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JEE Main 20202 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
2 September
Shift
Evening Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2020 · 2 September · Evening Shift · Mathematics · Inverse Trigonometric Functions

If y = ∑ _ k = 1 ^ 6 k ^ - 1 \ (3)/(5)coskx - (4)/(5)sinkx\ then...

If y=∑k=16kcos⁡−1{35coskx−45sinkx}y = \mathop{∑}_{k = 1}^{6}k\cos^{- 1}\left\{\dfrac{3}{5}coskx - \dfrac{4}{5}sinkx\right\} then dydx\dfrac{dy}{dx} at x=0x = 0is

Correct answer

91

Step-by-step solution

For, small values near 00, i.e., \left|x\right| < < . . 1</p><p></p><p>y = \mathop{∑}_{k = 1}^{6}k\cos^{- 1}\left(coskx \cdot cos\alpha - sinkx \cdot sin\alpha\right)

= \mathop{∑}_{k = 1}^{6}k\cos^{- 1}\left(\cos \left(kx + \alpha\right)\right)<br/><br />= \mathop{∑}_{k = 1}^{6}k \left(kx + \alpha\right)</p><p></p><p>= \mathop{∑}_{k = 1}^{6}\left({k}^{2}x + k\alpha\right)

\dfrac{dy}{dx} =\mathop{∑}_{k = 1}^{6}{k}^{2} = \dfrac{6\left(7\right)\left(13\right)}{6} = 91

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