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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 :

- Answer all the questions.
- 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^{-1}x

Question 9.

Find dy and ∆y of y = f(x) = x^{2} + x at x = 10 when ∆x = 0.1.

Solution:

Given y = f (x) = x^{2} + x

dy = f^{1}(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) – (x^{2} + x)

= x^{2} + 2x ∆x + x + ∆x – x^{2} – 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) = x^{2} + 4 in [-3, 3]

Solution:

Given y = f(x) = x^{2} + 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) = 3^{2} + 4 = 9 + 4 = 13

∴ f(-3) = f(3)

∴ f satisfies all the conditions of Roile’s theorem.

∴ there exists ct (-3, 3) such that f^{1}(c) = 0

f (x) = x^{2} + 4

⇒ f^{1}(x) = 2x

⇒ f^{1}(c) = 2c

⇒ 0 = 2c ⇒ c = 0t (-3, 3)

Hence Rolle’s theorem is verified.

Section – B

II. Short answer type questions.

- Attempt any five questions.
- 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 PA^{2} + PB^{2} = 2PC^{2} then find the equation of locus of P.

Solution:

Given A = (1, 2) B = (2, -3) and c = (-2, -3)

Let P(x_{1}, y_{1}) be any point on the locus

Given geometric condition is

PA^{2} + PB^{2} = 2PC^{2}

⇒ -6x_{1} + 2y_{1} + 18 = 8x_{1} – 12y_{1} + 26

⇒ 14x_{1} – 14y_{1} + 8 = 0

⇒ 7x_{1} – 7y_{1} + 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 3x^{2} + 10 xy + 3y^{2} = 9.

Solution:

Given equation is 3x^{2} + 10xy + 3y^{2} = 9 ………. (1)

⇒ 3 (x^{2} – 2xy + y^{2} ) + 10 (x^{2} – y^{2} ) + 3(x^{2} + 2xy + y^{2} ) = 18

⇒ 3x^{2} – 6xy + 3y^{2} + 10x^{2} – 10y^{2} + 3x^{2} + 6xy + 3y^{2} = 18

⇒ 16x^{2} – 4y^{2} = 18

⇒ 8x^{2} – 2y^{2} = 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 + t^{3}. 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 + t^{3}

differentiate w.r. to f on both sides, we have

v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 8(1) + 3t^{2} = 8 + 3t^{2}

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 + 3.2^{2}

= 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/sec^{2}.

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 – y_{1} = m(x – x_{1})

⇒ y – 5 = \(\frac{-5}{2}\)(x – 2)

⇒ 2y – 10 = -5x + 10

⇒ 5x + 2y – 20 = 0

Normal equation is y – y_{1} = \(\frac{-1}{m}\)(x – x_{1})

⇒ 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 :

- Attempt any five questions.
- 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 ax^{2} + 2hxy + by^{2} = 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

ax^{2} + 2hxy + by^{2} = 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 2x^{2} – 2xy + 3y^{2} + 2x – y – 1 = 0 and the line x + 2y = K are mutually perpendicular.

Solution:

Equation of the circle is x^{2} + y^{2} = a^{2} …….. (1)

Equation of AB is lx + my = 1 …….. (2)

Homogenising (1) with the help of (2)

Combined equation of OA, OB is

x^{2} + y^{2} = a^{2}.1^{2}

x^{2} + y^{2} = a^{2} (lx + my)^{2}

= a^{2}(l^{2}x^{2} + m^{2}y^{2} + 2lmxy)

= a^{2}l^{2}x^{2} + a^{2}m^{2}y^{2} + 2a^{2}lmxy

i.e., a^{2}l^{2}x^{2} + 2a^{2}lmxy + a^{2}m^{2}y^{2} – x^{2}-y^{2} = 0

(a^{2}l^{2} – 1) x^{2} + 2a^{2} lmxy + (a^{2}m^{2} – 1)y^{2} = 0

Since OA, OB are perpendicular

Co-efficient of x^{2} + co-efficient of y^{2} = 0

a^{2}l^{2} – 1 + a^{2}m^{2} – 1 = 0

a^{2}(l^{2} + m^{2}) = 2

This is the required condition.

Question 21.

Find the angle between the lines whose direction cosines satisfy the equations l + m + n = 0, l^{2} + m^{2} – n^{2} = 0.

Solution:

Given l + m + n = 0 ⇒ -m – n ……. (1)

l^{2} + m^{2} – n^{2} = 0 ……… (2)

Eliminate ‘l’ from (1) and (2)

(-m – n)^{2} + (m^{2} – n^{2}) = 0

⇒ (m + n)^{2} + (m^{2} – n^{2}) = 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, x^{2} + 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: