Let f: R -> R be such that f(xy) = f(x)f(y), for all x, y ∈ R and f(0) ≠ 0. Let g: [1, ∞) -> R be a differentiable function such that x^2 g(x) = ∫_1^x (t^2 f(t) - tg(t)) dt. Then g(2) is equal to :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q18 – JEE-MAIN 2026

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Home Maths Relations and FunctionsQuestion #12

Let

f : R \to R

be such that

f (x y) = f (x) f (y)

, for all

x, y \in R

and

f (0) \neq 0

. Let

g : [1, \infty) \to R

be a differentiable function such that

x^{2} g (x) = \int_{1}^{x} (t^{2} f (t) - t g (t)) d t .

Then

g (2)

is equal to :

Clue 1 of 3: Functional Equation Analysis

First, determine the nature of the function $f (x)$ using the given functional equation $f (x y) = f (x) f (y)$ and the condition $f (0) \neq 0$ .

🧠 Full Solution Path

● Video Walkthrough

Step 1: Determine the function

f (x)

✦ Active

Given the functional equation $f (x y) = f (x) f (y)$ for all $x, y \in R$ and $f (0) \neq 0$ . Substitute $y = 0$ into the equation:

f (x \cdot 0) = f (x) f (0) ⟹ f (0) = f (x) f (0)

Since $f (0) \neq 0$ , we can divide both sides by $f (0)$ , which implies $f (x) = 1$ for all $x \in R$ . This solution satisfies the given conditions.

Step 2: Formulate and solve the differential equation for

g (x)

○ Expand

Substitute $f (t) = 1$ into the given integral equation:

x^{2} g (x) = \int_{1}^{x} (t^{2} - t g (t)) d t

Differentiate both sides with respect to $x$ using Leibniz's integral rule:

2 x g (x) + x^{2} g^{'} (x) = x^{2} - x g (x)

Rearrange the terms to form a first-order linear differential equation:

x^{2} g^{'} (x) + 3 x g (x) = x^{2}

g^{'} (x) + \frac{3}{x} g (x) = 1

The integrating factor (IF) is $e^{\int \frac{3}{x} d x} = e^{3 \ln x} = x^{3}$ . Multiply the differential equation by the IF:

x^{3} g^{'} (x) + 3 x^{2} g (x) = x^{3}

\frac{d}{d x} (x^{3} g (x)) = x^{3}

Integrate both sides with respect to $x$ :

x^{3} g (x) = \int x^{3} d x = \frac{x^{4}}{4} + C

g (x) = \frac{x}{4} + \frac{C}{x^{3}}

💡 Teacher's Secret Hint

Remember to apply Leibniz's rule correctly for differentiating integrals with variable limits.

Step 3: Determine the constant

C

and calculate

g (2)

○ Expand

From the original integral equation, substitute $x = 1$ to find an initial condition for $g (x)$ :

1^{2} g (1) = \int_{1}^{1} (t^{2} - t g (t)) d t = 0

So, $g (1) = 0$ . Now, substitute $x = 1$ into the expression for $g (x)$ to find $C$ :

g (1) = \frac{1}{4} + \frac{C}{1^{3}} = 0 ⟹ C = - \frac{1}{4}

Thus, the function $g (x)$ is:

g (x) = \frac{x}{4} - \frac{1}{4 x^{3}}

Finally, calculate $g (2)$ :

g (2) = \frac{2}{4} - \frac{1}{4 (2^{3})} = \frac{1}{2} - \frac{1}{4 \cdot 8} = \frac{1}{2} - \frac{1}{32}

g (2) = \frac{16}{32} - \frac{1}{32} = \frac{15}{32}

💡 Teacher's Secret Hint

Always use the original integral equation to find the initial condition for the constant of integration, not the differentiated form.

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Maths Question 18 – JEE-MAIN 2026

Clue 1 of 3: Functional Equation Analysis

🧠 Full Solution Path