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MATHS ANSWER KEY
TYPE - A
Q.NO OPTION ANNSWER Q.NO OPTION ANSWER
1 1 2,1 21 4 1
2 4 1 22 3 e9,
3 3
23 4
X2 + y2 = a2
4 4 Xoy plane 24 2 8
5 1 (i), (iii), (iv) 25 4 2
6 3 No solution 26 2 Log
7 3 27 3 276!
8 2 3 28 1 -z
9 4
29 2
10 3 - tan x
30 3
11 1 xdx + ydy = 0 31 2 ( ) + ( . - q2 ) =0
12 3 Is a perpendicular bisector of the line joining z1 and z2 32 3 X2 – y2 -2xy = c
13 3 600
33 1 a=
14 4 An asymptote parallel to y –axis 34 3 Maximum height of the curve is
15 3 Has only trivial solution only if rank of the coefficient matrix is
equal to the number
35 1
16 2
36 1
( Z, . )
17 2 ( 6t2 , 8t ) 37 4 Fermat’s theorem
18 1 -16 38 2 ( 4, 4 ) , ( - 4 , - 4 )
19 3 If p and q are two statements then p q is a tautology 39 1 6
20 3 9 40 1 ( 1 , 1 , 2 )
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TYPE - B
Q.NO OPTIO
N ANSWER Q.NO OPTION ANSWER
1 3 Maximum height of the curve is
21 3
are parallel 2 3 No solution 22 3 If p and q are two statements then p q is a tautology
3 3 Has only trivial solution only if rank of the coefficient matrix is equal to the number
23 2 3
4 3
Is a perpendicular bisector of the line joining z1 and z2
24 4 2
5 3 X2 – y2 -2xy = c 25 3 9
6 1 ( Z, . ) 26 1
7 1 (i), (iii), (iv) 27 2 Log
8 2 ( 4, 4 ) , ( - 4 , - 4 ) 28 4 1 9 3 e9,
29 1 xdx + ydy = 0
10 2 ( 6t2 , 8t ) 30 1 -16
11 2
31 3 -tanx
12 1 2,1 32 1 6 13 3 276! 33 4 1 14 1 ( 1 , 1 , 2 ) 34 4 x2 + y2 = a2
15 4
35 4 Fermat’s theorem
16 4 Xoy plane 36 2
17 3
37 4 An asymptote parallel to y –axis
18 2 ( ) + ( . - q2 ) =0 38 1 a=
19 2 8 39 3
20 1 -z 40 3 600
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41.
|A| = -5 -6 = -11
[Aij ] =
Adj A= [ Aij]T =
A(adj A ) =
=
= -11
A(adj A) = |A| . I ………………….(1)
(adj A) A =
=
= -11
(adj A) A = |A| . I ………………….(2) From (1 ) & ( 2 ) A(adj A) = (adj A ) A= | A | . I Hence verified
42. 2x + 2y + z = 5 x - y + z = 1 3x + y + 2z = 4
=
= 0
x =
0
Since = 0 and x 0 (atleast one of the values of x , y , z non zero) The system is inconsistent and it has no solution
43.
(i) = 4 - 3 + , = 2 - 4 + 5 , = -
= - =- 2 - + 4 = | | =
= - =- + 3 - 5 = | | =
= - =- 3 + 2 - = | | =
+ =
+
=
= 35
The points form a right angled triangle (ii) x2 + y2 + z2 – 3x – 2y + 2z – 15 = 0 comparing with
x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0
u = -
, v = -1, w = 1
c = =
One end of the diameter A is (-1, 4, -3) Let B(x2, y2, z2) be the other end of the diameter
Mid point of AB is the centre
=
x2 = 4, y2 = -2, z2 = 1
The co – ordinates of B are (4, -2, 1) 44. let P(x) = anxn + an-1x
n-1 + ….. + a1x + a0 = 0 be a polynomial equation of degree n with real coefficients. Z be a root of P(x) = 0. is also a root of P(x) = 0 P(z) = anxn + an-1x
n-1 + ….. + a1x + a0 = 0 Taking the conjugate in both sides
=
=
+
+ ……+ + = 0
+ + ……. + + = 0
= and a0 , a1, a2 , ……… an are real numbers
an +an-1 + …. + a1 + a0 = 0 P( ) = 0 is also a root of P(x) = 0
