If e^{x+ ^{-1} x}≤ft({x²+2}{ ²≤ft( ^{-1} x)}) d x=e^{f(x)}+c, then ...

Correct Answer is : f(x)f(x)f(x) is strictly increasing on RRR.

f(x) is strictly decreasing on R.

f(x) is strictly increasing on R^{+}and strictly decreasing on R^{-}.

f(x) is strictly increasing on R.

f(x) is strictly decreasing on R^{+}and strictly increasing on R^{-}.

Correct Answer is : f(x)f(x)f(x) is strictly increasing on RRR.

Test Index

MHTCET 18 Apr 2026 Shift 2 Paper

Question : 128 of 150

Marks: +1, -0

\int e^{x+\tan ^{-1} x}\left(\frac{x^{2}+2}{\sec ^{2}\left(\tan ^{-1} x\right)}\right) d x=e^{f(x)}+c

, then ...

Solution:

Concept:

Use

\sec^2(\tan^{-1}x)=1+x^2

and express the integrand as

e^{g(x)}g'(x)

Explanation:

First, simplify the denominator:

\sec^2(\tan^{-1}x) = 1+\tan^2(\tan^{-1}x) = 1+x^2

So the integrand becomes:

e^{x+\tan^{-1}x}\left(\frac{x^2+2}{x^2+1}\right)

Let

g(x)=x+\tan^{-1}x

Then

g'(x)=1+\frac{1}{1+x^2}=\frac{x^2+2}{x^2+1}

Thus the integral is of the form

\int e^{g(x)}g'(x)\,dx = e^{g(x)}+c

Therefore,

\int e^{x+\tan^{-1}x}\left(\frac{x^2+2}{\sec^2(\tan^{-1}x)}\right)dx = e^{x+\tan^{-1}x}+c

Comparing with

e^{f(x)}+c

, we get:

f(x)=x+\tan^{-1}x

Now check monotonicity:

f'(x)=1+\frac{1}{1+x^2}

Since

1+x^2>0

for all real

x

, we have

f'(x)>0

for all

x\in\mathbb{R}

Thus

f(x)

is strictly increasing on

\mathbb{R}

Answer:

Option C:

f(x)

is strictly increasing on

\mathbb{R}

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