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Multiple Choice Questions
1. The value of $$3(sinx- cosx)^{4}+6(sinx+cosx)^{2}+4(sin^{6}x + cos^{6}x)$$ is
(A): 14
(B): 11
(C): 12
(D): 13
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By: anil on 05 May 2019 01.57 am
$$(sin^{6}x + cos^{6}x) = ((sin^2)^3x + (cos^2)^3x)$$
=$$ ((sin^2x)+ (cos^2x))((sin^4x)+ (cos^4x)+ sin^2x cos^2x$$
=$$(sin^4)+ (cos^4) + 2sin^2xcos^2 - 3sin^2xcos^2x$$
=$$(sin^2x + cos^2x)^2 - 3sin^2xcos^2x$$
=$$1 - 3sin^2xcos^2x$$
$$(sinx+cosx)^{2} $$= $$1 + sinxcosx$$
$$(sinx -cos x)^4 = (1-2sinxcosx)^2$$
$$= 1 + 4sin^2xcos^2x - 4sinxcosx$$
$$3(sinx- cosx)^{4}+6(sinx+cosx)^{2}+4(sin^{6}x + cos^{6}x)=3( 1+ 4sin^2xcos^2x - 4sinxcosx) + 6(1 +2sinx cosx) +4(1 -3sin^2xcos^2x)$$
=$$3 + 12sin^2xcos^2x - 12sinxcosx + 6 + 12sinxcosx + 4 -12 sin^2xcos^2 x$$
=$$13$$
Option D is the correct answer.
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SSC CGL 2013 Tier 1 19 May Morning
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=$$ ((sin^2x)+ (cos^2x))((sin^4x)+ (cos^4x)+ sin^2x cos^2x$$
=$$(sin^4)+ (cos^4) + 2sin^2xcos^2 - 3sin^2xcos^2x$$
=$$(sin^2x + cos^2x)^2 - 3sin^2xcos^2x$$
=$$1 - 3sin^2xcos^2x$$
$$(sinx+cosx)^{2} $$= $$1 + sinxcosx$$
$$(sinx -cos x)^4 = (1-2sinxcosx)^2$$
$$= 1 + 4sin^2xcos^2x - 4sinxcosx$$
$$3(sinx- cosx)^{4}+6(sinx+cosx)^{2}+4(sin^{6}x + cos^{6}x)=3( 1+ 4sin^2xcos^2x - 4sinxcosx) + 6(1 +2sinx cosx) +4(1 -3sin^2xcos^2x)$$
=$$3 + 12sin^2xcos^2x - 12sinxcosx + 6 + 12sinxcosx + 4 -12 sin^2xcos^2 x$$
=$$13$$
Option D is the correct answer.