CAT 2019

Slot 1

QA

Question & Solution

For $x <0$, $-x(6x^2+1) = 5x^2$

=> $(6x^2+1$) = $-5x$

=> $(6x^2 + 5x+ 1) = 0$ 

=>$(6x^2 + 3x+2x+ 1) = 0$

=> $(3x+1)(2x+1)=0$

$\Rightarrow x=\ -\frac{\ 1}{3}$  or$\quad x=\ -\frac{\ 1}{2}$

For $x=0, \quad LHS = RHS = 0$    (Hence, 1 solution)

For $x >0$, $x(6x^2+1) = 5x^2$

=> $(6x^2 - 5x+ 1) = 0$ 

$\Rightarrow(3x-1)(2x-1)=0$

$\Rightarrow x = \ \frac{\ 1}{3}$   or   $x=\ \frac{\ 1}{2}$

Hence, the total number of solutions = 5