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Q.
The function $f (x) = x^4 - 62x^2 + ax + 9$ attains its maximum value in the interval [0, 2] at x = 1. Then the value of a is:
Application of Derivatives
Solution:
Given $f (x) = x^4 - 62x^2 + ax + 9$
$f ' (x) = 4x^3 - 124x + a$
For maxima and minima:
$f ' (x) = 0 $
$ \Rightarrow \, 4x^3 - 124x + a = 0$
But given that the function f (x) attains its maximum value at x = 1.
Thus, $4 . 1^3 - 124 . 1 + a = 0 $
$ \Rightarrow \, a = 120$.