JEE MAIN - Mathematics (2024 - 31st January Morning Shift - No. 17)
Let $$a$$ be the sum of all coefficients in the expansion of $$\left(1-2 x+2 x^2\right)^{2023}\left(3-4 x^2+2 x^3\right)^{2024}$$ and $$b=\lim _\limits{x \rightarrow 0}\left(\frac{\int_0^x \frac{\log (1+t)}{t^{2024}+1} d t}{x^2}\right)$$. If the equation $$c x^2+d x+e=0$$ and $$2 b x^2+a x+4=0$$ have a common root, where $$c, d, e \in \mathbb{R}$$, then $$\mathrm{d}: \mathrm{c}:$$ e equals
$$2: 1: 4$$
$$1: 1: 4$$
$$1: 2: 4$$
$$4: 1: 4$$
Explanation
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, $$\mathrm{cx}^2+\mathrm{dx}+\mathrm{e}=0, \mathrm{x}^2+\mathrm{x}+4=0$$
$$(\mathrm{D}<0)$$
$$\therefore \frac{\mathrm{c}}{1}=\frac{\mathrm{d}}{1}=\frac{\mathrm{e}}{4}$$
Comments (0)
