Q33 / 105 · NTA Abhyas (JEE Main) 2020 · Mathematics · Circle

The least value of the quadratic polynomial, $f (x) = (2 p^{2} + 1) x^{2} + 2 (4 p^{2} - 1) x + 4 (2 p^{2} + 1)$ for real values of $p$ and $x$ is

Correct answer: 3

Solution

y = 2 p^{2} (x^{2} + 4x + 4) + x^{2} - 2 x + 1 + 3

y = 2 p^{2} {(x + 2)}^{2} + {(x - 1)}^{2} + 3

p = 0, x = 1

y_{min} = 3

Hence, the correct answer is 3.

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