Let a be the sum of all coefficients in the expansion of ( 1 - 2 x + 2 x 2 ) 2023 ( 3 - 4 x 2 + 2 x 3 ) 2024 and b = lim x 0 ( 0 x log ( 1 + t ) t 2024 + 1 ? d t x 2 ) . If the equations c x 2 + d x + e = 0 and 2 b x 2 + a x + 4 = 0 have a comm

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Solution

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

1 2 : 1 : 4

2 4 : 1 : 4

3 1 : 1 : 4

4 1 : 2 : 4

Ans.
(3)

$a = 1,$ $b = \lim_{x \to 0} (\frac{\int_{0}^{x} \frac{\log (1 + t)}{t^{2024} + 1}}{x^{2}})$

Using L-Hopital's Rule

$b = \lim_{x \to 0} \frac{\log (1 + x)}{2 x (x^{2024} + 1)} = \frac{1}{2} .$

Then, $x^{2} + x + 4 = 0$ (non-real) ...(i)

and $c x^{2} + d x + e = 0$ ...(ii)

Both equation (i) and (ii) have a common root

So, $\frac{c}{1} = \frac{d}{1} = \frac{e}{4}$

So, $d : c : e$ is equal to $1 : 1 : 4$

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