Download App

MATHS MCQ

COMPLEX NUMBERS AND QUADRATIC EQUATIONS · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 165 questions.

Questions · Page 1 of 4

MCQ

🎯

Test yourself on this topic

50 questions · timed · auto-graded

MCQ 11 Mark
If z is a comp lex num ber, then |3z – 1|= 3|z – 2| represents:
  • A
    y-axis
  • B
    a circle
  • C
    x-axis
  • ✓
    a line parallel to y-axis
Answer
Correct option: D.
a line parallel to y-axis
  1. a line parallel to y-axis
View full question & answer→
MCQ 21 Mark
If $z_1=2+3 i$ and $z_2=5+2 i$, then find sum of two complex numbers:
  • A
    4 + 8i
  • B
    3 - i
  • ✓
    7 + 5i
  • D
    7 - 5i
Answer
Correct option: C.
7 + 5i
  1. 7 + 5i
Solution:
In addition of two complex numbers, corresponding parts of two complex numbers are added i.e. real partsof both are added and imaginary parts of both are added.
So, sum = (2 + 5) + (3 + 2)i = 7 + 5i.
View full question & answer→
MCQ 31 Mark
Choose the correct answer. 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.
$\Rightarrow\frac{-3\text{i}\sin\alpha}{1+4\sin^2\alpha}=0$
$\Rightarrow-3\sin\alpha=0$
$\Rightarrow\sin\alpha=0$
$\Rightarrow\alpha=\text{n}\pi,\text{ n}\in\text{I}$
View full question & answer→
MCQ 41 Mark
What will be the sum of b + c if the equations $x^2 + bx + c = 0$ and $x^2 + 3x + 3 = 0$ have one common root:
  • A
    2
  • B
    4
  • ✓
    6
  • D
    8
Answer
Correct option: C.
6
  1. 6
Solution:
Comparing the coefficients of the above equation we get,
$\frac{1}{1}=\frac{\text{b}}{3}=\frac{\text{c}}{3}$
This means $\text{b}=3$ and $\text{c}=3$
$\therefore\text{b}+\text{c}=6$
View full question & answer→
MCQ 51 Mark
If the roots of $x^2− bx + c = 0$ are two consecutive integers, then $b^2 − 4 c$ is:
  • A
    0
  • ✓
    1
  • C
    2
  • D
    None of these.
Answer
Correct option: B.
1
  1. 1
Solution:
Given equation: $x^2 − bx + c = 0$
Let $\alpha$ and $\alpha+1$ be the two consecutive roots of the equation.
Sum of the roots $=\alpha+\alpha+1=2\alpha+1$
Product of the roots $=\alpha(\alpha+1)=\alpha^2+\alpha$
So, sum of the roots $=2\alpha+1=\frac{-\text{Coeffecient of x}}{\text{Coeffecient of x}^2}=\frac{\text{b}}{1}=\text{b}$
Product of the roots $=\alpha^2+\alpha=\frac{\text{Constant term}}{\text{Coeffecient of x}^2}=\frac{\text{c}}{1}=\text{c}$
Now,$\text{b}^2-4\text{c}=(2\alpha+1)^2-4(\alpha^2+\alpha)=4\alpha^2+4\alpha+1-4\alpha^2-4\alpha=1$
View full question & answer→
MCQ 61 Mark
If $\text{x}_1, \text{x}_2$ are real roots of $\text{ax}^2-\text{x}+\text{a}=0$ Then, find the set of all values of parameter ‘a’ for which $|\text{x}_1-\text{x}_1|<1$
  • ✓
    $\Big(\frac{1-5\text{a}}{\text{a}^2}\Big)<0$
  • B
    $\Big(\frac{1-5\text{a}}{\text{a}^2}\Big)=0$
  • C
    $\Big(\frac{1-5\text{a}}{\text{a}^2}\Big)>0$
  • D
    $\Big(\frac{1-5\text{a}}{\text{a}}\Big)<0$
Answer
Correct option: A.
