Mathematics - Limits Question with Solution | TestHub

Given: $a$ is the sum of all coefficients in ${\left(1-2x+2x^2\right)}^{2023}{\left(3-4x^2+2x^3\right)}^{2024}$

$\Rightarrow a={\left(1-2\times 1+2\times 1\right)}^{2023}{\left(3-4\times 1+2\times 1\right)}^{2024}$

$\Rightarrow a=1 ...\left(i\right)$

Now, $b=\underset{x\to 0}{\lim }\frac{{\int }_0^x{\displaystyle \frac{\log \left(1+t\right)}{t^{2024}+1}}dt}{x^2}$

Using L-Hospital's rule and Newton Leibnitz Theorem, we get

$\Rightarrow b=\underset{x\to 0}{\lim }\frac{\log \left(1+x\right)}{\left(x^{2024}+1\right)2x}$

$\Rightarrow b=\underset{x\to 0}{\lim }\frac{1}{2\left(x^{2024}+1\right)}$

$\Rightarrow b=\frac{1}{2} ...\left(ii\right)$

Also, $2b{x}^2+ax+4=0$

$\Rightarrow x^2+x+4=0$, which gives complex conjugates as roots.

Let $\alpha \text{and} \overline{\alpha }$ be those roots.

Then, $c{x}^2+dx+e=0$ will also have $\alpha \text{and} \overline{\alpha }$ as roots.

$\Rightarrow d:c:e=1:1:4$