Suppose that E₁ and E₂ are two events of a random experiment such that P(E₁) = 1/4, P(E₂/E₁) and P(E₁/E₂) = 1/4, observe the lists given below

        List I                        List II

(A)    P(E₂)                  (i) 1/4

(B)    $\mathrm{P}\left({\mathrm{E}}_1 \cup {\mathrm{E}}_2\right)$           (ii) 5/8

(C)   $\mathrm{P}\left(\overline{{\mathrm{E}}_1}/ \overline{{\mathrm{E}}_2}\right)$             (iii) 1/8

(D)   $\mathrm{P}\left({\mathrm{E}}_1/\overline{{\mathrm{E}}_2}\right)$              (iv) 3/8

                                  (v) 3/8

                                  (vi) 3/4

The correct matching of the List I from the List II is

The roots (x - a) (x - a - 1) + (x - a - 1) (x - a - 2) + (x - a) (x - a - 2) = 0, a $\in$ R are always

• equal

• imaginary

• real and distinct

• rational and equal

Let f(x) = x + ax + b, where a, b $\in$ R. If f(x) = 0 has all-its roots imaginary, then the roots of f(x) + f'(x) + f"(x) = 0 are

• real and distinct

• imaginary

• equal

• rational and equal

If $\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }$ are the roots of x³ + 4x + 1 = 0, then the equation whose roots are $\frac{{\mathrm{\alpha }}^3}{\mathrm{\beta } + \mathrm{\gamma }}, \frac{{\mathrm{\beta }}^2}{\mathrm{\gamma } + \mathrm{\alpha }}, \frac{{\mathrm{\gamma }}^2}{\mathrm{\alpha } + \mathrm{\beta }}$ is

• x³ - 4x - 1 = 0

• x³ - 4x + 1 = 0

• x³ + 4x - 1 = 0

• x³ + 4x + 1 = 0

$$\begin{gathered} \mathrm{If} \mathrm{f} : \left[2, 3\right]\to \mathrm{R} \mathrm{is} \mathrm{defined} \mathrm{by} \mathrm{f} \left(\mathrm{x}\right) = {\mathrm{x}}^3 + 3\mathrm{x} - 2, \mathrm{then} \mathrm{the} \mathrm{range} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \\ \mathrm{contained} \mathrm{in} \mathrm{the} \mathrm{interval} \end{gathered}$$

• [1, 12]

• [12, 34]

• [35, 50]

• [- 12, 12]

$\left\{\mathrm{x} \in \mathrm{R} :\hspace{0.17em}\frac{2\mathrm{x} - 1}{{\mathrm{x}}^3 +\hspace{0.17em}4{\mathrm{x}}^2 +\hspace{0.17em}3\mathrm{x}} \in \mathrm{R}\right\} \mathrm{equals}$

• R - {0}

• R - {0, 1, 3}

• R - {0, - 1, - 3}

• R - $\left\{0, - 1, - 3, +\frac{1}{2}\right\}$

$$\begin{gathered} \mathrm{Using} \mathrm{mathematical} \mathrm{induction}, \mathrm{the} \mathrm{numbers} {\mathrm{a}}_{\mathrm{n} }\text{'}\mathrm{s} \mathrm{are} \mathrm{defined} \mathrm{by}, \\ {\mathrm{a}}_0 = 1, {\mathrm{a}}_{\mathrm{n} + 1} = 3{\mathrm{n}}^2 + \mathrm{n} + {\mathrm{a}}_{\mathrm{n}}, \left(\mathrm{n} \ge 0\right) \\ \mathrm{Then} {\mathrm{a}}_{\mathrm{n}} = ? \end{gathered}$$

• n³ + n² + 1

• n³ - n² + 1

• n³ - n²

• n³ + n²

If $\mathrm{\alpha } \mathrm{and} \mathrm{\beta }$ are the roots of x² - 2x + 4 = 0, then the value of ${\mathrm{\alpha }}^6 + {\mathrm{\beta }}^6$ is

• 32

• 64

• 128

• 256

The locus of z satisfying the inequality $\left|\frac{\mathrm{z} + 2\mathrm{i}}{2\mathrm{z} + \mathrm{i}}\right| < 1$, where z = x + iy, is

• x² + y² < 1

• x² - y² < 1

• x² + y² > 1

• 2x² + 3y² < 1 

If n is an integer which leaves remainder one when divided by three, then ${\left(1 + \sqrt{3}\mathrm{i}\right)}^{\mathrm{n}} + {\left(1 - \sqrt{3}\mathrm{i}\right)}^{\mathrm{n}}$ equals

• - 2^{n + 1}

• 2^{n + 1}

• - (- 2)ⁿ

• - 2ⁿ