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JEE Mathematics 2025 Question with Solution

Let f(x)+2f(1x)=x2+5f(x) + 2f\left( \frac{1}{x} \right) = x^2 + 5 and 2g(x)−3g(12)=x,  x>0.  If  α=∫12f(x) dx,  β=∫12g(x) dx, then the value of 9α+β is:2g(x) - 3g\left( \frac{1}{2} \right) = x, \; x > 0. \; \text{If} \; \alpha = \int_{1}^{2} f(x) \, dx, \; \beta = \int_{1}^{2} g(x) \, dx, \text{ then the value of } 9\alpha + \beta \text{ is:}

  • A

    11

  • B

    00

  • C

    1010

  • D

    1111

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

f(x)+2f(1x)=x2+5f(x) + 2f\left( \frac{1}{x} \right) = x^2 + 5

and

2g(x)−3g(12)=x,x>02g(x) - 3g\left( \frac{1}{2} \right) = x, \quad x>0

Also,

α=∫12f(x) dx,β=∫12g(x) dx\alpha = \int_1^2 f(x) \, dx, \qquad \beta = \int_1^2 g(x) \, dx

Find: 9α+β9\alpha + \beta

Substitute x→1xx \to \frac{1}{x} in the first equation:

f(1x)+2f(x)=1x2+5f\left( \frac{1}{x} \right) + 2f(x) = \frac{1}{x^2} + 5

Now solve the pair

f(x)+2f(1x)=x2+5f(x) + 2f\left( \frac{1}{x} \right) = x^2 + 5 f(1x)+2f(x)=1x2+5f\left( \frac{1}{x} \right) + 2f(x) = \frac{1}{x^2} + 5

Using these, we obtain

f(x)=23x2−x23+53f(x) = \frac{2}{3x^2} - \frac{x^2}{3} + \frac{5}{3}

Therefore,

α=∫12(23x2−x23+53) dx=119\alpha = \int_1^2 \left( \frac{2}{3x^2} - \frac{x^2}{3} + \frac{5}{3} \right) \, dx = \frac{11}{9}

For g(x)g(x), the given equation is

2g(x)−3g(12)=x2g(x) - 3g\left( \frac{1}{2} \right) = x

From the working,

g(12)=−12g\left( \frac{1}{2} \right) = -\frac{1}{2}

So,

g(x)=x2−34g(x) = \frac{x}{2} - \frac{3}{4}

Hence,

β=∫12(x2−34) dx=0\beta = \int_1^2 \left( \frac{x}{2} - \frac{3}{4} \right) \, dx = 0

Finally,

9α+β=9(119)+0=119\alpha + \beta = 9\left( \frac{11}{9} \right) + 0 = 11

Therefore, the correct option is D.

Equation Solving Detail

Given:

f(x)+2f(1x)=x2+5f(x) + 2f\left( \frac{1}{x} \right) = x^2 + 5 2g(x)−3g(12)=x2g(x) - 3g\left( \frac{1}{2} \right) = x

Find: 9α+β9\alpha + \beta

From

f(x)+2f(1x)=x2+5f(x) + 2f\left( \frac{1}{x} \right) = x^2 + 5

replace xx by 1x\frac{1}{x} to get

f(1x)+2f(x)=1x2+5f\left( \frac{1}{x} \right) + 2f(x) = \frac{1}{x^2} + 5

Now multiply the first equation by 22:

2f(x)+4f(1x)=2x2+102f(x) + 4f\left( \frac{1}{x} \right) = 2x^2 + 10

Subtract

f(1x)+2f(x)=1x2+5f\left( \frac{1}{x} \right) + 2f(x) = \frac{1}{x^2} + 5

from it:

3f(1x)=2x2−1x2+53f\left( \frac{1}{x} \right) = 2x^2 - \frac{1}{x^2} + 5

This leads to the stated expression

f(x)=23x2−x23+53f(x) = \frac{2}{3x^2} - \frac{x^2}{3} + \frac{5}{3}

Then

α=∫12(23x2−x23+53) dx=119\alpha = \int_1^2 \left( \frac{2}{3x^2} - \frac{x^2}{3} + \frac{5}{3} \right) \, dx = \frac{11}{9}

For g(x)g(x),

2g(x)−3g(12)=x2g(x) - 3g\left( \frac{1}{2} \right) = x

Using the worked value

g(12)=−12g\left( \frac{1}{2} \right) = -\frac{1}{2}

we get

g(x)=x2−34g(x) = \frac{x}{2} - \frac{3}{4}

and hence

β=∫12(x2−34) dx=0\beta = \int_1^2 \left( \frac{x}{2} - \frac{3}{4} \right) \, dx = 0

Therefore,

9α+β=119\alpha + \beta = 11

So the correct option is D.

Common mistakes

  • A common mistake is not substituting x→1xx \to \frac{1}{x} in the first functional equation. Without the second equation, f(x)f(x) and f(1x)f\left( \frac{1}{x} \right) cannot be separated. Write both equations first and then solve the system.

  • Students may treat g(12)g\left( \frac{1}{2} \right) as an independent constant without evaluating it from the given relation. That gives an incomplete expression for g(x)g(x). First use the equation at x=12x=\frac{1}{2} to determine this value, then substitute back.

  • Another mistake is making sign errors while integrating 23x2\frac{2}{3x^2}. Since ∫x−2 dx=−x−1\int x^{-2} \, dx = -x^{-1}, the antiderivative must be handled carefully. Convert powers correctly before integration.

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