If p = ^ 2 x + ^ 4 x , then

p = \sin^{2} x + \cos^{4} x

, then

Options

A $\dfrac{3}{4} \le p \le 1$
B $\dfrac{3}{16} \le p \le \dfrac{1}{4}$
C $\dfrac{1}{4} \le p \le \dfrac{1}{2}$
DNone of these

Correct answer

A. $\dfrac{3}{4} \le p \le 1$

Step-by-step solution

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

$⇒ p = ( \sin^{2} x + \cos^{2} x ) - \sin^{2} x\cos^{2} x$

$\begin{array}{ll}⇒ p = 1 - \dfrac{1}{4}{( 2\sin x \cos x )}^{2} \\ ⇒ p = 1 - \dfrac{1}{4}\sin^{2}( 2x )\end{array}$

$\begin{array}{ll}Max . value = 1 \\ Min . value = \dfrac{3}{4}\end{array}$

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