Let the range of the function f(x) = 6 + 16 x ( 3 - x) ( 3 + x) 3x 6x, x R be [, ]. Then the distance of the point (,) from the line 3x + 4y + 12 = 0 is:

Given: f(x) = 6 + 16 x ( 3 - x) ( 3 + x) 3x 6x Find: The distance of the point (,), where [, ] is the range of f(x), from the line 3x + 4y + 12 = 0. Using the identity from the solution, x ( 3 - x) ( 3 + x) = 14 3x we get f(x) = 6 + 16 14 3x 3x 6x = 6 + 4 3x 3x 6x Now use 2 3x 3x = 6x so 4 3x 3x 6x = 2 6x 6x = 12x Hence, f(x) = 6 + 12x Since -1 12x 1 it follows that 5 f(x) 7 Therefore, [, ] = [5, 7] so the point is (5, 7). Distance from the line 3x + 4y + 12 = 0 is |3(5) + 4(7) + 12| 3^2 + 4^2 = |15 + 28 + 12| 9 + 16 = 555 = 11 Therefore, the distance is 11, so the correct option is A. The first approach in the solution contains inconsistent intermediate remarks such as [-10,10] and 225, but the second approach gives the coherent simplification and correct final result. Therefore, the answer is taken from the valid working.

Let the range of the function f(x) = 6 + 16 x ( 3 - x) ( 3 + x) 3… — Solution | JEE Mathematics

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Let the range of the function $f(x) = 6 + 16 \cos x \cdot \cos\left(\frac{\pi}{3} - x\right) \cdot \cos\left(\frac{\pi}{3} + x\right) \cdot \sin 3x \cdot \cos 6x, \quad x \in R$ be $[\alpha, \beta]$ . Then the distance of the point $(\alpha, \beta)$ from the line $3x + 4y + 12 = 0$ is:

A
$11$
B
$8$
C
$10$
D
$9$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

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

Find: The distance of the point $(\alpha, \beta)$ , where $[\alpha, \beta]$ is the range of $f(x)$ , from the line $3x + 4y + 12 = 0$ .

Using the identity from the solution,

\cos x \cos\left(\frac{\pi}{3} - x\right) \cos\left(\frac{\pi}{3} + x\right) = \frac{1}{4} \cos 3x

we get

f(x) = 6 + 16 \cdot \frac{1}{4} \cos 3x \cdot \sin 3x \cdot \cos 6x

= 6 + 4 \cos 3x \sin 3x \cos 6x

Now use

2 \sin 3x \cos 3x = \sin 6x

4 \cos 3x \sin 3x \cos 6x = 2 \sin 6x \cos 6x = \sin 12x

Hence,

f(x) = 6 + \sin 12x

Range and Distance Calculation

Since

-1 \le \sin 12x \le 1

it follows that

5 \le f(x) \le 7

Therefore,

[\alpha, \beta] = [5, 7]

so the point is $(5, 7)$ .

Distance from the line $3x + 4y + 12 = 0$ is

\frac{|3(5) + 4(7) + 12|}{\sqrt{3^2 + 4^2}}

= \frac{|15 + 28 + 12|}{\sqrt{9 + 16}}

= \frac{55}{5} = 11

Therefore, the distance is $11$ , so the correct option is A.

The first approach in the solution contains inconsistent intermediate remarks such as $[-10,10]$ and $\frac{22}{5}$ , but the second approach gives the coherent simplification and correct final result. Therefore, the answer is taken from the valid working.

Common mistakes

A common mistake is not using the identity for $\cos x \cos\left(\frac{\pi}{3}-x\right) \cos\left(\frac{\pi}{3}+x\right)$ correctly. This keeps the expression unnecessarily complicated. Instead, reduce this triple product first to simplify the range calculation.
Students may find the range of the trigonometric product directly without first showing that $f(x) = 6 + \sin 12x$ . That is wrong because the original product does not make the extrema obvious. Always convert to a single sine expression before taking the range.
Another mistake is using the point-line distance formula incorrectly by forgetting the modulus in the numerator or the square root in the denominator. Use $\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$ for the line $Ax + By + C = 0$ .

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