Question

The number of integral values of ${}^{\prime }{a}^{\prime }$ so that $f\left(x\right)=sgn(x^3-(a+1)x^2+(a+1)x-1)$ has exactly one point of non differentiablility for all real values of $x$ is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (C)

$4$

For $f\left(x\right)$ to have only one point of non-differentiability. There should be only one
root of equation $x^3-(a+1)x^2+(a+1)x-1=P\left(x\right)$ (say)

For this $P^{\prime }\left(x\right)>0\forall x\in R\Rightarrow 3x^2-2(a+1)x+(a+1)>0$

$\Rightarrow D<0\Rightarrow (a+1{)}^2-3(a+1)<0\Rightarrow (a+1\left)\right(a-3)<0$

$\Rightarrow a\in (-1,3)$. Check for $a=-1$. Clearly for this value of $a$
$P\left(x\right)=x^3-1$ which has only one real root. Thus $a\in [-1,3)$

Hence total number of integral points are $4$.