Question

Let $f(x) = x^2 + 2x + 2, g(x) = - x^2 + 2x - 1 $ and $a, b$ be the extreme values of $f(x), g(x)$ respectively. If $c$ is the extreme value of $\frac{f}{g} (x)$ (for x $\neq$ 1), then $a + 2b + 5c + 4$ =

• A. 2
• B. 1
• C. 4
• D. 3

Answer 1

Answer: (C)

Given, $f(x) =x^{2}+2 x+2 $
$=x^{2}+2 x+1+1=(x+1)^{2}+1 $
Here, $ f(x) \in[1, \infty) $ and $ g(x)=-x^{2}+2 x-1 $
$=-\left(x^{2}-2 x+1\right)=-(x-1)^{2}$
Here, $g(x) \in(-\infty, 0]$
Now, $\frac{f}{g}(x)=\frac{x^{2}+2 x+2}{-x^{2}+2 x-1}=y$
$\Rightarrow x^{2}+2 x+2=-y x^{2}+2 x y-y $
$ \Rightarrow x^{2}+y x^{2}+2 x-2 x y+2+y=0 $
$ \Rightarrow x^{2}(1+y)+(2-2 y) x+2+y=0 $
$ \because D \geq 0 $
$ \therefore (2-2 y)^{2}-4(2+y)(1+y) \geq 0 $
$ 4+4 y^{2}-8 y-4(2+y)(1+y) > 0$
$ \Rightarrow y \leq-\frac{1}{5} $
So, $ \frac{f}{g}(x) \in\left(-\infty,-\frac{1}{5}\right]$
So, $a=1, b=0$ and $c=-\frac{1}{5}$
Hence, $a+2 b+5 c+4$
$=1+0+5\left(-\frac{1}{5}\right)+4$
$=1-1+4=4$