Two numbers k_1 and k_2 are randomly chosen f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The value of $i^k$ for any natural number $k$ can only be one of the four values: $\{i, -1, -i, 1\}$.Since $k_1$ and $k_2$ are chosen randomly,the

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(C) The value of $i^k$ for any natural number $k$ can only be one of the four values: $\{i, -1, -i, 1\}$.Since $k_1$ and $k_2$ are chosen randomly,there are $4 	imes 4 = 16$ possible pairs of values for $(i^{k_1}, i^{k_2})$.We want to find the probability that $i^{k_1} + i^{k_2} 
eq 0$,which is equivalent to $1 - P(i^{k_1} + i^{k_2} = 0)$.The condition $i^{k_1} + i^{k_2} = 0$ implies $i^{k_1} = -i^{k_2}$.The possible pairs $(i^{k_1}, i^{k_2})$ that satisfy this are: $(i, -i), (-i, i), (1, -1), (-1, 1)$.There are $4$ such unfavorable cases.Thus,the number of favorable cases is $16 - 4 = 12$.The probability is $\frac{12}{16} = \frac{3}{4}$.

Two numbers $k_1$ and $k_2$ are randomly chosen from the set of natural numbers. Then,the probability that the value of $i^{k_1} + i^{k_2}$ (where $i = \sqrt{-1}$) is non-zero,equals:

Similar Questions

Let $z_1$ and $z_2$ be two complex numbers such that $z_1 + z_2 = 5$ and $z_1^3 + z_2^3 = 20 + 15i$. Then $|z_1^4 + z_2^4|$ equals-

If $z=x+iy$ and $z^{1/3}=p+iq$,where $x, y, p, q \in R$ and $i=\sqrt{-1}$,then the value of $\left(\frac{x}{p}+\frac{y}{q}\right)$ is

If $(2+i)$ is a root of the equation $x^3-5x^2+9x-5=0$,then the other roots are

Sum of the squares of the imaginary roots of the equation $z^8-20z^4+64=0$ is

If $\cos \alpha+3 \cos 3 \beta+5 \cos 5 \gamma=0$,$\sin \alpha+3 \sin 3 \beta+5 \sin 5 \gamma=0$ and $\cos 3 \alpha+27 \cos 9 \beta+125 \cos 15 \gamma=\left(\lambda^2-4\right) \cos (\alpha+3 \beta+5 \gamma)$,then $\lambda$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module