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
Solution
<p>Put $x=1$</p>
<p>$\therefore \mathrm{a}=1$</p>
<p>$$\mathrm{b}=\lim _\limits{\mathrm{x} \rightarrow 0} \frac{\int_\limits0^{\mathrm{x}} \frac{\ln (1+\mathrm{t})}{1+\mathrm{t}^{2024}} \mathrm{dt}}{\mathrm{x}^2}$$</p>
<p>Using L' HOPITAL Rule</p>
<p>$$\mathrm{b}=\lim _\limits{\mathrm{x} \rightarrow 0} \frac{\ln (1+\mathrm{x})}{\left(1+\mathrm{x}^{2024}\right)} \times \frac{1}{2 \mathrm{x}}=\frac{1}{2}$$</p>
<p>Now, $\mathrm{cx}^2+\mathrm{dx}+\mathrm{e}=0, \mathrm{x}^2+\mathrm{x}+4=0$</p>
<p>$(\mathrm{D}<0)$</p>
<p>$\therefore \frac{\mathrm{c}}{1}=\frac{\mathrm{d}}{1}=\frac{\mathrm{e}}{4}$</p>
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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