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 (where, ), then the value of \left|\sqrt{3}{f}^{'}\left(\dfrac{1}{2}\right)\right| is equal to (take )
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\}= \cos^{- 1} \left\{x\sqrt{x} - \sqrt{1 - {\left(\sqrt{x}\right)}^{2}}\sqrt{1 - {x}^{2}}\right\}= \cos^{- 1} x + \cos^{- 1} \sqrt{x}{f}^{′} \left(x\right) = \dfrac{- 1}{\sqrt{1 - {x}^{2}}} - \dfrac{1}{\sqrt{1 - x}} ⋅ \dfrac{1}{2\sqrt{x}}{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