45. x4 + 4 = 0 x4 = -4 = 4(-1)
x =
=
=
=
k = 0, 1, 2 3
The values are cis
, cis
, cis
, cis
46. y2 = 8x at t =
equations of the tangent at ‘t’ is yt = x + at2
t =
= x +
y = 2x + 1
P.A. godpag;gd; M.Sc.M.Phil.,B.Ed
KJfiy fzpj gl;ljhup Mrpupah; muR Mz;fs; Nky;epiyg;gs;sp
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2x – y + 1 = 0 Equations of the normal at t is y + tx = 2at +at3
y +
= 2 +
4y + 2x = 8 +1 2x + 4y - 9 = 0
47. f(x) = x3 – 3x + 1 f’(x) = 3x2 – 3 = 3(x2 – 1)
f’’(x) = 6x f’’(x) > 0 when x > 0 f’’(x) < 0 when x < 0
The curve is concave downward on (-, 0) and concave upward on (0, ). Since the curve changes from concave downward to concave upward when x = 0 The point of inflection (0, f(0)) is (0, 1)
48. u(tx, ty) = sin-1
tan
= sin-1
tan
== t0u(x, y) f is homogeneous of degree 0 by Euler theorem,
x
+ y
= nu
x
+ y
= 0
49. I5
In =
[ cos n-1 x . sin x ] +
In-2
I5 =
[ cos 4 x . sin x ] +
I3
I3 =
[ cos 2 x . sin x ] +
I1
I1 =
I5 =
( cos 4 x . sin x ) +
(
cos 2 x . sin x +
sinx )
=
cos 4 x . sin x +
cos 2 x . sin x +
sinx )
50. The characteristics equation is 2p2 + 5p + 2 = 0
P=
=
P=-
and -2
C.F = A
+ Be-2x
P.I =
Again PI =
=
the general solution y = C.F + P.I
= A
+ Be-2x +
51.
p q pq qp
T T T T
T F F T
F T T F
F F T T
The columns corresponding to pq and qp are not identical.
pq is not equivalent to qp 52.
(i) Let G be a group. Let e1 and e2 be identity elements in G e1 as an identity element we have e1 * e2 = e2 ……..(1) e2 as an identity element we have e1 * e2 = e1 ………(2) from (1) and (2) , e1 = e2
Identity element of a group is unique
(ii) a-1 G and hence (a-1)-1 G. a * a-1 = a-1 * a = e a-1 * (a-1)-1 = (a-1)-1 * a-1 = e a * a-1 = (a-1)-1 * a-1 a = (a-1)-1
53. s =
X 2 3 4 5 6 7 8 9 10 11 12
P(X =
P.A. godpag;gd; M.Sc.M.Phil.,B.Ed
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x)
E(X) = xipi =
+
+
+ ………. +
=
= 7
54. Let X denote the height of a student
= 64.5 inches , = 3.3 inches
P(-<z<c)=0.99
P(-<z<0)+p(0<z<c)=0.99 0.5+p(0<z<c)=0.99 P(0<z<c)=0.49 P(0<Z<2.33) = 0.49 c = 2.33
Z =
2.33 =
X = 72.19 inches 55.
(a) The required plane passes through the point A(4, -2, -5) and is
perpendicular to
= 4 - 2 - 5 and = = 4 - 2 - 5 . = .
.(4 - 2 - 5 ) = (4 - 2 - 5 ). (4 - 2 - 5 ) = 45
(x + y + z ) . (4 - 2 - 5 ) = 45 4x – 2y – 5z = 45
(b) f(1) = 10 and f’(x) 2, 1 x 4 f(x) is differentiable on (1,4) and continuous on [1,4].
By law of Mean there exists an element c (1,4) f(a) = f(1) = 10 f(b) = f(4)
f’(c) =
=
3f’(c) + 10 = f(4) (3 x 2) + 10 = f(4) 16 = f(4) f(4) must be atleast 16
56. The given system of equation can be written in matrix form as
The augmented matrix is
[A,B] =
~
~
~
Case (i) : = 0 Now (A) = 2 ; A, B ] = 2
= [ A, B ] = 2 <no of unknowns (3 )
The system is consistent and has infinitely many solutions for = 0
Case (ii) : 0 Now [ A B ] = 3 = number of unknowns
The system is consistent and has unique solutions for 57. Let the points P and Q on the unit circle with centre at the origin O.
