TYPICAL QUESTIONS & ANSWERS

PART – I

OBJECTIVE TYPES QUESTIONS

Each Question carries 2 marks.

Choose correct or the best alternative in the following:

Q.1 The co-ordinates of the middle points of the sides of a triangle are (4, 2) (3, 3) and (2, 2). Then the co-ordinates of the centroid are

(A) . (B) (3, 3).

(C) (4, 3). (D) (4, 7).

Ans: A

Coordinate of the Centroid is

Coordinate of the centroid =

Q.2 If x, 2x +2, 3x + 3 are first three terms of a G.P. then its 4^th term is

(A) 27. (B) -27.

(C) 13.5. (D) -13.5.

Ans: D

If x = -1 then three terms are

-1, 0, 0

If x = -4 then the first three terms are

-4, -6, -9

Therefore common ratio is

4^th terms = -13.5

Q.3 The angle made by any diameter of a circle at any point on the circumference is

(A) 90° (B) 180°

(C) 45° (D) 60°

Ans: D

Q.4 If then the value of r is

(A) 6. (B) 5.

(C) 4. (D) 7.

Ans:A

Or,

Or,

Q.5 is equal to

(A) . (B) .

(C) . (D) .

Ans: A

Q.6 If then is equal to

(A) 4896. (B) 816.

(C) 1632. (D) 408.

Ans: B

Q.7 is

(A) . (B)5.

(C) Not defined. (D) .

Ans: A

=

=

Q.8 is equal to

(A) . (B) .

(C) . (D) .

Ans: D

Q.9 The equation of the straight line which makes equal intercepts on the axes and passes through the point (1, 2) is

(A) x + y = 3 (B) x + 2y = 5

(C) x – y = 1 (D) 2x + y = 4

Ans: A

Straight line having equal intercepts on axes is x + y = a. If it passes through (1, 2), then a = 3. Hence required straight line x + y = 3.

Q.10 Area of the triangle whose vertices are (a, b) (a, a + b), (-a, -a + b) is

(A) a²b² (B) a² + b²

(C) a² (D) b²

Ans: C

Area of reqd.

= a²

Q.11

(A) 1 (B)

(C) (D) Zero

Ans: B

=

Q.12 The point on the curve y² = 4x at which the tangent to the curve is parallel to y = x is

(A) (0, 0) (B) (2,

(C) (4, 4) (D) (1, 2)

Ans: D

Here . If tangent is parallel to y = x,

Q.13 is equal to

(A) tan x – cot x (B) tan x + cot x

(C) sec x + cosec x (D) sec x - cosec x

Ans: C

Q.14 Sin³x dx is equal to

(A) (B)

(C) (D)

Ans: A

(By formula)

Q.15 Solution of differential equation is

(A) e^x + e^y = const (B) e^x – e^y = const

(C) e^x . e^y = const (D) e^x / e^y = const

Ans:

Q.16 Period of Sin (2x + 3) is

(A) 2π (B)

(C) π (D)

Ans: C

= = ……….Hence period is π.

Q.17 The value of Sin 105⁰ + Cos 105⁰ is

(A) (B)

(C) (D)

Ans: D

=

Q.18 If pth, (2p)^th and (3p)^th terms of a G.P. are x, y, z respectively, then x, y, z are in

(A) A.P. (B) H.P.

(C) G.P. (D) None of these

Ans: C

If a, ar, ar², ar³, ……….. be the G.P. then T_p = x = ar^p-1, T_2p = y ar^2p-1, T_3p = Z = ar^3p-1. Evidently y² = xz. Hence x, y, z are in G.P.

Q.19 Sum of the series is equal to

(A) 348551 (B) -1000

(C) 5151 (D) None of the above

Ans: C

Q.20 The value of tan is

(A) (B)

(C) (D)

Ans: A

Q.21 In a triangle ABC, let a = BC, b = CA and c = AB. If , then

(A) (B)

(C) (D) None of the above

Ans: C

Q.22 The circles and cut orthogonally if the

value of p is

(A) 3 (B) -2

(C) -3 (D) 1

Ans: A

Q.23 The eccentricity of the ellipse is

(A) 3 (B)

(C) 5 (D)

Ans: D

Q.24 The derivative of – cos (log x) is

(A) sin (log x) (B)

(C) – sin (log x) (D)

Ans: B

Q.25 The value of the is

(A) 0 (B) 1

(C) e (D) Does not exist

Ans: B

Q.26 The integral is equal to

(A) (B)

(C) 0 (D) 1

Ans: D

Q.27 The area under the curve between x = 0 and x = 1 is

(A) 1 (B)

(C) (D)

Ans: C

Q.28 The solution of is

(A) (B)

(C) (D)

Ans: A

Q.29 If one root of the equation is of the other root, then K is

(A) 2 (B) 8

(C) 10 (D) 12

Ans: D.

Q.30 The centroid of the triangle formed by the straight lines is

(A) (0, 0) (B) (1, 0)

(C) (0, 1) (D) (1, 1)

Ans: C.

The points of intersection of the

Given straight lines are A(3, 0), B(0, 3), C(-3, 0)

Q.31 The distance between the parallel lines 3x + 4y + 5 = 0 and 3x + 4y + 15 = 0 is

(A) 1 (B) 2

(C) 3 (D) 5

Ans: B.

The distance of origin from the line 3x + 4y +5 = 0 is

The distance of origin from the line 3x + 4y +15 = 0 is

\ distance between parallel lines =

Q.32 , where m n is equal to

(A) m (B) n

(C) m – n (D) m + n

Ans: C.

Q.33 If then is equal to

(A) 2 sin 4x (B) 4 sin 2x

(C) sin 4x (D) 2 sin 2x

Ans: A.

Q.34 is equal to

(A) (B) log

(C) tan x + sec x (D) tan x – sec x

Ans: D.

Q.35 is equal to

(A) (B)

(C) 1 (D) 0

Ans: B.

Q.36 The solution of the differential equation is

(A) (B)

(C) (D)

Ans: A.

Q.37 The value of is equal to

(A) 1 (B)

(C) (D) zero

Ans: C.

Q.38 The value of is

(A) (B)

(C) (D)

Ans: D.

Q.39 If , then n is equal to

(A) 8 (B) 12

(C ) 16 (D) 20

Ans: D

Or

(n - 8)(n - 9)(n - 10)(n - 11) = 12.11.10.9 = (20 - 8)(20 - 9)(20 - 10)(20 - 11)

Q.40 is equal to

(A) 0 (B) 1

(C) 2 (D) 3

Ans: B

=

= 1 x 1 = 1

Q.41 If the point P(x, y) is equidistant from the points and , then

(A) bx = ay (B) ax = by

(C) x = y (D) x + y = 0

Ans: A

=

Q.42 The area of the triangle formed by the lines y = a + x, y = a – x, y = 0, where a > 0, is

(A) 1 (B) a

(C) (D) zero

Ans: C

y = a + x

y = a – x

Solving, we get y = a, x = 0

Therefore vertex is (0, a) another vertices are (-a, 0), (a, 0).

