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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Differentiation

NTA Abhyas 2020 · Mathematics · Differentiation

If the function f(x) = ^ - 1 (x(3)/(2) - 1 - x - x ^ 2 + x ^ 3 )...

If the function f(x)=cos⁡−1(x32−1−x−x2+x3)f\left(x\right) = \cos^{- 1} \left(x\dfrac{3}{2} - \sqrt{1 - x - {x}^{2} + {x}^{3}}\right) (where, ∀0<x<1\forall 0 < x < 1), then the value of \left|\sqrt{3}{f}^{'}\left(\dfrac{1}{2}\right)\right| is equal to (take 3=1.73\sqrt{3} = 1.73)

Correct answer

3.73

Step-by-step solution

f\left(x\right) = \cos^{- 1} \left\{x\sqrt{x} - \sqrt{\left(1 - x\right)\left(1 - {x}^{2}\right)}\right\}<br/><br />= \cos^{- 1} \left\{x\sqrt{x} - \sqrt{1 - {\left(\sqrt{x}\right)}^{2}}\sqrt{1 - {x}^{2}}\right\}<br/><br />= \cos^{- 1} x + \cos^{- 1} \sqrt{x}<br/>Now,<br />Now,{f}^{′} \left(x\right) = \dfrac{- 1}{\sqrt{1 - {x}^{2}}} - \dfrac{1}{\sqrt{1 - x}} ⋅ \dfrac{1}{2\sqrt{x}}<br/><br />{f}^{'}\left(\dfrac{1}{2}\right) = \dfrac{- 1}{\sqrt{1 - \dfrac{1}{4}}} - \dfrac{1}{\sqrt{\dfrac{1}{2}}} \times \dfrac{1}{2\sqrt{\dfrac{1}{2}}} = - \dfrac{2}{\sqrt{3}} - 1</p><p>∣3f′(12)∣=∣−3(−23−1)∣=3.73</p><p>\left|\sqrt{3}{f}^{'}\left(\dfrac{1}{2}\right)\right| = \left|- \sqrt{3}\left(- \dfrac{2}{\sqrt{3}} - 1\right)\right| = 3 . 73

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