Mathematics Paper1

8/11/2019 Mathematics Paper1

1/34

MATHEMATICS GRAND TEST PAPER

1. If k1

ak(k 1)

=+

for k = 1, 2, 3, n then find

2n

kk 1

a=

=

1)n

n 1+ 2)

2

2

n

(n 1)+3)

4

4

n

(n 1)+ 4)

6

6

n

(n 1)+

2. ( )1/ 4

4f (x) 25 x= for 0 x 5< < then1

fof2

=

1)1

16 2)

1

83)

1

4 4)

1

2

3. 2 2 2 27 4

cos cos cos cos ...18 9 18 9

+ + + =

1) 0 2) 1

3) 1 4) 2

4. If tan A =a(a b c)

bc

+ +, tan B =

b(a b c)

ac

+ +and tan C =

c(a b c)

ab

+ +then A + B + C

=

1) /4 2) 3) /2 4) /3

5. The value of cosec 10 3 sec 10 is

1) 2 2) 4 3) 0 4) 1

6. If1

cos x cos y3

+ = and sin x + sin y =1

4then cot(x + y) =

1)24

25 2)

7

243)

16

24 4) None

7. The min value of1

3sin x 4cos x 7 +is

1)1

6 2)

1

83)

1

11 4)

1

12

s

ducatio

n.co

m

b(a

2)

sec 10

os x cos

1)24

www.sakshieducation.com



2/34

8. If sin(270 x) = cos 292 then the value of x lying between 0 and 360 are

1) 240, 120 2) 248, 112 3) 225, 115 4) none

9. If (cos x)y= (sin y)xthen dydx

=

1)log sin y y tan x

log cos x x cot y

+

2)

log sin y y tan x

log cos x x cot y

+ 3)

logsin y

logcosx 4) none

10. 1 14 2

tan cos tan5 3

+ =

1)6

17 2)

7

163)

17

6 4)

16

7

11. A solution of the equation 1 1tan (1 x) tan (1 x)2

+ + =

1) x = 1 2) x = 1 3) x = 0 4) x =

12. In ABC,cos A cos B cos C

csin B a sin C b sin A+ + =

1) R 2) 2R 3) 1/R 4) 2/R

13. If A A2= I then |I + A| =

1) 1 2) 1 3) 0 4) 2

14. In ABC, D, E are mid points of AB,AC then BE DC+ =

1)1

BC2

2)3

BC2

3) 2BC 4) 3BC

15. In ABC, if r1: r2: r3= 2 : 4 : 6, then

a : b : c =

1) 3 : 5 : 7 2) 1 : 2 : 3 3) 5 : 8 : 9 4) 1 : 8 : 9

16. If OA i j= , OB j k = then a unit vector perpendicular to plane AOB is

1)i j k

3

+ + 2)

i j k

3

+ 3)

i j k

3

4)

i j k

3

+

.

iu

tion.co

m e

4) x =

3) 0

po nts

2)3

, if r : r:

: c =

1) 3 : 5 ::

f




3/34

17. If u a b, v a b and| a | | b | 2= + = = = , then | u v | =

1) 216 (a b) 2) 216 (a b) 3) 4 216 (a b) 4) 3 216 (a b)

18. An open box is to be made out of give quantities of card board of area 432 sq.cm. If its

base is a square, then the measurements of box for maximum volume.

1) 12, 12, 6 2) 6, 6, 12 3) 12 3,12 3,6 3 4) 6 3,6 3,12 7

19. | a | 2,| b | 3,| c | 6= = = , a,b ,c are perpendi-cular to b c, c a, a b+ + + respectively, then

| a b c |+ + =

1) 7 2) 11 3) 121 4) 49

20. If three unit vectors a,b ,c are such that1

a(b c) b2 = , then the vector a makes with

b and c respectively the angles

1) 40, 80 2) 45, 45 3) 30, 60 4) 90, 60

21. Given P(a, 0) and Q(a, 0) and R is a variable point on one side of the line PQ such that

QPR RQP = 2. The locus of the point R is

1) x2 y2+ 2xy tan 2= a2 2) x2+ y2+ 2xy cot 2= a2

3) x2+ y

2 2xy tan 2= a2 4) x2 y2+ 2xy cot 2= a2

22. The transformed equation of 4xy 3x2= a2, if the axes are rotated through an angle

tan1

(2) is

1) X2 4Y2= a2 2) X2 + 4Y2= a2 3) X2 4Y2+ a2= 0 4) X2 + 4Y2+ a2= 0

23. If PL, PM be drawn perpendicular to YZ, ZX planes making at L, M from the point (a,

b, c) then the equation of plane OLM is

1) ax + by + cz = 0 2)x y z

0a b c

+ + =

3)x y z

1a b c

+ + = 4)x y z

0a b c

+ =

24. The distance from a point (, ) to a pair of lines passing through the origin is d. Thenthe equation to the pair of lines is

