10 Question MCQ (including Assertion)

If \(\mathrm{f}(x)=|x|+|x-1|\), then which of the following is correct?

Assertion (A): A finction must be continuous at a point if it is differentiable at that point.
Reason (R): If the limit of a function exists at a point, it must be differentiable at that point.

Derivative of \(\cos ^{-1}\left(\frac{\sin \mathrm{x}+\cos \mathrm{x}}{\sqrt{2}}\right),-\frac{\pi}{4}<\mathrm{x}<\frac{\pi}{4}\) with respect to x is :

If \(f(x)=\left\{\begin{array}{cc}\frac{x^2-4 x-5}{x+1}, & x \neq-1 \\ k, & x=-1\end{array}\right.\) is continuous at \(\mathrm{x}=-1\), then the value of k is :

Why is the function \(f(x)=\tan (x)\) discontinuous at \(x=\frac{\pi}{2}\) ?

The value of \(k\) for which the function \(f(x)= \begin{cases}x^2 \sin \frac{1}{x}, & x \neq 0 \\ k(x+1), & x=0\end{cases}\) is a continuous function, is :

If \(2 \cos ^{-1} x=y\), then \(\frac{d y}{d x}\) is :

Assertion (A): The function \(f(x)=\cos |x|\) is differentiable at \(x=0\).
Reason (R): The cosine function is an even function, so \(\cos (-x)=\cos (x)\).

For the function \(\mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{1}{x}, \mathrm{x} \in[1,3]\), the value of c for mean value theorem is