Limits, Continuity & Differentiability
Limit of Integral via L'Hôpital — Ratio of Coefficients
nta_pyq_2024_jan
Grade 12
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 $b=\displaystyle\lim_{x\to0}\left(\dfrac{\int_0^x\dfrac{\log(1+t)}{t^{2024}+1}\,dt}{x^2}\right)$. 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
Step-by-Step Solution
Key Concept: Sum of all coefficients: put $x=1$: $a=(1-2+2)^{2023}(3-4+2)^{2024}=1^{2023}\cdot1^{2024}=1$. For $b$: apply L'Hôpital: $b=\lim_{x\to0}\frac{\ln(1+x)/(1+x^{2024})}{2x}=\frac{1}{2}$. Second equation: $x^2+x+4=0$ (using $2b=1$, $a=1$). Common root means $cx^2+dx+e\propto x^2+x+4$, so $d:c:e=1:1:4$.
$a=1$, $b=1/2$. Common root: $cx^2+dx+e\propto x^2+x+4$. $d:c:e=1:1:4$.
Correct Answer: 4