Question

Let K be the set of all real values of x where the function f(x) = sin|x| - |x| + 2(x - π) cos|x| is not differentiable. Then the set K is equal to:
(A) Φ (en empty set) (B) {π}
(C) {0} (D) {0, π}

Answer 1

Answer:

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}}

-

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}}

Step-by-step explanation:

Answer 2

Answer:

123456

Step-by-step explanation: