JEE Advanced 2024 — Paper 2 — Question 13

Let

$$F(x)=\int_0^x\frac{f^2(t)}{1+t^2}\,dt,$$

so the given relation is

$$f(x)=(1+x^2)\big(1+F(x)\big).$$

Differentiate both sides (using $F'(x)=\dfrac{f^2(x)}{1+x^2}$):

$$f'(x)=2x(1+F(x))+(1+x^2)F'(x) =2x\frac{f(x)}{1+x^2}+\frac{f^2(x)}{1}\,.$$

Rearrange by setting

$$y(x)=\frac{f(x)}{1+x^2}.$$

Then

$$y'(x)=\frac{d}{dx}\!\left(\frac{f}{1+x^2}\right) =\frac{f'(1+x^2)-2x f}{(1+x^2)^2} =\frac{f^2}{1+x^2}=(1+x^2)y^2.$$

This is separable:

$$\frac{dy}{y^2}=(1+x^2)\,dx.$$

Integrate:

$$-\,\frac{1}{y}=x+\frac{x^3}{3}+C.$$

From $f(0)=(1+0^2)(1+0)=1$ we have $y(0)=1$, so

$$-1 = 0 + 0 + C \quad\Rightarrow\quad C=-1.$$

Thus

$$\frac{1}{y}=1-x-\frac{x^3}{3},\qquad y=\frac{1}{\,1-x-\dfrac{x^3}{3}\,}.$$

Hence

$$f(x)=(1+x^2)y=\frac{1+x^2}{\,1-x-\dfrac{x^3}{3}\,}.$$

Evaluate at $x=\tfrac12$:

$$f\!\Big(\tfrac12\Big) =\frac{1+\tfrac{1}{4}}{\,1-\tfrac12-\dfrac{1}{24}\,} =\frac{\tfrac{5}{4}}{\tfrac{11}{24}} =\frac{5}{4}\cdot\frac{24}{11}=\frac{30}{11}\approx2.72727\ldots$$

Therefore the greatest integer less than or equal to $f\!\big(\tfrac12\big)$ is

$$\left\lfloor f\!\Big(\tfrac12\Big)\right\rfloor=\boxed{2}.$$