If $$f\left( x \right)$$ is differentiable everywhere, then which one of the following is correct ?
A.
$$\left| f \right|$$ is differentiable everywhere
B.
$${\left| f \right|^2}$$ is differentiable everywhere
C.
$$f\left| f \right|$$ is not differentiable at some point
D.
none of the above
Answer :
$$f\left| f \right|$$ is not differentiable at some point
Solution :
If $$f\left( x \right)$$ is differential everywhere then $$\left| f \right|$$ is not differentiable at some point, so $$f\left| f \right|$$ is not differentiable at some point.
[ Example : $$f\left( x \right) = x$$ is differentiable everywhere but $$\left| {f\left( x \right)} \right| = \left| x \right|$$ is not differentiable at $$x = 0$$ ]
Releted MCQ Question on Calculus >> Differentiability and Differentiation
Releted Question 1
There exist a function $$f\left( x \right),$$ satisfying $$f\left( 0 \right) = 1,\,f'\left( 0 \right) = - 1,\,f\left( x \right) > 0$$ for all $$x,$$ and-
A.
$$f''\left( x \right) > 0$$ for all $$x$$
B.
$$ - 1 < f''\left( x \right) < 0$$ for all $$x$$
C.
$$ - 2 \leqslant f''\left( x \right) \leqslant - 1$$ for all $$x$$
If $$f\left( a \right) = 2,\,f'\left( a \right) = 1,\,g\left( a \right) = - 1,\,g'\left( a \right) = 2,$$ then the value of $$\mathop {\lim }\limits_{x \to a} \frac{{g\left( x \right)f\left( a \right) - g\left( a \right)f\left( x \right)}}{{x - a}}$$ is-
Let $$f:R \to R$$ be a differentiable function and $$f\left( 1 \right) = 4.$$ Then the value of $$\mathop {\lim }\limits_{x \to 1} \int\limits_4^{f\left( x \right)} {\frac{{2t}}{{x - 1}}} dt$$ is-