Area of the triangle is

=

Q.43 If is equal to

(A) 1 (B) 2

(C) (D)

Ans: A

thus

Q.44 is equal to

(A) sec x + cosec x (B)

(C) (D)

Ans: C

=

=

=

Q.45 The area bounded by the parabola and its latus rectum is

(A) (B)

(C) (D)

Ans: D

The parabola is symmetrical about the x – axis

Area =

=

=

Q.46 The solution of differential equation is

(A) (B)

(C) (D)

Ans: B

=

Q.47 Value of is

(A) (B)

(C) (D)

Ans: C

=

Q.48 Value of is

(A) 0 (B) 1

(C) (D)

Ans: A

=

=

Q.49 The number of terms in the sequence are

(A) 8 (B) 9

(C) 10 (D) 6

Ans: B

C.R = 2

Q.50 First three terms in the expansion of are

(A) (B)

(C) (D)

Ans: D

=

Q.51 Value of is

(A) (B)

(C) (D)

Ans: A

=

=

Q.52 If , then the value of cos 2A is

(A) (B)

(C) (D)

Ans: D

Q.53 The value of ‘x’ such that PQ = QR, where P, Q and R are (6, -1), (1, 3) and (x, 8) respectively is given by

(A) 5, –3 (B) 3, 5

(C) 2, 5 (D) 2, 3

Ans: A

Or

Or

Or

Q.54 Slope of the line passing through the points & is

(A) (B)

(C) (D)

Ans: A

Q.55 is equal to

(A) (B)

(C) (D)

Ans: C

Q.56 If then is equal to

(A) (B)

(C) (D)

Ans: A

=

Q.57 is equal to

(A) (B)

(C) (D)

Ans: C

=

=

Q.58 Order and degree of the differential equation is given by

(A) 3, 2 (B) 2, 3

(C) 1, 3 (D) 3, 1

Ans: D

Order – 3 [Power of higher directive]

Degree – 1

Q.59 Which term of the series 37+32+27+22+.............. is –103?

(A) 24^th (B) 30^th

(C) 15^th (D) 29^th

Ans: D

a = 37, d = -5

n = 29

Q.60 How many terms are there in the expansion of

(A) 4 (B) 6

(C) 16 (D) 10

Ans: C

The given expansion is

No. of terms in the expansion is = 15 + 1 = 16

(ONE more than the power of given expansion)

Q.61 If and , find the value of cot

(A) (B)

(C) (D)

Ans: B

, and

Q.62 Expansion of is equal to

(A) (B)

(C) (D)

Ans: A

=

=

Q.63 For what value of k do the points & lie on a straight line.

(A) 3 (B) 4

(C) 0 (D) 1

Ans: C

The points say A(-1, 4), B(-3, 8), C(-k + 1, 3k) lies on straight line if area of ABC = 0

Q.64 Mid point of the line joining (3, 5) and is given by

(A) (B) (1, 2)

(C) (2, 3) (D) (2, 1)

Ans: A

The midpoint of the line joining (3, 5) and (-7, -3) is

mid point.

Q.65 is equal to

(A) (B) 2

(C) (D)

Ans: B

Q.66 If y = x sin x, then is equal to

(A) cos x + sin x (B) cos x + x sin x

(C) x cos x + sin x (D) x cos x – sin x

Ans: C

If y = x sin x

Differentiating both side w.r. to x we have

Q.67 is equal to

(A) tan x + c (B)

(C) (D)

Ans: D

Q.68 The solution of the differential equation is

(A) (x + y) = k (1 – xy) (B) y – x = kxy

(C) (D)

Ans: A

Using variable separable method

Q.69 The square root of 12 – 5i is

(A) . (B) .

(C) . (D) .

Ans: C

Let

12-5i = =

,

=

Q.70 If be the roots of then the quadratic equation whose roots are is

(A) . (B) .

(C) . (D) .

Ans: A

We know that , . Since = , equation is

or

Q.71 The 12^th term in the binomial expansion of is

(A) . (B) .

(C) . (D) .

Ans: B

12^th term in the expansion of

is =

Q.72 The area of the triangle formed by the coordinate axes and the line 2x + 3y=6 is

(A) 3 sq. units. (B) 6 sq. units.

(C) 9 sq. units. (D) 12 sq. units.

Ans: A

Area of triangle is base ´ height = =3 sq. unit

Q.73 The eccentricity of the ellipse if its latus rectum is equal to one half of its minor axis is

(A) . (B) .

(C) ½ . (D) .

Ans: B

Eccentricity =

Q.74 In a triangle ABC, sin A – cos B = cos C, then angle B is

(A) . (B) .

(C) . (D) .

Ans: A

Given sin A-cos B=cos C Sin A= cos B+ cos C =

Q.75 is equal to

(A) e – 1. (B) e + 1.

(C) 0. (D) 1.

Ans: D

= =e-e+1 =1

Q.76 If , then is equal to

(A) . (B)

(C) . (D) .

Ans: C

Given

\ or

Q.77 The point is equidistant from points (7,6) and (-3, 4) if

(A) (B)

(C) (D)

Ans: A

Q.78 The value of is

(A) (B)

(C) 1 (D) 0

Ans: D

= -

= = 0

Q.79 The equation of a line through point (2, -3) and parallel to y-axis is

(A) y = -3 . (B) y = 2.

(C) x = 2. (D) x = -3.

Ans: C

The equation of line parallel to y-axis and at a distance 2 is

Q.80 The length of tangent from point (5,1) to the circle is

(A) 81. (B) 29.

(C) 7. (D) 21.

Ans: C

Here S =

= 49

length of tangent =

Q.81 The differential coefficient of log tan x is

(A) 2 sec2 x. (B) 2 cosec 2 x.

(C) 2 sec³ x. (D) 2 cosec³ x.

Ans: B

Let y = log tan x

Q.82 The expression where w is a cube root of unity, equals

(A) 16. (B) 16 w.

(C) 16 w² . (D) 0.

Ans: B

= = = = 16w

= 16w

Q.83 The complex number z = x +iy which satisfies the equation lie on

(A) The x-axis.

(B) The straight line y =5.

(C) A circle passing through the origin.

(D) The y-axis.

Ans: A

-10 y + 25 = 10 y + 25

y =0

Q.84 If , then the value of xyz is

(A) 0. (B) 1.

(C) 2. (D) 3.

Ans: B

Given

=()=

Q.85 The equation whose roots are the reciprocals of the roots of the equation is

(A) . (B) .

(C) . (D) .

Ans: D

We have --------(1)

Let are roots of (1), then

Again and .

Equation is Þ

Q.86 The smallest positive integer n for which is

(A) 2 (B) 4

(C) 3 (D) 5

Ans: B

We have

n is a multiple of 4

the smallest positive value of n is 4

Q.87 If then is equal to

(A) . (B) .

(C) . (D) .

Ans: A

Given

=,

Q.88 is equal to

(A) . (B) .

(C) . (D) .

Ans: B

Q.89 Integrating factor of the differential equation is

(A) 1 + x. (B) .

(C) . (D) .

Ans: C

I.F. = = =

Q.90 The distance between two parallel lines 3x + 4y = 5 and 6x + 8y = 35 is

(A) 1.0. (B) 1.5.

(C) 2.0. (D) 2.5.

Ans: D

Putting y=0 in we get

Thus ( lie on

The length of perpendicular from to 6x+8y =35 is

d= Hence, the distance between the given lines is 2.5

Q.91 The angle between the vectors and is

(A) . (B) .

(C) . (D) .

Ans: C

We know that :

= Þ

Q.92 The value of is

(A) 2. (B) 4.

(C) 6. (D) zero.

Ans: A

Q.93 is equal to

(A) . (B) .

(C) . (D) .

Ans: C

Q.94 If then x is equal to

(A) 3 (B) 27

(C) 9 (D) 15

Ans: B

Given = = =

Q.95 The value of is equal to

(A) 1. (B) 0.

(C) -1. (D) ½

Ans: A

==

Q.96 If are the roots of then is

(A) . (B) .

(C) . (D) .

Ans: B

Since are roots of

Now

Q.97 is equal to

(A) 1. (B) 0.

(C) e. (D) .

