Let a be the sum of all coefficients in the expansion of (1−2x+2x^2)^2023 (3−4x^2+2x^3)^2024

Question

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 \(b=\lim\limits _{\mathrm{x} \rightarrow 0}\left(\frac{\int_{0}^{\mathrm{x}} \frac{\log (1+\mathrm{t})}{\mathrm{t}^{2024}+1} \mathrm{dt}}{\mathrm{x}^{2}}\right)\). If the equations \(\mathrm{cx}^{2}+\mathrm{dx}+\mathrm{e}=0\) and \(2 \mathrm{bx}^{2}+\mathrm{ax}+4=0\) have \(\mathrm{a}\) common root, where \(\mathrm {c, d, e} \in \mathrm{R}\), then \(\mathrm {d : c : e}\) equals

(1) \(2: 1: 4\)

(2) \(4: 1: 4\)

(3) \(1: 2: 4\)

(4) \(1: 1: 4\)

Answer 1

Correct option is (4) \(1: 1: 4\)

Put \(\mathrm{x}=1\)

\(\therefore \mathrm{a}=1\)

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

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, 

\(\mathrm{cx}^{2}+\mathrm{dx}+\mathrm{e}=0, \mathrm{x}^{2}+\mathrm{x}+4=0\)

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

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