It is given that at $x = 1$, the function $x^4 - 62x^2 + ax + 9$ attains its maximum value in the interval [0,2]. Find the value of a.

120

The correct answer is Option (3) → 120

Let $f(x) = x^4- 62x^2+ax +9$   ...(i)

It is differentiable for all x in [0, 2].

Differentiating (i) w.r.t. x, we get

$f'(x) = 4x^3-62.2x + a = 4x^3 - 124x + a$

$∴ f'(1) = 4.1^3 – 124.1 + a = a − 120$.

Given that at $x = 1$, the function (i) has maximum value, therefore, $x = 1$ is a point of maxima

$⇒ x = 1$ is a critical point

$⇒ f'(1)=0⇒a-120=0⇒ a = 120$.