Let -f- be a real valued function defined on the interval -1-1 such that e - x - f x -2-0 x - t 4-1 dt - x -1-1 and let -g- be the inverse funciton of -f- Then g12-A- 3 B- 1-2 C- 1-3 D- 2

The correct option is C Differentiating given equation we get e^ x.f^1(x) e^ x.f(x)=√(1+x^4) Since (g of)(x)=x as 'g' is inverse of f ⇒ g[f(x)]=x ⇒ g^1[f(x) ...

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Standard XII

Mathematics

Term Independent of x

Let 'f' be a ...

Question

Let 'f' be a real valued function defined on the interval (-1, 1) such that $e^{- x} . f (x) = 2 + \int_{0}^{x} \sqrt{t^{4} + 1} d t \forall x ϵ (- 1, 1)$ and let 'g' be the inverse funciton of 'f'. Then $g^{1} (2) =$

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Solution

The correct option is C
Differentiating given equation we get
$e^{- x} . f^{1} (x) - e^{- x} . f (x) = \sqrt{1 + x^{4}} S i n c e (g o f) (x) = x a s^{'} g^{'} i s i n v e r s e o f f \Rightarrow g [f (x)] = x \Rightarrow g^{1} [f (x)] . f^{1} (x) = 1 \Rightarrow g^{1} [f (0)] = \frac{1}{f^{1} (0)} \Rightarrow g^{1} (2) = \frac{1}{f^{1} (0)}$
(Here f(0) = 2 observe from hypothesis)
Put x = 0 in (1) we get $f^{1} (0) = 3$

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Let 'f' be a real valued function defined on the interval (-1, 1) such that e−x.f(x)=2+∫x0√t4+1dt ∀xϵ(−1,1) and let 'g' be the inverse funciton of 'f'. Then g1(2)=

The correct option is C Differentiating given equation we get e−x.f1(x)−e−x.f(x)=√1+x4Since (g of)(x)=x as ′g′ is inverse of f⇒g[f(x)]=x⇒g1[f(x)].f1(x)=1⇒g1[f(0)]=1f1(0)⇒g1(2)=1f1(0) (Here f(0) = 2 observe from hypothesis) Put x = 0 in (1) we get f1(0)=3

Let 'f' be a real valued function defined on the interval (-1, 1) such that $e^{- x} . f (x) = 2 + \int_{0}^{x} \sqrt{t^{4} + 1} d t \forall x ϵ (- 1, 1)$ and let 'g' be the inverse funciton of 'f'. Then $g^{1} (2) =$