Sets, Relations & Functions
Range of Trigonometric Function
nta_pyq_2025_apr
Grade 11
Question:
Let the range of the function $f(x) = 6 + 16\cos x \cdot \cos\!\left(\frac{\pi}{3} - x\right) \cdot \cos\!\left(\frac{\pi}{3} + x\right) \cdot \sin 3x \cdot \cos 6x$, $x \in \mathbb{R}$, be $[\alpha, \beta]$. Then the distance of the point $(\alpha, \beta)$ from the line $3x + 4y + 12 = 0$ is:
Step-by-Step Solution
Key Concept: Use product-to-sum identity: $\cos(\pi/3-x)\cos(\pi/3+x) = \frac{1}{4}(1+\cos 2x)$... actually use $16\cos x \cdot \cos(\pi/3-x)\cos(\pi/3+x) = 4\cos 3x$, then $4\cos3x\cdot\sin3x\cdot\cos6x = \sin12x$.
$f(x) = 6 + \sin12x$. Range $= [5, 7]$, so $(\alpha,\beta)=(5,7)$. Distance from $3x+4y+12=0$: $\frac{|3(5)+4(7)+12|}{5} = \frac{|15+28+12|}{5} = \frac{55}{5} = 11$.
Correct Answer: 11