1) 2 2 2 2( x y) d (x y ) = + 2) 2 2 2 2( x y) d (x y ) + = +

3) 2 2 2 2( x y) d (x y ) = + 4) 2 2 2 2( x y) d (x y ) 0 + + =

a

a

n.c

spectively

, t en t ee

, 60 4))

iable poi

e point

) x y

4) x2

ation of 4

a

be dra

then the

1) ax +




4/34

25.1 1

3x 0

Tan x Sin xLt

x

=

1) 12

2) 12

3) 2 4) 2

26. If y =2n

1

2n

x 1Cos

x 1

+ then ( )2n

dy1 x

dx+ =

1) n2nx 2) n2nx 3) n2nx 4) n2nx

27. If f(x) =x 2

a| x 2 |

+

; x < 2; f(2) = a + b ;

| x 2 |f (x) b

(x 2)

= +

, x > 2, f(x) continuous at x = 2.

Then (a, b) =

1) (1, 1) 2) (1, 1) 3) (1, 1) 4) (1, 1)

28. The area (in square units) of the quadrilateral formed by the two pairs of lines

l2x

2 m

2y

2 n(lx + my) = 0 and l

2x

2 m

2y

2+ n(lx my) = 0 is

1)2n

2 | lm | 2)

2n

| lm|3)

n

2 | l m | 4)

2n

4 | l m |

29. Find the distance of the point (2, 3) from the line 2x 3y + 9 = 0 and along a line

making an angle of 45 with X-axis

1) 4 2 2) 4 3) 4 / 2 4) 2 2

30. If the d.c.s l, m, n of two lines are connected by the relation is l + m + n = 0, mn 2nl

2lm = 0 then d.r.s of the two lines

1) (1, 2, 1), (1, 1, 2) 2) (1, 1, 2), (1, 1, 2)

3) (1, 2, 1), (1, 1, 2) 4) (1, 2, 1), (1, 1, 2)

31. p a (b c),q b (c a), r c (a b)= = = then p, q, rare

1) Collinear 2) Coplanar 3) Non-coplanar 4) none

32. If sin cot cos tan4 4

=

then =

1)n

4 4

= + 2) n

2

= +

3) 3n

2

= + 4) n

2

=

on.

om x) contin

(1, 1)

for ed

(lx my)

m | 4)

(2, 3) frof

X-axis

4 3

of two li

.r.s of th

), (1, 1, 2,

, 2, 1), (1(

p a

1) C




5/34

33. x + 2y + 3z = 1, 2x + y + 3z = 2, 4x + 5y + 9z = 4 the system of equation has

solution

1) Unique 2) no solution 3) Infinite 4) two solutions

34. (r1 r) (r2+ r3) =

1) a2 2) b2 3) c2 4) none

35. If G is the centroid of ABC where

A = (a, 0), B = (a, 0) and C = (b, c) then (AB2+ BC

2+ CA

2) / (GA

2+ GB

2+ GC

2)

1) 1 2) 2 3) 3 4) 4

36. The value of sin h(ix) is

1) sin x 2) sin (ix) 3) i sin x 4) i sin x

37. The function f(x) is defined as 22

1 1f x x

x x

+ = +

when x 0 then the function is

1) f(x) = x2 2 2) f(x) = x

2+ 1 3) f(x) = x

2+ 2 4) f(x) =

1x

x+

38. If cosA .cos2A .cos 4A cos 2n1A = k sin 2nA, then k =

1) n2 sin A 2)n

1

2 sin A3)

n 1

1

2 sin A 4) n 12 cos A

39. The subnormal to the curve y = a1n

xn

is constant then n =

1) 2 2) 2 3) 1/2 4) 1/2

40. The base of a triangle is fixed the locus of the vertex when one base angle is double of

other

1) y2 3x2+ 2ax + a2= 0 2) y2+ 3x2+ 2ax + a2= 0

3) y2 3x

2 2ax + a

2= 0 4) y

2+ 3x

2 2ax a

2= 0

41. The equation whose roots are the squares of the roots of the equation x2 3x + 1 = 0

1) x2 7x + 1 = 0 2) x

2 7x 1 = 0

3) x2+ 7x + 1 = 0 4) x

2+ 7x 1 = 0

42.1

x iy1 cos i sin

+ =+ +

then x2=

1) 1 2) 3) 1/4 4) 4

.

n.c

+ GC2)

n x

hen x 0

= x + 2

k sin 2

n A3)

rve y = a

2) 2

triangle i

3x + 2a

y2 3x

22

. The e




6/34

43. The value of6

k 1

2k 2k sin i cos

7 7=

is equal to

1) i 2) I 3) 2i 4) 1

44. If xx

fxf =

13)(2 , then dxxf

2

1)( is equal to

(1) 2ln5

3 (2) )2ln1(

5

3+

(3) 2ln

5

3 (4) None of these

45. Number of ways of selecting 4 letters out of the letters of the word MATHEMATICS

is

1) 136 2) 11C4 3)8C4 4) none

46. A man speaks truth 3 out of 4 times. He throws a die and reports that it is a six.

Probability that it is actually six is

1)3

8 2)