Ans: C

Q.98 is equal to

(A) . (B) .

(C) . (D) .

Ans: A

Q.99 The maximum value of y = 2 cos 2x – cos 4x, is

(A) -1. (B) .

(C) . (D) 1.

Ans: C

Q.100 The equation of the line which is perpendicular to the line 3x – 4y +7=0 and passes through the point (-3, 2) is

(A) 4x + 3y + 5 = 0. (B) 4x + 3y –3 = 0.

(C) 4x + 3y + 6 = 0. (D) 3x – 4y + 6 = 0.

Ans: C

The equation of line perpendicular to

is

This passes through (-3,2)

== =

From (i), required equation is

=

PART – II

NUMERICALS

Q.1 If are the roots of the equation . Find the equation whose roots are and . (7)

Ans:

=

=

=

Therefore required equation is

Q.2 If the roots of the equation are equal, show that . (7)

Ans:

Q.3 In a show that . (7)

Ans:

R.H.S. =

=

=

=

Q.4 If then show that . (7)

Ans:

Q.5 Evaluate . (7)

Ans:

=

=

=

Q.6 Differentiate by the first principle. (7)

Ans:

=

=

= - sin 2x.

Q.7 Find the area bounded by the curve and the straight line .

(7)

Ans:

Area bounded by the curve x² = 4y and the straight line x = 4y – 2.

=

The above curve intersects at the points and (2, 1).

=

= Units.

Q.8 Find the equation of tangent to at , where and . (7)

Ans:

Equation of the given ellipse is

Tangent at the point (2, y₁) such as y₁ > 0.

Equation of tangent at point (x₁, y₁) is

Satisfies the point (2, y₁)

, y₁ > 0

The equation of tangent at is

Q.9 Find the equation of a line passing through and perpendicular to the line 3x –y +5 = 0. (7)

Ans:

Let the equation of line is y = wx + c ……………….(1)

because it is perpendicular to 3x – y + 5 = 0

Therefore (1) becomes

It is passing through the point (-2, -4) therefore

-12 -2 = 3c

~ -14 = 3c

required equation is

x + 3y + 14 = 0

Q.10 Find the equation of the circle whose centre lies on the line x – 4y = 1 and which passes through the points (3, 7) and (5,5). (7)

Ans:

Let the equation of the circle is

The centre lies on the line x – 4y = 1,

h – 4k = 1 …….(1)

Again the circle passes through (3, 7) and (5, 5)

h – k = -2 ……..(2)

Subtracting (1) from equation (2)

3k = -3 k = -1

and h = -3

putting the value of h and k , we have

required equation is

Q.11 Find the term independent of x in the expansion of . (7)

Ans:

Middle term is independent from x i.e.

=

=

= 8064.

Q.12 Evaluate . (7)

Ans:

=

Q.13 Using induction, prove that for all n. (7)

Ans:

By using mathematical induction method

2¹ > 1

This is true n = 1

Let 2^r > r

Now 2^r+1 = 2.2^r > 2r > r+1 if r > 1

Therefore on the statement is true for r = n

Hence it is true for all n.

Q.14 Solve . (7)

Ans:

Q.15 Evaluate . (7)

Ans:

Let

=

Q.16 Evaluate . (7)

Ans:

=

=

=

=

=

=

Q.17 Solve , given y =1 when x = 1. (7)

Ans:

=

When y = 1, x = 1

c = 2

Q.18 Find the differential equation of which is a solution. (7)

Ans:

Q.19 Find the term independent of x in the expansion of . (8)

Ans:

If n^th term is independent of x

12 – 2n = 0 i.e. n = 6

is independent of x and

Q.20 If the p^th, q^th and r^th terms of an A.P. are x, y, z respectively, show that x (q – r) + y (r – p) + z (p – q) = 0. (8)

Ans:

If a, a + d, a + 2d, ………… be A.P.,

T_p = x = a + (p – 1) d

T_q = y = a + (q – 1) d

T_r = z = a + (r – 1) d

=

=

Q.21 If A + B + C = π, show that

(8)

Ans:

A+B+C = π or

By cross multiplying

Dividing through out by , we get

Q.22 In any triangle ABC, show that

(8)

Ans:

In any triangle ABC,

A + B + C = π

and (say)

=

=

=

Hence

Q.23 Solve the equation . (8)

Ans:

Þ

I.F. =

Solution is

Q.24 Find the equation of a straight line when p is the length of perpendicular on it from the origin and the inclination of this perpendicular to the x – axis is . (8)

Ans:

Let ON = p be length of perpendicular from origin on st line AB and let ON make angle α with x-axis.

p = ON = OA cos α

= (OM + MA) cos α

= (x + MP tan α) cos α

= x cos α + y sin α

Hence required equation is

x cos α + y sin α = p

Q.25 Find the equation of the straight line which passes through the intersection of the straight lines 2x – 3y + 4 = 0 and 3x + 4y + 5 = 0 and is perpendicular to the straight line 6x – 7y + 8 = 0. (8)

Ans:

Any line through the intersection of two given lines in

2x – 3y + 4 + k(3x + 4y + 5) = 0

It is perpendicular to the line 6x – 7y + 8 = 0

Hence required straight line in

Or

Q.26 Show that x² + y² + 2gx + 2fy + c = 0 represents a circle. Find its centre and radius. (6)

Ans:

Given equation can be written as

Or . Comparing with which is a circle of centre (h, k) and radius a, we observe that given equation represents a circle with centre = (-g, -f), Radius =

Q.27 Find the vertex, focus, latus rectum and directrix of the parabola x² = 4x – y. (10)

Ans:

or

Or

Put x – 2 = X, y – 4 = Y

represents a parabola of the shape as shown below.

With vertex X = 0, Y = 0 i.e. x = 2, y = 4, axis x = 2,

LR = 4a = 1.

and Directrix is

Q.28 Evaluate , by using the fact that . (8)

Ans:

Put a^x – 1 = t or a^x = t + 1 or x = log_a(1 + t)

= because

Q.29 Differentiate with respect to x. (8)

Ans:

=

Hence

Q.30 Find the points at which the function y = 3 Sin²x + 4 Cos²x has maximum and minimum values in the interval (8)

Ans:

For Max or Min

points of maximum & minimum are

Hence x = 0 is a point of Maxima and Max. value is 4

is a point of Minima and Minimum value is 3.

Q.31 Evaluate , where a, b are not both zero. (8)

Ans:

put

=

=

=

Q.32 Find the area common to the circles x² + y² – 2ax = 0 and x² + y² – 2ay = 0. (10)

Ans:

Given circles and

intersect at (0, 0) and (a, a)

Common area = where and

=

=

Put x – a = t

=

Hence required common area = a².

Q.33 Evaluate . (6)

Ans:

put

=

Q.34 Solve following the differential equations

(i) ydx – xdy = . (8)

(ii) cos²x . (8)

Ans:

(i)

Or

Put y = vx,

Integrating,

Or

.

(ii)

Or

It is linear differential equation with

I.F =

Solution is

=

=

Hence required solution is

Q.35 Show that the sum to n terms of the series 1.3.5 + 3.5.7 + 5.7.9 + … is . (8)

Ans: The r^th term of the series is given by

\ s_n , the sum to n terms of the series is

Q.36 If are the roots of the quadratic equation and are the roots of the quadratic equation then show that . (8)

Ans:

We have a + b = -p, ab = 1

Therefore,

Q.37 If A + B + C = , prove that . (8)

Ans:

We have B+C-A = 180 – 2A. so that

L.H.S. = sin (180 – 2A) + sin(180 – 2B) + sin(180 – 2C)

= sin 2A + sin 2B + sin 2C

= 2 sinA cosA + 2 sin(B + C) cos (B - C)

= -2 sinA [cos (B + C)] + 2 sinA cos(B - C)

= 2 sinA [cos(B – C) – cos(B + C)]

= 2 sinA 2sinB sinC

= 4 sinA sinB sinC

Q.38 Show that sin is a root of the equation . (8)

Ans: It is sufficient to show that

Q.39 Find the value of such that the circles and touch each other. (8)

Ans:

and touch each other if the distance between their centre’s is equal to the sum or difference of their radii.

centres of circles is (-1, -1) (-1, -1)

distance between centres is 0

Q.40 For what values of k the points and are collinear?

