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JEE Main 202224 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
24 June
Shift
Morning Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2022 · 24 June · Morning Shift · Mathematics · Inverse Trigonometric Functions

The set of all values of k for which ( ^ - 1 x) ^ 3 + ( ^ - 1 x) ^...

The set of all values of kk for which (tan⁡−1x)3+(cot⁡−1x)3=kπ3,x∈R{\left(\tan^{- 1}x\right)}^{3} + {\left(\cot^{- 1}x\right)}^{3} = k{\pi}^{3}, x \in R, is the interval

Options

  1. A[132,78)\left[\dfrac{1}{32}, \dfrac{7}{8}\right)
  2. B(124,1316)\left(\dfrac{1}{24}, \dfrac{13}{16}\right)
  3. C[148,1316]\left[\dfrac{1}{48}, \dfrac{13}{16}\right]
  4. D[132,98)\left[\dfrac{1}{32}, \dfrac{9}{8}\right)

Correct answer

A. [132,78)\left[\dfrac{1}{32}, \dfrac{7}{8}\right)

Step-by-step solution

Let T=(tan⁡−1x)3+(cot⁡−1x)3T = {\left(\tan^{- 1}x\right)}^{3} + {\left(\cot^{- 1}x\right)}^{3}

=(tan⁡−1x+cot⁡−1x)3−3tan⁡−1x⋅cot⁡−1x(tan⁡−1x+cot⁡−1x)= {\left(\tan^{- 1}x + \cot^{- 1}x\right)}^{3} - 3\tan^{- 1}x \cdot \cot^{- 1}x\left(\tan^{- 1}x + \cot^{- 1}x\right)

=π38−3π2tan⁡−1x(π2−tan⁡−1x)= \dfrac{{\pi}^{3}}{8} - \dfrac{3\pi}{2}\tan^{- 1}x\left(\dfrac{\pi}{2} - \tan^{- 1}x\right)\left(∵ \tan^{- 1}x + \cot^{- 1}x = \dfrac{\pi}{2}\right)</p><p></p><p>= \dfrac{3\pi}{2}{\left(\tan^{- 1}x\right)}^{2} - \dfrac{3{\pi}^{2}}{4}\tan^{- 1}x + \dfrac{{\pi}^{3}}{8}</p><p></p><p>T = \dfrac{3\pi}{2}{\left(\tan^{- 1}x - \dfrac{\pi}{4}\right)}^{2} + \dfrac{{\pi}^{3}}{32}</p><p>Weknowthan</p><p>We know than\tan^{- 1}x \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)</p><p>Minimumvalueof</p><p>Minimum value ofTisatis at\tan^{- 1}x = \dfrac{\pi}{4} and maximum value at tan⁡−1x=−π2\tan^{- 1}x = - \dfrac{\pi}{2}

⇒π332≤T<78π3⇒ \dfrac{{\pi}^{3}}{32} \le T < \dfrac{7}{8}{\pi}^{3}

=π332≤kπ3<78π3= \dfrac{{\pi}^{3}}{32} \le k{\pi}^{3} < \dfrac{7}{8}{\pi}^{3}

⇒132≤k<78⇒ \dfrac{1}{32} \le k < \dfrac{7}{8}

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