M.C.Q (1 Marks) - Maths STD 11 Science Questions - Vidyadip

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M.C.Q (1 Marks)

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16 questions · timed · auto-graded

MCQ 11 Mark

The real value of $\alpha$ for which the expression $\frac{1-\text{i}\sin\alpha}{1+2\text{i}\sin\alpha}$ is purely real is:

A
$(\text{n}+1)\frac{\pi}{2}$
B
$(2\text{n}+1)\frac{\pi}{2}$
✓
$\text{n}\pi$
D
None of these, where $\text{n}\in\text{N}$

Answer

Correct option: C.

$\text{n}\pi$

$=\frac{(1-\text{i}\sin\alpha)(1-2\text{i}\sin\alpha)}{(1+2\text{i}\sin\alpha)(1-2\text{i}\sin\alpha)}$
$=\frac{1-\text{i}\sin\alpha-2\text{i}\sin\alpha+2\text{i}^2\sin^2\alpha}{1-4\text{i}^2\sin^2\alpha}$
$=\frac{1-3\text{i}\sin\alpha-2\sin^2\alpha}{1+4\sin^2\alpha}$
$=\frac{1-2\sin^2\alpha}{1+4\sin^2\alpha}-\frac{3\text{i}\sin\alpha}{1+4\sin^2\alpha}$
It is given that $z$ is a purely real.
$\Rightarrow\frac{-3\text{i}\sin\alpha}{1+4\sin^2\alpha}=0$
$\Rightarrow-3\sin\alpha=0$
$\Rightarrow\sin\alpha=0$
$\Rightarrow\alpha=\text{n}\pi,\text{ n}\in\text{I}$

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MCQ 21 Mark

The value of $(\text{z}+3)(\bar{\text{z}}+3)$ is equivalent to:

✓
$|z + 3|^2$
B
$|z - 3|$
C
$z^2 + 3$
D
None of these.

Answer

Correct option: A.

$|z + 3|^2$

$|z + 3|^{2}$
Let $z = x + iy$. Then
$(\text{z}+3)(\bar{\text{z}}+3)=(\text{x}+\text{iy}+3)(\text{x}-\text{iy}+3)$
$=(\text{x}+3)^2-(\text{iy})^2$
$=(\text{x}+3)^2+\text{y}^2$
$=|\text{x}+3+\text{iy}|^2$
$=|\text{z}+3|^2$

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MCQ 31 Mark

$\sin\text{x}+\text{i}\cos2\text{x}$ and $\cos\text{x}-\text{i}\sin2\text{x}$ are conjugate to each other for:

A
$\text{x}=\text{n}\pi$
B
$\text{x}=\Big(\text{n}+\frac{1}{2}\Big)\frac{\pi}{2}$
C
$\text{x}=0$
✓
no value of $x$

Answer

Correct option: D.

no value of $x$

$\sin\text{x}+\text{i}\cos2\text{x}$ and $\cos\text{x}-\text{i}\sin2\text{x}$ are conjugate to each other.
$\Rightarrow\overline{\sin\text{x}+\text{i}\cos2\text{x}}=\cos\text{x}-\text{i}\sin2\text{x}$
$\Rightarrow{\sin\text{x}-\text{i}\cos2\text{x}}=\cos\text{x}-\text{i}\sin2\text{x}$
On comparing real and imaginary parts of both the sides, we get
$\sin\text{x}=\cos\text{x}$ and $\cos2\text{x}=\sin2\text{x}$
$\Rightarrow\tan\text{x}=1$ and $\tan2\text{x}=1$
Now, $\tan2\text{x}=1$
$\Rightarrow\frac{2\tan\text{x}}{1-\tan^{2}\text{x}}=1,$
which is not satisfied by $\tan\text{x}=1$
Hence, no value of $x$ is possible.

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MCQ 41 Mark

$|z_1 + z_2| = |z_1| + |z_2|$ is possible if:

A
$\text{z}_2=\bar{\text{z}_1}$
B
$\text{z}_2=\frac{1}{\text{z}_1}$
✓
$\arg(\text{z}_1)=\arg(\text{z}_2)$
D
$|\text{z}_1|=|\text{z}_2|$

Answer

Correct option: C.

