Question

If $a{x}^2+bx+4$ attains its minimum value $-1$ at $x=1$, then the values of $a$ and $b$ are respectively

• A. $5,-10$
• B. $5,-5$
• C. $5,5$
• D. $10,-5$

Answer 1

The correct option is A $5,-10$

Let $f\left(x\right)=a{x}^2+bx+4$

On differentiating w.r.t. x, we get

$f^{\prime }\left(x\right)=2ax+b$

For minimum, put $f^{\prime }\left(x\right)=0$

$\Rightarrow x=-\frac{b}{2a}$

Since, it is given that at $x=1,$ minimum value is $-1$

$\therefore 1=-\frac{b}{2a}$

$\Rightarrow 2a+b=0$ ...(i)

and $f\left(1\right)=a+b+4=-1$

$\Rightarrow a+b+5=0$ ...(ii)

On solving Eq. (i) and (ii) we get $a=5,b=-10$