Question

Let ${\cos }^{-1}(4x^3-3x)=a+b{\cos }^{-1}x$. If $x\in (\frac{1}{2},1]$, then the value of $\underset{y\to a}{lim}b\cos \left(y\right)$ is

• A. $-\frac{1}{3}$
• B. $-3$
• C. $\frac{1}{3}$
• D. $3$

Answer 1

The correct option is D $3$

Let $x=\cos \theta$

Then, $\frac{1}{2}\le x\le 1$

$\frac{1}{2}\le \cos \theta \le 1$

${\cos }^{-1}\left(\frac{1}{2}\right)\ge \theta \ge {\cos }^{-1}\left(1\right)$

$\frac{\pi }{3}\ge \theta \ge 0$

Hence, $\theta ϵ[0,\frac{\pi }{3}]$

Hence, $\theta$ is in the first quadrant.

Now, ${\cos }^{-1}(4x^3-3x)$$={\cos }^{-1}(4{\cos }^3\theta -3\cos \theta )$

$={\cos }^{-1}\left(\cos 3\theta \right)$

Since $\theta ϵ[0,\frac{\pi }{3}]$,

Hence, $3\theta$ $=3{\cos }^{-1}\left(x\right)$

$=a+b{\cos }^{-1}\left(x\right)$

Hence, $a=0$ and $b=3$

Now, $li{m}_{y\to a}b\cos y$

$=bcos\left(0\right)$

$=b$
$=3$