Let a function f(x) be defined as f(x) = ∑_ k=1 ^ 8 cosec((kπ)/(3) + x) cosec(((k+1)π)/(3) + x) . The value of |f((π)/(12))| is equal to :

Correct answer: D. 4 The given function is f(x) = ∑_ k=1 ^ 8 (1)/( ((kπ)/(3) + x) (((k+1)π)/(3) + x)) . The difference between the angles in the denominator is (((k+1)π)/(3) + x) - ((kπ)/(3) + x) = (π)/(3) . Multiply and divide the expression by ((π)/(3)) : f(x) = (1)/( ((π)/(3))) ∑_ k=1 ^ 8 ( ((((k+1)π)/(3) + x) - ((kπ)/(3) + x)))/( ((kπ)/(3) + x) (((k+1)π)/(3) + x)) Using the identity (A - B) = A B - A B , each term in the sum splits into a difference of cotangents: f(x) = 2 3 ∑_ k=1 ^ 8 [ ((kπ)/(3) + x) - (((k+1)π)/(3) + x) ] This is a telescoping series, so all intermediate terms cancel out, leaving the first and last terms: f(x) = 2 3 [ ((π)/(3) + x) - ((9π)/(3) + x) ] = 2 3 [ ((π)/(3) + x) - (3π + x) ] Now, substitute x = (π)/(12) : f((π)/(12)) = 2 3 [ ((π)/(3) + (π)/(12)) - (3π + (π)/(12)) ] f((π)/(12)) = 2 3 [ ((5π)/(12)) - ((π)/(12)) ] Using standard trigonometric values, ((5π)/(12)) = (75^ ) = 2 - 3 and ((π)/(12)) = (15^ ) = 2 + 3 . f((π)/(12)) = 2 3 [ (2 - 3 ) - (2 + 3 ) ] = 2 3 (-2 3 ) = -4 Therefore, |f((π)/(12))| = |-4| = 4 . Answer: 4

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

Let a function f(x) be defined as f(x) = ∑_ k=1 ^ 8 cosec((kπ)/(3)...

Let a function

f(x)

be defined as

f(x) = \sum_{k=1}^{8} \text{cosec}\left(\dfrac{k\pi}{3} + x\right) \text{cosec}\left(\dfrac{(k+1)\pi}{3} + x\right)

. The value of

|f(\dfrac{\pi}{12})|

is equal to :

Options

A $2\sqrt{3}$
B $4\sqrt{3}$
C $\dfrac{8}{\sqrt{3}}$
D $4$

Correct answer

D. $4$

Step-by-step solution

The given function is

f(x) = \sum_{k=1}^{8} \dfrac{1}{\sin\left(\dfrac{k\pi}{3} + x\right) \sin\left(\dfrac{(k+1)\pi}{3} + x\right)}

.

The difference between the angles in the denominator is \left(\dfrac{(k+1)\pi}{3} + x\right) - \left(\dfrac{k\pi}{3} + x\right) = \dfrac{\pi}{3}

.<br><br>Multiply and divide the expression by

\sin\left(\dfrac{\pi}{3}\right)

:<br><br>

f(x) = \dfrac{1}{\sin\left(\dfrac{\pi}{3}\right)} \sum_{k=1}^{8} \dfrac{\sin\left(\left(\dfrac{(k+1)\pi}{3} + x\right) - \left(\dfrac{k\pi}{3} + x\right)\right)}{\sin\left(\dfrac{k\pi}{3} + x\right) \sin\left(\dfrac{(k+1)\pi}{3} + x\right)}

<br><br>Using the identity

\sin(A - B) = \sin A \cos B - \cos A \sin B

, each term in the sum splits into a difference of cotangents:<br><br>

f(x) = \dfrac{2}{\sqrt{3}} \sum_{k=1}^{8} \left[ \cot\left(\dfrac{k\pi}{3} + x\right) - \cot\left(\dfrac{(k+1)\pi}{3} + x\right) \right]

<br><br>This is a telescoping series, so all intermediate terms cancel out, leaving the first and last terms:<br><br>

f(x) = \dfrac{2}{\sqrt{3}} \left[ \cot\left(\dfrac{\pi}{3} + x\right) - \cot\left(\dfrac{9\pi}{3} + x\right) \right] = \dfrac{2}{\sqrt{3}} \left[ \cot\left(\dfrac{\pi}{3} + x\right) - \cot(3\pi + x) \right]

<br><br>Now, substitute

x = \dfrac{\pi}{12}

:<br><br>

f\left(\dfrac{\pi}{12}\right) = \dfrac{2}{\sqrt{3}} \left[ \cot\left(\dfrac{\pi}{3} + \dfrac{\pi}{12}\right) - \cot\left(3\pi + \dfrac{\pi}{12}\right) \right]

<br><br>

f\left(\dfrac{\pi}{12}\right) = \dfrac{2}{\sqrt{3}} \left[ \cot\left(\dfrac{5\pi}{12}\right) - \cot\left(\dfrac{\pi}{12}\right) \right]

<br><br>Using standard trigonometric values,

\cot\left(\dfrac{5\pi}{12}\right) = \cot(75^{\circ}) = 2 - \sqrt{3}

and

\cot\left(\dfrac{\pi}{12}\right) = \cot(15^{\circ}) = 2 + \sqrt{3}

.<br><br>

f\left(\dfrac{\pi}{12}\right) = \dfrac{2}{\sqrt{3}} \left[ (2 - \sqrt{3}) - (2 + \sqrt{3}) \right] = \dfrac{2}{\sqrt{3}} (-2\sqrt{3}) = -4

<br><br>Therefore, \left|f\left(\dfrac{\pi}{12}\right)\right| = |-4| = 4

.

Answer:

4

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