JEE Main 2022 — Mathematics Algebra

Since $p,r$p,r are roots of ${x}^{2}-8ax+2a=0$x2−8ax+2a=0 and

$q,s$q,s are the roots of ${x}^{2}+12bx+6b=0$x2+12bx+6b=0

So the equation $2a{x}^{2}-8ax+1=0$2ax2−8ax+1=0 has roots $\frac{1}{p},\frac{1}{r}$p1​,r1​ and

$6b{x}^{2}+12bx+1=0$6bx2+12bx+1=0 has roots $\frac{1}{q},\frac{1}{s}$q1​,s1​.

Given that $\frac{1}{p},\frac{1}{q},\frac{1}{r},\frac{1}{s}$p1​,q1​,r1​,s1​ are in A.P.

Let $\frac{1}{p}=\alpha -3\beta ,\frac{1}{q}=\alpha -\beta ,\frac{1}{r}=\alpha +\beta$p1​=α−3β,q1​=α−β,r1​=α+β and $\frac{1}{s}=\alpha +3\beta$s1​=α+3β

i.e. $\frac{1}{p}+\frac{1}{r}=2\alpha -2\beta =4$p1​+r1​=2α−2β=4 and

$$\frac{1}{q}+\frac{1}{s}=2\alpha +2\beta =-2$$q1​+s1​=2α+2β=−2

or $\alpha =\frac{1}{2},\beta =-\frac{3}{2}$α=21​,β=−23​

i.e. $\frac{1}{p}=5,\frac{1}{q}=2,\frac{1}{r}=-1,\frac{1}{s}=-4$p1​=5,q1​=2,r1​=−1,s1​=−4

Now

$\frac{1}{p}\times \frac{1}{r}=\frac{1}{2a}=-5\Rightarrow \frac{1}{a}=-10$p1​×r1​=2a1​=−5⇒a1​=−10 and $\frac{1}{q}\times \frac{1}{s}=\frac{1}{6b}=-8\Rightarrow \frac{1}{b}=-48$q1​×s1​=6b1​=−8⇒b1​=−48

Hence, ${a}^{-1}-{b}^{-1}=-10+48=38$a−1−b−1=−10+48=38