Ans: (8)

The points (-1, 4) (2, -2) and (-4 - k, 6 – 2k) are collinear iff

Q.41 Find the equation of the circle for which is a tangent and are normals. (8)

Ans:

Any two normals of a circle intersect at the centre of the circle. So, the centre is obtained by solving the equations of normals.

The point of intersection of the normals x + y = 0 and x – y + 4 = 0 is the point(-2, 2)

Now, the radius of the circle is the perpendicular distance from the centre of the circle to any target.

Hence,

Radius = perpendicular distance from (-2,2) to the target x –y – 1 = 0

Q.42 Find the values of a, b such that the line ax + by + 1 = 0 is tangent to the hyperbola and is parallel to the line y = 2x + 4. (8)

Ans:

The equation of the hyperbola is

The straight line y = mx + c is a target to the hyperbola

Since the straight line ax + by+1 = 0 is parallel to the line y = 2x + 4, thus m = 2.

Substituting in y = mx + c, we get y = 2x ± 1.

Thus, the required straight lines are y = 2x + 1 and y = 2x - 1

Or, 2x – y + 1 = 0 and –2x + y + 1 = 0

Hence, the values of a and b are: a = 2, b = -1 and a = -2, b = 1.

Q.43 Evaluate the limit. (8)

Ans:

Q.44 Consider the function . Find using first principle. Is continuous at x = 0? (8)

Ans:

Q.45 Find the local maximum and minimum values of in . (8)

Ans:

So, x = 3p/2 is a point of minimum

Q.46 Find the area of the region bounded by , and x = 1. (8)

Ans:

let f(x) = -x and g(x) = x² + 2.

Then f(x) £ g(x) and x in [0, 1]. Hence the required area is

Q.47 Evaluate the following integral . (8)

Ans:

Q.48 Evaluate the following definite integral . (8)

Ans:

Thus,

Q.49 Solve the differential equation . (8)

Ans:

Q.50 Solve the differential equation

. (8)

Ans:

Separating the variables by dividing by (x²+1) sin y, we get

or, ln (x²+1) + 2 ln |siny| = 2c

or, ln[(x²+1) sin²y] = 2c = ln k, say

or, (x²+1) sin²y = k, ------- (*)

where k is an arbitrary constant.

In dividing by (x² + 1) sin y, we assumed that sin y ¹ 0.

now, consider sin y = 0. These are given by y = np, n = 0, ±1, ±2,…. Writing the original differential equation in the derivative form, it is clear that y = np is a constant solution. Each of these constant solution is present in the solution (*). So, we have not lost any solution in the division process.

Q.51 Show that the coefficient of in the expansion of is double the coefficient of in the expansion of . (8)

Ans:

Q.52 If and , where then prove that . (8)

Ans:

Q.53 If A + B + C =, show that

. (8)

Ans:

LHS = sin 2A + sin 2B + sin 2C

= 2 sin(A+B) cos (A-B) + 2 sinC cosC (

= 2 sin(p-c) cos(A-B) + 2sinC cos(A+B)

= 2 sinC cos(A-B) – 2 sinC cos(A+B)

= 2 sinC[cos(A-B) – cos(A+B)]

= 2 sinC[cosA cosB + sinA sinB – cosA cosB + sinA sinB]

= 4 sinA sinB sinC = R.H.S.

Q.54 If a, b, c be the sides opposite to the angles A, B, C of a triangle ABC, show that . (8)

Ans:

Q.55 Derive the formula for finding the area of a triangle whose vertices are and . (8)

Ans:

Q.56 Find the equation of a straight line joining the point (3, 5) to the point of intersection of the lines 4x +y = 1 and 7x – 3 y = 35. (8)

Ans:

Any line passing through the point of intersection of the given lines is

4x + y – 1 + k(7x – 3y – 35) = 0 ----------(1)

Q.57 Find the equation of the circle which passes through the centre of the circle

and is concentric with the circle . (8)

Ans:

Any circle concentrate with the given circle is

Q.58 Find the focus, vertex, directrix and axis of the parabola . (8)

Ans:

The given parabola can be written as

The shape of the parabola is as shown in the figure

Q.59 Evaluate . (8)

Ans:

Q.60 Find , if . (8)

Ans:

Q.61 Derive the equation of the tangent and the normal to the curve at the point . (8)

Ans:

Q.62 Evaluate . (8)

Ans:

Q.63 Find the volume of the solid of revolution obtained by revolving the ellipse about x-axis. (8)

Ans:

Q.64 Evaluate , for any positive integer n. (8)

Ans:

Q.65 (i) .

(ii) .

(iii) . (16)

Ans: (i)

(ii)

(iii)

Q.66 The sum of first p terms of an A.P. is the same as the sum of its first q terms. Find the sum of its first (p + q) terms. (8)

Ans:

Or

Or 4a + (p + q – 2)d + (p + q)d = 0

Or 4a + (2p + 2q – 2)d = 0

Or 2[2a + (p+ q – 1)d] = 0

Or

Thus sum of (p + q) terms is 0.

Q.67 For what value of n are the coefficients of second, third and fourth terms in the expansion of in A.P.? (8)

Ans:

Since 2^nd, 3^rd, 4^th terms of (1 + x)ⁿ are in A.P. Thus

Or

n = 7 is only possible.

Q.68 Solve for the equation , where . (8)

Ans:

And

Q.69 If a, b, c be the sides opposite to the angles A, B, C for a triangle ABC, show that . (8)

Ans:

=

=

Q.70 Derive the formula for the angle between the straight lines and . (8)

Ans:

Let be the equation of line AC which makes an angle with x – axis, so and be the equation of line BC which makes an angle with x –axis so

The angle between the lines

or 180 – ()

Thus

=

=

=

Q.71 Find the equation of a straight line which is perpendicular to 2x – 5y = 30 and the sum of its intercepts on the coordinate axes is 7. (8)

Ans:

Let the equation of the line is

y = mx + C

It is perpendicular to 5y = 2x – 30

Equation of the line is

Or 2y + 5x = 2c

Its passing through (x, 0) therefore

Again it is passing through (0, 7-x)

Or 14 – 2x = 5 x 2 or x = 2

Equation of line is 2y + 5x = 10

Q.72 Find the equation of the circle concentric with the circle and having its area equal to . (8)

Ans:

Centre of the circle is

Or

Also area

Let the equation of the required circle is

Q.73 Find the centre, eccentricity, foci and length of the latus rectum of the ellipse . (8)

Ans:

Let x – 1 = X

y + 2 = Y,

thus

Centre (0, 0) X = 0, Y = 0

i.e. (1, -2)

Eccentricity

Foci

Or

Foci (4 , -2), (-2, -2)

Length = 4a = 4 x 3 = 12

Q.74 Differentiate from the first principle the function y = tan x. (8)

Ans:

y = tanx y = δy = tan(x + δx)

=

=

Q.75 Evaluate . (8)

Ans:

= (L-Hospital rule)

=

Q.76 Find the local maximum and minimum values of the function y = sin 3x – 3 sin x, . (8)

Ans:

y = sin3x – 3sinx

and

At

At x = 0, , .

