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Let K be the set of all real values of x where the function

$f(x)=\sin \left | x \right |-\left | x \right |+2(x-\pi )\cos \left | x \right |$ is not differentiable. Then the set K is equal to :
Option 1)
{0}
Option 2)
$\phi$ (an empty set)
Option 3)
{ ${\pi}$ }
Option 4)
{0, ${\pi}$ }

Answers (1)

best_answer

Condition for differentiability -

A function f(x) is said to be differentiable at $x=x_{\circ }$ if $Rf'(x_{\circ })\:\:and\:\:Lf'(x_{\circ })$ both exist and are equal otherwise non differentiable

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Differentiability -

Let f(x) be a real valued function defined on an open interval (a, b) and $x\epsilon$ (a, b).Then the function f(x) is said to be differentiable at $x_{\circ }$ if

$\lim_{h\rightarrow 0}\:\frac{f(x_{0}+h)-f(x_{0})}{(x_{0}+h)-x_{0}}$

$or\:\:\:\lim_{x\rightarrow x_{0}}\:\frac{f(x)-f(x_{0})}{x-x_{0}}$

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Checking differentiability at x=0

for x>0,

$f(x)=\sin x-x+2(x-\pi)\cos x$

$f{}'(x)=\cos x-1+2\cos x-2(x-\pi)\sin x$

RHD = $f{}'(0+)=1-1+2-2(-\pi)\cdot 0=2$

for x<0,

$f(x)=-\sin x+x+2(x-\pi)\cos x$

$f{}'(x)=-\cos x+1+2\cos x-2(x-\pi)\sin x$

LHD = $f{}'(0-)=-1+1+2-2(-\pi)\cdot 0=2$

$\because$ LHD = RHD

differentiable at x = 0 => differentiable everywhere

Option 1)

{0}

Option 2)
$\phi$ (an empty set)
Option 3)

{ ${\pi}$ }

Option 4)

{0, ${\pi}$ }

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