Question

If 'x' is real, then maximum value of $\frac{3x^2+9x+17}{3x^2+9x+7}$ is -

• A. $41$
• B. $1$
• C. $\frac{17}{7}$
• D. $-1$

Answer 1

The correct option is A $41$

Maximise: $\frac{3x^2+9x+17}{3x^2+9x+7}$

Now we can see that coefficient of $x^2$ and $x$ are same in $N^x$ and $D^x$ so

$\Rightarrow$ $\frac{3x^2+9x+10+7}{3x^2+9x+7}$

$y=1+\frac{10}{3x^2+9x+7}$

We want to maximise $y$ so we need to minimize $3x^2+3x+7$

$y=1+\frac{10}{min(3x^2+9x+7)}$

$y=1+40=41$ ($\because$ we know min value of quadratic is $\frac{-D}{ya}$)