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

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Given function is 
f(x) =x ^4 - 62x^2 + ax + 9
Function f(x) =x ^4 - 62x^2 + ax + 9 attains maximum value at x = 1 then x must one of the critical point of the given function that means
f^{'}(1)=0
f^{'}(x) = 4x^3-124x+a\\ f^{'}(1) = 4(1)^3-124(1)+a\\ f^{'}(1)=4-124+a = a - 120\\
Now,
f^{'}(1)=0\\ a - 120=0\\ a=120
Hence, the value of a is 120

Posted by

Gautam harsolia

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