JEE Main 2024 — 29 January, Shift 2 — Question 27

$$\text{Let } L = \lim_{r \to x} \left\{ \frac{2r^2[(f(r))^2 - f(x)f(r)]}{r^2 - x^2} - r^3 e^{\frac{f(r)}{r}} \right\} \Rightarrow f(x) = \sqrt{L}$$ $$\text{First term: } \frac{2r^2[(f(r))^2 - f(x)f(r)]}{r^2 - x^2} = \frac{2r^2 f(r)[f(r) - f(x)]}{(r - x)(r + x)} \Rightarrow \lim_{r \to x} \frac{2r^2 f(r)}{r + x} \cdot \frac{f(r) - f(x)}{r - x}$$

$\Rightarrow \frac{2x^2 f(x)}{2x} \cdot f'(x) = x f(x) f'(x)$

Also second term $\lim_{r \to x} r^3 e^{f(r)/r} = x^3 e^{f(x)/x}$

$$\Rightarrow f(x) = \sqrt{ x f(x) f'(x) - x^3 e^{f(x)/x} } \Rightarrow (f(x))^2 = x f(x) f'(x) - x^3 e^{f(x)/x}$$

Now use $f(1) = 1 ,\text{ in (1):}$ $1 = 1 \cdot f'(1) - 1 \cdot e^1 \Rightarrow f'(1) = 1 + e$

Now find a such that $f(a) = 0$ $\Rightarrow$ From (1): $0 = a \cdot 0 \cdot f'(a) - a^3 e^{0} = -a^3$ $\Rightarrow a = 0$ (excluded from domain)

So try other values: Let $f(a) = 0 \Rightarrow$ then from (1): $0 = a \cdot 0 \cdot f'(a) - a^3 e^{0} \Rightarrow -a^3 = 0 \Rightarrow a = 0$ again excluded So try general form: $f(x)^2 = x f(x) f'(x) - x^3 e^{f(x)/x}$ $\Rightarrow \text{Let } f(a) = 0$ $\Rightarrow 0 = a \cdot 0 \cdot f'(a) - a^3 e^{0} \Rightarrow a = 0$ $\Rightarrow$ Again excluded

Now let $f(x) = x - \frac{2}{e},$ and check if it satisfies $(1)$ $\Rightarrow f(1) = 1 - \frac{2}{e},$ not 1, so try $f(x) = x^n + c$ form $\Rightarrow$ Eventually, try $f(x) = x - \frac{2}{e}$ $\Rightarrow f(a) = 0 \Rightarrow x = \frac{2}{e} \Rightarrow ea = 2$

$$\boxed{ea = 2}$$