$\arg(\text{z}_1)=\arg(\text{z}_2)$

$\text{arg}(\text{z}_1)=\text{arg}(\text{z}_2)$
Let $\text{z}_1=\text{r}_1(\cos\theta_1+\text{i}\sin\theta_1)$ and $\text{z}_2=\text{r}_2(\cos\theta_2+\text{i}\sin\theta_2)$
Since $|\text{z}_1+\text{z}_2|=|\text{z}_1|+|\text{z}_2|$
$|\text{z}_1+\text{z}_2|=\text{r}_1\cos\theta_1+\text{i}\text{r}_1\sin\theta_1+\text{r}_2\cos\theta_2+\text{i}\text{r}_2\sin\theta_2$
$|\text{z}_1+\text{z}_2|=\sqrt{\text{r}^2_1\cos^2\theta_1+\text{r}^2_2\cos^2\theta_2+2\text{r}_1\text{r}_2\cos\theta_1\cos\theta_2\\+\text{r}^2_1\sin^2\theta_1+\text{r}^2_2\sin^2\theta_2+2\text{r}_1\text{r}_2\sin\theta_1\sin\theta_2}$
$=\sqrt{\text{r}^2_1+\text{r}^2_2+2\text{r}_1\text{r}_2\cos(\theta_1-\theta_2)}$
But $|\text{z}_1+\text{z}_2|=|\text{z}_1|+|\text{z}_2|$
So, $\sqrt{\text{r}^2_1+\text{r}^2_2+2\text{r}_1\text{r}_2\cos(\theta_1-\theta_2)}=\text{r}_1+\text{r}_2$
Squaring both sides, we get
$\text{r}^2_1+\text{r}^2_2+2\text{r}_1\text{r}_2\cos(\theta_1-\theta_2)=\text{r}^2_1+\text{r}^2_2+2\text{r}_1\text{r}_2$
$\Rightarrow2\text{r}_1\text{r}_2-2\text{r}_1\text{r}_2\cos(\theta_1-\theta_2)=1$
$\Rightarrow1-\cos(\theta_1-\theta_2)=0$
$\Rightarrow\cos(\theta_1-\theta_2)=1$
$\Rightarrow\theta_1-\theta_2=0$
$\Rightarrow\theta_1=\theta_2$
So, $\text{arg}(\text{z}_1)=\text{arg}(\text{z}_2)$

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MCQ 51 Mark

The complex number $z$ which satisfies the condition $\Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|=1$ lies on:

A
Circle $x^2+ y^2= 1$
✓
The $x-$axis
C
The $y-$axis
D
The line $x + y = 1$

Answer

Correct option: B.

The $x-$axis

Given that$, \Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|=1$
Let $\text{z}=\text{x}+\text{yi}$
$\therefore\ \Big|\frac{\text{i}+\text{x}+\text{yi}}{\text{i}-\text{x}-\text{yi}}\Big|=1$
$\Rightarrow\ \bigg|\frac{\text{x}-(\text{y}+1)\text{i}}{-\text{x}-(\text{y}-1)\text{i}}\bigg|=1$
$\Rightarrow|\text{x}+(\text{y}+1)\text{i}|=|-\text{x}-(\text{y}-1)\text{i}|$
$\Rightarrow\sqrt{\text{x}^2+(\text{y}+1)^2}=\sqrt{\text{x}^2+(\text{y}-1)^2}$
$\Rightarrow\ \text{x}^2+(\text{y}^2+1)^2=\text{x}^2+(\text{y}-1)^2$
$\Rightarrow(\text{y}+1)^2=(\text{y}-1)^2$
$\Rightarrow\text{y}^2+2\text{y}+1=\text{y}^2-2\text{y}+1$
$\Rightarrow2\text{y}=-2\text{y}$
$\Rightarrow2\text{y}+2\text{y}=0$
$\Rightarrow4\text{y}=0$
$\Rightarrow\text{y}=0$
$\Rightarrow\text{x-axis}$

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MCQ 61 Mark

If $\text{f(z)}=\frac{7-\text{z}}{1-\text{z}^2},$ where $z = 1 + 2i,$ then $|f(z)|$ is:

✓
$\frac{|\text{z}|}{2}$
B
$|\text{z}|$
C
$2|\text{z}|$
D
None of these

Answer

Correct option: A.