$\Big(\frac{1-5\text{a}}{\text{a}^2}\Big)<0$
$|\text{x}_1-\text{x}_1|<1$
$=(\text{x}_1-\text{x}_2)^2<1$
$=(\text{x}_1+\text{x}_2)^2-4\text{x}_1\text{x}_2-1<0$
$=\Big(\frac{1}{2}\Big)$
$=\frac{(1-5\text{a})}{\text{a}^2}<0.$
View full question & answer→
MCQ 71 Mark
The polar form of $(\text{i}^{25})^3$ is:
  • A
    $\cos\frac{\pi}{2}+\text{i}\sin\frac{\pi}{2}$
  • B
    $\cos\pi+\text{i}\sin\pi$
  • C
    $\cos\pi-\text{i}\sin\pi$
  • ✓
    $\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}$
Answer
Correct option: D.
$\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}$
$(\text{i}^{25})^3=(\text{i})^{75}$
$=(\text{i})^{4\times18+3}$
$=(\text{i})^3$
$=-\text{i} \ (\because\text{i}^4=1)$
Let $\text{z}=0-\text{i}$
Since, the point (0,−1) lies on the negative direction of imaginary axis.Therefore, $\text{arg(z)}=\frac{-\pi}{2}$
Modulus, $\text{r}=|\text{z}|=|1|=1$
$\therefore$ Polar form $=\text{r}(\cos\theta+\text{i}\sin\theta)$
$=\cos\Big(\frac{-\pi}{2}\Big)+\text{i}\sin\Big(\frac{-\pi}{2}\Big)$
$=\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}$
View full question & answer→
MCQ 81 Mark
The number of real roots of the equation $\left(x^2+2 x\right)^2-(x+1)^2-55=0$ :
  • ✓
    2
  • B
    1
  • C
    4
  • D
    None of these.
Answer
Correct option: A.
2
  1. 2
Solution:
$\left(x^2+2 x\right)^2-(x+1)^2-55=0$
$\Rightarrow\left(x^2+2 x+1-1\right)^2-(x+1)^2-55=0$
$\Rightarrow\left\{(x+1)^2-1\right\}^2-(x+1)^2-55=0$
$\Rightarrow\left\{(x+1)^2\right\}^2+1-3(x+1)^2-55=0$
$\Rightarrow\left\{(x+1)^2\right\}^2-3(x+1)^2-54=0$
Let $\mathrm{p}=(\mathrm{x}+1)^2$
$\Rightarrow p^2-3 p-54=0$
$\Rightarrow p^2-9 p+6 p-54=0$
$\Rightarrow(p+6)(p-9)=0$
$\Rightarrow p=9 \text { or } p=-6$
Rejecting $p=-6$
$\Rightarrow(x+1)^2=9$
$\Rightarrow x^2+2 x-8=0$
$\Rightarrow(x+4)(x-2)=0$
$\Rightarrow x=2, x=-4$
View full question & answer→
MCQ 91 Mark
A quadratic equation $\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0$ has two distinct real roots, if
  • A
    $a=0$
  • B
    $b^2-4 a c=0$
  • C
    $b^2-4 a c<0$
  • ✓
    $b^2-4 a c>0$
Answer
Correct option: D.
$b^2-4 a c>0$
  1. $b^2-4 a c>0$
Solutions:
If $\mathrm{a}=0$, it becomes linear equation.
If $b^2-4 a c=0$, then there will be real and equal roots.
If $b^2-4 a c<0$, then the roots will be unreal.
Only if $b^2-4 a c>0$, we will get two real distinct roots.
Option D is correct!
View full question & answer→
MCQ 101 Mark
What will be the product of b × c if the equations $x^2 + bx + c = 0$ and $x^2 + 3x + 3 = 0$ have one common root?
  • A
    3
  • B
    4
  • C
    6
  • ✓
    9
Answer
Correct option: D.
9
  1. 9
Solution:
Comparing the coefficients of the above equation we get,
$\frac{1}{1}=\frac{\text{b}}{3}=\frac{\text{c}}{3}$
This means $\text{b}=3$ and $\text{c}=3$
Therefore, $\text{b}\times\text{c}=9$
View full question & answer→
MCQ 111 Mark
Choose the correct answer. 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.
