KCET · Maths · Continuity and Differentiability
If \(f(x)=\left\{\begin{array}{l}2 a-x \text { when }-a < x < a \\ 3 x-2 a \text { when } a \leq x\end{array}\right.\).
Then, which of the following is true?
- A \(f(x)\) is not differentiable at \(x=a\)
- B \(f(x)\) is discontinuous at \(x=a\)
- C \(f(x)\) is continuous for all \(x < a\)
- D \(f(x)\) is differentiable for all \(x \geq a\)
Answer & Solution
Correct Answer
(A) \(f(x)\) is not differentiable at \(x=a\)
Step-by-step Solution
Detailed explanation
Given, \(f(x)=\left\{\begin{array}{l}2 a-x, \quad \text { when }-a < x < a \\ 3 x-2 a, \text { when } a \leq x\end{array}\right.\)
Now, \(f^{\prime}(x)=\left\{\begin{array}{cl}-1, & \text { when }-a < x < a \\ 3, & \text { when } a \leq x\end{array}\right.\) \(\because \quad L f^{\prime}(a)=-1\) and \(R f^{\prime}(a)=3\) i.e., \(\quad L f^{\prime}(a) \neq R f^{\prime}(a)\)
Hence, \(f(x)\) is not differentiable at \(x=a\).
Now, \(f^{\prime}(x)=\left\{\begin{array}{cl}-1, & \text { when }-a < x < a \\ 3, & \text { when } a \leq x\end{array}\right.\) \(\because \quad L f^{\prime}(a)=-1\) and \(R f^{\prime}(a)=3\) i.e., \(\quad L f^{\prime}(a) \neq R f^{\prime}(a)\)
Hence, \(f(x)\) is not differentiable at \(x=a\).
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