Q36 / 90 · KVPY 2013 (SB/SX) · Mathematics
Let $f(x)=\sqrt{2-x-x^{2}}$ and $g(x)=\cos x$. Which of the following statements are true?
(I) Domain of $f\left((g(x))^{2}\right)=$ Domain of $f(g(x))$
(II) Domain of $f(g(x))+g(f(x))=$ Domain of $g(f(x))$
(III) Domain of $f(g(x))=$ Domain of $f(g(x))$ (V) Domain of $g\left((f(x))^{3}\right)=$ Domain of $f(g(x))$
(IV) Domain of $g\left((f(\mathrm{x}))^{3}\right)=$ Domain of $f(\mathrm{~g}(\mathrm{x}))$
- AOnly (I)
- Correct answerBOnly (I) and (II)
- COnly (III) and (IV)
- DOnly (I) and (IV)
Solution
Domain of $f(g(x))$ is $R$
$$
\begin{array}{l}
\because 2-\cos x-\cos ^{2} x \geq 0 \\
(\cos x+2)(\cos x-1) \leq 0 \Rightarrow-2 \leq \cos x \leq 1 \\
x \in R
\end{array}
$$
Domain of $g(f(x))$ is $[-2,1]$
$$
\begin{array}{c}
\because \cos \left(\sqrt{2-x-x^{2}}\right) \\
2-x-x^{2} \geq 0
\end{array}
$$
Domain of $f\left(g(x)^{2}\right)$ is $R$
$$
\begin{array}{l}
\because 2-\cos ^{2} x-\cos ^{4} x \geq 0 \\
\left(\cos ^{2} x+2\right)\left(\cos ^{2} x-1\right) \leq 0 \\
-1 \leq \cos x \leq 1 \\
\quad x \in R
\end{array}
$$
Domain of $\mathrm{g}\left(\mathrm{f}^{\beta}(\mathrm{x})\right.$ is Domain of $\mathrm{g}(\mathrm{f}(\mathrm{x})$
i.e., $[-2,1]$
Hence, the correct answer is option: 'B'.