$|z+3|^2$
  1. $|z+3|^2$
Solution:
Let z = x + iy. Then
$(\text{z}+3)(\bar{\text{z}}+3)=(\text{x}+\text{iy}+3)(\text{x}-\text{iy}+3)$
$=(\text{x}+3)^2-(\text{iy})^2$
$=(\text{x}+3)^2+\text{y}^2$
$=|\text{x}+3+\text{iy}|^2=|\text{z}+3|^2$
View full question & answer→
MCQ 121 Mark
For any complex number z, the minimum value of |z| + |z – 2i| is equal to:
  • A
    0
  • B
    1
  • ✓
    2
  • D
    None of these
Answer
Correct option: C.
2
  1. 2
View full question & answer→
MCQ 131 Mark
What will be the value of f(x) if, 2A, A + B, C are integers and $f(x) = Ax^2 + Bx + C = 0$?
  • A
    Natural Number
  • B
    Unpredictable
  • ✓
    Integer
  • D
    Complex Number
Answer
Correct option: C.
Integer
  1. Integer
Solution:
$\text{f}(\text{x}) = A\text{x}^2 + \text{Bx} +\text{C} = 0$
So, $\text{f}(\text{x}) = \text{Ax}^2 + (\text{A} +\text{B})\text{x} – \text{Ax} + \text{C}$
$= \text{Ax}^2 – \text{Ax} + (\text{A} + \text{B}) \text{x} + \text{C}$
$=\frac{2\text{Ax} (\text{x }– 1)}{2 + (\text{A} + \text{B})\text{x} + \text{C}}$
$\therefore\text{f}(\text{x})$ is an integer.
View full question & answer→
MCQ 141 Mark
Convert -1 + i into polar form:
  • A
    $\sqrt{2},\frac{5\pi}{4}$
  • ✓
    $\sqrt{2},\frac{3\pi}{4}$
  • C
    $-\sqrt{2},\frac{\pi}{4}$
  • D
    $\sqrt{2},\frac{\pi}{4}$
Answer
Correct option: B.
$\sqrt{2},\frac{3\pi}{4}$
  1. $\sqrt{2},\frac{3\pi}{4}$
Solution:
$\text{r}=\sqrt{\text{x}^2+\text{y}^2}=\sqrt{(-1)^2+\text{1}^2}=\sqrt{1+1}=\sqrt{2}$
$\text{r}\cos\theta = -1$ and $\text{r}\sin\theta = 1$ so, $\theta$ is in $2^{nd}$ quadrant since sin is positive and cos is negative.
$\tan\theta = -1\Rightarrow \tan\theta =\frac{-\tan\pi}{4}$
$\Rightarrow\tan\theta=\tan(\frac{\pi-\pi}{4})=\frac{\tan3\pi}{4}$
$\Rightarrow\theta=\frac{3\pi}{4}$
View full question & answer→
MCQ 151 Mark
Choose the correct answer. $\sin\text{x}+\text{i}\cos2\text{x}$ 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.
View full question & answer→
MCQ 161 Mark
If, $(\text{a}+1)\text{x}^2+2(\text{a}+1)\text{x}+\text{a}-2=0$ then, for what parameter of ‘a’ the given equation have equal roots?
  • A
    $\Big(-\infty,-1\Big)$
  • B
    $\Big[-1,\infty\Big)$
  • C
    $\Big(0,1\Big)$
  • ✓
    Not possible
Answer
Correct option: D.
Not possible
For, equal roots, $\text{D}=0$
Where, $\text{D}=\text{b}^2-4\text{ac}$
In the equation, $(\text{a}+1)\text{x}^2+2(\text{a}+1)\text{x}+\text{a}-2=0$
$\text{D}=\Big[2(\text{a}+1)\Big]-4(\text{a}+1)(\text{a}-2)$
$=4\text{a}^2+4+8\text{a}-4(\text{a}^2-2\text{a}+\text{a}-2)$
$=4\text{a}^2+4+8\text{a}-4\text{a}^2+4\text{a}+8>0$
$\Rightarrow12\text{a}+12=0$
$\Rightarrow12\text{a}=-12$
$\Rightarrow\text{a}=-1$
So, from here it is clear that $\text{a}=-1$ is not possible because the equation is becoming linear.
View full question & answer→
MCQ 171 Mark
$\frac{1}{\text{Z}}$ is, _________________ for complex number z.