$\frac{|\text{z}|}{2}$

Given that, $\text{z}=1+2\text{i}$
$|\text{z}|=\sqrt{(1)^2+(2)^2}=\sqrt{5}$
Now, $\text{f(z)}=\frac{7-\text{z}}{1-\text{z}^2}$
$=\frac{7-(1+2\text{i})}{1-(1+2\text{i})^2}$
$=\frac{7-1-2\text{i}}{1-1-4\text{i}^2-4\text{i}}$
$=\frac{6-2\text{i}}{4-4\text{i}}=\frac{3-\text{i}}{2-2\text{i}}$
$=\frac{3-\text{i}}{2-2\text{i}}\times\frac{2+2\text{i}}{2+2\text{i}}$
$=\frac{6+6\text{i}-2\text{i}-2\text{i}^2}{4-4\text{i}}$
$=\frac{6+4\text{i}+2}{4+4}$
$=\frac{8+4\text{i}}{8}$
$=1+\frac{1}{2}\text{i}$
So, $|\text{f(z)}|=\sqrt{(1)+\Big(\frac{1}{2}\Big)^2}$
$=\sqrt{1+\frac{1}{4}}$
$=\frac{\sqrt{5}}{2}$
$=\frac{|\text{z}|}{2}$

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MCQ 71 Mark

The point represented by the complex number $2 - i$ is rotated about origin through an angle $\frac{\pi}{2}$ in the clockwise direction, the new position of point is:

A
$1 + 2i$
✓
$-1 - 2i$
C
$2 + i$
D
$-1 + 2i$

Answer

Correct option: B.

$-1 - 2i$

Given that, $\text{z}=2-\text{i}$
If $z$ rotated through an angle of $\frac{\pi}{2}$ about the origin in clockwise direction.
Then the new position $=\text{z}\cdot\text{e}^{-\big(\frac{\pi}{2}\big)}$
$=(2-\text{i})\text{e}^{-\big(\frac{\pi}{2}\big)}$
$=(2-\text{i})\Big[\cos\Big(\frac{-\pi}{2}\Big)+\text{i}\sin\Big(\frac{-\pi}{2}\Big)\Big]$
$=(2-\text{i})(0-\text{i})$
$=-1-2\text{i}$

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MCQ 81 Mark

If $z$ is $a$ complex number$,$ then:

A
$|\text{z}^2|>|\text{z}|^2$
✓
$|\text{z}^2|=|\text{z}|^2$
C
$|\text{z}^2|<|\text{z}|^2$
D
$|\text{z}^2|\geq|\text{z}|^2$

Answer

Correct option: B.

$|\text{z}^2|=|\text{z}|^2$

$|z|^2 = |z^2|$
Let $z = x + yi$
$|z| = |x + yi|$ and $|z|^2 = |x + yi|^2$
$\Rightarrow |z|^2 = x^2 + y^2 .....(i)$
Now$, z^2 = x^2 + y^2i^2 + 2xyi$
$z^2 = x^2 - y^2+ 2xyi$
$|\text{z}^2|=\sqrt{(\text{x}^2-\text{y}^2)^2+(2\text{xy})^2}$
$=\sqrt{\text{x}^4+\text{y}^4-2\text{x}^2\text{y}^2+4\text{x}^2\text{y}^2}$
$=\sqrt{\text{x}^4+\text{y}^4+2\text{x}^2\text{y}^2}$
$=\sqrt{(\text{x}^2+\text{y}^2)^2}$
So$, |\text{z}|^2=\text{x}^2+\text{y}^2=|\text{z}|^2$
So$, |\text{z}|^2=|\text{z}^2|$

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MCQ 91 Mark

If $z = x + iy$ lies in the third quadrant, then $\frac{\bar{\text{z}}}{\text{z}}$ also lies in the third quadrant if:

A
$x > y > 0$
B
$x < y < 0$
✓
$y < x < 0$
D
$y > x > 0$

Answer

Correct option: C.

$y < x < 0$

Since $\text{z}=\text{x}+\text{iy}$ lies in the third quadrant, we get
$\text{x}<0$ and $\text{y}<0$
Now, $\frac{\bar{\text{z}}}{\text{z}}=\frac{\text{x}-\text{iy}}{\text{x}+\text{iy}}=\frac{(\text{x}-\text{iy})(\text{x}-\text{iy})}{(\text{x}+\text{iy})(\text{x}-\text{iy})}$
$=\frac{\text{x}^2-\text{y}^2-2\text{ixy}}{\text{x}^2+\text{y}^2}=\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}-\frac{2\text{ixy}}{\text{x}^2+\text{y}^2}$
Since, $\frac{\bar{\text{z}}}{\text{z}}$ also lies in third quadrant, we get
$\frac{\text{x}^2-\text{y}^2}{\text{x}^2+\text{y}^2}<0$ and $\frac{-2\text{xy}}{\text{x}^2+\text{y}^2}<0$
$\Rightarrow\text{x}^2-\text{y}^2<0$ and $-2\text{xy}<0$
$\Rightarrow\text{x}^2<\text{y}^2$ and $\text{xy}>0$
But $\text{x, y}<0$
$\Rightarrow\text{y}<\text{x}<0$

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MCQ 101 Mark

Which of the following is correct for any two complex numbers $z_1$ and $z_2$ ?

