JEE MAIN - Mathematics (2020 - 9th January Evening Slot - No. 15)
Let a, b $$ \in $$ R, a $$ \ne $$ 0 be such that the equation,
ax2 – 2bx + 5 = 0 has a repeated root $$\alpha $$, which
is also a root of the equation, x2 – 2bx – 10 = 0.
If $$\beta $$ is the other root of this equation, then
$$\alpha $$2 + $$\beta $$2 is equal to :
28
24
26
25
Explanation
Roots of equation ax2
– 2bx + 5 = 0 are $$\alpha $$, $$\alpha $$.
$$ \therefore $$ $$\alpha $$ + $$\alpha $$ = $${{2b} \over a}$$
$$ \Rightarrow $$ 2$$\alpha $$ = $${{2b} \over a}$$
$$ \Rightarrow $$ $$\alpha $$ = $${{b} \over a}$$ ....(1)
and $$\alpha $$2 = $${5 \over a}$$ ....(2)
From (1) and (2), we get
$${{{b^2}} \over {{a^2}}} = {5 \over a}$$
$$ \Rightarrow $$ b2 = 5a
$$\alpha $$, $$\beta $$ are the roots of equation x2 – 2bx – 10 = 0
$$ \therefore $$ $$\alpha $$ + $$\beta $$ = 2b
and $$\alpha $$$$\beta $$ = -10
As $$\alpha $$ is a root of the equation x2 – 2bx – 10 = 0.
$$ \therefore $$ $$\alpha $$2 - 2b$$\alpha $$ - 10 = 0
$$ \Rightarrow $$ $${{{b^2}} \over {{a^2}}} - {{2{b^2}} \over a} - 10 = 0$$
$$ \Rightarrow $$ $${{5a} \over {{a^2}}} - {{10a} \over a} - 10 = 0$$
$$ \Rightarrow $$ $${5 \over a} - 10 - 10 = 0$$
$$ \Rightarrow $$ $$a$$ = $${1 \over 4}$$
$$ \Rightarrow $$ $${\alpha ^2}$$ = 20
$$\alpha $$$$\beta $$ = -10
$$ \Rightarrow $$$${\beta ^2}$$ = 5
$$ \therefore $$ $${\alpha ^2} + {\beta ^2}$$ = 20 + 5 = 25
$$ \therefore $$ $$\alpha $$ + $$\alpha $$ = $${{2b} \over a}$$
$$ \Rightarrow $$ 2$$\alpha $$ = $${{2b} \over a}$$
$$ \Rightarrow $$ $$\alpha $$ = $${{b} \over a}$$ ....(1)
and $$\alpha $$2 = $${5 \over a}$$ ....(2)
From (1) and (2), we get
$${{{b^2}} \over {{a^2}}} = {5 \over a}$$
$$ \Rightarrow $$ b2 = 5a
$$\alpha $$, $$\beta $$ are the roots of equation x2 – 2bx – 10 = 0
$$ \therefore $$ $$\alpha $$ + $$\beta $$ = 2b
and $$\alpha $$$$\beta $$ = -10
As $$\alpha $$ is a root of the equation x2 – 2bx – 10 = 0.
$$ \therefore $$ $$\alpha $$2 - 2b$$\alpha $$ - 10 = 0
$$ \Rightarrow $$ $${{{b^2}} \over {{a^2}}} - {{2{b^2}} \over a} - 10 = 0$$
$$ \Rightarrow $$ $${{5a} \over {{a^2}}} - {{10a} \over a} - 10 = 0$$
$$ \Rightarrow $$ $${5 \over a} - 10 - 10 = 0$$
$$ \Rightarrow $$ $$a$$ = $${1 \over 4}$$
$$ \Rightarrow $$ $${\alpha ^2}$$ = 20
$$\alpha $$$$\beta $$ = -10
$$ \Rightarrow $$$${\beta ^2}$$ = 5
$$ \therefore $$ $${\alpha ^2} + {\beta ^2}$$ = 20 + 5 = 25
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