Question

Let f be a differentiable function such that x² f(x) - x = \(4 \int^x_0tf(t)dt,f(1)=\frac{2}{3}. \;\text{Then 18f(3)is equal to}\)
 

• A. 150
• B. 160
• C. 180
• D. 210

Answer 1

The Correct Option is B

Given:

\( x^2 f(x) - x = 4 \int_0^x tf(t) \, dt \)

Step 1: Differentiate both sides with respect to x:

\( \frac{d}{dx} (x^2 f(x) - x) = \frac{d}{dx} \left( 4 \int_0^x tf(t) \, dt \right) \)

Using Leibniz rule:

\( x^2 f'(x) + 2x f(x) - 1 = 4x f(x) \)

Simplify:

\( x^2 f'(x) + 2x f(x) - 1 = 4x f(x) \)

\( x^2 f'(x) - 2x f(x) - 1 = 0 \)

Let \( y = f(x) \):

\( x^2 \frac{dy}{dx} - 2xy - 1 = 0 \)

Step 2: Rewrite the equation:

\( \frac{dy}{dx} - \frac{2}{x} y = \frac{1}{x^2} \)

This is a first-order linear differential equation. The integrating factor (I.F.) is:

\( \text{I.F.} = e^{\int -\frac{2}{x} \, dx} = e^{-2\ln x} = \frac{1}{x^2} \)

Step 3: Solve the differential equation:

\( \frac{y}{x^2} = \int \frac{1}{x^4} \, dx + C \)

\( \frac{y}{x^2} = -\frac{1}{3x^3} + C \)

Multiply through by \(x^2\):

\( y = -\frac{1}{3x} + Cx^2 \)

Step 4: Apply Initial Condition: Given \( f(1) = \frac{2}{3} \), substitute \( x = 1 \) and \( y = \frac{2}{3} \):

\( \frac{2}{3} = -\frac{1}{3} (1) + C(1^2) \)

\( \frac{2}{3} = -\frac{1}{3} + C \)

\( C = 1 \)

Thus:

\( y = -\frac{1}{3x} + x^2 \)

Step 5: Find \( f(3) \):

\( f(3) = -\frac{1}{3(3)} + 3^2 = -\frac{1}{9} + 9 = \frac{80}{9} \)

Step 6: Find \( 18 f(3) \):

\( 18 f(3) = 18 \times \frac{80}{9} = 160 \)

Final Answer: \( 18f(3) = 160 \).