Let a be the sum of all coefficients in the expansion of (1–2x+2x^2 )^2023(3-4x^2+2x^3 )^2024 and b=undersetx→ 0 lim( ∫ _0^x log(1+t) t^2024+1dt…

Given: $a$a is the sum of all coefficients in ${(1-2x+2{x}^{2})}^{2023}{(3-4{x}^{2}+2{x}^{3})}^{2024}$(1−2x+2x2)2023(3−4x2+2x3)2024

$$\Rightarrow a={(1-2\times 1+2\times 1)}^{2023}{(3-4\times 1+2\times 1)}^{2024}$$⇒a=(1−2×1+2×1)2023(3−4×1+2×1)2024

$$\Rightarrow a=1...(i)$$⇒a=1...(i)

Now, $\displaystyle b=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{{\int }_{0}^{x}\frac{\mathrm{log}(1+t)}{{t}^{2024}+1}dt}{{x}^{2}}$b=x→0lim​x2∫0x​t2024+1log(1+t)​dt​

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

$$\Rightarrow b=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{\mathrm{log}(1+x)}{({x}^{2024}+1)2x}$$⇒b=x→0lim​(x2024+1)2xlog(1+x)​

$$\Rightarrow b=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{1}{2({x}^{2024}+1)}$$⇒b=x→0lim​2(x2024+1)1​

$$\Rightarrow b=\frac{1}{2}...(ii)$$⇒b=21​...(ii)

Also, $2b{x}^{2}+ax+4=0$2bx2+ax+4=0

$\Rightarrow {x}^{2}+x+4=0$⇒x2+x+4=0, which gives complex conjugates as roots.

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

Then, $c{x}^{2}+dx+e=0$cx2+dx+e=0 will also have $\alpha \text{and}\alpha$αandα as roots.

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