Let $f(x)=x^2+2 x+2, g(x)=-x^2+2 x-1$ 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 $x
eq 1$ ), then $a+2 b+5 c+4=$

Question

Let $f(x)=x^2+2 x+2, g(x)=-x^2+2 x-1$ 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 $x 
eq 1$ ), then $a+2 b+5 c+4=$

Aicharya · Accepted Answer

Given, $ \begin{aligned} f(x) & =x^2+2 x+2 \ & =x^2+2 x+1+1=(x+1)^2+1 \end{aligned} $ Here, $f(x)$ $ \begin{aligned} & \in[1, \infty) 	ext { and } g(x)=-x^2+2 x-1 \ & =-\left(x^2-2 x+1ight)=-(x-1)^2 \end{aligned} $ Here, $g(x) \in(-\infty, 0]$ Now, $\frac{f}{g}(x)=\frac{x^2+2 x+2}{-x^2+2 x-1}=y$ $ \begin{array}{ll} \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 \quad & D \geq 0 \ 	herefore \quad & (2-2 y)^2-4(2+y)(1+y) \geq 0 \ & 4+4 y^2-8 y-4(2+y)(1+y) \geq 0 \ & 1+y^2-2 y-(2+y)(1+y) \geq 0 \ & 1+y^2-2 y-\left(2+2 y+y+y^2ight) \geq 0 \ & 1+y^2-2 y-2-3 y-y^2 \geq 0 \ & -5 y-1 \geq 0 \ & y \leq-\frac{1}{5} \end{array} $ So, $\quad \frac{f}{g}(x) \in\left(-\infty,-\frac{1}{5}ight]$ So, $\quad a=1, b=0$ and $c=-\frac{1}{5}$ Hence, $ \begin{aligned} a+2 b+5 c+4 & \ & =1+0+5\left(-\frac{1}{5}ight)+4 \ & =1-1+4=4 \end{aligned} $

Let $f(x)=x^2+2 x+2, g(x)=-x^2+2 x-1$ and $a, b$ be the extreme values of $f(x), g(x)$ respectively. If $c$…

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