logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

The function $f(x)$ = $(x^2-1)\;$ $|x^2-3x+2|$ + $\cos(|x|)$ is not differentiable at

$(a)\;-1\qquad(b)\;0\qquad(c)\;1\qquad(d)\;2$

Can you answer this question?
 
 

1 Answer

0 votes
We have $|x|=\left\{\begin{array}{1 1}-x&if\;x < 0\\x&if\;x\geq 0\end{array}\right.$
Also $|x^2-3x+2|=|(x-1)(x-2)|$
$\Rightarrow \left\{\begin{array}{1 1}(1-x)(2-x)&if\;x < 1\\(x-1)(2-x)&if\;1\leq x\leq 2\\(x-1)(x-2)&if\;x\geq 2\end{array}\right.$
As $\cos(-\theta)=\cos\theta\Rightarrow \cos |x|=\cos x$
Given function can be written as
$\Rightarrow f(x)= \left\{\begin{array}{1 1}(x^2-1)(x-1)(x-2)+\cos x&if\;x \leq 1 \\-(x^2-1)(x-1)(x-2)+\cos x&if\;1\leq x < 2\\(x^2-1)(x-1)(x-2)+\cos x&if\;x\geq 2\end{array}\right.$
This function is differentiable at all points except possibly at $x=1$ and $x=2$
$Lf'(1)=\{\large\frac{d}{dx}$$\{(x^2-1)(x-1)(x-2)+\cos x\}_{x=1}$
$\Rightarrow -\sin 1$
$Lf'(1)=Rf'(1)$
$f$ is differentiable at $x=1$
$Lf'(2)=\{\large\frac{d}{dx}$$\{-(x^2-1)(x-1)(x-2)+\cos x\}_{x=2}$
$\Rightarrow -3-\sin 2$
$Rf'(2)=\{\large\frac{d}{dx}$$\{(x^2-1)(x-1)(x-2)+\cos x\}_{x=2}$
$\Rightarrow 3-\sin 2$
$Lf'(2)\neq Rf'(2)$
$\therefore f$ is not differentiable at $x=2$
Hence (d) is the correct answer.
answered Jan 3, 2014 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...