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JEE Main 202429 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
29 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Differentiation

JEE Main 2024 · 29 January · Morning Shift · Mathematics · Differentiation

Suppose f(x) = ( 2 ^ x + 2 ^ - x ) x ^ - 1 ( x ^ 2 - x + 1) (7 x ^...

Suppose f(x)=(2x+2−x)tan⁡xtan⁡−1(x2−x+1)(7x2+3x+1)3f\left(x\right) = \dfrac{\left({2}^{x} + {2}^{- x}\right)\tan x\sqrt{\tan^{- 1}\left({x}^{2} - x + 1\right)}}{{\left(7{x}^{2} + 3x + 1\right)}^{3}}. Then the value of f′(0){f}^{'}\left(0\right) is equal to

Options

  1. Aπ\pi
  2. B00
  3. Cπ\sqrt{\pi}
  4. Dπ2\dfrac{\pi}{2}

Correct answer

C. π\sqrt{\pi}

Step-by-step solution

Given: f(x)=(2x+2−x)tan⁡xtan⁡−1(x2−x+1)(7x2+3x+1)3f\left(x\right) = \dfrac{\left({2}^{x} + {2}^{- x}\right)\tan x\sqrt{\tan^{- 1}\left({x}^{2} - x + 1\right)}}{{\left(7{x}^{2} + 3x + 1\right)}^{3}}

We know that, f′(0)=lim⁡h→0f(h)−f(0)h{f}^{'}\left(0\right) = \lim_{h \to 0}\dfrac{f\left(h\right) - f\left(0\right)}{h}

⇒f′(0)=lim⁡h→0(2h+2−h)tanhtan⁡−1(h2−h+1)(7h2+3h+1)3−(20+20)tan⁡0tan⁡−1(0−0+1)(0+0+1)3h⇒ {f}^{'}\left(0\right) = \lim_{h \to 0}\dfrac{\dfrac{\left({2}^{h} + {2}^{- h}\right)tanh\sqrt{\tan^{- 1}\left({h}^{2} - h + 1\right)}}{{\left(7{h}^{2} + 3h + 1\right)}^{3}} - \dfrac{\left({2}^{0} + {2}^{0}\right)\tan 0\sqrt{\tan^{- 1}\left(0 - 0 + 1\right)}}{{\left(0 + 0 + 1\right)}^{3}}}{h}

⇒f′(0)=lim⁡h→0(2h+2−h)tanhtan⁡−1(h2−h+1)h(7h2+3h+1)3⇒ {f}^{'}\left(0\right) = \lim_{h \to 0}\dfrac{\left({2}^{h} + {2}^{- h}\right)tanh\sqrt{\tan^{- 1}\left({h}^{2} - h + 1\right)}}{h{\left(7{h}^{2} + 3h + 1\right)}^{3}}

⇒f′(0)=lim⁡h→0(2h+2−h)tan⁡−1(h2−h+1)(7h2+3h+1)3⇒ {f}^{'}\left(0\right) = \lim_{h \to 0}\dfrac{\left({2}^{h} + {2}^{- h}\right)\sqrt{\tan^{- 1}\left({h}^{2} - h + 1\right)}}{{\left(7{h}^{2} + 3h + 1\right)}^{3}}

⇒f′(0)=(20+20)tan⁡−1(1)(0+1)3⇒ {f}^{'}\left(0\right) = \dfrac{\left({2}^{0} + {2}^{0}\right)\sqrt{\tan^{- 1}\left(1\right)}}{{\left(0 + 1\right)}^{3}}

⇒f′(0)=(2)π4⇒ {f}^{'}\left(0\right) = \left(2\right)\sqrt{\dfrac{\pi}{4}}

⇒f′(0)=π⇒ {f}^{'}\left(0\right) = \sqrt{\pi}

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