Binomial Theorem

$\mathrm{The} \mathrm{value} \mathrm{of} \left\{\mathrm{x} \in \mathrm{R} ⋮ \left[\log \left(\log {\left(1.6\right)}^{1 - {\mathrm{x}}^2} - {\left(0.625\right)}^{6\left(1 + \mathrm{x}\right)}\right)\right] \in \mathrm{R}\right\} \mathrm{is}$

• $\left(- \infty , - 1\right) \cup \left(7, 0\right)$

• (- 1, 5)

• (1, 7)

• (- 1, 7)

$\frac{1}{2 . 3} + \frac{1}{4 . 5} + \frac{1}{6 . 7} + \frac{1}{8 . 9} + ... \infty = ?$

• $\log \left(\frac{2}{\mathrm{e}}\right)$

• $\log \left(\frac{\mathrm{e}}{2}\right)$

• $\log \left(2\mathrm{e}\right)$

• e - 1

If the harmonic mean between the roots of (5 + $\sqrt{2}$) x² - bx + (8 + 2$\sqrt{5}$) = 0 is 4, then the value of b is

• 2

• 3

• 4 - $\sqrt{5}$

• $4 + \sqrt{5}$

The set of solutions satisfying both x² + 5x + 6 $\ge$ 0 and x² + 3x - 4 < 0 is

• (- 4, 1)

• $\left(- 4, - 3\right) \cup (- 2, 1)$

• $\left(- 4, - 3\right) \cup \left(- 2, 1\right)$

• $\left(- 4, - 3\right) \cup \left( - 2, 1\right)$

The relation between pressure p and volume V is given by pV^1/4 constant. If the percentage decrease in volume is 1/2, then the percentage increase in pressure is

• $- \frac{1}{8}$

• $\frac{1}{16}$

• $\frac{1}{8}$

• $\frac{1}{2}$

The least positive integer n for which (1 + i)ⁿ = (1 - i)ⁿ , is

• 8

• 2

• 4

• 6

If the mean and variance of a binomial variate X are 8 and 4 respectively, then P(X < 3) equals to 

• $\frac{265}{2^{15}}$

• $\frac{137}{2^{14}}$

• $\frac{137}{2^{16}}$

• $\frac{265}{2^{16}}$

If ${\mathrm{x}}_1, {\mathrm{x}}_2, ... , {\mathrm{x}}_{\mathrm{n}}$ are n observations such that $\sum _{\mathrm{i} = 1}^{\mathrm{n}}{{\mathrm{x}}_1}^2 = 400 \mathrm{and} \sum _{\mathrm{i} = 1}^{\mathrm{n}}{\mathrm{x}}_1 = 80$, then the least value of n is 

• 18

• 12

• 15

• 16

If $\left|\mathrm{x}\right|$ < 1, then the coefficient of x⁵ in the expansion of $\frac{3\mathrm{x}}{\left(\mathrm{x} - 2\right)\left(\mathrm{x} + 1\right)}$ is

• $\frac{33}{32}$

• $\frac{- 33}{32}$

• $\frac{31}{32}$

• $\frac{- 31}{32}$

If the coefficients of x⁹, x1⁰ and x¹¹ in the expansion of (1 + x)ⁿ are in arithmetic progression, then n² - 41n is equal to

• 399

• 298

• - 398

• 198