If the middle term in the expansion of (1 + x)²ⁿ is the greatest term, then x lies in the interval

• $\left(\frac{\mathrm{n}}{\mathrm{n}\hspace{0.17em}+ 1}, \frac{\mathrm{n} + 1}{\mathrm{n}}\right)$

• $\left(\frac{\mathrm{n} + 1}{\mathrm{n}}, \frac{\mathrm{n}}{\mathrm{n} + 1}\right)$

• (n - 2, n)

• (n - 1, n)

To find the coefficient of x⁴ in the expansion of $\frac{3\mathrm{x}}{\left(\mathrm{x} - 2\right)\left(\mathrm{x} - 1\right)}$, the interval in which the expansion is valid, is

• $- 2 < \mathrm{x} < \infty$

• $- \frac{1}{2} < \mathrm{x} < \frac{1}{2}$

• $- 1 < \mathrm{x} < 1$

• $- \infty < \mathrm{x} < \infty$

$$\begin{gathered} \mathrm{Let} \mathrm{\alpha } > 0, \mathrm{\beta } > 0 \mathrm{be} \mathrm{such} \mathrm{that} {\mathrm{\alpha }}^3 + {\mathrm{\beta }}^2 = 4. \mathrm{If} \mathrm{the} \mathrm{maximum} \mathrm{value} \mathrm{of} \mathrm{the} \mathrm{term} \\ \mathrm{independent} \mathrm{of} \mathrm{x} \mathrm{in} \mathrm{the} \mathrm{binomial} \mathrm{expansion} \mathrm{of} {\left({\mathrm{\alpha x}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$9$}\right.} + {\mathrm{\beta x}}^{- \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\right)}^{10} \mathrm{is} 10\mathrm{k}, \mathrm{then} \mathrm{k} \mathrm{is} \mathrm{equal} \mathrm{to} : \end{gathered}$$

• 176

• 336

• 352

• 84

If the term independent of x in the expansion of ${\left(\frac{3}{2}{\mathrm{x}}^2 - \frac{1}{3\mathrm{x}}\right)}^9$, then 18k is equal to:

• 11

• 5

• 9

• 7

If for some positive integer n, the coefficients of three consecutive terms in the binomial expansion of

(1 + x)^{n + 5} are in the ratio 5 : 10 : 14, then the largest coefficient in the expansion is :

• 330

• 252

• 792

• 462

The natural number m, for which the coefficient of x in the binomial expansion of ${\left({\mathrm{x}}^{\mathrm{m}} + \frac{1}{{\mathrm{x}}^2}\right)}^{22} \mathrm{is} 1540, \mathrm{is}$

107.

If the constant term in the binomial expansion of ${\left(\sqrt{\mathrm{x}} - \frac{\mathrm{k}}{{\mathrm{x}}^2}\right)}^{10} \mathrm{is} 405, \mathrm{then} \left|\mathrm{k}\right| = ?$

• 3

• 2

• 1

• 9

If $8!\left(\frac{1}{3!} + \frac{5}{4!}\right) = {}_9\mathrm{P}_{\mathrm{r}}$, then the value of r is

• 4

• 5

• 3

• 2