If fx-x- gx-ex-1- and -fogx d x-A cogx-B tan -1fogx-C- then A-B is equal toA- 2B- 1C- 3D- 0

The correct option is D 0fog(x) = √(e^x 1) ∴ I = ∫√(e^x 1)dx = ∫2t^2/t^2 + 1dt { where t= √(e^x 1)} = 2t 2 tan^ 1 t+C = 2 √(e^x 1) 2tan^ 1 (√(e^x ...

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Standard XII

Mathematics

Standard Formulae - 2

If fx=√x, gx=...

Question

If $f (x) = \sqrt{x}, g (x) = e^{x - 1}, a n d \int f o g (x) d x = A f o g (x) + B t a n^{- 1} (f o g (x)) + C$ , then A + B is equal to

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Solution

$f o g (x) = \sqrt{e^{x} - 1} ∴ I = \int \sqrt{e^{x} - 1} d x = \int \frac{2 t^{2}}{t^{2} + 1} d t {w h e r e t = \sqrt{e^{x} - 1}} = 2 t - 2 t a n^{- 1} t + C = 2 \sqrt{e^{x} - 1} - 2 t a n^{- 1} (\sqrt{e^{x} - 1}) + C = 2 f o g (x) - 2 t a n^{- 1} (f o g (x)) + C ∴ A + B = 2 + (- 2) = 0$

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Similar questions

Q. If

f (x) = \sqrt{x}, g (x) = e^{x - 1}, a n d \int f o g (x) d x = A f o g (x) + B t a n^{- 1} (f o g (x)) + C

, then A + B is equal to

Q. If

f (x) = \sqrt{x}, g (x) = e^{x - 1}, a n d \int f o g (x) d x = A f o g (x) + B t a n^{- 1} (f o g (x)) + C

, then A + B is equal to

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f (x) = \sqrt{x}, g (x) = e^{x} - 1 \forall x \in (0, \infty)

and

\int f o g (x) d x = A f o g (x) + B {tan}^{- 1} (f o g (x)) + C

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f (x) = \sqrt{x}, g (x) = e^{x} - 1 \forall x \in (0, \infty)

and

\int f o g (x) d x = A f o g (x) + B {tan}^{- 1} (f o g (x)) + C

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A + B

$I f f (x) = x^{n} l n x, x > 0 = 0 x = 0 a n d R o l l e^{'} s t h e o r e m i s a p p l i c a l b e t o f (x) f o r x ϵ [0, 1] t h e n n m a y b e e q u a l t o$

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If f(x)=√x,g(x)=ex−1,and∫fog(x)dx=Afog(x)+Btan−1(fog(x))+C, then A + B is equal to

fog(x)=√ex−1∴I=∫√ex−1dx=∫2t2t2+1dt{where t=√ex−1}=2t−2tan−1t+C=2√ex−1−2tan−1(√ex−1)+C=2fog(x)−2tan−1(fog(x))+C∴A+B=2+(−2)=0

If $f (x) = \sqrt{x}, g (x) = e^{x - 1}, a n d \int f o g (x) d x = A f o g (x) + B t a n^{- 1} (f o g (x)) + C$ , then A + B is equal to

$f o g (x) = \sqrt{e^{x} - 1} ∴ I = \int \sqrt{e^{x} - 1} d x = \int \frac{2 t^{2}}{t^{2} + 1} d t {w h e r e t = \sqrt{e^{x} - 1}} = 2 t - 2 t a n^{- 1} t + C = 2 \sqrt{e^{x} - 1} - 2 t a n^{- 1} (\sqrt{e^{x} - 1}) + C = 2 f o g (x) - 2 t a n^{- 1} (f o g (x)) + C ∴ A + B = 2 + (- 2) = 0$