Question

Let $-1\le p\le 1.$ Then the equation ${4x}^3-3x-p=0$ has a unique root in the interval $[\frac{1}{2},1]$ if

• A. $x=\cos \left\{\frac{1}{3}{\cos }^{-1}2p\right\}$
• B. $x=\cos \left\{\frac{1}{5}{\cos }^{-1}p\right\}$
• C. $x=\cos \left\{\frac{1}{4}{\cos }^{-1}p\right\}$
• D. $x=cos\left\{\frac{1}{3}{\cos }^{-1}p\right\}$

Answer 1

The correct option is D $x=cos\left\{\frac{1}{3}{\cos }^{-1}p\right\}$

Given equation is ${4x}^3-3x-p=0$

Put $x=\cos \theta$

$\Rightarrow \cos 3\theta =p$

$\Rightarrow \theta =\frac{1}{3}{\cos }^{-1}p$

So, $x=\cos \left(\frac{1}{3}{\cos }^{-1}p\right)$

Also, cosine function is one-one and range of $\cos \theta$ is $[-1,1]$

Hence, the given equation has a unique solution given by $x=\cos \left(\frac{1}{3}{\cos }^{-1}p\right)$

Since, $\frac{1}{2}\le x\le 1$
$\Rightarrow \frac{1}{2}\le \cos \theta \le 1$

$\Rightarrow 0\le \theta \le \frac{\pi }{3}$

$\Rightarrow 0\le 3\theta \le \pi$