Let f(x) = cos x and g(x) = [x + 1], where [.] denotes the greatest integer function. They, (gof)' (π/2), is
Answer & explanation
Correct answer: option 4
We have, f(x) = cos x and g(x) = [x + 2]
∴ gof(x) = g(f(x)) = [cos x + 2]
$\Rightarrow gof(x)= \begin{cases}2, & \text { if } 0<x \leq \pi / 2 \\ 1, & \text { if } \pi / 2 \leq x<\pi\end{cases}$
We have,
$\lim\limits_{x \rightarrow \pi / 2^{-}} gof(x)=2$ and $\lim\limits_{x \rightarrow \pi / 2^{+}} gof(x)=1$
Clearly, gof(x) is not continuous at $x=\frac{\pi}{2}$.
Hence, it is not differentiable at x = π/ 2.
Consequently, (gof)'(π/2) does not exist.