JEE MAIN - Mathematics (2018 - 15th April Morning Slot - No. 6)
If $$\lambda $$ $$ \in $$ R is such that the sum of the cubes of the roots of the equation,
x2 + (2 $$-$$ $$\lambda $$) x + (10 $$-$$ $$\lambda $$) = 0 is minimum, then the magnitude of the difference of the roots of this equation is :
x2 + (2 $$-$$ $$\lambda $$) x + (10 $$-$$ $$\lambda $$) = 0 is minimum, then the magnitude of the difference of the roots of this equation is :
$$4\sqrt 2 $$
$$2\sqrt 5 $$
$$2\sqrt 7 $$
20
Explanation
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)
$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $$\lambda $$ $$-$$ 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)
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