It is given that at $x = 1$, the function ${x^4} - 63{x^2} + ax + 9$ attains its maximum value, on the interval $\left[ {0,2} \right]$. Find the value of $a$.

$f(x) = {x^4} - 62{x^2} + ax + 9,0 \le x \le 2$
$\Rightarrow f'(x) = 4{x^3} - 124x + a$

As $f(x)$ attains maximum value at $x = 1 \in [0,2]$,
Therefore we must have $f'(1) = 0 \Rightarrow 4 - 124 + a = 0 \Rightarrow a = 120$

Thus, we have$f(x) = {x^4} - 62{x^2} + 120x + 9$

$f(0) = 9,f(1) = 1 - 62 + 120 + 9 = 68$
and $f(2) = {2^4} - 62 \times {2^2} + 120 \times 2 + 9 = 17$

Hence we can say that $f(1)$ is maximum and $a = 120$.