QJEEM18Maa2e1: JEE Main 2018 Mathematics - Class 11 JEEM18Maa2e1 Solution: If R is such that the sum of the cubes of the roots of the…

Let $\alpha$ , $\beta$ are the roots of the equation,

$\therefore$ $\alpha$ + $\beta$ = $\lambda$ $-$ 2 and $\alpha$ $\beta$ = 10 $-$ $\lambda$

${\alpha ^3} + {\beta ^3}$ = ( $\alpha$ + $\beta$ )³ $-$ 3 $\alpha$ $\beta$ ( $\alpha$ + $\beta$ )

= ( $\lambda$ $-$ 2)³ $-$ 3(10 $-$ $\lambda$ )( $\lambda$ $-$ 2)

= $\lambda ^3$ $-$ 3 $\lambda ^2$ $-$ 24 $\lambda$ + 52

Let $f(\lambda$ ) = $\lambda ^3$ $-$ 3 $\lambda ^2$ $-$ 24 $\lambda$ + 52

$\therefore$ ${{df(\lambda )} \over {d\lambda }}$ = 3 $\lambda ^2$ $-$ 6 $\lambda$ $-$ 24

$\therefore$ at maximum of minimum ${{df(\lambda )} \over {d\lambda }}$ = 0

$\therefore$ $\lambda ^2$ $-$ 2 $\lambda$ $-$ 8 = 0

$\Rightarrow$ ( $\lambda$ + 2) ( $\lambda$ $-$ 4) = 0

$\Rightarrow$ $\lambda$ = $-$ 2, 4

${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$ = 2 $\lambda$ $-$ 2

When $\lambda$ = $-$ 2

${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$ = $-$ 6 < 0

$\therefore$ at $\lambda$ = $-$ 2, f( $\lambda$ ) has maximum value.

When $\lambda$ = 4

${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$ = 6 > 0

$\therefore$ at $\lambda$ = 4, f( $\lambda$ ) has minimum value.

$\therefore$ When $\lambda$ = 4 equation is,

x² $-$ 2x + 6 = 0

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

$\Rightarrow$ x² $-$ 4 $\times$ 6

= $-$ 20

$\Rightarrow$ ( $\alpha$ $-$ $\beta$ ) = $2\sqrt 5 i$

$\Rightarrow$ $\left| {\alpha - \beta } \right|$ = $2\sqrt 5$ (ans)