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TS EAMCET · Maths · Continuity and Differentiability

If \(f(x)= \begin{cases}\frac{8}{x^3}-6 x, & \text { if } 0 \lt x \leq 1 \ \frac{x-1}{\sqrt{x}-1}, & \text { if } x\gt1\end{cases}\) is a real valued function, then at \(x=1, f\) is

  1. A continuous and differentiable
  2. B continuous but not differentiable
  3. C neither continuous nor differentiable
  4. D differentiable but not continuous
Verified Solution

Answer & Solution

Correct Answer

(B) continuous but not differentiable

Step-by-step Solution

Detailed explanation

\(\lim _{x \rightarrow 1^{-}} f(x)=8-6=2 ; \lim _{x \rightarrow 1^{+}} f(x)=\sqrt{x}+1=1+1=2\) Hence, \(f(x)\) is continuous. \(\mathrm{LHD}=\frac{-24}{\mathrm{x}^4}-6 ;\) LHD at \(x=1\) is -30 . For…