Question

If ${\cos }^{-1}x=\{\begin{array}{c}a\pi -b{\cos }^{-1}(2x^2-1),if-1\le x<0\\ c{\cos }^{-1}(2x^2-1),if0\le x\le 1\end{array}$.
Find the value of $a+b+c$.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

Let ${\cos }^{-1}x=\theta$, where $\theta ϵ[0,\pi ]$ or $\cos \theta =x$

Now,
${\cos }^{-1}(2x^2-1)={\cos }^{-1}(2{\cos }^2\theta -1)={\cos }^{-1}\left(\cos 2\theta \right)={\cos }^{-1}\left(\cos \theta \right)$, where $\alpha ϵ[0,2\pi ]$

Refer to the graph of $y={\cos }^{-1}\left(\cos \alpha \right)$, $\alpha ϵ[0,2\pi ]$

From the graph, we have
${\cos }^{-1}(2x^2-1)=\{\begin{array}{c}\alpha ,if0\le \alpha <\pi \\ 2\pi -\alpha ,if\pi \le \alpha \le 2\pi \end{array}$

$=\{\begin{array}{c}2{\cos }^{-1}x,if0\le 2{\cos }^{-1}x\le \pi \\ 2\pi -{\cos }^{-1}x,if\pi <2{\cos }^{-1}x\le 2\pi \end{array}$

$=\{\begin{array}{c}2{\cos }^{-1}x,if0\le {\cos }^{-1}x\le \left(\pi /2\right)\\ 2\pi -{\cos }^{-1}x,if\left(\pi /2\right)<{\cos }^{-1}x\le \pi \end{array}$

$=\{\begin{array}{c}2{\cos }^{-1}x,if0\le x\le 1\\ 2\pi -{\cos }^{-1}x,if-1\le x<0\end{array}$

$\Rightarrow$ $2{\cos }^{-1}x=\{\begin{array}{c}2\pi -{\cos }^{-1}(2x^2-1),if-1\le x<0\\ {\cos }^{-1}(2x^2-1),if0\le x\le 1\end{array}$

$a=2,b=1,c=1$