1

5 3)

3

4 4) none

47. If the letters of the work REGULATION be arranged at random then the probability

that there will be exactly four letters between R and E is

1) 1/2 2) 1/5 3) 1/9 4) 1/10

48. The number of irrational terms in the expansion of ( )100

54 5 4+ is

1) 50 2) 95 3) 94 4) 49

49. X is a poisson variate such that P(X = 2) = 9P(X = 4) + 90 P(X = 6) then mean

1)1

2 2) 1 3) 2 4)

3

2

50. Six + and four signs can be arranged in a row so that no two signs are together

is

1) 35 2) 7p4 3) 7! 4) 1

51. The number of partial fractions of

2

2 4

3x 1

(x 1)

+

+ is

1) 1 2) 2 3) 3 4) 4

.

t

.c THEM

and repo

4) non

be arrana

etween R

/9

ms n t e

95

ate suche c

and fou

1) 35)




7/34

52. A has 3 shares in a lottery containing 3 prizes and 9 blanks. B has 2 shares in a lottery

containing 2 prizes and 6 blanks. Compare their chances of success

1)34 15

:55 28 2)32 13

:55 28 3)

32 15:55 28 4)

34 13:55 28

53. Radical axes of two circles with centres

(a, c), (b, c) is y-axis radius of first circle is r, radius of second circle is

1) 2 2 2r a b+ + 2) 2 2 2r a b + 3) 2 2 2r a b 4) 2 2 2r a b+

54. Out of (2n + 1) tickets consecutively numbered three are drawn at random. The

probability that the numbers on them are in AP is

1)2

2

3n

4n 1 2)

2

n

4n 1 3)

2

3n

4n 1 4)

2

6n

4n 1

55. If the normal at P on y2= 4ax cuts the axis of the parabola in G and S is the focus then

SG =

1) SP 2) 2SP 3)1

SP2

4) SP

56. The equation to the asymptotes of the hyperbola 2x2+ 5xy + 2y

2 11x 7y + k=0 is

1) 2x2+ 5xy + 2y

2 11x 7y = 3 2) 2x

2+ 5xy + 2y

2 11x 7y = 5

3) 2x2+ 5xy + 2y

2 11x 7y + 3 = 0 4) 2x

2+ 5xy + 2y

2 11x 7y + 5 = 0

57.dx

1 sin x cos x=

+ +

1)1

log(1 tan x / 2) c2

+ +

2) log(1 cot x / 2) c+ +

3) log(1 tan x / 2) c+ + 4) none of these

58. I) Equation of the ellipse whose foci (5, 0) and5

e8

= is2 2x y

764 39

+ = .

II) The eccentricity of the ellipse

2 2x y116 25+ = is

5

3 .

Which of the above is correct?

1) Only I 2) only II

3) Both I and II 4) neither I nor II

. ak

c

.

2a

n at ran

4))6n

n

arabola i

SP2

he hyper

3 2)

y + 3 = 0

an x

(1 an x

. I) Equa




8/34

59.1 1

1 1

20 0

1Tan dx K Tan x dx

1 x x

= + +

, then K =

1) 1 2) 0 3) 2 4) none

60. Area of the region bounded by y = ex, y = e

xand x = 2 is square units

1)

21

ee

+

2)

21

ee

3)

21

ee

4)

21

ee

+

61. The length of the perpendicular from the focus S of the parabola y2 = 4ax to any

tangent at any point P is

1) 2 2OS SP+ 2) OS SP 3) OS + SP 4) OS SP

62. sin x dxcos(x a)

=

1) sin a log cos(x a) + x cot a + c 2) cos a log cos(x a) + x sin a + c

3) sin a log sin(x a) + x cos a + c 4) cos a log sin(x a) + x sin a + c

63.n

1 1 1 1Lt ...

3n 1 3n 2 3n 3 6n

+ + + +

+ + + =

1) log 6 2) log 3 3) log 2 4) 0

64. The differential equation of 3x 5xy ae be= where a, b are arbitrary constants is

1) y2 8y1+ 15y = 0 2) y2+ 8y1+ 15y = 0

3) y2+ 2y1 15y = 0 4) y2 2y1+ 15y = 0

65. Let nxxx ,....,, 21 be nobservations such that 4002 = ix and 80= ix . Then a possible value

of namong the following is

(a) 9 (b) 12 (c) 15 (d) 18

66.a ib

tan i loga ib

=

+

1)2 2

ab

a b+ 2)

2 2

ab

a b 3)

2 2

2ab

a b 4)

2 2

2ab

a b+

d

tion.c

2 = 4ax

g cos(x

a log sin(

3)

f y

0

e nob

ong the fo

9

6. ana




9/34

67. If2 23 2x 3x 2

(20) (40 5) = , then x =

1)13

12 2)

12

13

3)4

5 4)

