Question

Let $a$ be a real number such that the function $f\left(x\right)=a{x}^2+6x-15,$ $x\in R$ is increasing in $(-\infty ,\frac{3}{4})$ and decreasing in $(\frac{3}{4},\infty )$. Then the function $g\left(x\right)=a{x}^2-6x+15,$ $x\in R$ has a

• A. Local maximum at $x=\frac{3}{4}$
• B. Local maximum at $x=-\frac{3}{4}$
• C. Local minimum at $x=-\frac{3}{4}$
• D. Local minimum at $x=\frac{3}{4}$

Answer 1

The correct option is B Local maximum at $x=-\frac{3}{4}$

$\because f\left(x\right)=a{x}^2+6x-15$
$f\left(x\right)$ increases in $(-\infty ,\frac{3}{4})$ and decreases in $(\frac{3}{4},\infty )$
$\therefore f\left(x\right)$ is stationary at $x=\frac{3}{4}$
$\Rightarrow f^{\prime }\left(x\right)=0\Rightarrow x=-\frac{3}{a}=\frac{3}{4}\Rightarrow a=-4$
Now, $g\left(x\right)=-4x^2-6x+15$
$g^{\prime }\left(x\right)=-8x-6=-8(x+\frac{6}{8})$
$\therefore g\left(x\right)$ has local maximum at $x=-\frac{3}{4}$