AP EAPCET 2025 - 21st May Evening Shift | Binomial Theorem Question 22 | Mathematics

The coefficient of $x^3$ in the expansion of $\frac{x^4+1}{\left(x^2+1\right)(x-1)}$ when it is expressed in terms of positive integral powers of $x$, is

AP EAPCET 2025 - 21st May Morning Shift

MCQ (Single Correct Answer)

-0

If $(1+x)^n=\sum_{r=0}^n C, x^r$, then the value of $C_0+\left(C_0+C_1\right)+\left(C_0+C_1+C_2\right)+\ldots+ \left(C_0+C_1+C_2+\ldots+C_n\right)$ is

$n R^{n-1}$

$2^n+n$

$(n+2) 2^n$

$(n+2) 2^{n-1}$

AP EAPCET 2025 - 21st May Morning Shift

MCQ (Single Correct Answer)

-0

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

$\frac{18}{7}$

$\frac{7}{18}$

$-\frac{7}{18}$

$-\frac{18}{7}$

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