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  • If <img alt="\mathrm{x^2 f(x)=\sqrt{1+\cos 2 x}+|f(x)|}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2%20f%28x%29%3D%5Csqrt%7B1&plus;%5Ccos%202%20x%7D&plus;%7Cf%28x

If \mathrm{x^2 f(x)=\sqrt{1+\cos 2 x}+|f(x)|} then the set of points within the domain of f where f(x) is not differentiable is

Option: 1

\mathrm{\{x \in R: x=(2 n+1) \pi / 2, n \in I\}}


Option: 2

\mathrm{\{1,-1\}}


Option: 3

\mathrm{\{x \in R: x=(2 n+1) \pi / 2\} \cup\{1,-1\}}


Option: 4

\mathrm{\{x \in R: x=n \pi\} \cup\{1,-1\}}


Answers (1)

best_answer

\mathrm{ x^2 f(x)=\sqrt{2}|\cos x|+|f(x)| }

RHS is non-negative, so \mathrm{f(x) \geq 0}

\mathrm{ \therefore f(x)=\frac{\sqrt{2}|\cos x|}{x^2-1} }

f is not defined at \pm 1
so f(x) is not differentiable at \mathrm{x=(2 n+1)^{\frac{\pi}{2}}.}

Posted by

Devendra Khairwa

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