5

4

68. The solution of x(y x)dy = y(x + y)dx

1) y / xe cxy= 2) x / ycxy e 1= 3)y / xcxy e 1= 4) xy

cye

x=

69. Write the decreasing order of the product of perpendicular from any point on

following hyperbolas to its asymptotes is

A)2 2

(x 3) (y 2)1

16 9

=

B)

2 2(x 3) (y 2)

15 6

=

C)2 2

(x 1) (y 2)1

6 9

=

D)

2 2(x 3) (y 2)

12 3

=

1) A, B, D, C 2) A, C, B, D 3) A, B, C, D 4) A, C, D, B

70. Tn denotes the number of triangles which can be formed with the vertices of regular

polygon of n sides. If n 1 nT T 21+ = then n =

1) 5 2) 7 3) 6 4) 4

71. PQ is a double ordinate of the parabola axy 42 = . The locus of the points of trisection

of PQ is

(1) axy 492 = (2) ayx 49

2 = (3) 0492 =+ axy (4) 049

2 =+ ayx

72. The area of the region between y = sin x, y = cos x and y-axis is 0 x2

<


10/34

75. P is any point on the ellipse2 2

2 2

x y1

a b+ = with foci S and Sthen maximum area of SS

is

1) ab 2) 2ab 3) abe 4) abe2A

76. circle of radius r passes through the origin O and cuts the axes at A and B. Locus of

the centroid of triangle OAB is

1) x2+ y2= 4r2 2) (x2+ y2) = 4r2 3) x2+ y2= r2 4) x2+ y2= 3r2

77. x2+ y

2+ 2x 3y + 1 = 0, x

2+ y

2+ 6x 5y + 9 = 0 are two circles. Find equation of circle

with common chord as diameter.

1) 5x2+ 5y

2 12x 16y 9 = 0 2) 5x

2+ 5y

2 12x 16y + 9 = 0

3) 5x2+ 5y2+ 12x + 16y + 9 = 0 4) 5x2+ 5y2+ 12x 16y + 9 = 0

78. If1 1 3 1 3 5

x ...5 5 10 5 10 15

= + + +

, then 3x

2+ 6x =

1) 0 2) 1 3) 2 4) 1

79. Area of region enclosed by ay = 3(x2 a

2) and the X-axis is

1) a2 2) 4a

23) 2

4a

3 4) 2a

2

80. If x2+ 6x 27 > 0, x

2 3x 4 < 0 then x lies in the interval

1) (3, 4) 2) [3, 4] 3) (, 3] [4, ) 4) (9, 4)

www

.sa

a

.

y = 3r

d equatio

y + 9 = 0

16y + 9

4) 1

a ) and th

2a3

4 < 0 the

[3, 4]




11/34

HINTS AND SOLUTIONS

1. (2)

Sol. k1 1 1

ak(k 1) k k 1

= = + +

2n

kk 1

1 1 1 1 1a 1 ...

2 2 3 n n 1=

= + +

+

2 2

2

1 n1

n 1 (n 1)

= =

+ +

Hence (2) is correct choice.

2. (4)

Sol.1 1

fof f f 2 2

=

1/ 4 1/ 4

1/ 4 1/ 4

1 399f 25 f

16 16

399 1 125

16 16 2

= =

= = =


3. (4)

Sol. cos210 + cos

220 + cos

270 + cos

280 = 2

If A + B = 90 cos2A + cos2B = 1

Hence (4) is correct choice.(2)

4. (2)

Sol. tan A=a(a b c)

bc

+ +, tan B =

b(a b c)

ac

+ +and tan C =

c(a b c)

ab

+ +

.s

hied

ucatio

n.co

m

2

oice..

+ cos 20

A + B = 9=

Hence




12/34

By observation

Tan A + tan B + tan C = tan A tan B tan C

A + B + C =


5. (2)

Sol. The value of cosec 10 3 sec 10

1 3 cos10 3 sin10

sin10 cos10 sin10 cos10

1 32 cos10 sin10

2 2

1(2sin10 cos10 )

2

4sin(30 10 )4

sin 20

=

=

= =

6. (2)

Sol.sin x sin y 3 x y 3

tancos x cos y 4 2 4

+ + = =

+

32

244tan(x y) 9 71

16

+ = =

7cot(x y)

24 + =

7. (4)

Sol. The min. value of1

3sin x 4cos x 7 +

1 1

max(3sin x 4 cos x 7) 7 25

1 1

7 5 12

= = + +

= =+

w.saks

hied

ucatio

n.co

m

4

24=

. The min..




