JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 8)
The value of $$\lambda $$ such that sum of the squares of the roots of the quadratic equation, x2 + (3 – $$\lambda $$)x + 2 = $$\lambda $$ has the least value is -
1
2
$${{15} \over 8}$$
$${4 \over 9}$$
Explanation
$$\alpha $$ + $$\beta $$ = $$\lambda $$ $$-$$ 3
$$\alpha $$$$\beta $$ = 2 $$-$$ $$\lambda $$
$$\alpha $$2 + $$\beta $$2 = ($$\alpha $$ + $$\beta $$)2 $$-$$ 2$$\alpha $$$$\beta $$ = ($$\lambda $$ $$-$$ 3)2 $$-$$ 2$$\left( {2 - \lambda } \right)$$
= $$\lambda $$2 + 9 $$-$$ 6$$\lambda $$ $$-$$ 4 + 2$$\lambda $$
= $$\lambda $$2 $$-$$ 4$$\lambda $$ + 5
= ($$\lambda $$ $$-$$ 2)2 + 1
$$ \therefore $$ $$\lambda $$ = 2
$$\alpha $$$$\beta $$ = 2 $$-$$ $$\lambda $$
$$\alpha $$2 + $$\beta $$2 = ($$\alpha $$ + $$\beta $$)2 $$-$$ 2$$\alpha $$$$\beta $$ = ($$\lambda $$ $$-$$ 3)2 $$-$$ 2$$\left( {2 - \lambda } \right)$$
= $$\lambda $$2 + 9 $$-$$ 6$$\lambda $$ $$-$$ 4 + 2$$\lambda $$
= $$\lambda $$2 $$-$$ 4$$\lambda $$ + 5
= ($$\lambda $$ $$-$$ 2)2 + 1
$$ \therefore $$ $$\lambda $$ = 2
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