Q. If $f:R \rightarrow R$ is defined by $f(x) = \sin x$ and $g:(1,\ \infty) \rightarrow R$ is defined by $g(x) = \sqrt{x^{2} - 1}$, then $gof(x)$ is
A) $\sqrt{\sin{(x^{2} - 1)}}$
B) $\sin\sqrt{x^{2} - 1}$
C) $\cos x$
D) Not defined
Explanation
(d)
Given, $f(x) = \sin x$
And $g(x) = \sqrt{x^{2} - 1}$
$\therefore$ Range of $f = \lbrack - 1,\ 1\rbrack \notin$ domain of $g = (1,\ \infty)$
$\therefore gof$ is not defined.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.