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 cx2+dx+e=0 and 2bx2+ax+4=0 have a common root, where \[c, d, e \in \mathbb{R}\], then \[\mathrm{d}: \mathrm{c}:\] e equals
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Official 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, \[\mathrm{cx}^2+\mathrm{dx}+\mathrm{e}=0, \mathrm{x}^2+\mathrm{x}+4=0\]