Question

If $f\left(x\right)=0$ then $a^2$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{1}{5}$
• C. $\frac{5}{9}$
• D. $\frac{11}{5}$

Answer 1

Answer: (C)

$\frac{5}{9}$

$f\left(x\right)=0\Rightarrow \sin \left\{{\cot }^{-1}(x+1)\right\}=\cos \left({\tan }^{-1}x\right)\Rightarrow \sin {\sin }^{-1}\frac{1}{\sqrt{1+{(x+1)}^2}}=\cos {\cos }^{-1}\frac{1}{\sqrt{1+x^2}}\Rightarrow \frac{1}{\sqrt{1+{(x+1)}^2}}=\frac{1}{\sqrt{1+x^2}}\Rightarrow 1+x^2=2+x^2+2x\Rightarrow x=-\frac{1}{2}a=\cos {\tan }^{-1}\sin {\cot }^{-1}x=\cos {\tan }^{-1}\sin {\sin }^{-1}\left(\frac{1}{\sqrt{1+{\left(x\right)}^2}}\right)=\cos {\tan }^{-1}\sin {\sin }^{-1}\left(\frac{1}{\sqrt{x^2+1}}\right)=\cos {\tan }^{-1}\left(\frac{1}{\sqrt{x^2+1}}\right)=\cos {\cos }^{-1}\left(\frac{1}{\sqrt{1+{\left(\frac{1}{\sqrt{x^2+1}}\right)}^2}}\right)=\cos {\cos }^{-1}\left(\sqrt{\frac{x^2+1}{x^2+2}}\right)$

At $x=-\frac{1}{2}$ we get $a^2=\frac{5}{9}$