Given that at x=1, the function $x^4-62x^2+ax+90$ attains its maximum value on the interval [0, 2]. The value of 'a' is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 120
$y=x^4-62x^2+ax+90$
$\frac{dy}{dx}=4x^3-124x+a$
so $\left.\frac{dy}{dx}\right]_{x=1}=0$
so $4-124+a=0$
$a=120$