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## AP Inter 2nd Year Maths 2B Question Paper March 2019

Time : 3 Hours

Max. Marks : 75

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

Section – A (10 × 2 = 20)

I. Very Short Answer Type Questions.

- Attempt all questions.
- Each question carries two marks.

Question 1.

Find the equation of the circle for which the points (4, 2), (1, 5) are the end points of a diameter.

Solution:

The equation of the circle for which the points (4, 2), (1, 5) are the end points of a diameter is

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

⇒ (x – 4) (x – 1) + (y – 2) (y – 5) = 0

⇒ x^{2} – x – 4x + 4 + y^{2} – 5y – 2y + 10 = 0

⇒ x^{2} + y^{2} – 5x + 7y + 14 = 0.

Question 2.

Find the value of k if the points (4, 2) and (k, -3) are conjugate points with respect to the circle x^{2} + y^{2} – 5x + 8y + 6 = 0.

Solution:

Let S ≡ x^{2} + y^{2} – 5x + 8y + 6 = 0.

Since the points (4, 2) and (k, -3) are conjugate with respect to the circle s = 0.

∴ S_{12} = 0.

⇒ x_{1}x_{2} + y_{1}y_{2} – \(\frac{5}{2}\)(x_{1} + x_{2}) + 4 (y_{1} + y_{2}) + 6 = 0.

⇒ 4k – 6 – \(\frac{5}{2}\) (4 + k) + 4 (2 – 3) + 6 = 0.

⇒ 4k – 6 – 10 – \(\frac{5}{2}\) k – 4 + 6 = 0.

⇒ \(\frac{3}{2}\) k – 14 = 0

⇒ \(\frac{3}{2}\) k = 14 ⇒ 3k = 28

⇒ k = \(\frac{28}{3}\).

Question 3.

Find the equation of the radical axis of the circles x^{2} + y^{2} – 4x + 6y – 7 = 0, 4(x^{2} + y^{2}) + 8x + 12y – 9 = 0.

Solution:

Let s ≡ x^{2} + y^{2} + 4x + 6y – 7 = 0.

S’ ≡ x^{2} + y^{2} + 2x + 3y – \(\frac{9}{4}\) = 0.

The equation of the radical axis of the circles s = 0, s’ = 0 is s – s’ =0

⇒ (x^{2} + y^{2} + 4x + 6y – 7) – (x^{2} + y^{2} + 2x + 3y – \(\frac{9}{4}\) = 0.

⇒ x^{2} + y^{2} + 4x + 6y – 7 – x^{2} – y^{2} – 2x – 3y + \(\frac{9}{4}\) = 0.

⇒ 2x + 3y – (7 – \(\frac{9}{4}\)) = 0

⇒ 2x + 3y – \(\left(\frac{28-9}{4}\right)\) = 0

⇒ 2x + 3y – \(\frac{19}{4}\) = 0

⇒ 8x + 12y – 19 = 0.

Question 4.

Find the equation of the tangent to the parabola x^{2} – 4x – 8y + 12 = 0 at (4, \(\frac{3}{2}\))

Solution:

Let s ≡ x^{2} – 4x – 8y + 12 = 0.

The equation of the tangent to the parabola s = 0 at (4, \(\frac{3}{2}\)) is s_{1} = 0.

⇒ xx_{1} – 2 (x + x_{1}) – 4 (y + y_{1}) + 12 = 0.

⇒ x (4) – 2 (x + 4) – 4 (y + \(\frac{3}{2}\)) + 12 = 0

⇒ 4x – 2x – 8 – 4y – 6 + 12 = 0.

⇒ 2x – 4y – 2 = 0

⇒ x – 2y – 1 = 0.

Question 5.

Find the product of lengths of the perpendiculars from any point on the hyperbola \(\frac{x^2}{16}\) – \(\frac{y^2}{9}\) = 1 to its asymptotes.

Solution:

Given \(\frac{x^2}{16}\) – \(\frac{y^2}{9}\) = 1

Here a^{2} = 16, b^{2} = 9.

The product or lengths of the perpendiculars from any point on the hyperbola \(\frac{x^2}{16}\) – \(\frac{y^2}{9}\) = 1 to its asymptotes is \(\frac{a^2 b^2}{a^2+b}\) = \(\frac{144}{25}\) = \(\frac{16(9)}{16+9}\)

Question 6.

