Consider the function f (x) = ta n ^ - 1 \ (3x - 2)/(3 + 2x)\ , x ≥ 0 . If g(x) is the inverse function of f (x), then the value of g ^ ′ ((π)/(4)) is equal to

Correct answer: 26 f(x) = ta n ^ - 1 \ (x - (2)/(3))/(1 + x . (2)/(3))\ = ta n ^ - 1 x - ta n ^ - 1 (2)/(3) ⇒ f ^ ' (x) = 1 1 + x ^ 2 . . . (1) Now, g ^ ' (f(x)) = 1 f ^ ' (x) \ ∵ g(f(x)) = x\ Putting x = 5 , we get, g ^ ' (f(5)) = 1 f ^ ' (5) ⇒ g ^ ' ( ^ - 1 ((3(5) - 2)/(3 + 2(5)))) = 1 ( 1 1 + 5 ^ 2 ) ⇒ g ^ ' ( ^ - 1 (1)) = 26 ⇒ g ^ ' ( π 4 ) = 26

NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Differentiation

Consider the function f (x) = ta n ^ - 1 \ (3x - 2)/(3 + 2x)\ , x ≥...

Consider the function

f

\left(x\right) = ta{n}^{- 1}

\left\{\dfrac{3x - 2}{3 + 2x}\right\}, \forall x \ge 0 .

g\left(x\right)

is the inverse function of

f

\left(x\right),

then the value of

{g}^{′}

\left(\dfrac{\pi}{4}\right)

is equal to

Correct answer

Step-by-step solution

f\left(x\right) = ta{n}^{- 1}\left\{\dfrac{x - \dfrac{2}{3}}{1 + x . \dfrac{2}{3}}\right\} = ta{n}^{- 1}x - ta{n}^{- 1}\dfrac{2}{3}

⇒ {f}^{'}\left(x\right) = \dfrac{1}{1 + {x}^{2}} . . . \left(1\right)

Now,

{g}^{'}\left(f\left(x\right)\right) = \dfrac{1}{{f}^{'}\left(x\right)}\left\{∵ g\left(f\left(x\right)\right) = x\right\}

}
Putting

x = 5

, we get,

{g}^{'}\left(f\left(5\right)\right) = \dfrac{1}{{f}^{'}\left(5\right)} ⇒ {g}^{'}\left({\text\tan}^{- 1}\left(\dfrac{3\left(5\right) - 2}{3 + 2\left(5\right)}\right)\right) = \dfrac{1}{\left(\dfrac{1}{1 + {5}^{2}}\right)}

⇒ {g}^{'}\left({\text\tan}^{- 1}\left(1\right)\right) = 26 ⇒ {g}^{'}\left(\dfrac{\text{\pi}}{4}\right) = 26

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