MCQ
$2 + i 0$ is point on $...........?$
  • $x -$ axis
  • B
    $y -$ axis
  • C
    $z -$ axis
  • D
    any axis

Answer

Correct option: A.
$x -$ axis
Since imaginary part of complex number is zero.
So, it is plotted on real axis
i.e. $x -$ axis.
$2 + i 0$ is point on $x -$ axis.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The number of distinct solutions of the equation $\frac{5}{4} \cos ^2 2 x+\cos ^4 x+\sin ^4 x+\cos ^6 x+\sin ^6 x=2$ in the interval $[0,2 \pi]$ is
$\lim _{n \rightarrow \infty} \tan \left\{\sum_{r=1}^{n} \tan ^{-1}\left(\frac{1}{1+r+r^{2}}\right)\right\}$ is equal to..........
If the tangent on the point $(2\sec \phi ,\;3\tan \phi )$ of the hyperbola $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{9} = 1$ is parallel to $3x - y + 4 = 0$, then the value of $\phi$ is ............ $^o$
In a $\triangle A B C$, points $X$ and $Y$ are on $A B$ and $A C$, respectively, such that $X Y$ is parallel to $B C$. Which of the two following equalities always hold? (Here $[P Q R]$ denotes the area of $\triangle P Q R)$.

$I$. $[B C X]=[B C Y]$

$II$. $[A C X] \cdot[A B Y]=[A X Y] \cdot[A B C]$

Let $f(x)=\lim _{n \rightarrow \infty}\left(\frac{n^n(x+n)\left(x+\frac{n}{2}\right) \cdots\left(x+\frac{n}{n}\right)}{n!\left(x^2+n^2\right)\left(x^2+\frac{n^2}{4}\right) \cdots\left(x^2+\frac{n^2}{n^2}\right)}\right)^{\frac{x}{n}}$, for all $x>0$. Then

($A$) $f\left(\frac{1}{2}\right) \geq f(1)$

($B$) $f\left(\frac{1}{3}\right) \leq f\left(\frac{2}{3}\right)$

($C$) $f^{\prime}(2) \leq 0$

($D$) $\frac{f^{\prime}(3)}{f(3)} \geq \frac{f^{\prime}(2)}{f(2)}$

If $f(x) = \frac{{x - |x|}}{{|x|}}$, then $f( - 1) = $
It is given that the number $43361$ can be written as a product of $two$ distinct prime number $p_1, p_2$. Further, assume that there are $42900$ numbers which are less than $43361$ and are coprime to it. Then, $p_1+p_2$ is
If $\sin\alpha+\sin\beta=\text{a}$ and $\cos\alpha-\cos\beta=\text{b}$ then $\tan\frac{\alpha-\beta}{2}=$
Circles ${x^2} + {y^2} - 2x - 4y = 0$ and ${x^2} + {y^2} - 8y - 4 = 0$
$n^2+ 3n$ is always divisible by which number, provided $n$ is an integer?