Functions
Functional equation; integer count satisfying inequality
MMTS_Full_Test_12
Grade 12
Question:
Let $f(x)=2\tan^{-1}x$ and $g(x)$ differentiable with $g\!\left(\dfrac{x+2y}{3}\right)=\dfrac{g(x)+2g(y)}{3}$, $g'(0)=1$, $g(0)=2$. Number of integers $x$ in $(-10,20)$ satisfying $f^2(g(x))-5f(g(x))+4>0$ is
Step-by-Step Solution
Key Concept: Functional equation gives $g$ linear: $g(x)=x+2$. Inequality: $(f(g(x))-1)(f(g(x))-4)>0$. $f=2\tan^{-1}$: $f<1$ or $f>4$. $f>4$ impossible (range$=(-\pi,\pi)$). $f<1\Rightarrow 2\tan^{-1}(x+2)<1\Rightarrow x+2<\tan(1/2)\approx0.546\Rightarrow x<-1.45$. Count integers in $(-10,20)\cap(-\infty,-1.45)$: $\{-9,-8,...,-2\}=8$.
8 integers.
Correct Answer: (D) 8