Continuity and Differentiability — Class 12 Maths Solution

exemplar objective MCQ NCERT Exemp. Ex.5.3 ,Q.90,Page 114
Question

If $f(x) = |\sin x|,$ then

  • (a) $f$ is everywhere differentiable
  • (b) $f$ is everywhere continuous but not differentiable at $x = n\pi ,$ $n \in Z$ ✓ Correct
  • (c) $f$ is everywhere continuous but not differentiable at $x = (2n + 1)\frac{\pi }{2},$ $n \in Z$
  • (d) None of the above
Step-by-step Solution
Correct answer: option (b)

We have, $f(x) = |\sin x|$

Let $f(x) = {\mathop{\rm vou}\nolimits} (x) = v[u(x)]$ [where, $u(x) = \sin x$ and $v(x) = |x|$]
$= v(\sin x) = |\sin x|$

where, $u(x)$ and $v(x)$ are both continuous.

Hence, $f(x) =$ $vo\,\,u(x)$ is also a continuous function but $v(x)$ is not differentiable at $x = 0$.

So, $f(x)$ is not differentiable where $\sin x = 0 \Rightarrow x = n\pi ,n \in Z$

Hence, $f(x)$ is continuous everywhere but not differentiable at $x = n\pi ,n \in Z$.

NCERT & Exemplar solution for CBSE Class 12 Mathematics, Continuity and Differentiability. Curated by Sachin Sharma. Free for all students.