Let $$fk\left( x \right) = {1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)$$ where $$x \in R$$ and $$k \ge \,.$$
Then $${f_4}\left( x \right) - {f_6}\left( x \right)\,\,$$ equals
A
$${1 \over 4}$$
B
$${1 \over 12}$$
C
$${1 \over 6}$$
D
$${1 \over 3}$$
Explanation
Let $${f_k}\left( x \right) = {1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}.x} \right)$$
Consider
$${f_4}\left( x \right) - {f_6}\left( x \right) $$
$$ABCD$$ is a trapezium such that $$AB$$ and $$CD$$ are parallel and $$BC \bot CD.$$ If $$\angle ADB = \theta ,\,BC = p$$ and $$CD = q,$$ then AB is equal to:
In a $$\Delta PQR,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} $$ If $$3{\mkern 1mu} \sin {\mkern 1mu} P + 4{\mkern 1mu} \cos {\mkern 1mu} Q = 6$$ and $$4\sin Q + 3\cos P = 1,$$ then the angle R is equal to :
A
$${{5\pi } \over 6}$$
B
$${{\pi } \over 6}$$
C
$${{\pi } \over 4}$$
D
$${{3\pi } \over 4}$$
Explanation
Given $$3$$ $$\sin \,P + 4\cos Q = 6$$ $$\,\,\,\,\,\,\,\,...\left( i \right)$$
$$4\sin Q + 3\cos P = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$