A
$|\text{z}_1\text{z}_2|=|\text{z}_1||\text{z}_2|$
B
$\arg(\text{z}_1\text{z}_2)=\arg(\text{z}_1)\cdot\arg(\text{z}_2)$
✓
$|\text{z}_1+\text{z}_2|=|\text{z}_1|+|\text{z}_2|$
D
$|\text{z}_1+\text{z}_2|\geq|\text{z}_1|-|\text{z}_2|$

Answer

Correct option: C.

$|\text{z}_1+\text{z}_2|=|\text{z}_1|+|\text{z}_2|$

$|\text{z}_1\text{z}_2|=|\text{z}_1||\text{z}_2|$
Let $\text{z}_1=\text{r}_1(\cos\theta_1+\text{i}\sin\theta_2)$
$\therefore\ |\text{z}_1|=\text{r}_1$
and $\text{z}_2=\text{r}_2(\cos\theta_1+\text{i}\sin\theta_2)$
$\therefore\ |\text{z}_2|=\text{r}_2$
$\text{z}_1\text{z}_2=\text{r}_1(\cos\theta_1+\text{i}\sin\theta_1)\cdot\text{r}_2(\cos\theta_2+\text{i}\sin\theta_2)$
$=\text{r}_1\text{r}_2(\cos\theta_1+\text{i}\sin\theta_1)\cdot(\cos\theta_2+\text{i}\sin\theta_2)$
$=\text{r}_1\text{r}_2(\cos\theta_1\cos\theta_2+\text{i}\sin\theta_2\cos\theta_1+\text{i}\sin\theta_1\cos\theta_2+\text{i}_2\sin\theta_1\sin\theta_2)$
$=\text{r}_1\text{r}_2\big[(\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2)+\text{i}(\sin\theta_1\cos\theta_2+\cos\theta_1+\sin\theta_2)\big]$
$=\text{r}_1\text{r}_2\big[\cos(\theta_1+\theta_2)+\text{i}\sin(\theta_1+\theta_2)\big]$
$\therefore\ |\text{z}_1\text{z}_2|=|\text{z}_1||\text{z}_2|$

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MCQ 111 Mark

If $\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^{\text{x}}=1,$ then:

A
$x = 2n + 1$
✓
$x = 4n$
C
$x = 2n$
D
$x = 4n + 1$, where $n \in N$

Answer

Correct option: B.

$x = 4n$

$\Rightarrow\Big[\frac{(1+\text{i})(1+\text{i})}{(1-\text{i})(1+\text{i})}\Big]^{\text{x}}=1$
$\Rightarrow\Big[\frac{1+2\text{i}+\text{i}^2}{1-\text{i}^2}\Big]^{\text{x}}=1$
$\Rightarrow\Big[\frac{2\text{i}}{1+1}\Big]^{\text{x}}=1$
$\Rightarrow\text{i}^{\text{x}}=1$
$\Rightarrow\text{x}=4\text{n},\text{n}\in\text{N}$

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MCQ 121 Mark

If $a + ib = c + id,$ then:

A
$a^2 + c^2 = 0$
B
$b^2 + c^2 = 0$
C
$b^2+ d^2 = 0$
✓
$a^2 + b^2 = c^2 + d^2$

Answer

Correct option: D.

$a^2 + b^2 = c^2 + d^2$

$a^2 + b^2 = c^2 + d^2$
Given that $\text{a}+\text{ib}=\text{c}+\text{id}$
$\Rightarrow|\text{a}+\text{ib}|=|\text{c}+\text{id}|$
$\Rightarrow\sqrt{\text{a}^2+\text{b}^2}$
$=\sqrt{\text{c}^2+\text{d}^2}$
Squaring both sides$,$ we get ${a^2 + b^2 = c^2 + d^2}$

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MCQ 131 Mark

The value of $\text{arg(x),}$ when $x < 0$ is:

A
$0$
B
$\frac{\pi}{2}$
✓
$\pi$
D
None of these

Answer

Correct option: C.

$\pi$

Let $z = -x + 0i$ and $x < 0$
$\therefore\ |\text{z}|=\sqrt{(-1)^2+(0)^2}=1,\text{ x}<0$
Since, the point $(-x, 0)$ lies on the negative side of the real axis
$(\because\text{x}<0)$
$\therefore$ Principle argument $(\text{z})=\pi$

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MCQ 141 Mark

A real value of $x$ satisfies the equation $\Big(\frac{3-4\text{ix}}{3+4\text{ix}}\Big)=\alpha-\text{i}\beta(\alpha,\beta\in\text{R})$ if $\alpha^2+\beta^2=$

✓
$1$
B
$-1$
C
$2$
D
$-2$

Answer

Correct option: A.

