If ∫ex+e−x+ex−e−xsin⁡x1+cos⁡xdx=f(x)tan⁡x2+c then which of the following is false ?

I=∫ex+e−x+ex−e−x2sin⁡x/2cos⁡x/22cos2⁡x/2dx=∫ex+e−x2sec2⁡x/2dx+∫ex−e−xtan⁡x2dx=ex+e−xtan⁡x2−∫ex−e−xtan⁡x2dx+∫ex−e−xtan⁡x2dx=ex+e−xtan⁡x2+C f(x)=ex+e−x

If ∫ex+e−x+ex−e−xsin⁡x1+cos⁡xdx=f(x)tan⁡x2+c then which of the following is false ?

I=∫ex+e−x+ex−e−x2sin⁡x/2cos⁡x/22cos2⁡x/2dx=∫ex+e−x2sec2⁡x/2dx+∫ex−e−xtan⁡x2dx=ex+e−xtan⁡x2−∫ex−e−xtan⁡x2dx+∫ex−e−xtan⁡x2dx=ex+e−xtan⁡x2+C f(x)=ex+e−x

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If $\int \frac{(e^{x} + e^{- x}) + (e^{x} - e^{- x}) \sin x}{1 + \cos x} dx = f (x) \tan (\frac{x}{2}) + c$ then which of the following is false ?

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f (x) is even

f (x) is differentiable

range of f (x) is [2, $\infty$ )

f (x) is always increasing on R

answer is B.

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Detailed Solution

$\begin{array}{l} I = \int \frac{(e^{x} + e^{- x}) + (e^{x} - e^{- x}) 2 \sin x / 2 \cos x / 2}{2 \cos^{2} x / 2} dx \\ = \int \frac{(e^{x} + e^{- x})}{2} \sec^{2} x / 2 dx + \int (e^{x} - e^{- x}) \tan \frac{x}{2} dx \\ = (e^{x} + e^{- x}) \tan (\frac{x}{2}) - \int (e^{x} - e^{- x}) \tan \frac{x}{2} dx + \int (e^{x} - e^{- x}) \tan \frac{x}{2} dx \\ = (e^{x} + e^{- x}) \tan (\frac{x}{2}) + C f (x) = e^{x} + e^{- x} \end{array}$