Question

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

• A. $[\frac{1}{2},1]$
• B. $[-1,1]$
• C. $[0,3]$
• D. $[0,\frac{1}{2}]$

Answer 1

The correct option is A $[\frac{1}{2},1]$

Let $x=cos\theta ,$ we get $4co{s}^3\theta -3cos\theta =p$
$\Rightarrow p=cos3\theta$
$\Rightarrow \theta =\frac{1}{3}co{s}^{-1}\left(p\right)\&x=cos\theta$
$\Rightarrow x=cos\left(\frac{1}{3}co{s}^{-1}\right(p\left)\right)$ ...(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
$\Rightarrow \frac{1}{2}\le x\le 1$ $\Rightarrow xϵ[\frac{1}{2},1]$