Thus maximum is obtained at and maximum value is 4.

The minimum is obtained at and minimum value is -4.

Q.77 Evaluate . (8)

Ans:

=

= Let

Also let

=

=

=

Q.78 Find the area bounded by the curve and the coordinate axes. (8)

Ans:

Let x = 0 y = a

y = 0 x = a

=

=

=

=

Q.79 Evaluate . (8)

Ans:

=

=

=

= log2 – log1 – log3 + log2

= 2log2 – log3

=

Q.80 Solve any TWO of the following differential equations:- (24)

(i) .

(ii) .

(iii) .

Ans:

(i)

Or

Or

Or

Or

(ii)

Thus , therefore eq is exact. Hence solution is

Or .

(iii)

Or

I.F =

=

=

=

.

Q.81 If 5 times the 5^th term of an A.P. is equal to the 10 times the 10^th term, find the 15^th term of the A.P. (8)

Ans:

5^th term of an A.P = a + 4d

10^th term of an A.P = a + 9d

Here 5(a + 4d) = 10(a + 9d)

a + 4d = 2a + 18d

a = -14d

= a – a = 0

Q.82 If denotes the sum of n terms of a G.P., prove that . (8)

Ans:

=

=

=

=

=

=

L.H.S = R.H.S

Q.83 Show that . (8)

Ans:

L.H.S =

=

=

=

= cot 2a = R.H.S

Q.84 If in the triangle ABC, A = , prove that . (8)

Ans:

To prove

This is true since .

Q.85 Find the equation of the straight line which passes through the intersection of the lines

x + y – 3 = 0 and 2x – y = 0 and is inclined at an angle of with x-axis. (8)

Ans:

Point of intersection is (1, 2)

x + y = 3

, x = 1, y = 2

Let the equation of the line is

y = wx + c

Here w = tan 45⁰ = 1.

And became the line passing through (1, 2) therefore

2 = 1 + C

Therefore the equation of required line is y = x + 1 i.e. x – y + 1 = 0

Q.86 Show that represents an ellipse. Find its centre, vertices, foci, eccentricity, directrices, latusrectum and equations of major and minor axis. (8)

Ans:

Let x – 3 = X, y – 7 = Y

…………….(1)

Center of the ellipse = (0, 0)

X = 0 x – 3 = 0 x = 3, Y = 0 y – 7 = 0 y = 7

Center = (3, 7)

About major axis: - x = a x – 3 = 2 x = 5; x = 5. Also x = -a x – 3 = 2

x = (5, 7)(1, 7)

Y = 0 y – 7 = 0 y = 7

About minor axis: - X = 0 x – 3 = 0 x = 3

Y = b y – 7 = 3 y = 10

Y = -b y – 7 = -3 y = 4

For y = b y – 7 = 3 y = 10,

y – 7 = 3 y = 4

foci (3, 10)(3, 4)

Eccentricity

=

Directories y = b y – 7 = 3 y – 10 = 0, y – 4 = 0

Latus rectum 4a = 4 x 2 = 8

Equation x = a x = 5

X = -a x = 1

Minor Axis y = b y = 10

y = -b y = 4

Q.87 Find the equation of the circle which passes (4, 1) & (6, 5) and having centre on the line 4x+y =16. (8)

Ans:

Equation of the circle, …………….(*)

Which passes through (4, 1) and (6, 5)

16 + 1 + 8g + 2f + c = 0

8g + 2f + c + 17 = 0 ………………………. (1)

36 + 25 + 12g + 10f + c = 0

12g + 10f + c + 61 = 0 ……………………. (2)

Since centre lies on line 4x + y = 16, thus

-4g –f -16 = 0

4g + f + 16 = 0 ……………………………. (3)

Equation (2) – (1)

4g + 8f + 44 = 0

g + 2f + 11 = 0 ……………………………. (4)

8g + 2f + 32 = 0 …………………………... (5)

Equation (5) – (4)

7g + 21 = 0

g = -3

Putting the value of g in Equation (3)

-12 + f + 16 = 0

g = -3, f = -4

From Equation (1)

-24 – 8 + c + 17 = 0

-32 + 17 + c = 0

-15 + c = 0

c = 15

Thus the Equation of circle is: -

Q.88 Find the value of (8)

Ans:

form 0/0

Using L-Hospital rule.

Q.89 Differentiate y = tan x w.r.t. ‘x’ from first principle. (6)

Ans:

y = tanx

=

=

=

=

=

Q.90 Differentiate y = w.r.t ‘x’. (10)

Ans:

Let

Let

Q.91 Prove that straight line touches the curve at the point where the curve crosses the axis of y. (8)

Ans:

The point where the curve crosses the axis is given by put .

Equation of tangent at the point (0, b)

Hence Proved.

Q.92 Find the volume generated by revolving the ellipse about x-axis. (8)

Ans:

Required value =

Q.93 Prove that . (10)

Ans:

=

=

Q.94 Solve . (6)

Ans:

Let

Q.95 Solve . (8)

Ans:

Let

Q.96 Solve subject to the initial condition y(0) = 0. (8)

Ans:

I.F =

x = 0, y = 0

Q.97 How many terms are there in a finite AP whose first and fifth terms are respectively –14 & 2 and the sum of terms is 40. (8)

Ans:

Let first term in AP be ‘a’ and ‘d’ be the common difference.

According to the given condition

First term a = -14

Fifth term a + 4d = 2

-14 + 4d = 2

4d = 16

d = 4

According to another condition

=

=

Neglecting n = -2 because no of terms cannot be negative

The only possibility is n = 10.

Q.98 The sum of three numbers in G.P. is and their product is –1. Find the numbers. (8)

Ans:

Let the three numbers in G.P be , a, ar

Then according to the first condition

…………………………. (1)

According to the second condition

a = -1 ……………………………………… (2)

Substituting the value of a in equation (1)

=

r =

r =

a = -1, r =

Then the three numbers be

ie

When a = -1, r =

Then the three numbers be

ie

Q.99 If , prove that

(8)

Ans:

Given A + B + C = 180

To Prove that

L.H.S.

------------------------- (1)

=

=

= ------------- (2)

=

=

=

=

=

=

=

= = R.H.S.

Q.100 In any triangle ABC, prove that (8)

Ans:

L.H.S.

= = R.H.S.

Q.101 Find the vertex, axis, focus, latus rectum and directrix of the parabola . (8)

Ans:

The given equation is

------------------------------ (2)

--------------------------------------------- (1)

Comparing it units

Vertex

Axis

Focus

L.R. 2

Q.102 Find the equation of the circle which passes through the points (1, 1) & (2, 2) & whose radius is 1. (8)

Ans:

(1)

equation (1) passes through the point (1, 1)

2g + 2f + c = -2 (2)

equation (1) passes through the point (2, 2)

4g + 4f + c = -8 (3)

Also radius = 1

(4)

Solving equation (2) and (3)

-2g -2f = 6

g + f = -3 (5)

Solving equation (3) and (4)

4g + 4f + c = -8

(6)

Solving equation (5) and (6)

f = -1 f = -2

g = -3 + 1 = -2 g = -3 + 2 = -1

4 + 1 – c = 1 1 + 4 – c =1

c = 4 c = 4

Thus the required equation of the circle is

&

Q.103 Find the equation of the straight line perpendicular to 7x + 9y – 3 = 0 and passing through (3, 8) (8)

Ans:

Equation of straight line perpendicular to 7x+ 9y + 3 = 0 is 9x – 7y + k = 0

It passes through (3, 8)

Any line perpendicular to ax + by + c = 0 is given by bx + ay + k = 0

9(3) – 7(8) + k = 0

27 – 56 + k = 0

k = 29

Thus the required equation be

9x – 7y + 29 = 0

Q.104 Differentiate from the first principle the function y = sin 3x. (8)

Ans:

If f(x) = y = sin 3x

Using first principle

=

=

=

=

= 3 cos 3x.