Evaluate: \(\int \frac{e^x(1+x)}{\cos ^2\left(x e^x\right)} d x\) on I ⊂ R \ {x ∈ R : cos (xe’) = 0}

Solution:

I = \(\int \frac{e^x(1+x)}{\cos ^2\left(x e^x\right)}\)dx

Let xe^{x} = t

(xe^{x} + e^{x.1}) dx = dt

e^{x} (x + 1) dx = dt

∴ I = \(\int \frac{1}{\cos ^2 t}\) dt

= ∫Sec^{2}t dt

= tant + c

= tan (xe^{x}) + c

∴ \(\int \frac{e^x(1+x)}{\cos ^2\left(x e^x\right)}\)dx = tan (xe^{x}) + c.

Question 7.

Evaluate : ∫\(\frac{d x}{(x+1)(x+2)}\).

Solution:

Question 8.

Evaluate : \(\int_0^1 \frac{d x}{\sqrt{3-2 x}}\).

Solution:

I = \(\int_0^1 \frac{d x}{\sqrt{3-2 x}} d x\)

Let 3 – 2x = t

o – 2dx = dt

dx = \(\frac{-1}{2}\)dt

L.L : x = 0 ⇒ t = 3

U.L : x = 1 ⇒ t = 1

Question 9.

Evaluate : \(\int_0^{\frac{x}{2}}\)sin^{6}cos^{4} x dx

Solution:

Question 10.

Form the differential equation corresponding to y = cx – 2c^{2}, where c is a parameter.

Solution:

Given y = cx – 2c^{2}

differentiating with respect to ‘x on both sides, we have \(\frac{d y}{d x}\) = c

substitute ‘c’ value in (1)

∴ y = x\(\frac{d y}{d x}\) – 2\(\left(\frac{d y}{d x}\right)^2\) is the required differential equation.

II. Short Answer Type Questions.

- Answer any five questions.
- Each question carries four marks.

Question 11.

If a point P is moving such that the lengths of tangents drawn from P to the circle x^{2} + y^{2} – 4x – 6y – 12 = 0 and x^{2} + y^{2} – 6x – 18y + 26 = 0 are in the ratio 2 : 3, then find the equation of the locus of P.

Solution:

Question 12.

Find the equation of the circle passing through the points of intersection of the circles x^{2} + y^{2} – 8x – 6y – 21 = 0 x^{2} + y^{2} – 2x – 15 = 0 and (1, 2).

Solution:

Let S ≡ x^{2} + y^{2} – 8y – 6y + 21 = 0

S’ ≡ x^{2} + y^{2} – 2x – 15 = 0

S – S’ = x^{2} + y^{2} – 8x – 6y + 21 – x^{2} – y^{2} + 2x + 15

= 6x – 6y + 36

The equation of the circle passing through the points of intersection of the circles s = 0, S’ = 0 is

s + λ(S – S’) = 0

⇒ (x^{2} + y^{2} – 8x – 6y + 21) + λ(-6x – 6y + 36) = 0

If this circle passing (1, 2) then

(1 + 4 – 8 – 12 + 21) + λ(-6 – 12 + 36) = 0

⇒ 6 + 18λ = 0

⇒ 18λ = – 6

⇒ λ = \(\frac{-1}{3}\)

Hence required circle equation is

(x^{2} + y^{2} – 8x – 6y + 21) – \(\frac{1}{3}\) (-6x – 6y + 36) = 0

⇒ 3x^{2} + 3y^{2} – 24x – 18y + 63 + 6x + 6y – 36 = 0

⇒ 3x^{2} + 3y^{2} – 18x – 12y + 27 = 0

⇒ x^{2} + y^{2} – 6x – 4y + 9 = 0

Question 13.

Find the equation of the ellipse referred to its major and minor axes as the co-ordinate axes X, Y – respectively with latus rectum of length 4 and distance between foci 4\(\sqrt{2}\).

