VITEEE 2018 Paper

Let f(x) = x4−62x2 + ax + 9
⇒ f '(x) = 4x3 - 124x + a
It is given that function f attains its maximum value on the interval [0,2] at x = 1.
.∴ f'(l) = 0 ⇒ 4 x 13 - 124 x 1 + 1 = 0
⇒ 4 - 124 + a = 0 ⇒ a = 120
Hence, the value of a is 120.