Q.105 Evaluate . (8)

Ans:

Form

Q.106 Find the points of maxima or minima values of the function . (8)

Ans:

Differentiating both sides w.r.t ‘x’

--------------------------------- (1)

Put

x = 8, 4

Differentiating (1) w.r.t x both side

At x = 4,

is a point of maxima and maximum value

= 64 – 18(16) + 384

= 64 -288 + 384 = 160

At x = 8,

is a point of minima and minimum value

= 512 – 1152 + 768

= 1280 – 1152

= 128

Q.107 Evaluate. (8)

Ans:

Put

Differentiating both side w.r.t ‘x’

=

= …..(1)

…...(2)

Q.108 Evaluate (8)

Ans:

Put

x = 0, θ = 0

x = 1, (1)

Let

(1)

Using property

=

=

=

=

=

=

Q.109 Find the area enclosed by the ellipse . (8)

Ans:

The equation of the curve is

The curve is symmetrical about the axis

Area enclosed by the ellipses

= 4 (area enclosed by the ellipse and coordinate axes in first quadrant)

Required area =

=

=

=

=

= sq units

Q.110 Solve . (8)

Ans:

Let y = vx (homogenous form)

Differentiating both side w.r.t x

Integrating both side

(2)

Taking antilog on both sides

Q.111 Solve . (8)

Ans:

Comparing the above equation with

I.F = (1)

I.F = (2)

Required solution

(3)

Q.112 If Show that (7)

Ans:

We have

=

Q.113 Put the following in the form , where r is a positive real number and . (7)

Ans:

Let r

, ,

Q.114 A two-digit number is four times the sum and three times the product of the digits. Find the number. (7)

Ans:

Let the number is where is tens digit and y is unit digit.

Given (1)

and (2)

From (1),we get

Using this in (2), 10x + 2x = 3x(2x) or 12x = 6x² or x² -2x = 0 ,
Þ x = 0, x = 2.

If x = 0, then y = 0 which is inadmissible. If x = 2 then y = 4, hence the required number is 10(2) +4=24

Q.115 Solve the simultaneous equations: . (7)

Ans:

We have

.....................(1)

x + y = 10 ..........................(2)

(1) , using (2).

Thus, the given system of equations is

x + y = 10, xy = 16 y = 10 – x and x (10- x) = 16

x² – 10x + 16 = 0 x = 2, 8

If x = 2, y = 8. And if x= 8, y = 2.

Hence roots are x = 2, y = 8 and x = 8. y = 2

Q.116 The diagonal of a square lies along the line and one vertex of the square is (1, 2). Find the equations of the sides of the square. (7)

Ans:

Let ABCD be a square such that the diagonal AC is 8x – 15y = 0 and the vertex B is (1,2). We have to find the sides passing through B clearly, sides BA and BC pass through B(1,2) and are inclined at an angle of to the diagonal AC. So, the equations of BA and BC are

where m is the slope of the line

or

........(3)

D C

A B

Coordinates of A, C are ,

other two sides are parallel to the sides (3)

hence are

These respectively pass through C and A. We can find by using this condition.

Q.117 Find the centroid and incentre of the triangle whose sides have the equations . (7)

Ans:

Let ABC be the triangle whose sides BC,CA and AB have the equations

y -15 = 0, 3x - 4y = 0, 5x +12y = 0 respectively. Solving these equations pair wise we can obtain the coordinates of the vertices A,B,C as A(0,0), B(-36,15), C(20,15) respectively

(0,0)

A

A

(20,15)

C B

Centroid:

The coordinates of centroid are

For Incentre:

We have

a =BC=

b=CA=

c=AB=

\Coordinates of incentre are

=(-1,8)

Q.118 (i) Find the equation of the circle which touches both the axes and whose radius is 5.

(ii) Find the coordinates of the centre and radius of the circle

. (7)

Ans:

(i) The equation of circles which touch both the axes are

and

Here and radius equals 5, Therefore circles are

and

and

(ii) In the given equation the coefficients of and one not unity.

We have to re-write the equation to make the coefficients of and unity. We have

The coordinates of centre areand radius=

=

Q.119 Find the equation of a circle passing through the points (1, 2) and (3, 0) and cutting an intercept 4 on the x-axis. (7)

Ans:

Let the equation of the circle be

(1)

Since it passes though the points

(1,2)and (3,0)

1 + 4 + 2 g + 4 f + c = 0

2 g +4 f + c = -5 (2)

and 9+6g+c=0 6g+c=-9 (3)

Also the length of x-intercept is 4

(4)

From (3) and(4)

g = -1, -5

From (3),if g = -1, c = -3

if g = -5, c = 21

Also from (2) if g = -1, c = 3 then f = 0

and If g = 5, c = 21 then f = -4

Equations are

Q.120 Find the equation of the parabola whose focus is (3, 0) and the directrix is 3x +4y = 1. (7)

Ans:

Let P be any point on the parabola whose focus is S and the directrix

Draw PM perpendicular to . Then, by definition for parabola

SP=PM SP= PM

Z

M P(x,y)

3x+4y=1 S(3,0)

or

or is the required equation of parabola.

Q.121 Find the equation of an ellipse whose foci are at (± 3, 0) and which passes through (4, 1). (7)

Ans:

Let the equation of ellipse be

. The coordinates of foci are

( a e,o). But .............(1)

Also the ellipse passes though (4,1)

=or =. Substituting in (1)

or

or

or

or

If

If (not possible)

Equation of ellipse is

Q.122 If , prove that . (7)

Ans:

Given

Diff. w.r to x

Þ

or

or

Q.123 (i) A man 2 metres high walks at a uniform speed of 6 metres per minute away from a lamp post, 5 metres high. Find the rate at which the length of his shadow increases.

(ii) Use differentials to find the approximate value of . (7)

Ans:

(i)	Let AB be the lamp-post. Let at any time t, the man CD be at a distance x meters from the lamp-post and y meters be the length of his shadow CE. Then = 6 meters / minute (given) Now, triangle ABE and CDE are similar, therefore B D x y A C E Thus, the shadow increases at the rate of 4 meters/minute.
(ii)	Let y = f (x) = x = 0.040 and x + ∆x = 0.037 then ∆x = -0.003. For x = .040, y = .2 Let dx = ∆x = -0.003 Now, . By using we get Now, ∆y is the approximately equal to dy, so . Hence

Q.124 A square piece of tin of side 24 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. What should be the side of the square to be cut off so that the volume of the box is maximum. (7)

Ans:

x 24-2x

Let x cm be the length of a side of the square which is cut-off from each corner of the plate. Then sides of the box as shown in fig. above are 24 - 2x, 24 - 2x and x.

Let V be the volume of the box .Then

For maximum or minimum V,

But x = 12 is not possible , thus x = 4

Now,

= - 96 <0

Thus, V is maximum

when x = 4

Hence, the volume of the box is maximum when the side of the square cut off is 4 cm.

Q.125 Evaluate the following integrals

(i) (ii) . (7)

Ans:

(i)	= = = --------------------(1) I = Let Þ Þ I = = From (1)
(ii)	= = tan x – sec x +c

Q.126 Draw the rough sketch of area enclosed by curves Also find this area. (7)

Ans:

The point of intersections of

are (2,1) and (2,-1).

Required area is shaded area in the figure

. Area=Sq units

Q.127 Using integration, show that the volume of a sphere of radius a is . (7)

Ans:

The sphere is generated by the revolution of a semi circular area about its bounding diameter. The equation of the generating circle of radius ‘a’ with centre at origin is

Let A be the bounding diameter about which the semi-circle revolves

The required volume of the sphere

Q.128 Solve the following differential equations

(i) . (ii)

(iii) . (14)

Ans:

(i)	Separating the variables or , On integration, we have , c arbitrary, as the general solution.
(ii)	costan x Þ This is linear differential equation I.F = = Solution is +C Let tan x = t, then and integral on r.h.s. becomes s Solution is or
(iii)	------------(1) Let is the solution of (1), then auxiliary equation is Þ m = 2, 3 The general solution of differential equation is y = CF + PI . Where

Q.129 (i) Find a unit vector perpendicular to both the vectors and .

(ii) If are unit vectors inclined at an angle , then prove that

(iii) Find the moment of the couple formed by the forces and acting at the points and respectively. (14)

Ans:

(i)	Let the unit vector perpendicular to both the vectors is Let , and are perpendicular to each other ------------------(1) Also and are perpendicular -------------------(2) from (1) and (2) unit normal vector
(ii)	Now
(iii)	Here Now

Q.130 Find the term independent of x in the expansion of

(7)

Ans:

Given . Let (r+1)^th term be independent of x.

Now

For this term be independent of x, we must have

15-3r = 0, So, 6^th term is independent of x.

Q.131 If prove that

(7)

Ans:

Given

Q.132 If where are complex cube roots of unity show that . (7)

Ans:

Given

and

Let

Now

Q.133 If the roots of the equation be equal prove that either . (7)

Ans:

Given that the roots of

are equal.

The discriminant of the equation is zero

or

Q.134 Find the derivative of from the first principles. (7)

Ans:

let Then

=

Q.135 Take A semicircle with a rectangle on its diameter as shown in the figure below. If the perimeter of the figure is 20 feet, find its dimension in order that its area may be maximum.

(7)

Ans:

Let ABCD consists of a rectangle and let the semi-circle be described on side AB as diameter. Let AB=2x and AC = 2y. Let P be the perimeter and A be the area of fig. then

------------------(1)

------------------(2)

, ,

For maxima or minima, , Thus

Also for all values of x. Thus, A is maximum when

. From(1),

. So, dimensions of rectangle are
and semicircle top has radius

Q.136 Evaluate . (6)

Ans:

Divide by x Ans.

Q.137 The rectangular co-ordinates of a point on the curve are . Find the equation of the normal at any point on the curve and show that at the point with , the normal passes through the origin. (8)

Ans:

Here ,

Equation of normal is

,

,

The equation of normal passes through origin,

Q.138 Show that the curves and intersect each other at point (a, 2 a) at an angle . (7)

Ans:

Solving for (x,y)

,, and

A point of intersection is (a,2a)

Now,

Here slopes at point are and

Q.139 Differentiate with respect to . (7)

Ans:

Let ,

Let , ,

Q.140 Prove that the straight line joining the mid-points of two non-parallel sides of a trapezium is parallel to the parallel sides. (7)

Ans:

Let ABCD be the given trapezium. Let the position vectors of A,B,C and D with reference to some origin O be respectively.

Let P and Q be the mid-points of AD and BC respectively. Then, the position vectors of P and Q are

respectively we have , and

Since is parallel to , Therefore there exists a scalar

such that

-------------(1)

Now position vector of Q-position vector of P

= =

----------------(2)

This shows that PQ is parallel, to AB. But, AB is parallel to CD,

Therefore PQ is parallel to CD

Q.141 Find a unit vector that is perpendicular to both the vectors

(7)

Ans:

4i+3j+k.

2i-j+2k

=ijk =7i-6j-10k

Q.142 Find the square root of 12-6i. (7)

Ans:

Let z be the square root of 12-6i

then or

Q.143 Evaluate the integral

(7)

Ans:

Putting we get B=

Putting we get C=

Comparing coefficients of on both sides of the identity. we get

A+C=0

=

=log -

Q.144 Evaluate the definite integral

(6)

Ans:

Let

I=--------------(1)

I=

=-------------(2)

Adding 1 and 2, we get

2I= =

I=

When

I= = =

Q.145 Find the area bounded by the parabola and the curve , where a > 0. (8)

Ans:

The curveis symmetrical about y-axis. Equating to zero the coefficient of the highest power of x in the given equation, we find that y=0 i.e x-axis is an asymptote of the curve. Also this curve cuts the

y-axis at (0, 2a). Solving the two given equations we get their points of intersection as

Now the required area OBACO

= area OAC (By symmetry) [area OACE –area OCE]

Q.146 Solve the differential equation

. (7)

Ans:

Given

It is homogeneous differential equation

Putting y=

Integrating both sides, we get

log

or or or

Q.147 Find , where . (7)

Ans:

We have

y

Q.148 Solve the differential equation

(7)

Ans:

------------(1)

Homogeneous differential equation

Let

(1)becomes

Separating the variables

on integration

Let

, dv = dt

Q.149 Two stones are thrown up from the ground simultaneously. The equation of motion for the first stone is s= 19.6 t – 4.9 t² and for the second stone it is s = 9.8 t – 4.9 t² . What is the height of the second stone from the ground, when the height of the first stone is maximum. (7)

Ans:

,

Þ

or

Since

S is maximum when t = 2sec.

Then after 2 sec. the height of the second stone from the ground is

and the maximum height of the first stone is

Q.150 Find real values of x and y if and are complex conjugate to each other. (7)

Ans:

Since

are complex conjugates, therefore

Q.151 Evaluate . (7)

Ans:

But the given expression is positive hence

Q.152 Show that the coefficient of in the expansion of is double the coefficient of in the expansion of . (7)

Ans:

Let A and B be the coefficients of in the binomial expansions of and respectively, Then

Q.153 Resolve into partial fractions , assuming a, b, c and d are distinct. (7)

Ans:

Q.154 Find the general solution of the equation . (7)

Ans:

Q.155 If A, B and C are the angles of a triangle, show that . (7)

Ans:

=

=

=

=

=

Q.156 Find the area of a triangle whose angular points are , and . Find for what value of K, these points will be collinear. (7)

Ans:

Here

Area of Triangle

Three points are collinear if Area of Triangle is zero.

Q.157 If p is the length of perpendicular from the origin on a straight line whose intercepts on the axes of x and y are a and b respectively, show that . (7)

Ans:

The given line is

P = length of perpendicular from the origin to (1)

=

Q.158 Find the equation of the circle which passes through the points and and has its centre on the line x = 2y. (7)

Ans:

Let the equation of the required circles be

It passes through (-1,2) and (3,-2)

5-2g+4f+c=0........................(2)

13+6g-4f+c=0......................(3)

The centre (-g, -f) of (i) lies on x=2y

Solving (2), (3) and (4), we get

From (1), equation is

Q.159 Find the vertex, the axis, the focus and latus rectum of the parabola . (7)

Ans:

The given equation is

Shifting the origin to the point (-1,2) without rotating the axes and denoting the coordinates with respect to new axes by X and Y, we have

.............(ii)

Using these relations in equation (i) it reduces to

......................................(iii)

Here

Vertex: The coordinates of vertex with new axes are X=0, Y=0

so, coordinates of the vertex with respect to old axes are (-1,2)

Focus: The coordinates of the focus w.r. to new axes are

X=1, Y=0

So, Coordinates of the focus w.r. to old axes are (0,2)

Axis: Equation of the axis of the parabola w.r. to new axes is Y=0

So, equation of axis w.r. to old axes is y=2

Latus rectum:

The length of latus rectum =4

Q.160 If and find such that is perpendicular to . (7)

Ans:

Given

Q.161 Find a unit vector normal to the plane of the vectors and . (7)

Ans:

Given and , Unit normal vector

=

= =

Q.162 If y = . Show that . (7)

Ans:

Given

Putting

y = =

===

=

y =Þ

Q.163 Show that for all values of n, the curve touches the straight line at the point . (7)

Ans:

Differentiate both sides

Q.164 Find the maximum and minimum values of . (7)

Ans:

For maxima and minima

=

.

