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Definite Integration question

2023 · Shift 2 · Q25

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JEE Advanced
Mathematics
Definite Integration
2023 · Shift 2 · Q25

Definite Integration question

2023 · Shift 2 · Q25

JEE AdvancedMathematicsDefinite IntegrationNumerical+4 / −1

For

x \in \mathbb{R}

, let

\tan ^{-1}(x) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

. Then the minimum value of the function

f: \mathbb{R} \rightarrow \mathbb{R}

defined by

f(x)=\int\limits_0^{x \tan ^{-1} x} \frac{e^{(t-\cos t)}}{1+t^{2023}} d t

is :

Numerical answer

View written solutionFree

Correct answer: 0

We are given

f(x)=\int_0^{x\tan^{-1}x} \frac{e^{(t-\cos t)}}{1+t^{2023}}\,dt

and we need the minimum value of $f(x)$ for $x\in\mathbb R$ .

First, study the upper limit:

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

Since $\tan^{-1}x$ has the same sign as $x$ , we get

if $x>0$ , then $\tan^{-1}x>0$ , so $g(x)>0$ ,
if $x<0$ , then $\tan^{-1}x<0$ , so $g(x)>0$ ,
if $x=0$ , then $g(0)=0$ .

Hence,

x\tan^{-1}x\ge 0 \quad \text{for all } x\in\mathbb R.

So the upper limit of integration is always nonnegative.

Now examine the integrand:

\phi(t)=\frac{e^{(t-\cos t)}}{1+t^{2023}}.

For the integral to be defined over $[0,\,x\tan^{-1}x]$ , note that $t\ge 0$ on this interval.

For $t\ge 0$ :

$e^{(t-\cos t)}>0$ always,
$1+t^{2023}>0$ .

Therefore,

\phi(t)>0 \quad \text{for all } t\ge 0.

Since the integrand is strictly positive on $[0, a]$ for every $a>0$ , we have

\int_0^a \phi(t)\,dt >0 \quad \text{for } a>0,

and of course

\int_0^0 \phi(t)\,dt=0.

Thus,

f(x)\ge 0 \quad \text{for all } x\in\mathbb R.

Equality occurs when the upper limit is $0$ , i.e. when

x\tan^{-1}x=0.

This happens only at

x=0.

Then

f(0)=\int_0^0 \frac{e^{(t-\cos t)}}{1+t^{2023}}dt=0.

Therefore, the minimum value of $f(x)$ is

\boxed{0}.

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