fx=1−1−x2 , if −1≤x 1 is

f1=1−1−1=1f1−=Ltx→1+1−1−x2=1f1+=Ltx→1−1+log1x=1f11 +=Lth→0f1+h−f1h=−1f11 −=Lth→0f1+h−f1−h=∞ Hence f(x) is non - differentiable at x=1

fx=1−1−x2 , if −1≤x 1 is

f1=1−1−1=1f1−=Ltx→1+1−1−x2=1f1+=Ltx→1−1+log1x=1f11 +=Lth→0f1+h−f1h=−1f11 −=Lth→0f1+h−f1−h=∞ Hence f(x) is non - differentiable at x=1

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$f (x) = \{\begin{cases} 1 - \sqrt{1 - x^{2}}, if - 1 \leq x < 1 \\ 1 + \log \frac{1}{x}, if x > 1 \end{cases} is$

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Continuous and differentiable at $x = 1$

Neither continuous not differentiable at $x = 1$

None of these

Continuous but not differentiable at $x = 1$

answer is B.

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Detailed Solution

$f (1) = 1 - \sqrt{1 - 1} = 1$

$\begin{array}{l} f (1^{-}) = \underset{x \to 1 +}{L t} (1 - \sqrt{1 - x^{2}}) = 1 \\ f (1^{+}) = \underset{x \to 1 -}{L t} (1 + \log \frac{1}{x}) = 1 \end{array}$

$\begin{array}{l} f^{1} (1 {}^{+}) = \underset{h \to 0}{L t} \frac{f (1 + h) - f (1)}{h} = - 1 \\ f^{1} (1 {}^{-}) = \underset{h \to 0}{L t} \frac{f (1 + h) - f (1)}{- h} = \infty \end{array}$