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 $b=\lim _{x \rightarrow 0}\left(\frac{\int_{0}^{x} \frac{\log (1+t)}{t^{2024}+1} d t}{x^{2}}\right)$. If the equations $c x^{2}+d x+e=0$ and $2 b x^{2}+a x+4=0$ have a common root, where $\mathrm{c}, \mathrm{d}, \mathrm{e} \in \mathrm{R}$, then $\mathrm{d}: \mathrm{c}: \mathrm{e}$ equals

Quadratic Equations and Inequalities — JEE Main · 2024 · 31 Jan 2024 · Shift 1 Question with Solution

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{l o g ( 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

Answer: (D)

a

= sum of coefficients of

(1 - 2 x + 2 x^{2})^{2023} (3 - 4 x^{2} + 2 x^{3})^{2024}

, obtained by setting

x = 1

a = (1 - 2 + 2)^{2023} (3 - 4 + 2)^{2024} = (1)^{2023} (1)^{2024} = 1

For

b

, apply L'Hopital's rule to the

0/0

limit:

b = x \to 0 lim \frac{\int _{0}^{x} \frac{ln ( 1 + t )}{t ^{2024} + 1} d t}{x ^{2}} = x \to 0 lim \frac{ln ( 1 + x ) / ( x ^{2024} + 1 )}{2 x} = x \to 0 lim \frac{ln ( 1 + x )}{x} \cdot \frac{1}{2 ( x ^{2024} + 1 )} = \frac{1}{2}

So the second equation

2 b x^{2} + a x + 4 = 0

becomes

x^{2} + x + 4 = 0

, whose discriminant is

1 - 16 < 0

— its roots are a complex-conjugate pair.

For

c x^{2} + d x + e = 0

(with real

c, d, e

) to share a root with a quadratic having non-real roots, it must share both conjugate roots, i.e. it must be proportional to

x^{2} + x + 4

. So

c : d : e = 1 : 1 : 4 ⟹ d : c : e = 1 : 1 : 4

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option