AP EAMCET · Maths · Continuity and Differentiability
\(f(x)=\left\{\begin{array}{lc}4 & ,-\infty < x < -\sqrt{5} \\ x^2-1, & -\sqrt{5} < x \leq \sqrt{5} \\ 4, & \sqrt{5} \leq x < \infty\end{array}\right.\)
If \(k\) is the number of points where \(f(x)\) is not differentiable, then \(k-2=\)
- A \(2\)
- B \(1\)
- C \(0\)
- D \(3\)
Answer & Solution
Correct Answer
(C) \(0\)
Step-by-step Solution
Detailed explanation
Points to check, \(x=-\sqrt{5}, x=\sqrt{5}\) \[ f(x)=\left\{\begin{array}{l} 4,-\infty < x < -\sqrt{5} \\ x^2-1,-\sqrt{5} \leq x \leq \sqrt{5} \\ 4, \sqrt{5} \leq x < \infty \end{array}\right. \] At \(x=-\sqrt{5}\), \(\mathrm{LHL}=\mathrm{RHL}=f(-\sqrt{5})=4\)…
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