13/34

8. (2)

Sol. sin(270 x) = cos 292

sin(270 x) = cos(270 + 22) = sin 22

270 x = 22, 180 22

x = 248, x = 270 158 = 112

9. (1)

Sol. y x(cos x) (sin y)=

y log cos x x log sin y

dy y( tan x) logsin y

dx log cos x x cot y

y tan x log sin y

log cos x x cot y

=

=

+=

10. (3)

Sol. 1 14 2

tan cos tan5 3

+ =

1 1

3 23 2 4 3tan tan tan

3 24 3

1 4 3

9 8 17

12 6 6

+

+ =

+= =

11. (3)

Sol. 1 1tan (1 x) tan (1 x)2

+ + =

Option (3) is satisfied by verification

12. (3)

Sol.cos A cos B cos C

csin B a sin C bsin A+ + =

1 cos A cos B cosC

2R sin B sin C sin Csin A sin A sin B

+ +

.sak

hied

ucatio

n.co

m

3 2

4 3

x

tion (3) is




14/34

1[sinA cos A

2sin Asin Bsin C

sinB cosB sinCcosC]

=

+ +

1[sin 2A sin 2B sin 2C]

4RsinAsinBsinC= + +

4sin Asin Bsin C 1.

4R sin A sin B sin C R = =

13. (3)

Sol. A2= I

A2 I = 0 (A + I)(A I) = 0

|A + I||A I| = 0

|A + I| = 0

Since |A I| 0

14. (2)

Sol.

OA a,OB b,OC c= = =

BE DC OE OB OC OD

a c a b 3(c b)b c

2 2 2

3BC

2

+ = +

+ + = + =

=

15. (3)

Sol.2 3 3 1 1 2

1 1 1 1 1 1a : b : c : :

r r r r r r = + + +

B C

E

A

D

.ks

hied

ucatio

n.co

m

,

a

2

3

cc

=




15/34

1 1 1 1 1 1: : 10 :16 ::18

4 6 6 2 2 4

5:8:9.

= + + + =

=

16. (1)

Sol. Unit vector perpendicular to plane AOB

OA OB i j k

| OA OB | 3

+ += =

17. (3)

Sol. | u v | 2 | a b | =

2 2 2

2

4 a b (a b)

4 16 (a b)

=

=

18. (1)

Sol. a2= 432

a 12 3=

Measurements of box are

a a a, , 12,12,6

3 3 2 3

= .

19. (1)

Sol. a b c, b (c a), c (a b) + + +

a b b c c a 0 + + =

2 2 2 2| a b c | | a | | b | | c |

4 9 36 49.

+ + = + +

= + + =

20. (4)

Sol.1

a (b a) b2

=

kshi

educ

ation.co

m

c a c

c c a

a c

=

(4)




16/34

1(a c)b (a b)c b

2

1a c a b 0

2

(a, c) (a, b)3 2

=

= =

= =

21. (4)

Sol. QPR RQP = 2

1 2= 2

tan1 tan2= tan 2(1 + tan 1tan 2)

2

1 1 12 2

1 1 1

y y ytan 2 1x a x a x a

= +

+

2 2 21 1 1 12x y tan 2 (x y a ) =

Locus is x2 y2+ 2xy cot 2= a2.

22. (1)

Sol. f(X cos Y sin , X sin+ y cos) = 0

Tan = 2, cos=1

5, sin=

2

5

2

2X 2Y 2X Y X 2Y4 3 a5 5 5

+ =

i.e., X2 4Y

2= a

2

23. (4)

Sol. PL perpendicular YZ then L(0, b, c)

PM perpendicular ZX then M(a, 0, c)

1

R(x1,y1

2

Q(a,0) P(a,0)

.

iedu

catio

n.co

m

= a .

, X sin

os =

i.e.,. .,




17/34

x y z

0 b c 0

a 0 c

=

bcx + acy abz = 0

24. (3)

Sol. 2 2 2 2( x y) d (x y ) = +

25. (2)

Sol.1 1

3x 0

Tan x Sin xLt

x

2 2

2x 0

1 11 x 1 x

Lt3x

+ =

2 2 2 3/ 2

x 0

2x x

(1 x ) (1 x )Lt

6x

+ =

2 2 3/ 2

x 0

2 1

2 1 11 x (1 x )Lt

6 6 2

+ = = =

26. (2)

Sol. y =2n

1

2n

x 1Cos

x 1

+

Put xn= tan

2n 2

2n 2

x 1 tan 1cos2

x 1 tan 1

cos( 2 )

= =

+ +

= +

1 n

y 2 2Tan (x )

= + = +

n 1

2n

dy 12 nx

dx 1 x

= +

. ak

educ

ation.co

m

1 1

6 2=

an

x

an

= coss




18/34

2n n 1dy(1 x ) 2nxdx

+ =

27. (3)

Sol.x 2

f (x) a, x 2| x 2 |

= +

f(x) is continuous

x 2 x 2

Lt f (x) f (2) Lt f (x)

1 a a b 1 b

a 1,b 1

+ = =

+ = + = +

= =

28. (1)

Sol. The given equations represents

lx + my + n= 0, lx my n = 0

lx + my = 0, lx my = 0

Area of parallelogram = 1 1 2 2

1 2 2 1

(c d )(c d )

a b a b

2(n 0)( n 0) n

l( m) l(m) 2 | lm |

= =

29. (1)

Sol. 1 1ax by c

ra cos bsin

+ +=

+

2(2) 3(3) 9 4

1 1 12 3

2 2 2

+ + = =

4 2=

s

ducatio

n.co

m

)

2n

2 | l m |

c1

acoso b

2 2=




19/34

30. (3)

Sol. Let l = (m + n)

mn + 2n(m + n) + 2m(m + n) = 0

lm2+ 2m2+ 5mn = 0

let l = 1, m = 1 n

2 + 2n2+ 4n + 2n2 5n 5n2= 0

n2 n + 2 = 0

n2+ n 2 = 0

(n + 2)(n 1) = 0

n = 2, 1

n = 1, m = 2

(1, 2, 1)

n = 2, m = 1

(1, 1, 2)

31. (2)

Sol. p q r 0+ + =

p, q, r are coplanar.

