Question 3 [JEE adv, 2021]

Concepts

Consider the sign of \(x^2 - 1\) to get two quadratic equations.

Solution

Step 1

\[ \begin{eqnarray} x^2 - 1 > 0&\implies& x\in (-\infty, -1)\cup(1,\infty)\\ x^2 - 1 < 0&\implies& x\in (-1, 1) \end{eqnarray} \]

Step 2

Case 1: \(x^2 - 1>0\)

This gives the quadratic equation \[ \begin{eqnarray} 3x^2 -4(x^2 - 1) + x - 1 &=& 0\\ \implies x^2-x -3 &=& 0\\ \implies x &=& \frac{1\pm \sqrt{13}}{2} \end{eqnarray} \] Both the roots are in the allowed interval \(x\in (-\infty, -1)\cup(1,\infty)\).

Case 2: \(x^2 - 1<0\)

This gives the quadratic equation \[ \begin{eqnarray} 3x^2 +4(x^2 - 1) + x - 1 &=& 0\\ \implies 7x^2+x -5 &=& 0\\ \implies x &=& \frac{-1\pm \sqrt{141}}{14} \end{eqnarray} \] Both the roots are in the allowed interval \(x\in (-1,1)\).

Step 3

There are two roots each of the two quadratic equations. Hence, the total number of roots is 4.