If the function $f(x)=x^{4}-62 x^{2}+a x+9$ attains a local maximum at $x=1$, then $a=$ _______________
It is given that, the function $f(x)=x^{4}-62 x^{2}+a x+9$ attains a local maximum at $x=1$
$\therefore f^{\prime}(x)=0$ at $x=1$
$f(x)=x^{4}-62 x^{2}+a x+9$
Differentiating both sides with respect to x, we get
$f^{\prime}(x)=4 x^{3}-124 x+a$
Now,
$f^{\prime}(1)=0$
$\Rightarrow 4 \times(1)^{3}-124 \times 1+a=0$
$\Rightarrow a=124-4=120$
Thus, the value of a is 120.
At $x=1$, we have
$f^{\prime \prime}(1)=12 \times(1)^{2}-124=12-124=-112<0$
So, x = 1 is the point of local maximum of f(x).
If the function $f(x)=x^{4}-62 x^{2}+a x+9$ attains a local maximum at $x=1$, then $a=$ ___120____.
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