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JEE Previous Year Question

$The minimum value of the sum of the squares of the roots of x^{2} + (3 - a)x + 1 = 2a is:$

Topic: JEE PYQ

Answer: C. Full step-by-step solution below.

A $4$
B $5$
C $6$
D $8$

Correct answer: C

Solution

(c) α^{2} + β^{2} = (α + β)^{2} - 2 αβ Let f (a) = (3 - a)^{2} - 2 (1 - 2 a) f (a) = a^{2} - 2 a + 7 f (a) = (a - 1)^{2} + 6 f (a)_{min .} = 6

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