Solution:

Given Length of the latus rectum = 4

⇒ \(\frac{2 b^2}{a}\) = 4

⇒ \(\frac{a^2\left(1-e^2\right)}{a}\) = 2

⇒ a(1 – e^{2}) = 2 ………. (1)

Given distance between foci = 4\(\sqrt{2}\)

⇒ 2ae = 4\(\sqrt{2}\)

⇒ ae = 2\(\sqrt{2}\)

⇒ e = \(\frac{2 \sqrt{2}}{a}\) ………. (2)

From (1)

a(1 – \(\frac{8}{a^2}\)) = 2

⇒ a\(\left(\frac{a^2-8}{a^2}\right)\) = 2

⇒ \(\frac{a^2-8}{a}\) = 2

⇒ a^{2} – 8 = 2a

⇒ a^{2} – 2a – 8 = 0

⇒ (a – 4) (a + 2) = 0

Since a > 0

∴ a = 4

From (2) e = \(\frac{2 \sqrt{2}}{4}\) = \(\frac{1}{\sqrt{2}}\)

b^{2} = a^{2}(1 – e^{2}) = 16(1 – \(\frac{1}{2}\)) = 16(\(\frac{1}{2}\)) = 8.

∴ Required ellipse equation is \(\frac{x^2}{16}\) + \(\frac{y^2}{8}\) = 1.

Question 14.

Show that the locus of the feet of the perpendicular’s drawn from foci to any tangent of the ellipse is the auxiliary circle.

Solution:

Equation of the ellipse is \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1

Equation of the tangent to the ellipse is y = mx ± \(\sqrt{a^2 m^2+b^2}\)

⇒ y – mx ± \(\sqrt{a^2 m^2+b^2}\) ……….. (1)

Equation to the perpendicular from either focus (± ae, 0) on this tangent is

y – 0 = \(\frac{-1}{m}\) (x ± ae)

⇒ my = – (x ± ae)

⇒ x + my = ± ……. (2)

squaring and adding (1) and (2)

∴ (y – mx)^{2} + (x + mx)^{2} = a^{2}m^{2} + b^{2} + a^{2}e^{2}

⇒ y^{2} – 2mxy + m^{2}x^{2} + x + 2mxy + m^{2}y^{2} = a^{2}m^{2} + a^{2} – a^{2}e^{2} + a^{2}e^{2}

⇒ (1 + m^{2})x^{2} + (1 + m^{2})y^{2} = (1 + m^{2})

⇒ x^{2} + y^{2} = a^{2}

∴ The locus is the auxiliary circle concentric with the ellipse.

Question 15.

Find the equations of the tangents to the hyperbola x^{2} – 4y^{2} = 4 which are

(i) Parallel

(ii) Perpendicular to the line x + 2y = 0.

Solution:

Given hyperbola equation is x^{2} – 4y^{2} = 4

⇒ \(\frac{x^2}{4}\) + \(\frac{y^2}{1}\) = 1 ……… (1)

Here a^{2} = 4, b2 = 1

Given line equation is x + 2y = 0 …….. (2)

(i) Equation of a line parallel to (2) is x + 2y + k = 0

⇒ 2y = – x – k

⇒ y = –\(\frac{1}{2}\)x – \(\frac{k}{2}\)

If this line is a tangent to the hyperbola (1) then

c^{2} = a^{2}m^{2} – b^{2}

\(\left(\frac{-k}{2}\right)^2\) = 4\(\left(\frac{-1}{2}\right)^2\) – 1

\(\frac{k^2}{4}\) = 1 – 1

k^{2} = 0 ⇒ k = 0

∴ Equation to the required tangent is x + 2y = 0

ii) Equation of a line perpendicular to (2) is 2x – y + k = 0

⇒ y = 2x + k

If this line is a tangent to the hyperbola (1) then

c^{2} = a^{2} m^{2} – b^{2}

k^{2} = 4 (2)^{2} – 1

k^{2} = 16 – 1 = 15

k = ± \(\sqrt{15}\)

∴ Equation to the required tangent is 2x – y ± \(\sqrt{15}\) = 0.

Question 16.

Find the area, of one of the curvilinear triangles bounded by y = sin x, y = cos x and X-axis.

Solution:

Question 17.

Solve : x(x – 1)\(\frac{d y}{d x}\) – y = x^{3} (x – 1)^{3}.

Solution:

Section – C

III. Long Answer Type Questions.

- Answer any five questions.
- Each question carries seven marks.

