Functions
Min and max of composed function; range of product
MJMT_Full_Test_01
Grade 12
Question:
Let $f(x)=\frac{1}{2}\begin{vmatrix}1&\tan x&1\\-\tan x&1&\tan x\\-1&-\tan x&1\end{vmatrix}+\begin{vmatrix}\cot\frac{\pi}{2}&\sec(x+\frac{\pi}{3})&\sec(x+\frac{\pi}{12})\\\csc(x-\frac{\pi}{6})&\sin2024\pi&e^{i2024\pi}\\\csc(x-\frac{5\pi}{12})&e^{2025\pi}&\tan(2025\pi)\end{vmatrix}$ and $g(x)=\sqrt{f(x)-1}+\sqrt{f(2025\pi/2-x)-1}$ on $(0,\pi/2)$. Let $m$ be minimum of $f(x)$ and $M$ minimum of $g(x)$. Range of $h(x)=(x-m)(x-M)$ on $[0,3]$ is
$[-\frac{1}{4},\infty)$
$[-\frac{1}{4},2]$
$[-2,4]$
$[\frac{3}{2},4]$
Step-by-Step Solution
Key Concept: $f(x)=\sec^2x$ (first det$=1+\tan^2x$; second det$=0$). $g(x)=|\tan x|+|\cot x|\geq2$: min $M=2$. $m=\min f=1$. $h(x)=(x-1)(x-2)$.
Range of $h(x)=[-1/4,2]$.
Correct Answer: 2