Question

Let a, b, c $ϵ$ R such that a + b + c = 0 and a + b - c = 0, then the polynomial function $f\left(x\right)=a{x}^2+bx+c$; $(a>0)$ attains its least value at 'x' equal to

• A. 0
• B. 1
• C. $\frac{1}{2}$
• D. not possible

Answer 1

Answer: (C)

$\frac{1}{2}$

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

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

$f^{\prime }\left(x\right)=0$

$\Rightarrow 2ax=-b$

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

$\Rightarrow f\left(\frac{-b}{2a}\right)=a\times \frac{b^2}{4a^2}-b\times \frac{b}{2a}+c=\frac{b^2}{4a}-\frac{b^2}{2a}+c=-\frac{b^2}{4a}+c=c-\frac{b^2}{4a}$

Now,

$a+b+c=0$ .....................(1)

$a+b-c=0$ .....................(2)

Adding (1) and (2);

$a+b=0,$ $a=-b$ $c=0$

$\Rightarrow f\left(\frac{-b}{2a}\right)=0-\frac{b^2}{4(-b)}=\frac{b}{4}$

$\Rightarrow x=\frac{-b}{2a}=\frac{-b}{2(-b)}=\frac{1}{2}$