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JEE Mathematics 2025 Question with Solution

Let (a,b)(a, b) be the point of intersection of the curve x2=2yx^2 = 2y and the straight line y=2x−6y = 2x - 6 in the second quadrant. Then the integral I=∫ab9x21+5x3 dxI = \int_a^b \frac{9x^2}{1 + 5x^3} \, dx is equal to:

  • A

    2424

  • B

    2727

  • C

    1818

  • D

    2121

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The curves are x2=2yx^2 = 2y and y=2x−6y = 2x - 6, and I=∫ab9x21+5x3 dxI = \int_a^b \frac{9x^2}{1 + 5x^3} \, dx.

Find: The value of the definite integral.

From the solution, the intersection is obtained by solving the equations simultaneously. Using the line as written in the working,

y=2x+6y = 2x + 6

and substituting into x2=2yx^2 = 2y,

x2=2(2x+6)x^2 = 2(2x + 6) x2=4x+12x^2 = 4x + 12 x2−4x−12=0x^2 - 4x - 12 = 0 (x−6)(x+2)=0(x - 6)(x + 2) = 0

So the intersection points are (6,18)(6, 18) and (−2,2)(-2, 2).

The point in the second quadrant is (−2,2)(-2, 2). Therefore,

a=−2,b=2a = -2, \quad b = 2

The solution then evaluates the integral as

I=∫−229x21+5x dxI = \int_{-2}^{2} \frac{9x^2}{1 + 5^x} \, dx

Let

f(x)=9x21+5xf(x) = \frac{9x^2}{1 + 5^x}

Then

f(−x)=9(−x)21+5−x=9x2⋅5x1+5xf(-x) = \frac{9(-x)^2}{1 + 5^{-x}} = \frac{9x^2 \cdot 5^x}{1 + 5^x}

Hence,

f(x)+f(−x)=9x21+5x+9x2⋅5x1+5x=9x2f(x) + f(-x) = \frac{9x^2}{1 + 5^x} + \frac{9x^2 \cdot 5^x}{1 + 5^x} = 9x^2

Using the symmetric-limit property,

I=∫029x2 dxI = \int_0^2 9x^2 \, dx

Now,

I=9∫02x2 dxI = 9 \int_0^2 x^2 \, dx I=9[x33]02I = 9 \left[ \frac{x^3}{3} \right]_0^2 I=3[x3]02I = 3[x^3]_0^2 I=3(23−03)=24I = 3(2^3 - 0^3) = 24

Therefore, the value of the integral is 2424, so the correct option is A.

Note: The provided question text and the solution contain inconsistencies in the line equation and integrand, but the solution explicitly concludes option A.

Using symmetry of the integrand

Given: The bounds come from the second-quadrant intersection point, giving a=−2a = -2 and b=2b = 2 according to the solution.

Find: Evaluate the definite integral using the symmetry relation shown in the working.

For symmetric limits,

∫−ccf(x) dx=∫0c[f(x)+f(−x)] dx\int_{-c}^{c} f(x) \, dx = \int_0^c [f(x) + f(-x)] \, dx

The working uses

f(x)=9x21+5xf(x) = \frac{9x^2}{1 + 5^x}

so

f(−x)=9x2⋅5x1+5xf(-x) = \frac{9x^2 \cdot 5^x}{1 + 5^x}

Adding,

f(x)+f(−x)=9x2f(x) + f(-x) = 9x^2

Thus the integral reduces immediately to

I=∫029x2 dxI = \int_0^2 9x^2 \, dx

which is a standard power-function integral:

I=9[x33]02=3(8)=24I = 9 \left[ \frac{x^3}{3} \right]_0^2 = 3(8) = 24

Hence the correct option is A.

Common mistakes

  • Using the wrong quadrant while choosing the intersection point. The second quadrant requires x<0x < 0 and y>0y > 0, so one must select (−2,2)(-2, 2), not (6,18)(6, 18).

  • Missing the symmetric-limit identity. If the interval is [−2,2][-2, 2], then checking f(x)+f(−x)f(x) + f(-x) can simplify the integral drastically; integrating the original form directly is unnecessary here.

  • Not noticing the discrepancy between the given question text and the solution. The working uses a different line equation and integrand form, so the answer must be derived from the solution.

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