Let $f(x)=(x-a)(x-b)-\left(\frac{a+b}{2}\right)$. If $f(x)=0$ has both non-negative roots, then the minimum…
Let $f(x)=(x-a)(x-b)-\left(\frac{a+b}{2}\right)$. If $f(x)=0$ has both non-negative roots, then the minimum value of $f(x)$.
- $=\left(\frac{a+b}{4}\right)$
- $\geq \frac{(a+b)^2}{4}$
- $\geq \frac{-(a+b)^2}{4}$
- $\leq \frac{-(a+b)^2}{4}$
Solution
Let $f(x)=(x-a)(x-b)-\left(\frac{a+b}{2}\right)$
Now, $\quad f^{\prime}(x)=(x-b)+(x-a)=2 x-b-a$
Now, $\quad f^{\prime}(x)=0$
$
\begin{aligned}
x-b+x-a & =0 \Rightarrow 2 x=a+b \\
\Rightarrow \quad x & =\frac{a+b}{2}
\end{aligned}
$
Now, $\quad f^{\prime \prime}(x)=2>0$
So, at $x=\frac{a+b}{2}, f(x)$ has minimum value.
Now, at $x=\frac{a+b}{2}$
$
\begin{aligned}
f(x) & =\left(\frac{a+b}{2}-a\right)\left(\frac{a+b}{2}-b\right)-\left(\frac{a+b}{2}\right) \\
& =\frac{-(a-b)^2}{4}-\left(\frac{a+b}{2}\right) \\
& =-\frac{1}{2}\left\lfloor\frac{a^2+b^2-2 a b+2 a+2 b}{2}\right] \\
& =-\frac{1}{2}\left\lfloor\frac{a^2+b^2+2 a b-4 a b+2(a+b)}{2}\right]
\end{aligned}
$
$
\begin{aligned}
& \text { Given, } \quad(x-a)(x-b)=\frac{a+b}{2} \\
& \Rightarrow x^2-(a+b) x+\left(\frac{2 a b-a-b}{2}\right)=0
\end{aligned}
$
Now, roots are non-negative, so $(a+b)>0$
$
\text { and } \frac{2 a b-(a+b)}{2}>0 \Rightarrow \frac{4 a b-2(a+b)}{4}>0
$
So, from Eq. (i)
$
f(x)=-\frac{(a+b)^2}{4} .
$
Asked in: AP EAMCET 2018 (23 Apr Shift 1)
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