Limit Continuity And Differentiability Ques 131
- If $f(x)=-\cos x, \quad-\frac{\pi}{2}<x \leq 0$, then
$$ x-1, \quad 0<x \leq 1 $$
$$ \ln x, \quad x>1 $$
(a) $f(x)$ is continuous at $x=\frac{\pi}{2}$
(2011)
(b) $f(x)$ is not differentiable at $x=0$
(c) $f(x)$ is differentiable at $x=1$
(d) $f(x)$ is differentiable at $x=-\frac{3}{2}$
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Answer:
Correct Answer: 131.True
Solution:
- Given, $h(x)=[f(x)]^{2}+[g(x)]^{2}$
$$ \begin{alignedat} \Rightarrow \quad h^{\prime} x & =2 f(x) \cdot f^{\prime}(x)+2 g(x) \cdot g^{\prime}(x) \\ & =2[f(x) \cdot g(x)-g(x) \cdot f(x)] \\ & =0 \quad\left[\because f^{\prime}(x)=g(x) \text { and } g^{\prime}(x)=-f(x)\right] \end{aligned} $$
$\therefore h(x)$ is constant.
$\Rightarrow h(10)=h(5)=11$
BETA
