Question

Let \( a \) be the sum of all coefficients in the expansion of \( (1 - 2x + 2x^2)^{2023} (3 - 4x^2 + 2x^3)^{2024} \). and \( b = \lim_{x \to 0} \frac{\int_0^x \frac{\log(1 + t)}{t^{2024} + 1} \, dt}{x^2} \).If the equations \( cx^2 + dx + e = 0 \) and \( 2bx^2 + ax + 4 = 0 \) have a common root, where \( c, d, e \in \mathbb{R} \), then \( d : c : e \) equals

• A. \( 2 : 1 : 4 \)
• B. \( 4 : 1 : 4 \)
• C. \( 1 : 2 : 4 \)
• D. \( 1 : 1 : 4 \)

Answer 1

The Correct Option is D

• Substitute $x = 1$: 
$\therefore a = 1$

• Consider:

$b = \lim_{x \to 0} \left( \frac{x \int_0^x \frac{\log(1+t)}{1+t^2} dt}{x^2} \right)$

• Using L'Hôpital's Rule:

$b = \lim_{x \to 0} \left( \frac{\frac{d}{dx} \left( \int_0^x \frac{\log(1+t)}{1+t^2} dt \right)}{\frac{d}{dx}(x^2)} \right) = \lim_{x \to 0} \left( \frac{\log(1+x)}{2x} \right) = \lim_{x \to 0} \frac{1}{2(1+x)} = \frac{1}{2}$

• Now, for the equations $cx^2 + dx + e = 0$ and $2bx^2 + ax + 4 = 0$ to have a common root:

$cx^2 + dx + e = 0$, $2bx^2 + ax + 4 = 0$

Since $D < 0$ (where $D$ denotes the discriminant of the equation), we find:

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