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# Let $f\left( x \right) = {\sin ^4}x + {\cos ^4}x.$ Then f is an increasing function in the interval:$a.{\text{ }}\left[ {\dfrac{{5\pi }}{8},{\text{ }}\dfrac{{3\pi }}{4}} \right] \\ b.{\text{ }}\left[ {\dfrac{\pi }{2},{\text{ }}\dfrac{{5\pi }}{8}} \right] \\ c.{\text{ }}\left[ {\dfrac{\pi }{4},{\text{ }}\dfrac{\pi }{2}} \right] \\ d.{\text{ }}\left[ {0,{\text{ }}\dfrac{\pi }{4}} \right] \\$ Verified
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Hint: Check the graph of first derivative of the given function

Given equation is $f\left( x \right) = {\sin ^4}x + {\cos ^4}x.................\left( 1 \right)$
We know the function is increasing if its differentiation is greater than or equal to zero.
I.e.$f'\left( x \right) \geqslant 0$ so, differentiate equation 1 w.r.t.$x$
$\Rightarrow f'\left( x \right) = 4{\sin ^3}x\dfrac{d}{{dx}}\sin x + 4{\cos ^3}x\dfrac{d}{{dx}}\cos x \\ \Rightarrow f'\left( x \right) = 4{\sin ^3}x\left( {\cos x} \right) + 4{\cos ^3}x\left( { - \sin x} \right) \\ \Rightarrow f'\left( x \right) = 4\sin x\cos x\left( {{{\sin }^2}x - {{\cos }^2}x} \right) \\$
As we know$2\sin x\cos x = \sin 2x$, and${\cos ^2}x - {\sin ^2}x = \cos 2x$, so apply this
$\Rightarrow f'\left( x \right) = - 2\sin 2x\cos 2x = - \sin 4x$
But for increasing function $f'\left( x \right) \geqslant 0$
$\Rightarrow - \sin 4x \geqslant 0 \\ \Rightarrow \sin 4x \leqslant 0 \\$
As we know $\sin x$is zero at $\left( {0,{\text{ }}\pi ,{\text{ }}2\pi } \right),$in the interval between $\left[ {0,2\pi } \right]$

So, in $\sin x$graph $\sin x$is less than or equal to zero in between $\left[ {\pi ,2\pi } \right]$
$\Rightarrow 4x \in \left[ {\pi ,2\pi } \right] \\ \Rightarrow x \in \left[ {\dfrac{\pi }{4},\dfrac{\pi }{2}} \right] \\$
Hence, option $c$ is correct.

Note: - In such a type of question the key concept we have to remember is that for increasing function the differentiation of function w.r.t. the variable is always greater than or equal to zero, then simplify this we will get the required answer and the required answer is the shaded region in the figure.