32. (1)

Sol. sin cot cos tan4 4

=

2 2

cot tan4 4 2

1tan 2

tan

1 tan 2 tan (1 tan ) 01 tan 0 tan 1

n4

= + =

+ =

+ = = = =

= +

.ks

hied

ucatio

n.co

m

cos

4 4

1

cot t




20/34

33. (3)

Sol.

1 2 3 1

2 1 3 2

4 5 9 4

R2 2R1

R3 4R1

1 2 3 1

0 3 3 0

0 3 3 0

R3 R2

1 2 3 1

0 3 3 0

0 0 0 0

r = r1= 2

the system has infinite solution.

34. (1)

Sol. 1 2 3(r r)(r r ) +

2

22

2

s a s s b s c

s s a s c s b

s(s a) (s b)(s c)

[a a]a

+

+ +

=

35. (3)

Sol.2 2 2

2 2 2

AB BC CA

AG BG CG

+ +

+ +

2 2 2

2 2 2

3(AG BG CG )3

AG BG CG

+ += =

+ +.

w.

shi

educ

ation.co

m

a

bs

s c

)

ol.AB

AG




21/34

36. (3)

Sol.ix ixe e

sinh ix2

=

[ ]1

(cos x i sin x) (cos x isin x)2

1[2i sin x ] i sin x .

2

= +

= =

37. (1)

Sol. 22

1 1f x x

x x

+ = +

=

21

x 2x

+

2

f (x) x 2= .

38. (2)

Sol. cosA .cos2A .cos 4A cos 2n1A

n

n n

sin(2 A) 1;k

2 sin A 2 sin A=

39. (4)

Sol. y = a1nxn

1 n n 1

1 n n 1 n n 1

2 2n 2n 1

dy dyna x , S.N. ydx dx

a x n a x

na x

= =

=

=

SN is constant 2n 1 = 0.

40. (3)

Sol.

B(a,0)A(a,0)

C(, )

11

O

2

w

.shi

educ

ation.co

m

n n

n n

dydxx

a x

x

onstant 2n2

(3)

l.




22/34

Let C(, ); AO = OB = a

1= 22

21 2

2

2tantan1 tan

=

2

2a

a1

a

=

+

is 2 2 23 2a a 0 + =

locus is y2 3x2 2ax + a2 = 0.

41. (1)

Sol. The equation whose roots are squares of roots of f(x) = 0 is

2

2

f ( x ) 0

x 3 x 1 0

(x 1) 9x

x 7x 1 0

=

+ =

+ =

+ =

42. (3)

Sol. Put = 90

1 1 ix iy

1 i 2

+ = =

+

2 1x4

=

43. (2)

Sol.

6

k 12k 2k sin icos7 7=

6

k 1

2k 2k i cos isin

7 7

i( 1) i

=

= +

= =

. aks

hied

ation.co

m

f(x) = 0 i

1 i

i 2=

1

4=

. (2).




23/34

44. (2)

xx

fxf =

13)(2 (i)

Replacingxby

x

1 in (i), we getx

xfx

f1

)(31

2 =

..(ii)

Eliminating

xf

1 from (i) and (ii), we get

3

3232)(5

2 +=+=

x

xxxf

+=

x

xxf

5

32)(

2

=2

1)( dxxf 21

22

1

2

]log3[5

1

5

32xxdx

x

xe+=

+

[ ]2ln15

3]2log1[5

3 +=+= e .

45. (1)

Sol. M 2 4 are different =8C4= 70

A 4 2 same 2 different = 37C2= 63

T 2 2 same 2 same = 3 = 3

H, E, I, C, S 136

46. (1)

Sol. A is number on the dice

E is event of man says truth

AP(E)P

E EP

A AAP(E)P P(E)P

E E

= +

3 1

34 63 1 1 5 8

4 6 4 6

=

+

.k

ducatio

n.co

m

= 63

3

e dice

man says t

=




24/34

47. (3)

Sol. R and E can be arranged in following position

R E1 place 6 place

2 7

3 8

4 9

5 10

R and E can be interchanged in 2 ways remaining letters are arranged in 8! ways.

Probability =5 2 8! 1

10! 9

=

48. (2)

Sol. Number of rational terms in ( )100

54 5 4+

1001 5 1 6

20

= + = + =

Number of irrational terms = total terms no.of rational term

= 101 6 = 95

49. (2)

Sol.2 4 6

e e9e 90

2! 4! 6!

