Question

If $f\left(x\right)=x|x^2-3x|e^{|x-3|}+[\frac{3+sgn(x^2-1-5x)}{4+x^2}]+\{\frac{1}{3}+[\frac{x^2-2x}{|x|+1}]\}$ is non differentiable at $x=x_1,x_2,x_3.....x_n$ then ${\sum }_{i=1}^n{x}_i$ equals
( where $[.],\{.\},sgn(.)$, denotes greatest integer function, fractional part function and signum function respectively)

• A. \N
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is B 3

$sgn(x^2-1-5x)=0,-1,+1$
$\Rightarrow 3+sgn(x^2-1-5x)=3,2,4$
$\therefore \left[\frac{3+sgn(x^2-1-5x)}{4+x^2}\right]=0\text{and}\{\frac{1}{3}+\left[\frac{x^2-2x}{|x|+1}\right]\}=\frac{1}{3}$
$\therefore f\left(x\right)=x|x^2-3x|e^{|x-3|}+\frac{1}{3}$
So we have to check differentiability of the function $f\left(x\right)$ at $x=0\text{and}3$.
At $x=0$
For LHD
$f\left(x\right)=(x^3-3x^2)e^{3-x}\therefore f^{\prime }\left(x\right)=(x^3-3x^2)e^{3-x}(-1)+e^{3-x}(3x^2-6x)f^{\prime }\left(0\right)=0$
For RHD
$f\left(x\right)=(-x^3+3x^2)e^{3-x}\therefore f^{\prime }\left(x\right)=(-x^3+3x^2)e^{3-x}(-1)+e^{3-x}(-3x^2+6x)f^{\prime }\left(0\right)=0$

At $x=3$
For LHD
$f\left(x\right)=(-x^3+3x^2)e^{3-x}\therefore f^{\prime }\left(x\right)=(-x^3+3x^2)e^{3-x}(-1)+e^{3-x}(-3x^2+6x)f^{\prime }\left(0\right)=-9$
For RHD
$f\left(x\right)=(x^3-3x^2)e^{x-3}\therefore f^{\prime }\left(x\right)=(x^3-3x^2)e^{x-3}+e^{x-3}(3x^2-6x)f^{\prime }\left(0\right)=9$
$f\left(x\right)$ is differentiable at $x=0$ but not differentiable at $x=3.\therefore \sum x_i=3$