  • A
    additive inverse
  • B
    additive identity element
  • C
    multiplicative identity element
  • ✓
    multiplicative inverse
Answer
Correct option: D.
multiplicative inverse
On multiplying reciprocal of complex number $\frac{1}{\text{z}}$ to complex number z, we get multiplying inverse one i.e. z × 1 = z.
View full question & answer→
MCQ 181 Mark
If $\alpha$ and $\beta$ are the roots of $4x^2+ 3x + 7 = 0$, then the value of $\frac{1}{\alpha}+\frac{1}{\beta}$ is:
  • A
    $\frac{4}{7}$
  • ✓
    $-\frac{3}{7}$
  • C
    $\frac{3}{7}$
  • D
    $-\frac{3}{4}$
Answer
Correct option: B.
$-\frac{3}{7}$
  1. $-\frac{3}{7}$
Solution:
Given equation: $4x^2 + 3x + 7 = 0$
Also, $\alpha$ and $\beta$ are the roots of the equation.
Then, sum of the roots = $\alpha$ + $\beta$ $=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}=-\frac{3}{4}$
Product of the roots $=\alpha\beta=\frac{\text{constant term}}{\text{Coefficient of x}^2}=\frac{7}{4}$
$\therefore\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{-\frac{3}{4}}{\frac{7}{4}}=-\frac{3}{7}$
View full question & answer→
MCQ 191 Mark
The argument of $\frac{1-\text{i}\sqrt{3}}{1+\text{i}\sqrt{3}}$ is:
  • A
    $60^\circ$
  • B
    $120^\circ$
  • C
    $210^\circ$
  • ✓
    $240^\circ$
Answer
Correct option: D.
$240^\circ$
$\frac{1-\text{i}\sqrt{3}}{1+\text{i}\sqrt{3}}$
Rationalising the denominator,
$\frac{1-\text{i}\sqrt{3}}{1+\text{i}\sqrt{3}}\times\frac{1-\text{i}\sqrt{3}}{1-\text{i}\sqrt{3}}$
$=\frac{1+3\text{i}^2-2\sqrt{3}\text{i}}{1-3\text{i}^2}$
$=\frac{-2-2\sqrt{3}\text{i}}{4} \ (\because\text{i}^2=-1)$
$=\frac{-1}{2}-\text{i}\frac{\sqrt{3}}{2}$
Then, $\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=\sqrt{3}$
$\Rightarrow\alpha=60^\circ$
Since the points $(\frac{-1}{2},-\frac{-\sqrt{3}}{2})$ lie in the third quadrant, the argument is given by:
$\theta=180^\circ+60^\circ$
$=240^\circ$
View full question & answer→
MCQ 201 Mark
If z and w are two non - zero complex numbers such that |zw|= 1 and arg(z) – arg(w) $=\frac{\pi}{2}$ then $\overline{\text{z}}$ w is equal to:
  • A
    1
  • ✓
    -i
  • C
    i
  • D
    -1
Answer
Correct option: B.
-i
  1. -i
View full question & answer→
MCQ 211 Mark
If $|\text{z}+4|\leq3,$ then the great est and the least value of $|\text{z}+1|$ are:
  • A
    6 – 6
  • ✓
    6, 0
  • C
    7, 2
  • D
    0, -1
Answer
Correct option: B.
6, 0
  1. 6, 0
View full question & answer→
MCQ 221 Mark
If $\text{a}=\cos\theta+\text{i}\sin\theta,$ then $\frac{1+\text{a}}{1-\text{a}}=$
  • A
    $\cot\frac{\theta}{2}$
  • B
    $\cot\theta$
  • ✓
    $\text{i}\cot\frac{\theta}{2}$
  • D
    $\text{i}\tan\frac{\theta}{2}$
Answer
Correct option: C.
