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Bernoulli Equation from an Integral Equation | JEE

JEE Maths question with a full step-by-step solution.

Question
A continuous function f:R→Rf: R \to R satisfies f(x)=(1+x2)[1+∫0xf2(t)1+t2 dt]f(x) = (1+x^2)\left[1 + \displaystyle\int_0^x \dfrac{f^2(t)}{1+t^2}\,dt\right]. Then the value of f(−2)f(-2) is:
A00
B1715\dfrac{17}{15}
C−1715-\dfrac{17}{15}
D1517\dfrac{15}{17}correct
Solution
Step 1: At x=0x = 0 the integral vanishes, so f(0)=1f(0) = 1. Write f(x)1+x2=1+∫0xf2(t)1+t2dt\dfrac{f(x)}{1+x^2} = 1 + \displaystyle\int_0^x\frac{f^2(t)}{1+t^2}dt and differentiate both sides:
(1+x2)f′(x)−2x f(x)(1+x2)2=f2(x)1+x2.\frac{(1+x^2)f'(x) - 2x\,f(x)}{(1+x^2)^2} = \frac{f^2(x)}{1+x^2}.
 Step 2: Multiply by (1+x2)(1+x^2) and write y=f(x)y = f(x):
dydx−2x1+x2 y=y2(Bernoulli).\frac{dy}{dx} - \frac{2x}{1+x^2}\,y = y^2 \quad(\text{Bernoulli}).
 Step 3: Divide by y2y^2 and put t=−1yt = -\dfrac{1}{y}, so dtdx=1y2dydx\dfrac{dt}{dx} = \dfrac{1}{y^2}\dfrac{dy}{dx}:
dtdx+2x1+x2 t=1,IF=e∫2x1+x2dx=1+x2.\frac{dt}{dx} + \frac{2x}{1+x^2}\,t = 1, \qquad \text{IF} = e^{\int \frac{2x}{1+x^2}dx} = 1+x^2.
 Step 4: Solve:
t(1+x2)=∫(1+x2) dx=x+x33+c⇒t=−1f=x+x33+c1+x2.t(1+x^2) = \int (1+x^2)\,dx = x + \frac{x^3}{3} + c \Rightarrow t = -\frac{1}{f} = \frac{x + \frac{x^3}{3} + c}{1+x^2}.
∴f(x)=−1+x2x33+x+c=−3(1+x2)x3+3x+3c.\therefore f(x) = -\frac{1+x^2}{\frac{x^3}{3} + x + c} = -\frac{3(1+x^2)}{x^3 + 3x + 3c}.
 Step 5: f(0)=1⇒1=−33c⇒c=−1f(0) = 1 \Rightarrow 1 = -\dfrac{3}{3c} \Rightarrow c = -1, so
f(x)=−3(1+x2)x3+3x−3,f(−2)=−3(5)−8−6−3=−15−17=1517.f(x) = -\frac{3(1+x^2)}{x^3 + 3x - 3}, \qquad f(-2) = -\frac{3(5)}{-8 - 6 - 3} = \frac{-15}{-17} = \frac{15}{17}.
 Correct answer: (4)
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