Question

The minimum value of $f\left( x \right) = {\sin ^4}x + {\cos ^4}x,0 \leqslant x \leqslant \dfrac{\pi }{2}$ is
1. $\dfrac{1}{{2\sqrt 2 }}$
2. $\dfrac{1}{4}$
3. $\dfrac{{ - 1}}{2}$
4. $\dfrac{1}{2}$

Answer 1

Hint:In this problem, we have to find the minimum value of $f\left( x \right) = {\sin ^4}x + {\cos ^4}x$ where $0 \leqslant x \leqslant \dfrac{\pi }{2}$. First, start using trigonometric identities to make the given function easier. Now using the range of $\sin x$calculate the range of ${\sin ^2}2x$ and use that condition to know the minimum value.

Formula Used:
Trigonometric identities –
${\sin ^2}\theta + {\cos ^2}\theta = 1$
$\sin 2A = 2\sin A\cos A$

Complete step by step Solution:
Given that,
$f\left( x \right) = {\sin ^4}x + {\cos ^4}x$
$ = {\left( {{{\sin }^2}x} \right)^2} + {\left( {{{\cos }^2}x} \right)^2}$
$ = \left( {{{\sin }^2}x + {{\cos }^2}x} \right) - 2{\sin ^2}x{\cos ^2}x\left( {\dfrac{2}{2}} \right)$
$ = 1 - \dfrac{{{{\sin }^2}2x}}{2}$
As we know that, the range of $\sin x$ is $\left[ { - 1,1} \right]$
Therefore, the range of ${\sin ^2}2x$ is $\left[ {0,1} \right]$
It implies that, $0 \leqslant {\sin ^2}2x \leqslant 1$
$0 \geqslant - {\sin ^2}2x \geqslant - 1$
$0 \geqslant - \dfrac{{{{\sin }^2}2x}}{2} \geqslant - \dfrac{1}{2}$
$1 + 0 \geqslant 1 + \left( { - \dfrac{{{{\sin }^2}2x}}{2}} \right) \geqslant 1\left( { - \dfrac{1}{2}} \right)$
$1 \geqslant 1 - \dfrac{{{{\sin }^2}2x}}{2} \geqslant \dfrac{1}{2}$
$1 \geqslant f\left( x \right) \geqslant \dfrac{1}{2}$
Hence, the minimum value of $f\left( x \right)$ is $\dfrac{1}{2}$.

Hence, the correct option is 4.

Note: In such questions, always try to apply to find the solution using the range of any trigonometric function and try to make the function the same as the given function by applying any operation with the term all sides. Also, while multiplying and dividing any negative term with the inequality don’t forget to change the sign of inequality.