solution:
The distance s and the time are connected by the relation
s = f(t) = 8t + t³
differentiate w.r. to f on both sides, we have
v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 8(1) + 3t² = 8 + 3t²
again differentiate w.r. to ‘t’ on both sides, we have
a = \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = \(\frac{d^2 s}{d t^2}\) = 0 + 3(2t) = 6t

i) Velocity at t = 2 sec
V = \(\left(\frac{d s}{d t}\right)_{t=2}\) = 8 + 3.2²
= 8 + 12
= 20 cm/sec

ii) initial velocity (t = 0)
v = \(\left(\frac{d s}{d t}\right)_{t=0}\) = 8 + 3.0^{2</sup = 8 cm/sec}

iii) Acceleration at t = 2 sec
a = \(\left(\frac{d^2 s}{d t^2}\right)_{t=2}\) = 6(2)
= 12 cm/sec².

Question 17.
Find the equations of tangent and normal to the curve xy = 10 at (2, 5).
Solution:
Given curve equation is xy = 10
differentiate w.r. to ‘x’ on both sides, we have
x\(\frac{d y}{d x}\) + y.1 = 10
⇒ x\(\frac{d y}{d x}\) = -y
⇒ \(\frac{d y}{d x}\) = \(\frac{-y}{x}\)
slope m = \(\left(\frac{d y}{d x}\right)_{(2,5)}\) = \(\frac{-5}{2}\)
Tangent equation is
y – y₁ = m(x – x₁)
⇒ y – 5 = \(\frac{-5}{2}\)(x – 2)
⇒ 2y – 10 = -5x + 10
⇒ 5x + 2y – 20 = 0
Normal equation is y – y₁ = \(\frac{-1}{m}\)(x – x₁)
⇒ y – 5 = \(\frac{-1}{-5 / 2}\)(x – 2)
⇒ y – 5 = \(\frac{2}{5}\)(x – 2)
⇒ 5y – 25 = 2x – 4
⇒ 2x – 5y + 21 = 0

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 (-2, 3), (2, -1) and (4, 0).
Solution:
Let A = (-2, 3)
B = (2, -1)
C = (4, 0)
Let S(α, β) be the circumcenter of the ∆ABC

Question 19.
Show that the area of the triangle formed by the lines ax² + 2hxy + by² = 0, lx + my + n = 0 is \(\left|\frac{n^2 \sqrt{h^2-a b}}{a m^2-2 h l m+b l^2}\right|\)
Solution:
Let \(\overleftrightarrow{\mathrm{OA}}\) and \(\overleftrightarrow{\mathrm{OB}}\) be the pair of straight lines represented by the equation
ax² + 2hxy + by² = 0 (see figure)
and \(\stackrel{\leftrightarrow}{\mathrm{AB}}\) be the line lx + my + n = 0

Question 20.
Find the values of K, if the lines joining the origin to the points of intersection of the curve 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 + 2a²lmxy + a²m²y² – 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 ⇒ -m – n ……. (1)
l² + m² – n² = 0 ……… (2)
Eliminate ‘l’ from (1) and (2)
(-m – n)² + (m² – n²) = 0
⇒ (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

∴ Direction ratios of the lines are (-1, 0, 1) and (0, -1, 1)
If ‘θ’ is the acute angle between the lines then

Question 22.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\), if y = (sin x)log x + x sinx.
Solution:
Given y = (sin x)^logx + x^sinx
Let u = (sinx)^logx
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.
Find the angle between the curves xy = 2, x² + 4y = 0.
Solution:

Question 24.
A wire of length l is cut into two parts which are bent respectively in the form of a square and a circle. What are the lengths of the pieces of the wire respectively so that the sum of the areas is the least?
Solution: