AP EAPCET 2025 - 21st May Morning Shift | Binomial Theorem Question 24 | Mathematics

If $x$ is so large that terms containing $x^{-3}, x^{-4}, x^{-5}, \ldots$ can be neglected, then the approximate value of $\left(\frac{3 x-5}{4 x^2+3}\right)^{-1 / 5}$ is

$\left(\frac{3}{4 x}\right)^{4 / 5}\left(1-\frac{4}{3 x}-\frac{7}{5 x^2}\right)$

$\left(\frac{4 x}{3}\right)^{4 / 5}\left(1+\frac{4}{3 x}+\frac{13}{5 x^2}\right)$

$\left(\frac{4 x}{3}\right)^{4 / 5}\left(1+\frac{4}{3 x}-\frac{13}{5 x^2}\right)$

$\left(\frac{3}{4 x}\right)^{4 / 5}\left(1-\frac{4}{3 x}+\frac{7}{5 x^2}\right)$

AP EAPCET 2024 - 23th May Morning Shift

MCQ (Single Correct Answer)

-0

The independent term in the expansion of $\left(1+x+2 x^2\right)\left(\frac{3 x^2}{2}-\frac{1}{3 x}\right)^9$ is

$\frac{18}{7}$

$\frac{7}{18}$

$-\frac{7}{18}$

$-\frac{18}{7}$

AP EAPCET 2024 - 23th May Morning Shift

MCQ (Single Correct Answer)

-0

For $|x|<\frac{1}{\sqrt{2}}$, the coefficient of $x$ in the expansion of $\frac{(1-4 x)^2\left(1-2 x^2\right)^{1 / 2}}{(4-x)^{3 / 2}}$ is

$\frac{61}{64}$

$-\frac{61}{64}$

$\frac{69}{64}$

$-\frac{69}{64}$

AP EAPCET 2024 - 22th May Evening Shift

MCQ (Single Correct Answer)

-0

If $P$ is the greatest divisor of $49^n+16 n-1$ for all $n \in N$, then the number of factors of $P$ is

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