= +

2 4

4

19 90

2 7202

= +

4 2

2 2

3 4 0

( 4)( 1) 0

+ =

+ =

= 1

50. (1)

Sol. Six + signs are arranged in 6!/6! ways among 7 gaps between 6 + signs 4 signs are

arranged in

.ks

ducatio

com

ed in 8! wn

l terms

e

7

(




25/34

47p 7 6 5 4 354! 24

= =

51. (2)

Sol. x2+ 1 = y x2= y - 1

2

2 4 4 4 3 4

3x 1 3(y 1) 1 3y 2 3 2

(x 1) y y y y

+ + = = =

+

No.of partial fractions = 2

52. (4)

Sol.9

312

3

C 21 34P(A) 1 1

55 55C= = =

62

82

C 15 13P(B) 1 1

28 28C= = =

53. (2)

Sol. (x a)2+ (y 1)

2= r

2

(x b)2+ (y c)

2= r1

2

Radical axis is (x a)2 (x b)

2= r

2 r1

2

but x = 0 is radical axis.

2 2 2 21

2 2 21

a b r r

r r a b

=

= +

54. (3)

Sol. If n = 2

Number of tickets = 5

n(s) =5C3= 10

E = {(1, 2, 3) (2, 3, 4) (3, 4, 5) (1, 3, 5)}

n(E) = 4

Probability =4 2

10 5= .

.saks

hdu

catio

n.co

m

r r

ber of tic

n(s) = C

E =




26/34

55. (1)

Sol. Equation of normal at t is

y + xt = 2at + a + 3It cuts axis at G y = 0

= 2a + at2

G(2a + at2, 0)

SG = a + at2= a(1 + t2)

SP = 2 2 2(at a) (2at) +

( )2 2 2

2 2 2

2

a (t 1) 4t

a ( t 1)

a(t 1)

SG SP

= +

= +

= +

=

56. (4)

Sol. 2x2+ 5xy + 2y

2 11x 7y + k = 0

a h g

h b f 0

g f c

=

5 112

2 2

5 72 0

2 2

1 7k

2 2

=

4 5 11

5 4 7 0

11 7 2k

=

.saks

hied

ucatio

n.co

m

70

2

k2

4

5




27/34

4(8k 49) 5(10k 77) 11( 35 44) 0

32k 196 50k 385 385 484 0

18k 680 770 0

18k 90

k 5

+ =

+ + =

+ =

=

=

57. (2)

Sol.dx

1 sin x cos x+ +

2

dx

2 cos x / 2 2sin x / 2 cos x / 2=

+

2sec x / 2dx log(1 tan x / 2) c2(1 tan x / 2)= = + ++

58. (4)

Sol. I) focus (5, 0), e = 5/8

ae = 5, e = 5/8 a = 8

b2=25

64 1 3964

=

2 2x y1

64 39 + =

II)2 2x y 25 16 3

1,e16 25 25 5

+ = = =

Both are wrong.

59. (3)

Sol.1 1

1 1

20 0

1Tan dx K Tan x dx

1 x x

= + +

.shi

educ

ation.co

m

25 16

25=

rong.

. Tan




28/34

1

0

1 11 1

0 0

1 11 1

0 0

1 x xTan dx

1 (1 x)x

Tan xdx Tan dx

Tan (1 x)dx Tan xdx

+=

= +

= +

1 11 1

0 0

Tan x Tan x = +

11

0

2 Tan xdx=

k = 2.

60. (3)

Sol. y = ex, y = ex

A =2

x x x x 20

0

(e e )dx (e e ) = +

22 2 1 2 1

e e 2 (e e ) ee

= + = =

61. (2)

Sol. Let P(a, 2a) is a point on parabola equation of tangent at P : x y + a = 0

Length of perpendicular from focus to tangent =| a 0 a |

, 2a2

+

a(2a) OS SP= =

62. (2)

Sol.sin x

dxcos(x a)

Put x a = t x = t + a

dx = dt

.s

iedu

catio

n.co

m

e

t on parabn r

endiculara

. (2).




29/34

sin t cosa cos t s in aI dt

cost

+=

(cosa tan t sina)dt= +

cosa log(sect) ts ina= +

cos a log sec(x a) sin a(x a)= +

cos a log cos(x a) x sin a c= + +

63. (3)

Sol.n

1 1 1 1Lt ...

3n 1 3n 2 3n 3 3n 3n

+ + + +

+ + + +

3n 3n

n nr 1 n 1

1 1 1

Lt Lt r3n r n3

n

= ==+ +

[ ]3

3

00

1dx log(3 x)

3 x= +

+

log 6 log 3 log 2= = .

64. (4)

Sol. 3x 5xy ae be=

2 1y 2y 15y 0 =

65. (4) Since, root mean square arithmetic mean

1680400112

=

== nnnn

x

n

x

n

i

i

n

i

i

Hence, possible value of n= 18.

