The function f (x) = x4 - 62x2 + ax + 9 attains its maximum value in the interval [0, 2] at x = 1. Then the value of a is:

Question

The function f (x) = x4 - 62x2 + ax + 9 attains its maximum value in the interval [0, 2] at x = 1. Then the value of a is: (A) 120 (B) -120 (C) 52 (D) 102

Tardigrade Admin · Accepted Answer

Given f (x) = x4 - 62x2 + ax + 9 f ' (x) = 4x3 - 124x + a For maxima and minima: f ' (x) = 0 ⇒ 4x3 - 124x + a = 0 But given that the function f (x) attains its maximum value at x = 1. Thus, 4 . 13 - 124 . 1 + a = 0 ⇒ a = 120.

Q. The function $f (x) = x^{4} - 62 x^{2} + a x + 9$ attains its maximum value in the interval [0, 2] at x = 1. Then the value of a is:

120

-120

52

102

Given $f (x) = x^{4} - 62 x^{2} + a x + 9$
$f^{'} (x) = 4 x^{3} - 124 x + a$
For maxima and minima:
$f^{'} (x) = 0$
$\Rightarrow 4 x^{3} - 124 x + a = 0$
But given that the function f (x) attains its maximum value at x = 1.
Thus, $4. 1^{3} - 124.1 + a = 0$
$\Rightarrow a = 120$ .

Q. The function f(x)=x4−62x2+ax+9 attains its maximum value in the interval [0, 2] at x = 1. Then the value of a is:

120

-120

52

102

Given f(x)=x4−62x2+ax+9 f′(x)=4x3−124x+a For maxima and minima: f′(x)=0 ⇒4x3−124x+a=0 But given that the function f (x) attains its maximum value at x = 1. Thus, 4.13−124.1+a=0 ⇒a=120.

Q. The function $f (x) = x^{4} - 62 x^{2} + a x + 9$ attains its maximum value in the interval [0, 2] at x = 1. Then the value of a is:

Given $f (x) = x^{4} - 62 x^{2} + a x + 9$
$f^{'} (x) = 4 x^{3} - 124 x + a$
For maxima and minima:
$f^{'} (x) = 0$
$\Rightarrow 4 x^{3} - 124 x + a = 0$
But given that the function f (x) attains its maximum value at x = 1.
Thus, $4. 1^{3} - 124.1 + a = 0$
$\Rightarrow a = 120$ .