MCQMediumJEE 2025Integration Techniques (Substitution, Parts, Partial Fractions)

JEE Mathematics 2025 Question with Solution

Let f(x)f(x) be a positive function and I1=1212xf(2x(12x))dxI_1 = \int_{-\frac{1}{2}}^1 2x \, f\left(2x(1-2x)\right) \, dx and I2=12f(x(1x))dx.I_2 = \int_{-1}^2 f\left(x(1-x)\right) \, dx. Then the value of I2I1\frac{I_2}{I_1} is equal to _____

  • A

    44

  • B

    66

  • C

    1212

  • D

    99

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

I1=1212xf(2x(12x))dx,I_1 = \int_{-\frac{1}{2}}^1 2x \, f\left(2x(1-2x)\right) \, dx, I2=12f(x(1x))dxI_2 = \int_{-1}^2 f\left(x(1-x)\right) \, dx

Find: The value of I2I1\frac{I_2}{I_1}.

From the solution, the conclusion stated is that the correct option is A and hence I2I1=4\frac{I_2}{I_1} = 4.

Solution Working

Given:

I1=1212xf(2x(12x))dx,I_1 = \int_{-\frac{1}{2}}^1 2x \, f\left(2x(1-2x)\right) \, dx, I2=12f(x(1x))dxI_2 = \int_{-1}^2 f\left(x(1-x)\right) \, dx

Find: The value of I2I1\frac{I_2}{I_1}.

The solution proceeds by considering substitutions for both integrals.

For I1I_1, it takes

u=2x(12x)u = 2x(1-2x)

and writes

du=2(14x)dx.du = 2(1-4x) \, dx.

It then tracks the limits:

  • at x=12x = -\frac{1}{2}, the working gives u=0u = 0,
  • at x=1x = 1, the working gives u=2u = -2.

Using this, the extracted working states

I1=1202f(u)du=1220f(u)du=1220f(u)du.I_1 = \frac{1}{2} \int_{0}^{-2} f(u) \, du = -\frac{1}{2} \int_{-2}^{0} f(u) \, du = \frac{1}{2} \int_{-2}^{0} f(u) \, du.

For I2I_2, the working takes

u=x(1x)u = x(1-x)

with

dudx=12x.\frac{du}{dx} = 1-2x.

It evaluates the endpoint values as

  • at x=1x = -1, u=2u = -2,
  • at x=2x = 2, u=2u = -2.

The working then appeals to symmetry and states

I2=20f(u)du\therefore I_2 = \int_{-2}^{0} f(u) \, du

and hence

I2=4(1220f(u)du)=4I1.I_2 = 4\left(\frac{1}{2}\int_{-2}^{0} f(u) \, du\right) = 4I_1.

Therefore,

I2I1=4.\frac{I_2}{I_1} = 4.

So the correct option is A.

Note: The intermediate derivation in the provided the solution is not fully rigorous and contains inconsistencies, but its final conclusion explicitly gives option A, so the extracted answer is 44.

Shortcut from the solution

Given: The same integrals I1I_1 and I2I_2. Find: I2I1\frac{I_2}{I_1}.

The second approach in the solution tests the constant function

f(x)=1.f(x) = 1.

Then

I1=1212xdx=x2121=114=34I_1 = \int_{-\frac{1}{2}}^1 2x \, dx = \left.x^2\right|_{-\frac{1}{2}}^1 = 1 - \frac{1}{4} = \frac{3}{4}

and

I2=12dx=3.I_2 = \int_{-1}^2 dx = 3.

So

I2I1=334=4.\frac{I_2}{I_1} = \frac{3}{\frac{3}{4}} = 4.

Thus the correct option is A.

Why this shortcut works in the extracted solution: the working uses this test to match the constant ratio claimed by the main solution and identify the correct option.

Common mistakes

  • Assuming a substitution is complete without transforming the differential correctly. In I1I_1, replacing only the inside expression and ignoring the derivative structure can produce an invalid integral. Always check how dxdx and every multiplicative factor transform together.

  • Using symmetry too loosely for x(1x)x(1-x). Although the expression is symmetric about x=12x = \frac{1}{2}, that does not mean the integral can be rewritten arbitrarily. First split the interval carefully, then map each subinterval with a valid substitution.

  • Trusting endpoint substitution alone for I2I_2. Since x=1x = -1 and x=2x = 2 both give the same transformed value, one may incorrectly conclude the integral is zero. That is wrong because a many-to-one substitution requires interval splitting or monotonic pieces.

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