Logo
Jeeify
MCQMediumJEE 2025Definite Integrals

JEE Mathematics 2025 Question with Solution

The value of ∫−11(1+∣x∣−x)ex+(∣x∣−x)e−xex+e−x dx\int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx is equal to

  • A

    3−2233 - \frac{2\sqrt{2}}{3}

  • B

    2+2232 + \frac{2\sqrt{2}}{3}

  • C

    1−2231 - \frac{2\sqrt{2}}{3}

  • D

    1+2231 + \frac{2\sqrt{2}}{3}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

I=∫−11(1+∣x∣−x)ex+(∣x∣−x)e−xex+e−x dxI = \int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx

Find: The value of II.

Because of the term ∣x∣|x|, split the analysis according to the sign of xx.

For x≥0x \ge 0,

∣x∣−x=x−x=0|x| - x = x - x = 0

so

∣x∣−x=0\sqrt{|x| - x} = 0

and hence

f(x)=exex+e−x.f(x) = \frac{e^x}{e^x + e^{-x}}.

For x<0x < 0,

∣x∣=−x⇒∣x∣−x=−x−x=−2x|x| = -x \Rightarrow |x| - x = -x - x = -2x

so

∣x∣−x=−2x.\sqrt{|x| - x} = \sqrt{-2x}.

Now use the substitution x=−tx = -t in the integral over [−1,0][-1,0]. Then

I=∫01(f(x)+f(−x)) dx.I = \int_0^1 \bigl(f(x) + f(-x)\bigr) \, dx.

For x∈[0,1]x \in [0,1],

f(x)=exex+e−xf(x) = \frac{e^x}{e^x + e^{-x}}

and

f(−x)=(1+2x)e−x+(2x)exex+e−x.f(-x) = \frac{(1 + \sqrt{2x})e^{-x} + (\sqrt{2x})e^x}{e^x + e^{-x}}.

Therefore,

f(x)+f(−x)=ex+(1+2x)e−x+2xexex+e−x=(1+2x)(ex+e−x)ex+e−x=1+2x.f(x) + f(-x) = \frac{e^x + (1+\sqrt{2x})e^{-x} + \sqrt{2x}e^x}{e^x + e^{-x}} = \frac{(1+\sqrt{2x})(e^x + e^{-x})}{e^x + e^{-x}} = 1 + \sqrt{2x}.

So,

I=∫01(1+2x) dx.I = \int_0^1 (1 + \sqrt{2x}) \, dx.

Evaluate the integral:

I=∫011 dx+∫012x dxI = \int_0^1 1 \, dx + \int_0^1 \sqrt{2x} \, dx =[x]01+2∫01x1/2 dx= [x]_0^1 + \sqrt{2}\int_0^1 x^{1/2} \, dx =1+2[23x3/2]01= 1 + \sqrt{2}\left[\frac{2}{3}x^{3/2}\right]_0^1 =1+223.= 1 + \frac{2\sqrt{2}}{3}.

Therefore, the value of the integral is 1+2231 + \frac{2\sqrt{2}}{3}. The correct option is D.

Using symmetry after splitting at $$x = 0$$

Given:

I=∫−11(1+∣x∣−x)ex+(∣x∣−x)e−xex+e−x dxI = \int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx

Find: The exact value of the integral.

The key observation is that the absolute value changes form at x=0x = 0. Hence, the interval must be treated in two parts.

For the positive side,

∣x∣=x(x≥0)|x| = x \quad (x \ge 0)

therefore

∣x∣−x=0,|x| - x = 0,

which gives

∣x∣−x=0.\sqrt{|x| - x} = 0.

So the integrand becomes

exex+e−x.\frac{e^x}{e^x + e^{-x}}.

For the negative side,

∣x∣=−x(x<0)|x| = -x \quad (x < 0)

so

∣x∣−x=−x−x=−2x,|x| - x = -x - x = -2x,

and hence

∣x∣−x=−2x.\sqrt{|x| - x} = \sqrt{-2x}.

Thus the negative-side expression is

(1+−2x)ex+(−2x)e−xex+e−x.\frac{(1 + \sqrt{-2x})e^x + (\sqrt{-2x})e^{-x}}{e^x + e^{-x}}.

Now write

I=∫−10f(x) dx+∫01f(x) dx.I = \int_{-1}^{0} f(x) \, dx + \int_{0}^{1} f(x) \, dx.

In the first integral, substitute

x=−t,dx=−dt.x = -t, \qquad dx = -dt.

Then, after changing limits,

∫−10f(x) dx=∫01f(−t) dt.\int_{-1}^{0} f(x) \, dx = \int_0^1 f(-t) \, dt.

Renaming tt as xx,

I=∫01(f(x)+f(−x)) dx.I = \int_0^1 \bigl(f(x) + f(-x)\bigr) \, dx.

Next compute the sum explicitly for x>0x > 0:

f(x)=exex+e−x,f(x) = \frac{e^x}{e^x + e^{-x}}, f(−x)=(1+2x)e−x+(2x)exex+e−x.f(-x) = \frac{(1 + \sqrt{2x})e^{-x} + (\sqrt{2x})e^{x}}{e^x + e^{-x}}.

So,

f(x)+f(−x)=ex+(1+2x)e−x+2xexex+e−x=(1+2x)(ex+e−x)ex+e−x=1+2x.\begin{aligned} f(x) + f(-x) &= \frac{e^x + (1+\sqrt{2x})e^{-x} + \sqrt{2x}e^x}{e^x + e^{-x}} \\ &= \frac{(1+\sqrt{2x})(e^x + e^{-x})}{e^x + e^{-x}} \\ &= 1 + \sqrt{2x}. \end{aligned}

Therefore,

I=∫01(1+2x) dx.I = \int_0^1 (1 + \sqrt{2x}) \, dx.

Now integrate termwise:

∫011 dx=1,\int_0^1 1 \, dx = 1,

and

∫012x dx=2∫01x1/2 dx=2[23x3/2]01=223.\int_0^1 \sqrt{2x} \, dx = \sqrt{2}\int_0^1 x^{1/2} \, dx = \sqrt{2}\left[\frac{2}{3}x^{3/2}\right]_0^1 = \frac{2\sqrt{2}}{3}.

Hence,

I=1+223.I = 1 + \frac{2\sqrt{2}}{3}.

Therefore, the value of the given integral is 1+2231 + \frac{2\sqrt{2}}{3}, so the correct option is D.

Common mistakes

  • Treating ∣x∣|x| as xx over the entire interval is incorrect because the interval includes negative values. You must split at x=0x = 0 and use ∣x∣=x|x| = x for x≥0x \ge 0 and ∣x∣=−x|x| = -x for x<0x < 0.

  • Forgetting that ∣x∣−x=−2x|x| - x = -2x when x<0x < 0 leads to a wrong square-root term. The correct expression is ∣x∣−x=−2x\sqrt{|x| - x} = \sqrt{-2x} on the negative side.

  • Using symmetry incorrectly by assuming the integrand is even or odd is not valid here. Instead, compute f(x)+f(−x)f(x) + f(-x) explicitly and then simplify the sum.

Practice more Definite Integrals questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Definite Integrals

Related questions