Question

The function $f\left(x\right)=|x^2-3x+2|+cos|x|$ is not differentiable at x is equal to -

• A. - 1
• B. 1
• C. 2

Answer 1

Answer: (B), (C)

The correct options are
B 1
C 2

$\because f\left(x\right)=|x^2-3x+2|+cos|x|=|(x-1)||x-2|+cos|x|f\left(x\right)=\{\begin{array}{cc}{x}^2-3x+2+cosx,& x<0\\ x^2-3x+2+cosx,& 0\le x<1\\ -x^2+3x-2+cosx,& 1\le x<2\\ x^2-3x+2+cosx,& x>2\end{array}\therefore f^{{}^{\prime }}\left(x\right)=\{\begin{array}{cc}2x-3-sinx,& x<0\\ 2x-3-sinx,& 0\le x<1\\ -2x+3-sinx,& 1\le x<2\\ 2x-3-sinx,& x>2\end{array}$
It is clear f(x) is not differentiable at x = 1and x = 2.
$\therefore f^{{}^{\prime }}\left(1^{-}\right)=-1-sin1and{f}^{{}^{\prime }}\left(1^{+}\right)=1-sin1$
$\therefore f^{{}^{\prime }}\left(2^{-}\right)=-1-sin2and{f}^{{}^{\prime }}\left(2^{+}\right)=1-sin2$