Assume that OP and OQ make angles A and B with x – axis respectively POQ = POX - = A – B The coordinates of P and Q are ( cos A, sinA ) and ( cos B , sin B ) Take the unit vectors and along x and y axis
= + = cos + sin A
= + = cos B + sin B
= ( cos A + sin A ) . (cos B + sin B ) ----------1
P.A. godpag;gd; M.Sc.M.Phil.,B.Ed
KJfiy fzpj gl;ljhup Mrpupah; muR Mz;fs; Nky;epiyg;gs;sp
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cos A cos B + sin A sin B
definition , = | | | | cos ( A – B ) = cos ( A – B ) -------2 (1 ) and ( 2) cos ( A – B ) = cos A cos B + sin A sin B 58. The points ( x1, y1, z1 ) = ( 1, 2 , 3 )
( x2, y2, z2 ) = ( 2, 3 , 1 ) ( l1, m1, n1 ) = ( 3, -2, 4 )
Vector equation
= ( 1 – s ) + s + t or = + s ( - ) + t
= ( 1 –s ) ( + 2 + 3 ) + s ( 2 + 3 + ) + t ( 3 - 2 +4 ) (or)
= ( + 2 + 3 ) + s ( + t( 3 - 2 4 ) Cartesian form :
(x1, y1, z1 ) is ( 1, 2, 3 ) ; ( x2,y2 , z2) is ( 2, 3 , 1 ) and ( l1 , m1 , n1) is ( 3,-2,4 )
The equation of the plane is
= 0
= 0
(x-1) (4-4) – (y-2) (4+6) + (z-3) (-2-3) = 0
59. Let z = x+ iy
=
x
Re
= 1
= 1
x2 + y2 – x + y = x2 + y2 + 2y + 1 Locus of P is x + y + 1 = 0
60. Consider the suspension bridge to be open upwards i.e. x2 = 4ay From the given data , the vertex of the bridge lies 5 meter above
the roadway . The span of the bridge being 40 meters. The point A(20, 50) lies on the parabola
400 = 4a (50)
a= 2 The equation is x2 = 8y The point Q is ( x1 , 25 ) lies on the parabola x1
2 = 8 ( 25 ) = 200
x1 = 10
PQ = 2x1 = 20 mts is the length of the support 61. 4x2 - 9y2 + 8x -36 y – 68 = 0
4x2 + 8x - 9y2 – 36 y = 68 4[x2 + 2x ] – 9 [ y2 + 4y ] = 68 4 { [x +1 ]2 -1 } -9 { (y + 2 )2 – 4 } = 68 4 (x +1 ) 2 - 9 ( y + 2 )2 = 68 + 4 – 36
-
= 1
-
= 1
( h, k) = ( -1 , - 2 ) Length of semi major axis a = 3 Length of minor axis 2b = 2 b=1 (iv) The equation major axis x = -1
Equation of ellipse
+
Eccentricity
e=
=
=
=
e =
62. One of the asymptotes is x +2y – 5 =0 The other asymptotes is of the form 2x – y + k = 0 Equation of the R.H. is
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( x + 2y – 5 ) ( 2x - y + k ) + c = 0 It passes through ( 6 , 0 ) ( 6 – 5 ) ( 12 + k ) + c = 0 k + c = -12 ---------(1) It also passes through ( -3 , 0 )
(-3 -5 ) ( -6 + k) + c = 0 (-8 ) ( -6 + k ) + c = 0 48 – 8k + c = 0 -8k+ c = - 48 -------- (2) Solving ( 1 ) – ( 2 ) k + c = - 12 -8 k + c = - 48 (+) (-) (+) 9k = 36 k = 4 substitute k = 4 in (1) , we get 4 + c = -12 c = -12 - 4 c = -16
Equation of the R.H is ( x + 2y - 5 ) ( 2x – y + 4 ) - 16 = 0 63. Pick a point P and draw the tangent line at P intersecting again at Q. Let a
be the x – co-ordinate of P so that P is the point ( a, a3 ) We have y = x3
The slope at P is
m =
Now Q is the point of intersection of the tangent at P and the curve y = x3 -----(1)
Equation of this tangent line at P is
y – a 3 = 3a2 ( x – a ) y – a3 = 3a2 x – 3a3 or y = 3a2x – 2a3 ------(2) Solving (1) and (2) x3 = 3a2 x – 2a3 or x3 – 3a2 x + 2a3 = 0 that is (x – a)2 ( x +2a ) = 0 x =a or x = -2a Now a is the x – coordinate of P -2a must be the coordinate of Q.
