JEE Main 2024 — Binomial Theorem Question with Solution

Put $$x=1$$

$$\therefore \mathrm{a}=1$$

\mathrm{b}=\lim _\limits{\mathrm{x} \rightarrow 0} \frac{\int_\limits0^{\mathrm{x}} \frac{\ln (1+\mathrm{t})}{1+\mathrm{t}^{2024}} \mathrm{dt}}{\mathrm{x}^2}

Using L' HOPITAL Rule

\mathrm{b}=\lim _\limits{\mathrm{x} \rightarrow 0} \frac{\ln (1+\mathrm{x})}{\left(1+\mathrm{x}^{2024}\right)} \times \frac{1}{2 \mathrm{x}}=\frac{1}{2}

Now, $${\mathrm{cx}}^2+\mathrm{dx}+\mathrm{e}=0,{\mathrm{x}}^2+\mathrm{x}+4=0$$

$$(\mathrm{D}<0)$$

$$\therefore \frac{\mathrm{c}}{1}=\frac{\mathrm{d}}{1}=\frac{\mathrm{e}}{4}$$