Again At

==

and minimum value is

At

==

At , f(x) is maximum and maximum value is

Q.165 Integrate the following:

(i) .

(ii) . (3 + 4)

Ans:

(i)	Given =
	=
(ii)	Given
	Let ,
	= = =
	==
	==

Q.166 Find the area enclosed by the parabolas and . (7)

Ans:

The equations of the given curves are

The points of intersection of (i) and (ii) are

So, the two curves intersect at (0,0) and (4a,4a)

The region whose area we have to find is the shaded region. Here we slice this region into vertical strips. We observe that all vertical strips have lower end on the parabola and the upper end on the parabola , For the approximating rectangle shown in fig, we have width and the area

Since the approximating rectangle can move between and=4a,

Thus required area = =

== sq. units.

Q.167 Find the volume of the solid of revolution obtained by revolving the ellipse about its major axis. (7)

Ans:

Volume of solid

Q.168 Solve the following equations :-

(i) .

(ii) .

(iii) . (4+5+5)

Ans:

(i)
	Separating the variables
	or
	Integrating both sides
	or
	or is the required solution.
(ii)	Given
	ÞÞÞ
	Solution is
	Let ,
(iii)
	= is the general solution of differential equation.

Q.169 Prove that 7 divides for all positive integers n. (7)

Ans:

Let, For

which is divisible by 7 . Let ,where k is a positive integer

i.e. is divisible by 7. We have to show that this relation is true for

Here is divisible by 7 and 7 itself divisible by 7. Thus P(k+1) is divisible by 7. Hence result is true for ,But it is true for . Thus it is true for every positive integer

Q.170 Find the condition that the roots of equation are equal. (7)

Ans:

Let

and

Now

Q.171 Evaluate . (6)

Ans:

=

Q.172 If prove that . (8)

Ans:

Cubic both sides

Þ

or Þ

Q.173 If a, b, c are lengths of sides opposite to angles A, B, C in a triangle ABC, then show that . (7)

Ans:

When DABC is an acute angled triangle.Draw perpendicular CD from C on AB

In D CAD, we have

In DCBD, we have

In DCBD,

Q.174 Show that in a triangle ABC,

a sin (B – C) + b sin (C – A) + c sin (A – B) = 0,

where a , b, c are lengths of sides opposite to angles A, B, C. (7)

Ans:

Let

L.H.S.

a sin(B-C)+ b sin (C-A)+c sin(A-B)

= K sin A sin(B-C)+K sin B sin(C-A)+K sin C sin(A-B)

= K[sin (B+C) sin (B-C)+sin(C+A) sin(C-A)+sin(A+B) sin(A-B)]

= K(sin² B-sin²C+ sin²C -sin²A+sin²A-sin²B]

= K(0)

= 0 = RHS

Q.175 Find the condition that the points (1, 1), (3, 5) and (a, b) are collinear. (7)

Ans:

Let A= (1,1) ,B= (3,5) ,C= (a,b)

The given points are collinear if x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)=0

Q.176 Find equations of lines which pass through the point (4, 5) and make an angle with the line 2x + y +1 = 0. (7)

Ans:

A line through point (4,5) is This makes angle 45⁰ with the line

whose slope is -2. Therefore.

tan 45⁰ = or

The required lines are

Q.177 Find the equation of the circle concentric with the circle

and which passes through (-4, 5). (7)

Ans:

Given circle is x²+ y²-4x-6y-9=0. Its center is (-f,-g) =(2,3)

The equation of circle whose center is (2,3) and radius r is (x-2)²+(y-3)²= r²

It passes through (-4,5) Þ(-4-2)²+(5-3)² =r² Þ 36+4 = r² Þ r² = 40

Required Circle is (x-2)²+(y-3)²=40

Q.178 Show that represents a parabola. Find its focus, vertex and directrix. (7)

Ans:

y²-8y-x+19=0 Þ (y-4)² =(x-3) ..............(1)

Let Y = y-4, X= x-3 (1) becomes Y²=X, which is a parabola.
Here 4a=1 Þ

Vertex: Vertex = (X=0, Y=0) Þ (x-3=0, y-4=0) Þ (x=3, y=4) So, Vertex = (3,4)

Focus: (X=a, Y=0) Þ Þ

Directrix: Equation of directorix is X= -a Þ x-3= - Þ x =

Q.179 Find . (6)

Ans:

= = 3.1 = 3

Q.180 Examine the continuity of the function f(x) = [x], where [x] is greatest integer , x being any real number. (8)

Ans:

Let a be any real number, then there exists an integer k such that k-1£ a £ k,

Case1: a ¹k-1

(LHL at x=a) =

(RHL at x=a) =

and f(a)=k-1. Thusso, f(x) is continuous at x=a.

Case2: a=k-1

Now while

Thus f(x) is not continuous at point a=k-1. Thus f(x) continuous at all points x¹ an

integer while it is discontinuous at integer points.

Q.181 Show that the semi verticle angle of a cone of maximum volume and a given slant height is . (7)

Ans:

Let be the semi-vertical angle of a cone of given slant height . Then, CO=Cos, OA= sin. Let V be the volume of the cone.
Then

For maximum or minimum V,

Thus V is maximum when tan

Q.182 Find the equation of tangent and normal to the curve at the point where it intersects the positive x-axis. (7)

Ans:

The equation of given curve is y=x²-9..............(1)

This cuts the x-axis at the point where y=0 Þ x²-9=0 Þ x=3

Point of contact = (3,0) Differentiating (1) w.r. to x, we get

Equation of tangent at (3,0) is y-0 =6 (x-3) Þ y -6x+18=0

Evaluation of normal at (3,0) is y-0=

Q.183 Find a reduction formula for the integral . (7)

Ans:

Let I_n = =

=

=

=

=

Q.184 Evaluate . (7)

Ans:

Let I=………………….. (1)

= =……(2)

Now

2.I = = = Þ I =

Q.185 Find the area bounded by and its latus rectum. (7)

Ans:

A rough sketch of the parobola is shown in Fig.

Let S(a,o) be the focus and , be the latus rectum of the parabola y²=4ax. The required area is Since the curve is symmetric about x-axis. So, required area = 2 area ()

Here, we slice the area into vertical strips. For the approximating rectangle shown in fig. we have length =y, width = Dx

Area = y Dx =

Since the approximating rectangle can move between x=0 and x=a

\Required area = 2 Area =2 =

Q.186 Find the volume of the solid obtained by revolving the ellipse , about its major axis. (7)

Ans:

Volume of solid

Q.187 Solve the equation . (6)

Ans:

This is homogeneous equation

Let y=vx

= v+x

Þ\v+x Þ v+x Þ x

= =

Separating the variables

Integrating both sides

. Let 1-2v-v² = t on LHS Þ (-2-2v)dv=dt

Þ (1+v)dv=

\ Þ Þ

= (1-2v-v²)^-½ =

Þ