The number of points at which the function $$f\left( x \right) = \left| {x - 0.5} \right| + \left| {x - 1} \right| + \tan \,x$$ does not have a derivative in the interval $$\left( {0,\,2} \right)$$ is :
A.
0
B.
1
C.
2
D.
3
Answer :
3
Solution :
$$\left| {x - a} \right|$$ is not differentiable at $$x = a.$$
Also $$\tan \,x$$ is not differentiable if $$x = \left( {2k + 1} \right)\frac{\pi }{2},\,k\, \in \,I$$
$$\therefore $$ In the interval $$\left( {0,\,2} \right),\,f\left( x \right)$$ is not derivable at $$x = 0.5,\,x = 1$$ and $$x = \frac{\pi }{2}.$$
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-