Hence slope at Q =
3(-2a)2 = 3 x 4a2 = 4 (3 a2)
That is the slope at Q = 4 times the slope at P.
64. (i) Domain , extent , intercepts and origin The function is defined for all real values of x and hence the domain is ( - The horizontal extent is - and the vertical extent is - . Clearly it passes through the origin since ( 0, 0 ) satisfies the equation
(ii)symmetry
It is symmetrical about the origin (iii) Asymptotes
The curve does not admit any asymptotes (iv) Monotonicity
Since y’ 0 for all x , the curve is increasing in ( - , ) (v) Special points
Since y’’ = 6x the curve is concave upward in ( 0, ) and convex upward in ( - , 0 0 y’’ =0 for x = 0 yield ( 0, 0 ) as the point of inflection
65. For x – axis 4y2 = 9x gives the maximum area 3y2 = 16y gives the minimum area
P.A. godpag;gd; M.Sc.M.Phil.,B.Ed
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Thus w.r. to x – axis 4y2 -9x is the first curve and 3y2= 16y is the second curve . But if we take y – axis as axis of boundedness 3y2 = 16y is the first curve and 4y2 = 9x is the second curve
=
= ( 8 – 4 ) – 0 = 4 sq . units
66. x = a cos3t, y = a sin3t is the parametric form of a given asteroid, where 0
1 2
= - 3a cos2t sin t
= 3a sin2t cos t
= = 3a sin t cos t
Since the curve is symmetrical about both axes, the total length of the curve is 4 times the length of the first quadrant
But t varies from 0 to
in the first quadrant
Length of the entire curve = 4
dt
= 4
cos t dt = 6a
= 6a. –
= -3a[cos - cos 0]
= - 3a [-1 – 1] = 6a
67.
= -
Put y = vx
L.H.S = v + x
; R.H.S = -
= -
v + x
= -
x
= -
= -
dv
Integrating we have 4 log x = - log(v4 + 6v2 + 1) + log c log[x4(v4 + 6v2 + 1)] = log c i.e. x4 (v4 + 6v2 + 1) = c or y4 + 6x2y2 + x4 = c
68. Let A be the population at time t
A
= kA
A = cekt Take the year 1960 as the initial time i.e. t = 0 When t = 0, A = 130000 130000 = ce0 = c A = 130000ekt When the year 1990 i.e. when t = 30, A = 160,000
160,000 = 130000 x e30k e30k =
Find A, when the year 2020 i.e., when t = 60
A = 130000 x e60k = 130, 000 x
~ 197000
The approximate population in 2020 is 197000
69. a) for number of incoming buses per minute = 0.9
for number of incoming buses per 5 minutes = 0.9 x 5 = 4.5
P exactly 9 incoming buses during 5 minutes =
i.e P(X = 9) =
Fewer than 10 incoming buses during a period of 8 minutes = P(X < 10)
Here = 0.9 x 8 = 7.2
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Required probability =
P atleast 14 incoming buses during a period of 11 minutes = P(X 14) = 1 – P(X < 14)
Here = 11 x 0.9 = 9.9
Required probability = 1 –
70. (a) a is the radius of the sphere
let x be the base radius of the cone and h is the height of the cone
volume V =
h
=
(a + y)
where OC = y so that height h = a + y From the diagram x2 + y2 = a2
Using (2) in (1) we have
V=
( ) (a+y)
For the volume to be maximum v’=0
[ a2 – 2ay – 3y2 ] = 0
y=
and y = -a is not possible
now V’’ =
(a + 3y) <0 at y =
The volume is maximum when y =
and the maximum volume is
x
( a +
=
(
)
=
( volume of the sphere)
(b) Let Zn = { [ 0 ] , [ 1 ] , [ 2] , …..[ n -1 ] } be the set of all congruence
classes modulo n. and let [l] , [ m ] , Zn 0 l , m , < n
(i) Closure axiom : By definition
[l] +n [ m ] =
where l + m =q. n + r 0 r < n
In both the cases , [ l + m ] Zn and [ r ] Zn \ Closure axiom is true (ii) Addition modulo n is always associative in the set of congruence classes modulo n (iii) The identity element [0] Zn and it satisfies the identity axiom (iv) The inverse of [l] Zn and [l] + n [ n – l ] = [ 0 ] [ n – l ] + n[l] = [0] The inverse axiom is also true . Hence ( Zn , +n ) is a group
P.A. godpag;gd;
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