Question

The number of integral values of $k$ for which $h\left(x\right)=\text{sgn}(x^2-2kx+\text{sgn}(k+2\left)\right)$ is continuous for all $x\in R$ is
[Note : $\text{sgn}\left(y\right)$ denotes the signum function of $y$.]

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

For the signum function to be continuous for all $x\in R$,
$x^2-2kx+\text{sgn}(k+2)>0$
$D<0\Rightarrow 4k^2-4\text{sgn}(k+2)<0$
$\Rightarrow k^2-\text{sgn}(k+2)<0$

Case $1:k+2>0\Rightarrow k>-2$,
$\Rightarrow k^2-1<0\Rightarrow k\in (-1,1)$
Which is possible.

Case $2:k+2<0\Rightarrow k<-2$
$\Rightarrow k^2+1<0$
Which is not possible.

Case $3:k+2=0\Rightarrow k=-2$
$\Rightarrow k^2<0$
Which is not possible.

$\therefore k\in (-1,1)$
Hence, the number of integral value of $k$ is $1.$