Question

Let $y^2=\frac{(x+5)(x-3)}{x-1},$ then all the real values of x, for which y has non-zero real values, are:

• A. -5 < x < 1
• B. x > 15
• C. -5 < x < 1 or x > 3
• D. None of these

Answer 1

The correct option is C -5 < x < 1 or x > 3

If y is real and non-zero $y^2>0,i.e.,\frac{(x+5)(x-3)}{(x-1)}>0$
Case (i) : If numerator and denominator are > 0
$\Rightarrow x>3orx<-5andx>1$
But with x < -5 , $y^2$ is negative which is not possible.
Case (ii): If numerator and denominator are < 0
$\Rightarrow -5<x<3andx<1.$