JEE MAIN - Mathematics (2020 - 2nd September Evening Slot - No. 4)
Let f(x) be a quadratic polynomial such that
f(–1) + f(2) = 0. If one of the roots of f(x) = 0
is 3, then its other root lies in :
f(–1) + f(2) = 0. If one of the roots of f(x) = 0
is 3, then its other root lies in :
(–3, –1)
(1, 3)
(–1, 0)
(0, 1)
Explanation
Let the other root is $$\alpha $$.
$$ \therefore $$ f(x) = a(x – 3) (x – $$\alpha $$)
f(2) = a($$\alpha $$– 2)
f(–1) = 4a(1 + $$\alpha $$)
Given f(–1) + f(2) = 0
$$ \Rightarrow $$a($$\alpha $$ – 2 + 4 + 4$$\alpha $$) = 0
$$ \Rightarrow $$ 5$$\alpha $$ = -2 As a $$ \ne $$ 0
$$ \Rightarrow $$ $$\alpha $$ = $$-\frac{2}{5} $$ = - 0.4
$$ \therefore $$ $$\alpha $$ $$ \in $$ (–1, 0)
$$ \therefore $$ f(x) = a(x – 3) (x – $$\alpha $$)
f(2) = a($$\alpha $$– 2)
f(–1) = 4a(1 + $$\alpha $$)
Given f(–1) + f(2) = 0
$$ \Rightarrow $$a($$\alpha $$ – 2 + 4 + 4$$\alpha $$) = 0
$$ \Rightarrow $$ 5$$\alpha $$ = -2 As a $$ \ne $$ 0
$$ \Rightarrow $$ $$\alpha $$ = $$-\frac{2}{5} $$ = - 0.4
$$ \therefore $$ $$\alpha $$ $$ \in $$ (–1, 0)
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