JEE MAIN - Mathematics (2019 - 9th April Evening Slot - No. 12)
If m is chosen in the quadratic equation
(m2 + 1) x2 – 3x + (m2 + 1)2 = 0
such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is :-
(m2 + 1) x2 – 3x + (m2 + 1)2 = 0
such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is :-
$$4\sqrt 3 $$
$$8\sqrt 3 $$
$$8\sqrt 5 $$
$$10\sqrt 5 $$
Explanation
Given quadratic equation
(m2 + 1) x2 – 3x + (m2 + 1)2 = 0
Let roots of the equation $$\alpha $$ and $$\beta $$.
$$ \therefore $$ Sum of roots = $$\alpha $$ + $$\beta $$ = $${3 \over {{m^2} + 1}}$$
Product of roots = $$\alpha $$$$\beta $$ = m2 + 1
$${3 \over {{m^2} + 1}}$$ is maximum when m = 0
Hence equation becomes x2 – 3x + 1 = 0
$$\alpha + \beta = 3$$, $$\alpha \beta = 1$$
$$\left| {\alpha - \beta } \right|$$ = $$\sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } $$ = $$\sqrt {{{\left( 3 \right)}^2} - 4.1} $$ = $$\sqrt 5 $$
$$\left| {{\alpha ^3} - {\beta ^3}} \right| = \left| {(\alpha - \beta )({\alpha ^2} + {\beta ^2} + \alpha \beta )} \right| $$
= $$\sqrt 5 \left| {{{\left( {\alpha + \beta } \right)}^2} - \alpha \beta } \right|$$
$$= \sqrt 5 (9 - 1) = 8\sqrt 5 $$
(m2 + 1) x2 – 3x + (m2 + 1)2 = 0
Let roots of the equation $$\alpha $$ and $$\beta $$.
$$ \therefore $$ Sum of roots = $$\alpha $$ + $$\beta $$ = $${3 \over {{m^2} + 1}}$$
Product of roots = $$\alpha $$$$\beta $$ = m2 + 1
$${3 \over {{m^2} + 1}}$$ is maximum when m = 0
Hence equation becomes x2 – 3x + 1 = 0
$$\alpha + \beta = 3$$, $$\alpha \beta = 1$$
$$\left| {\alpha - \beta } \right|$$ = $$\sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } $$ = $$\sqrt {{{\left( 3 \right)}^2} - 4.1} $$ = $$\sqrt 5 $$
$$\left| {{\alpha ^3} - {\beta ^3}} \right| = \left| {(\alpha - \beta )({\alpha ^2} + {\beta ^2} + \alpha \beta )} \right| $$
= $$\sqrt 5 \left| {{{\left( {\alpha + \beta } \right)}^2} - \alpha \beta } \right|$$
$$= \sqrt 5 (9 - 1) = 8\sqrt 5 $$
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