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

Options

A. [-1,0]
B. `[0,1/2]`
C. `[-1,-1/2]`
D. None of these

Answer

C

Explanation

f(x)=0
`rArr (1-3 ) sin^2 x cos^2 x)+k[1-2 sin^2 x cos^2 x]=0`
`rArr K+1 =(sin ^2 x cos^2 x)/(1-2 sin^2x cos^2x)`
`rArrk=(3sin^2xcos^2x-1)/(1-2sin^2xcos^2x)`
`=-3/2(1-2sin^2xcos^2x-1/3)/(1-2sin^2xcos^2x)`
`=-3/2(1-(1/3)/(1-2sin^2xcos^2x))`
minimum of `sin^2xcos^2x " is 0 at " x=0,pi//2`
Maximum of `sin^2xcos^2x" is "1//4" at " x=pi//2`
Hence, k in[1,-1/2]