Question 18.

Show that the following four points (1, 1), (-6, 0), (-2, 2), (-2, -8) are concyclic and find the equation of the circle on which they lie.

Solution:

Suppose the equation of the required circle is

x^{2} + y^{2} + 2gx + 2fy + c = 0 ………. (i)

This circle passes through A (1, 1)

1 + 1 + 2g + 2f + c = 0

⇒ 2g + 2f + c = 2 ………… (ii)

This circle passes through B (-6, 0)

36 + 0 + 12g + 0 + c = 0

-12g + c = -36 ……… (iii)

This circle passes through C (-2, 2)

4 + 4 + 4g + 4f + c = 0

-4g + 4f + c = -8 ……….. (iv)

(iii) – (iv) gives – 8g – 4f = 0

⇒ 2g + f = 7

(i) – (ii) gives 14g + 2f = 34

7g + f = 17 …….. (v)

Solving (iv) and (v) we get g = 2, f = 3

Putting g = 2, f = 3 in

⇒ 4 + 6 + c = -2

c = – 12

Required circle is x^{2} + y^{2} + 4x + 6y – 12 = 0

Substituting (-2, -8) in this above equation, we get 4 + 64 – 8 – 48 – 12 = 68 – 68 = 0

(-2, -8) satisfies the above equation

∴ A, B, C, D are concyclic.

Equation of the circle is x^{2} + y^{2} + 4x + 6y – 12 = 0.

Question 19.

Find the transverse common tangents of the circles

x^{2} + y^{2} – 4x – 10y + 28 = 0 and x^{2} + y^{2} + 4x – 6y + 4 = 0.

Solution:

= [1 + \(\frac{81}{4}\) – 4 – 10 × \(\frac{9}{2}\) + 28]

(-2x – y + 7)^{2} = (x^{2} + y^{2} – 4x – 10y + 28)

4x^{2} + y^{2} + 4xy – 28x – 14y + 4 = x^{2} + y^{2} – 4x – 10y + 28

3x^{2} + 4xy – 24x – 4y + 21 = 0

(3x + 4y – 21); (x – 1) = 0

3x + 4y – 21 = 0; x – 1 = 0.

Question 20.

Find the equation of the parabola whose focus is S (3, 5) and vertex is A (1, 3).

Solution:

Given s = (3, 5) and vertex A = (1, 3)

Since A is the mid point or SZ

∴ Z = 2A – S

= (2 – 3, 6 – 5)

= (-1, 1)

Slope of SZ = \(\frac{5-1}{3+1}=\) = \(\frac{4}{4}\) = 1

Slope of directrix = \(\frac{-1}{1}\) = -1

Equation of directrix is

y – 1 = – 1 (x + 1)

y – 1 = -x – 1

Question 21.

Evaluate : ∫\(\frac{\cos x+3 \sin x+7}{\cos x+\sin x+1}\) dx.

Solution:

Let I = \(\int \frac{\cos x+3 \sin x+7}{\cos x+\sin x+1}\) dx

Let cos x + 3 sin x + 7 = A \(\frac{d}{d x}\) (cos x + sin x + 1) + B (cos x + sin x + 1) + k

⇒ cos x + 3 sin x + 7 = A (- sin x + cos x) + B (cos x + sin x + 1) + k

– A + B = 3 ⇒ A – B + 3 = 0 ………. (1)

A + B = 1 ⇒ A + B – 1 = 0 ……… (2)

B + K = 7 ……….. (3)

(1) + (2) ⇒ 2 A + 2 = 0

⇒ A + 1 = 0

⇒ A = – 1

From (2)

– 1 + B – 1 = 0

⇒ B = + 2

From (3)

2 + k = 7

⇒ k = 5

∴ A = – 1, B = 2, k = 5

Question 22.

Obtain the reduction formula for I_{n} = ∫cosec^{n}x dx, n being a positive integer, n ≥ 2 and deduce the value of ∫cosec^{5}x dx.

Solution:

Question 23.

Evaluate : \(\int_0^{\frac{\pi}{4}}\) log (1 + tan x) dx.

Solution:

Question 24.

Solve : \(\frac{d y}{d x}\) = \(\frac{y^2-2 x y}{x^2-x y}\)

Solution:

From (2)