Functions
Domain & Range
MMTS_Full_Test_04
Grade 12
Question:
Let $f:A\to B$ be a function defined by $f(x)=\sqrt{3}\sin x+\cos x+4$. If $f$ is both one-one and onto, then
$A=[-\pi/6,5\pi/6]$, $B=[2,6]$
$A=[-\pi/3,2\pi/3]$, $B=[2,6]$
$A=[0,\pi]$, $B=[2,6]$
$A=[-\pi/6,5\pi/6]$, $B=[2,6]$
Step-by-Step Solution
Key Concept: $f(x)=2(\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x)+4=2\sin(x+\pi/6)+4$; range $[2,6]$ on suitable domain
$f=2\sin(x+\pi/6)+4$. For bijection: $x+\pi/6\in[-\pi/2,\pi/2]\Rightarrow x\in[-2\pi/3,\pi/3]$... or $x+\pi/6\in[-\pi/2,3\pi/2]$: full period. For $[-\pi/6,5\pi/6]$: $x+\pi/6\in[0,\pi]$: one arch; range$=[2,6]$. ✓
Correct Answer: 1