$\text{i}\cot\frac{\theta}{2}$
$\text{a}=\cos\theta+\text{i}\sin\theta$ (given)
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{1+\cos\theta+\text{i}\sin\theta}{1-\cos\theta-\text{i}\sin\theta}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{1+\cos\theta+\text{i}\sin\theta}{1-\cos\theta-\text{i}\sin\theta}\times\frac{1-\cos\theta+\text{i}\sin\theta}{1-\cos\theta+\text{i}\sin\theta}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{(1+\text{i}\sin\theta)^2-\cos^2\theta}{(1-\cos\theta)^2-(\text{i}\sin\theta)^2}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{1-\text{i}\sin^2\theta+2\text{i}\sin\theta-\cos^2\theta}{1+\cos^2\theta-2\cos\theta+\sin^2\theta}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{1-(\sin^2\theta+\cos^2\theta)+2\text{i}\sin\theta}{1+(\sin^2\theta+\cos^2\theta)-2\cos\theta}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{2\text{i}\sin\theta}{2(1-\cos\theta)}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{2\text{i}\sin\frac{\theta}{2}-\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\frac{\text{i}\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}$
$\Rightarrow\frac{1+\text{a}}{1-\text{a}}=\text{i}\cot\frac{\theta}{2}$
View full question & answer→
MCQ 231 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
  1. The x-axis
Solution:
$\Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|=1$
$\Rightarrow \Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|^2=1^2$
$\Rightarrow\Big(\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big)\overline{\Big(\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big)}=1$
$\Rightarrow\Big(\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big)\Big(\frac{-\text{i}+\overline{\text{z}}}{-\text{i}-\overline{\text{z}}}\Big)=1$
$\Rightarrow\Big(\frac{-\text{i}^2-\text{zi}+\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}}{-\text{i}^2+\text{z}\text{i}-\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}}\Big)=1$
$\Rightarrow-\text{i}^2-\text{zi}+\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}=-\text{i}^2+\text{zi}-\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}$
$\Rightarrow-\text{zi}+\overline{\text{z}}\text{i}=\text{zi}-\overline{\text{z}}\text{i}$
$\Rightarrow\overline{\text{z}}\text{i}+\overline{\text{z}}\text{i}=\text{zi}-\text{zi}$
$\Rightarrow2\overline{\text{z}}\text{i}=2\text{zi}$
$\Rightarrow\overline{\text{z}}=\text{z}$
$ \Rightarrow\text{z}$ is purely real.
View full question & answer→
MCQ 241 Mark
If $z=2-3 i$ then $z^2-4 z+13=$
  • ✓
    0
  • B
    1
  • C
    2
  • D
    3
Answer
Correct option: A.
0
  1. 0
Solutions:
$z=2-3 i$
$z^2=2^2-3^2-12 i$
$=-5-12 i$
$\therefore z^2-4 z+13$
$=(-5-12 i)-4(2-3 i)+13$
$=-5-12 i-8+12 i+13$
$=-13+13$
$=0$
View full question & answer→
MCQ 251 Mark
If one root of the equation $x^2+p x+12=0$, is 4 , while the equation $x^2+p x+q=0$ has equal roots, the value of $q$ is:
  • ✓
    $\frac{49}{4}$
  • B
    $\frac{4}{49}$
  • C
    4
  • D
    None of these.
Answer
Correct option: A.
$\frac{49}{4}$
  1. $\frac{49}{4}$
Solution:
It is given that, 4 is the root of the equation $\mathrm{x}^2+\mathrm{px}+12=0$.
$\therefore 16+4 p+12=0$
$\Rightarrow p=-7$
It is also given that, the equation $x^2+p x+q=0$ has equal roots. So, the discriminant of:
$x^2+p x+q=0 \text { will be zero. }$
$\therefore p^2-4 q=0$
$\Rightarrow 4 q=(-7)^2=49$
$\Rightarrow q=\frac{49}{4}$
View full question & answer→
MCQ 261 Mark
Which one is the complete set of values of x satisfying log $\text{x}^2(\text{x+1})>0$
  • A
    $(1,\infty)$
  • B
    $(-1,\text{ 0})-(0)$
  • C
    $(-1,\text{ 1})-(0)$
  • ✓
    $(1,\text{ 0})\cup(1,\infty)$
Answer
Correct option: D.
$(1,\text{ 0})\cup(1,\infty)$
  1. $(1,\text{ 0})\cup(1,\infty)$
Solution:
If, $\text{x}^2>1,$ then $\text{x}+1>0$
So, $\text{x}>0$
$\text{x}\in(1,\infty)$
If, $0<\text{x}<1,$ the $0<\text{x}+1<1$
$\text{x}\in(-1,\text{ 0})$
Thus, $\text{x}\in(1,\infty)\cup(1,\infty)$
View full question & answer→
MCQ 271 Mark
In polar representation of a complex number $(\text{r, } 2\pi)$ lies on, ____________?
