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Let f(x)=sin^6x+cos^6x+k(sin^4x+cos^4x) ...

Let `f(x)=sin^6x+cos^6x+k(sin^4x+cos^4x)` for some real number `kdot` Value of `k` for which `f(x)` is constant for all values of `x` is `-1/2` (b) `1/2` (c) `1/4` (d) `-3/2` All real numbers `k` for which the equation `f(x)=0` has solution lie in `[-1,0]` (b) `[0,1/2]` (c) `[-1,-1/2]` (d) none of these Number of values of `k` for which `f(x)=0` is an identity is (a) 0 (b) 1 (c) infinite (d) none of these

A

`-1//2`

B

`1//2`

C

`1//4`

D

`-3//2`

Text Solution

Verified by Experts

The correct Answer is:
D
f(x)`=sin^(6)x+cos^(6)x+k (sin^4x+cos(4)x)`
f(x)` =sin^6x+cos^6x+k(sin^4x+cos^4 x)`
`=(sin ^2x)^3+cos^2 x)^3+k (sin^2 x)+(cos^2 )^2]`
`=(sin ^2x)^3+(cos ^2 x)^3-3 sin^2x.cos^2x(sin^2 x+ cos^2)`
`+ k[ sin^x +cos^2 x)^2-2sin^2 x. cos^2x]`
`=(1-3 sin^2 x cos^2x)+ k[1-2 sin^2 x cos^2 x]`
f(x) is constant if k = -3/2.
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