Let f x be a real valued function defined on -1-1 and e - x f x -2-0 x - t 4-1 dt- Then the value of f 10 isA-B- 0C- 1D- 3

The correct option is D 3e^ xf(x)=2+∫_0^x√(t^4+1)dt e^ x(f^1(x) f(x))=√(x^4+1)⇒ f^1(0)=3 ...

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Let f x be a ...

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Let f(x) be a real valued function defined on (-1, 1) and $e^{- x} f (x) = 2 + \int_{0}^{x} \sqrt{t^{4} + 1} d t$ . Then the value of $f^{1} (0)$ is

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\frac{1}{3}

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Let f(x) be a real valued function defined on (-1,1) and $e^{- x} f (x) = 2 + \int_{0}^{x} \sqrt{t^{4} + d t}$ . Then the value of $f^{'} (0)$ is

f

(- 1, 1)

e^{- x} f (x) = 2 + \int_{0}^{x} \sqrt{t^{4} + 1} d t

x \in (- 1, 1)

f^{- 1}

f

(f^{- 1})^{'} (2)

e^{- x} . f (x) = 2 + \int_{0}^{x} \sqrt{t^{4} + 1} d t \forall x ϵ (- 1, 1)

g^{1} (2) =

e^{- x} \cdot f (x) = 2 + \int_{0}^{x} \sqrt{t^{4} + 1} d t \forall ϵ (- 1, 1)

f^{- 1}

(f^{- 1})^{'} (2)

f

(0, \infty)

f (x) = ln x + \int_{0}^{x} \sqrt{1 + sin t} d t

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