QJEEM21Macfdd: JEE Main 2021 Mathematics - Class 11 JEEM21Macfdd Solution: The numbers of pairs (a, b) of real numbers, such that whenever …

Consider the equation x² + ax + b = 0

If has two roots (not necessarily real $\alpha$ & $\beta$ )

Either $\alpha$ = $\beta$ or $\alpha$ $\ne$ $\beta$

Case (1) If $\alpha$ = $\beta$ , then it is repeated root. Given that $\alpha$ ² $-$ 2 is also a root

So, $\alpha$ = $\alpha$ ² $-$ 2 $\Rightarrow$ ( $\alpha$ + 1)( $\alpha$ $-$ 2) = 0

$\Rightarrow$ $\alpha$ = $-$ 1 or $\alpha$ = 2

When $\alpha$ = $-$ 1 then (a, b) = (2, 1)

$\alpha$ = 2 then (a, b) = ( $-$ 4, 4)

Case (2) If $\alpha$ $\ne$ $\beta$

Then

(I) $\alpha$ = $\alpha$ ² $-$ 2 and $\beta$ = $\beta$ ² $-$ 2

Hence, (a, b) = ( $-$ ( $\alpha$ + $\beta$ ), $\alpha$ $\beta$ )

( $-$ 1, $-$ 2)

(II) $\alpha$ = $\beta$ ² $-$ 2 and $\beta$ = $\alpha$ ² $-$ 2

Then $\alpha$ $-$ $\beta$ = $\beta$ ² $-$ $\alpha$ ² = ( $\beta$ $-$ $\alpha$ ) ( $\beta$ + $\alpha$ )

Since $\alpha$ $\ne$ $\beta$ we get $\alpha$ + $\beta$ = $\beta$ ² + $\alpha$ ² $-$ 4

$\alpha$ + $\beta$ = ( $\alpha$ + $\beta$ )² $-$ 2 $\alpha$ $\beta$ $-$ 4

Thus $-$ 1 = 1 $-$ 2 $\alpha$ $\beta$ $-$ 4 which implies

$\alpha$ $\beta$ = $-$ 1 Therefore (a, b) = ( $-$ ( $\alpha$ + $\beta$ ), $\alpha$ $\beta$ )

= (1, $-$ 1)

(III) $\alpha$ = $\alpha$ ² $-$ 2 = $\beta$ ² $-$ 2 and $\alpha$ $\ne$ $\beta$

$\Rightarrow$ $\alpha$ = $-$ $\beta$

Thus $\alpha$ = 2, $\beta$ = $-$ 2

$\alpha$ = $-$ 1, $\beta$ = 1

Therefore (a, b) = (0, $-$ 4) & (0, 1)

(IV) $\beta$ = $\alpha$ ² $-$ 2 = $\beta$ ² $-$ 2 and $\alpha$ $\ne$ $\beta$ is same as (III) Therefore we get 6 pairs of (a, b)

Which are (2, 1), ( $-$ 4, 4), ( $-$ 1, $-$ 2), (1, $-$ 1), (0, $-$ 4)

Option (a)