Question

If $pϵ[-1,1]$, then the value of x for which $4x^3-3x-p=0$ has a root lies in

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$Letx=cos\theta ,weget4co{s}^3\theta -3cos\theta =p$
$\Rightarrow p=cos3\theta$
$\Rightarrow \theta =co{s}^{-1}\left(p\right)andx=cos\theta$
$\Rightarrow x=cos\left(\frac{1}{3}co{s}^{-1}\right(p\left)\right)$ ...(i)
$Since,-1\le p\le 1\Rightarrow 0\le co{s}^{-1}p\le \pi$
$or0\le \frac{1}{3}co{s}^{-1}p\le \frac{\pi }{3}$ ,as we know cos x is dereasing. .
$\therefore cos0\ge cos\left(\frac{1}{3}co{s}^{-1}\right(p\left)\right)\ge cos\frac{\pi }{3}$ ....(ii)
From Eqs. (i) and (ii), we get
$\Rightarrow \frac{1}{2}\le x\le 1$ $\Rightarrow xϵ[\frac{1}{2},1]$