$1$

Given that, $\Big(\frac{3-4\text{ix}}{3+4\text{ix}}\Big)=\alpha-\text{i}\beta$
$\Rightarrow8\Big(\frac{3-4\text{ix}}{3+4\text{ix}}\times\frac{3-4\text{ix}}{3-4\text{ix}}\Big)=\alpha-\text{i}\beta$
$\Rightarrow\frac{9-12\text{ix}-12\text{ix}+16\text{i}^2\text{x}^2}{9-16\text{i}^2\text{x}^2}=\alpha-\text{i}\beta$
$\Rightarrow\frac{9-24\text{ix}-16\text{x}^2}{9+16\text{x}^2}=\alpha-\text{i}\beta$
$\Rightarrow\frac{9-16\text{x}^2}{9+16\text{x}^2}-\frac{24\text{x}}{9+16\text{x}^2}\text{i}=\alpha+\text{i}\beta\ .....(\text{i})$
$\Rightarrow\frac{9-16\text{x}^2}{9+16\text{x}^2}+\frac{24\text{x}}{9+16\text{x}^2}\text{i}=\alpha+\text{i}\beta\ .....(\text{ii})$
Multiplying eq. $(i)$ and $(ii)$ we get
$\Big(\frac{9-16\text{x}^2}{9+16\text{x}^2}\Big)^2+\Big(\frac{24\text{x}}{9+16\text{x}^2}\Big)^2=\alpha^2+\beta^2$
$\Rightarrow\frac{(9-16\text{x}^2)^2+(24\text{x})^2}{(9+16\text{x}^2)^2}=\alpha^2+\beta^2$
$\Rightarrow\frac{81+256\text{x}^4-288\text{x}^2+576\text{x}^2}{(9+16\text{x}^2)^2}=\alpha^2+\beta^2$
$\Rightarrow\frac{81+256\text{x}^4+288\text{x}^2}{(9+16\text{x}^2)^2}=\alpha^2+\beta^2$
$\Rightarrow\frac{(9+16\text{x}^2)}{(9+16\text{x}^2)^2}=\alpha^2+\beta^2$
So, $\alpha^2+\beta^2=1$

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MCQ 151 Mark

Let $x, y \in R,$ then $x + iy$ is a non real complex number if:

A
$x = 0$
B
$y = 0$
C
$x \neq 0$
✓
$y \neq 0$

Answer

Correct option: D.

$y \neq 0$

$x + yi$ is a non$-$real complex number if $y \neq 0.$ If $x, y \in R$

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MCQ 161 Mark

The real value of $\theta$ for which the expression $\frac{1+\text{i}\cos\theta}{1-2\text{i}\cos\theta}$ is a real number is:

A
$\text{n}\pi+\frac{\pi}{4}$
B
$\text{n}\pi+(-1)^{\text{n}}\frac{\pi}{4}$
✓
$2\text{n}\pi\pm\frac{\pi}{2}$
D
None of these.

Answer

Correct option: C.

$2\text{n}\pi\pm\frac{\pi}{2}$

Let $\text{z}=\frac{1+\text{i}\cos\theta}{1-2\text{i}\cos\theta}=\frac{1+\text{i}\cos\theta}{1-2\text{i}\cos\theta}\times\frac{1+2\text{i}\cos\theta}{1+2\text{i}\cos\theta}$
$=\frac{1+2\text{i}\cos\theta+\text{i}\cos\theta+2\text{i}^2\cos^2\theta}{1-4\text{i}^2\cos^2\theta}$
$=\frac{1-3\text{i}\cos\theta-2\cos^2\theta}{1+4\cos^2\theta}$
$=\frac{1-2\cos^2\theta}{1+4\cos^2\theta}+\frac{3\cos\theta}{1+4\cos^2\theta}\text{i}$
If $z$ is real number, then
$\frac{3\cos\theta}{1+4\cos^2\theta}=0$
$\Rightarrow3\cos\theta=0$
$\Rightarrow\cos\theta=0$
$\therefore\ \theta=(2\text{n}+1)\frac{\pi}{2},\text{ n}\in\text{N}$

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M.C.Q (1 Marks) - Maths STD 11 Science Questions - Vidyadip