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CBSE Sample Paper Class 11 MathsSet 2 SolutionANSWERSSection A1. Solution:3 2i 3 2i 1 i .1 i1 i 1 i3 2i 3i 2i 2 1 5 i1 i22 23 2i1 55 3(1 2i )i ( 2 i) ( i ) i1 i2 22 22. Solution:Domain [ 1,1] Range [0, π ]3. Solution :0 cos 1 x πsin in this interval is positive and hence y is positive4. Solution: 6π π 1 sin 1 sin sin sin π 7 7 π sin 1 sin 7 ππ2 π7 π275. Solution: ( a, 2a ) , ( a, 2a )6. Solution:x 7 10x 3x y 8y 5Section B7. Solution:

1 cos 2 x 1 cos 4 x 122cos2x cos4x 02 cos3x cos x 0Cos3x 0x π π6 3Cosx 0x π2n πk π6 π3n n is integer8. Solution:i 30 i 40 i 60 (i 4 )7 .i 2 (i 4 )10 (i 4 )15i 4 1 1 1 1 19. Solution:Substituting the points (0, 0) and (5, 5) on the given linex y–8 00 0 – 8 -85 5–8 2Since the signs of the resulting numbers are different the given points lie on opposite sides ofthe given line.10. Solution :tan 1 x Atan A xcot 1 x Bcot B xπtan( B) x2tan 1 x π2 1tan x AA π2 B BA B π2tan 1 x cot 1 x π2

11. Solution:11n 2 122 n 1 is divisible by 133n 1113 123 (11 12)(112 11.12 122 ) 23.133Let it be true for k11k 2 122 k 1 is divisible by 133For k k 111k 3 122 k 3 11.11k 2 122.122 k 1 11.11k 2 144.122 k 1 11.11k 2 133.12 2 k 1 11.122 k 1 11.11k 2 11.12 2 k 1 133.122 k 1Is divisible by 133 since 11k 2 122 k 1 is divisible by 13312. Solution:n( A′ B′) n( A B)′ n(U ) n( A B ) n(U ) [n( A) n( B) n( A B)]800 [200 300 100] 40013. Solution: α β bαβ cα 2 β 2 (α β ) 2 2αβ b 2 2c14. Solution:( x a )n P Q( x a)n P Q( P Q )( P Q ) ( x a)( x a)P 2 Q 2 ( x 2 a 2 )n15. Solution:Discriminant of numerator 9 – 24 andCoefficient of x 2 is positive. Hence Numerator is always positiveHence dividing by the numerator on both sides of

The equality does not change the sign of the inequalityHence we need only considerx 1 03x 4 434x ( , )316. Solution:cos( A 15) sin( A 15) sin( A 15) cos( A 15)cos( A 15) cos( A 15) sin( A 15) sin( A 15) sin( A 15) cos( A 15)cos 2 A 11(sin 2 A )222 cos 2 A 1sin 2 A 24 cos 2 A 1 2sin 2 Acot( A 15) tan( A 15) 17. Solution:4 x2 0x2 4 0Domain of x [ 2, 2]y2 4 x2x2 4 y 2x 4 y24 y2 0y2 4 0y [ 2, 2]Also for all values of x [ 2, 2]y 4 x2 0Range y [0, 2]18. Solution:

cos θ 1 tan 21 tan2θ2θ21 4 3cos θ 1 4 52 tan θ 22.2 4sin θ 4 512 θ1 tan21111 2 cos θ sin θ 2 3 4 55 519. Solution:sin 5 x5sin 5 x lim3x 0 x xx 0 5 x (1 x 2 )5sin 5 x1 limlimx 05 x x 0 (1 x 2 ) 5.1.1lim 5Section C20. Solution :y log10 xx 10 ylog e x y log e 10y log e xlog e 10dy 1 1 dx log e 10 x21. Solution:There are 3 even numbers 2, 4, 6So the units place, 10th places can be filled in 3 p2 waysRemaining 5 digits can be used to fill 4 places in 5 p4 ways.Hence the total numbers satisfying the above condition is 3 p2 5 p4 72022. Solution:Let the origin be shifted to (h, k )

x x′ hy y′ kThen( x′ h) 2 ( y′ k )2 4( x′ h) 6( y′ k ) 36x′2 2hx′ h 2 y′2 2ky′ k 2 4( x′ h) 6( y′ k ) 36x′2 y′2 x′(2h 4) y′(2k 6) h 2 k 2 4h 6k 36 02h – 4 0h 22k 6 0k -3x′2 y′2 22 ( 3) 2 8 18 36 0x′2 y′2 13 62 0x′2 y′2 4923. Solution:2 4 12 14 11 x y 8743 x y 56x y 1322 42 122 142 112 x 2 y 2 ( mean) 2 1974 16 144 196 121 x 2 y 2 64 197481 x 2 y 2 837481 x 2 y 2 581x 2 y 2 100( x y ) 2 ( x y )2 2( x 2 y 2 )169 ( x y ) 2 200( x y )2 31x y 5.57x y 13x 9.285y 3.71524. Solution:

1 log b alog a b1 log b 2alog 2 a b1 log b 4alog 4 a blog b a log b 4a log b (2a ) 2 22log 2a 2 b2 log b 2aThus,111is, the, AM, between,log a b log 4 a blog 2 a b25. Solution:Probability of surviving 910Required to find out the probability of 4 are safe or 5 are safe 9 Probability of 5 is safe 10 54 9 1Probability of 4 is safe C4 10 10554 9 9 1 45927Required Probability 5 10 10 10 500026. Solution:T2 r 1 40 C2 rTr 2 40 Cr 140C2 r 40 Cr 12r r 1 403r 39r 13

CBSE Sample Paper Class 11 Maths Set 2 Solution. 1 cos2 1 cos4 1 2 2 x x cos2x cos4x 0 2 cos3x cos x 0 Cos3x 0 6 3 0 2 6 3 x n Cosx x k n n is integer π π π π π π 8. Solution: 30 40 60 4 7 2 4 10 4 15 4 ( ) . ( ) ( ) 1 1 1 1 1 i i i i i i i i 9. Solution: Substituting the points (0, 0) and (5, 5) on the given line x y – 8 0 0 .

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