JEE-Mains 2014 (28/30)

Geometry Level 2

Let $f_k(x)=\frac{1}{k} (\sin^kx+\cos^kx$ ) where $x \in \mathbb{R}$ and $k \geq 1$ . What is the value of $f_4(x)-f_6(x)$ ?

\frac{1}{12}

\frac{1}{3}

\frac{1}{4}

\frac{1}{6}

1 solution

Aarsh Srivastava
Mar 7, 2018

We all are familiar with the fact that $\sin^{2}\ x + \cos^{2}\ x = 1$ ;

So,

$(\sin^{2}\ x + \cos^{2}\ x)^{2} = \sin^{4}\ x + \cos^{4}\ x + 2\sin^{2}\ \cos^{2}\ x =1$

According to question;

$f_{4} (x) - f_{6} (x) = \frac {\sin^{4}\ x + \cos^{4}\ x} {4} - \frac {\sin^{6}\ x + \cos^{6}\ x} {6}$

Now; $a^{3}+b^{3} = (a+b)(a^{2} - ab + b^{2})$ . Similarly,

$\sin^{6}\ x + \cos^{6}\ x = (1)(\sin^{4}\ x + \cos^{4}\ x - \sin^{2}\ \cos^{2}\ x)$

So,

$f_{4} (x) - f_{6} (x) = \frac {\sin^{4}\ x + \cos^{4}\ x} {4} - (\frac {\sin^{4}\ x + \cos^{4}\ x - \sin^{2}\ \cos^{2}\ x)} {6})$

$= \frac {\sin^{4}\ x + \cos^{4}\ x } {12} + \frac {2\sin^{2}\ \cos^{2}\ x} {12}$

$= \frac {\sin^{4}\ x + \cos^{4}\ x + 2\sin^{2}\ \cos^{2}\ x} {12}$

$=\frac{1} {12}$