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)

4

Answer 2

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, $a+2 b+5 c+4$ $=1+0+5\left(-\frac{1}{5}\right)+4$ $=1-1+4=4$