Limits Question 8
Question 8 - 2024 (31 Jan Shift 1)
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 $b=\lim {x \rightarrow 0}\left(\frac{\int{0}^{x} \frac{\log (1+t)}{t^{2024}+1}}{x^{2}}\right)$. If the equations $\mathrm{cx}^{2}+\mathrm{dx}+\mathrm{e}=0$ and $2 \mathrm{bx}^{2}+\mathrm{ax}+4=0$ have a common root, where $c, d, e \in R$, then $d: c$ : e equals
(1) $2: 1: 4$
(2) $4: 1: 4$
(3) $1: 2: 4$
(4) $1: 1: 4$
Show Answer
Answer (4)
Solution
Put $x=1$
$\therefore a=1$
$b=\lim {x \rightarrow 0} \frac{\int{0}^{x} \frac{\ln (1+t)}{1+t^{2024}} d t}{x^{2}}$
Using L’ HOPITAL Rule
$b=\lim _{x \rightarrow 0} \frac{\ln (1+x)}{\left(1+x^{2024}\right)} \times \frac{1}{2 x}=\frac{1}{2}$
Now, $\mathrm{cx}^{2}+\mathrm{dx}+\mathrm{e}=0, \mathrm{x}^{2}+\mathrm{x}+4=0$
$(\mathrm{D}<0)$
$\therefore \frac{\mathrm{c}}{1}=\frac{\mathrm{d}}{1}=\frac{\mathrm{e}}{4}$
