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 $\ne$ 1), then $a+2b+5c+4$ =

• A. 2
• B. 1
• C. 4
• D. 3

Answer 1

Answer: (C)

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