If the function

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____.