66. (3)

Sol. tan (i x) = i tan hx

a ib a ibtan i log i tanh log

a ib a ib

=

+ +

.shi

educ

ation.co

m

an square

n

n

ii

Hence, pe

. (3)

an




30/34

a ib a ibi

a ib a ib

a ib a ib

a ib a ib

+

+ = +

+

+

2 2

2 2

(a ib) (a ib)i

(a ib) (a ib)

+=

+ +

2 2 2 2

4abi 2abi

2(a b ) a b

= =

67. (2)

Sol. ( ) ( )2 23 2x 3x 2

2 3(2 5) (2 5)

=

2 26 4x 9x 6(2 5) (2 5) =

12 = 13x2 12

x13

=

68. (2)

Sol. x(y x)dy y(x y)dx = +

dy y(x y)

dx x(y x)

+=

Put y = vx,dy dv

v xdx dx

= +

2

2

2 2 2

2

dv x v(1 v)v x

dx x (1 v)

dv v(1 v)x v

dx 1 v

v v v v 2v

1 v 1 v

1 v dx2

xv

++ =

+=

+=

=

kshi

educ

ation.co

m

dvx

dx

v(1 v)

x (1x 1

v(1 v

dx 1

v v




31/34

2

2

x / y

x / y

1 1 dxdv 2

v xv

1

log v 2log x log cv

1log(cx v)

v

xlog(cxy)

y

e cxy

cxy e 1

=

= +

=

=

=

69. (2)

Sol. Product of perpendicular from any point on hyperbola to asymptotes =2 2

2 2

a b

a b+

A)2 2

2 2

a b 144

25a b=

+ B)

2 2

2 2

a b 30

11a b=

+

C)2 2

2 2

a b 54

15a b=

+ D)

2 2

2 2

a b 6

5a b=

+

A, C, B, D

70. (2)

Sol. n 1 nT T 21+ =

(n + 1)C3 nC3= 21

2

(n 1)n(n 1) n(n 1)(n 2)21

6 6

n(n 1)(3) 126

n n 42 0 n 7

+ =

=

= =

71. (1) Required locus is axy 4)3( 2 =

axy 49 2 = .

(x1,y1)

O

Q

P

X

Y (x1,3y1)

aksh

educ

a

n.co

m

asymptote

21

(n

n

n n




32/34

72. (2)

Sol.

y = sin x (1)

y = cos x (2)

required area =/ 4

0

(cosx sin x)dx

/ 4

0

1 1sin x cos x

2 2

= + = +

(1)

2 1=

73. (1)

Sol.2

x

2

1 xe dx

(1 x)

+

+

2x

2

x 1 2e dx

(1 x)

+

+

x

2

x 1 2e dx

x 1 (x 1)

= +

+ +

2

x

x 1 2f (x) , f (x)

x 1 (x 1)

x 1e c

x 1

= =

+ +

= +

+

74. (3)

Sol. SP = 6, a = 2

PSQ is focal chord1 1 1

SP SP a + =

XBAMO

(1)Q

(2)

kshi

educ

ation.co

m

dx

x

x

1,

1= e




33/34

1 1 1

6 SQ 2

1 1 1 1

SQ 2 6 3

+ =

= =

SQ = 3

75. (3)

Sol. S(ae, 0), S(ae, 0), P(a cos, b sin )

Area of triangle PSS=

a cos bsin 11

ae 0 12

ae 0 1

1( bsin )(2ae) abe | sin |

2 =

Maximum mean |sin | = 1

76. (2)

Sol. AB = 2r

a2+ b2= 4r2

centroid 1 1

a b

, (x , y )3 3

= =

a = 3x1, b = 3y1

a2+ b

2= 4r

2 2 2 21 19x 9y 4r + =

Locus is 9(x2+ y

2) = 4r

2

77. (4)

Sol. Radical axis is 2x y + 4 = 0

x2+ y

2+ 2x 3y + 1 + (2x y + 4) = 0

centre3

1,2

+

lines on Radical axis.

3 12 2 4 0

2 5

+ + = =

a shi

educ

ation.co

m

9x

9(x2+ y

. Radical

x +




34/34

circle is 5x2+ 5y2+ 12x 16y + 9 = 0.

78. (2)

Sol. 1 1 3 1 3 5x ...5 5 10 5 10 15

= + + +

2

1/ 2

2 2

1 1 3 11 x 1 1 ...

5 2! 5

2 51 squaring

5 3

51 x 2x 3x 6x 2

4

+ = + + +

= =

+ + = + =

79. (2)

Sol. A(a, 0), B(a, 0)

2 2ay 3(x a )

y 0 x a

=

= =

Required area =a

2 2

a

3(x a )dx

a

a3

2

0

3 33 2

3 x2 a x

a 3

6 a 6 2aa 4a

a 3 a 3

=

= = =

80. (1)

Sol. x2+ 6x 27 > 0

(x + 9)(x 3) > 0

2

x ( , 9) (3, )

x 3x 4 0

(x 4)(x 1) 0

x ( 1,4) x (3,4)

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