1.

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

 

A. (A) (B) (C) (D)	(i) (ii) (iii) (vi) (i)
B. (A) (B) (C) (D)	(ii) (iv) (v) (vi) (i)
C. (A) (B) (C) (D)	(iii) (iv) (ii) (vi) (i)
D. (A) (B) (C) (D)	(iv) (i) (ii) (iii) (iv)

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