Question

Consider a function $f(x)=\frac{3x+a}{x^2+3}$ which has greatest value equal to $\frac{3}{2}$.
The minimum value of $f(x)$ is equal to

• A. $\tan \left(\frac{-\pi }{3}\right)$
• B. $\sin \left(\frac{-\pi }{6}\right)$
• C. $\cos \left(\frac{-\pi }{3}\right)$
• D. $\cot \left(\frac{\pi }{2}\right)$

Answer 1

Answer: (B)

$y=f(x)=\frac{3x+3}{x^2+3}\Rightarrow x^{2}y+3y=3x+3\Rightarrow x^{2}y-3x+3y-3=0$
As $x\in R$, so $D\ge 0$
$\Rightarrow 9-4y(3y-3)\ge 0\Rightarrow 12y^2-12y-9\le 0\Rightarrow 4y^2-4y-3\le 0$
$\Rightarrow 4y^2-6y+2y-3\le 0\Rightarrow (2y-3)(2y+1)\le 0$
Hence $y\in \left[\frac{-1}{2},\frac{3}{2}\right]$
Hence minimum value of $f(x)$ is $\frac{-1}{2}=\sin \left(\frac{-\pi }{6}\right)$