Explanation

(a)
Let $f(x) = ax^{2} + bx + 4$
On differentiating w. r. t., we get
$f^{'}(x) = 2ax + b$
For minimum, put $f^{'}(x) = 0\ \ \Rightarrow \ \ \ x = - \frac{b}{2a}$
Since, it is given that at $x = 1$ minimum value is -1
$\therefore\ \ \ \ \ 1 = - \frac{b}{2a}\ \ \ \Rightarrow \ \ \ \ \ 2a + b = 0$ …(i)
And $f(1) = a + b + 4 = - 1$
$\Rightarrow \ \ \ \ \ \ a + b + 5 = 0$ …(ii)
On solving Eqs. (i) and (ii), we get $a = 5,\ b = - 10$