JEE Mathematics 2024 Question with Solution

Given: $a$ is the sum of all coefficients of $\left(1-2x+2x^2\right)^{2023}\left(3-4x^2+2x^3\right)^{2024}$. The equations $cx^2+dx+e=0$ and $2bx^2+ax+4=0$ have a common root.

Find: The ratio $c:d:e$.

From the solution, the sum of coefficients is obtained by substituting $x=1$:

$$\left(1-2\cdot1+2\cdot1^2\right)^{2023}\left(3-4\cdot1^2+2\cdot1^3\right)^{2024}=1^{2023}\cdot1^{2024}=1$$

Hence, $a=1$.

The solution then concludes that the correct option is D and states the required ratio as $1:1:4$.

There is inconsistency in the extracted working for $b$ and in the intermediate justification, but the solution explicitly identifies D as the correct option.

Therefore, the correct option is D, so $c:d:e=1:1:4$.