JEE MAIN - Mathematics (2014 (Offline) - No. 19)
Let $$f_k\left( x \right) = {1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)$$ where $$x \in R$$ and $$k \ge \,1.$$
Then $${f_4}\left( x \right) - {f_6}\left( x \right)\,\,$$ equals :
Then $${f_4}\left( x \right) - {f_6}\left( x \right)\,\,$$ equals :
$${1 \over 4}$$
$${1 \over 12}$$
$${1 \over 6}$$
$${1 \over 3}$$
Explanation
Let $${f_k}\left( x \right) = {1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}.x} \right)$$
Consider
$${f_4}\left( x \right) - {f_6}\left( x \right) $$
$$=$$ $${1 \over 4}\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - {1 \over 6}\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$$
$$ = {1 \over 4}\left[ {1 - 2{{\sin }^2}x{{\cos }^2}x} \right] - {1 \over 6}\left[ {1 - 3{{\sin }^2}x{{\cos }^2}x} \right]$$
$$ = {1 \over 4} - {1 \over 6} = {1 \over {12}}$$
Consider
$${f_4}\left( x \right) - {f_6}\left( x \right) $$
$$=$$ $${1 \over 4}\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - {1 \over 6}\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$$
$$ = {1 \over 4}\left[ {1 - 2{{\sin }^2}x{{\cos }^2}x} \right] - {1 \over 6}\left[ {1 - 3{{\sin }^2}x{{\cos }^2}x} \right]$$
$$ = {1 \over 4} - {1 \over 6} = {1 \over {12}}$$
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