Exams Joint Entrance Examination Mathematics Limits, Continuity and Differentiability Question 222 Question 222 of 401 Medium MCQ
If f ( x ) = sin − 1 ( 2 ⋅ 3 x 1 + 9 x ) f(x)=\sin^{-1}\left(\dfrac{2\cdot3^x}{1+9^x}\right) f ( x ) = sin − 1 ( 1 + 9 x 2 ⋅ 3 x ) , then f ′ ( − 1 2 ) f'\left(-\dfrac12\right) f ′ ( − 2 1 ) equals.
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Let 3 x = tan t 3^x=\tan t 3 x = tan t . Then f ( x ) = sin − 1 ( sin 2 t ) = 2 t = 2 tan − 1 ( 3 x ) f(x)=\sin^{-1}(\sin2t)=2t=2\tan^{-1}(3^x) f ( x ) = sin − 1 ( sin 2 t ) = 2 t = 2 tan − 1 ( 3 x ) .
\[f'(x)=\frac{2}{1+9^x}\cdot3^x\ln3\]
At x = − 1 2 x=-\dfrac12 x = − 2 1 : f ′ ( − 1 2 ) = 2 ⋅ 3 − 1 / 2 ln 3 1 + 3 − 1 = 2 ln 3 / 3 4 / 3 = 3 2 ln 3 = 3 ln 3 f'\left(-\dfrac12\right)=\dfrac{2\cdot3^{-1/2}\ln3}{1+3^{-1}}=\dfrac{2\ln3/\sqrt{3}}{4/3}=\dfrac{\sqrt{3}}{2}\ln3=\sqrt{3}\ln\sqrt{3} f ′ ( − 2 1 ) = 1 + 3 − 1 2 ⋅ 3 − 1/2 ln 3 = 4/3 2 ln 3/ 3 = 2 3 ln 3 = 3 ln 3 . Answer the question to unlock