  • ✓
    x - axis
  • B
    y - axis
  • C
    z - axis
  • D
    any axis
Answer
Correct option: A.
x - axis
To convert polar representation $(\text{r, }\theta)$ into argand plane (x, y), substitute $\text{x}=\text{r}\cos\theta$ and $\text{y}=\text{r}\sin\theta$
$\text{x}=\text{r}\cos2\pi=\text{r}$ and $\text{y}=\text{r}\sin2\pi=0$ Argand plane representation is (r, 0).
Since imaginary part is zero, so it lies on real axis i.e.x - axis.
View full question & answer→
MCQ 281 Mark
The values of x satisfying $\log _3\left(x^2+4 x+12\right)=2$ are:
  • A
    2, −4
  • B
    1, −3
  • C
    −1, 3
  • ✓
    −1, −3
Answer
Correct option: D.
−1, −3
  1. −1, −3
Solution:
The given equation is $\log _3\left(x^2+4 x+12\right)=2$
$\Rightarrow x^2+4 x+12=3^2=9$
$\Rightarrow x^2+4 x+3=0$
$\Rightarrow(x+1)(x+3)=0$
$\Rightarrow x=-1,-3$
View full question & answer→
MCQ 291 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.
$|\text{z}|^2=|\text{z}|^2$
It is obvious that, for any complex number z,
$|\text{z}|^2=|\text{z}|^2$
View full question & answer→
MCQ 301 Mark
The sum of two complex numbers a + ib and c + id is purely imaginary if
  • ✓
    a + c = 0
  • B
    a + d = 0
  • C
    b + d = 0
  • D
    b + c = 0
Answer
Correct option: A.
a + c = 0
  1. a + c = 0
Solutions:
It is given that
$z_1=a+i b \text { and }$
$z_2=c+i d$
$z_1+z_2=(a+c)+i(b+d)$
$z_1+z_2$ is purely imaginary. (Given)
Then the real part has to be 0 .
Hence
$a+c=0 \text {. }$
View full question & answer→
MCQ 311 Mark
The complete set of values of $k$, for which the quadratic equation $x^2-k x+k+2=0$ has equal roots, consists of:
  • A
    $2+\sqrt{2}$
  • ✓
    $2\pm\sqrt{12}$
  • C
    $2-\sqrt{12}$
  • D
    $-2-\sqrt{12}$
Answer
Correct option: B.
$2\pm\sqrt{12}$
  1. $2\pm\sqrt{12}$
Solution:
Since the equation has real roots.
$\Rightarrow D=0$
$\Rightarrow b^2-4 a c=0$
$\Rightarrow K^2-4(1)(K+2)=0$
$\Rightarrow K^2-4 K-8=0$
$\Rightarrow\text{K}=\frac{4\pm\sqrt{16-4(1)(-8)}}{2(1)}$
$\Rightarrow\text{K}=\frac{4\pm2\sqrt{12}}{2}$
$\Rightarrow\text{K}=2\pm\sqrt{12}$
View full question & answer→
MCQ 321 Mark
If $\text{z}=1-\cos\theta+\text{i}\sin\theta,$ then $|\text{z}|=$
  • A
    $2\sin\frac{\theta}{2}$
  • B
    $2\cos\frac{\theta}{2}$
  • ✓
    $2\Big|\sin\frac{\theta}{2}\Big|$
  • D
    $2\Big|\cos\frac{\theta}{2}\Big|$
Answer
Correct option: C.
$2\Big|\sin\frac{\theta}{2}\Big|$
$\therefore\text{z}=1-\cos\theta+\text{i}\sin\theta$
$\Rightarrow|\text{z}|=\sqrt{(1-\cos\theta)^2+\sin^2\theta}$
$\Rightarrow|\text{z}|=\sqrt{1+\cos^2\theta-2\cos\theta+\sin^2\theta}$
$\Rightarrow|\text{z}|=\sqrt{1+1-2\cos\theta}$
$\Rightarrow|\text{z}|=\sqrt{2(1-2\cos\theta)}$
$\Rightarrow|\text{z}|=\sqrt{4\sin^2\frac{\theta}{2}}$
$\Rightarrow|\text{z}|=2\Big|\sin\frac{\theta}{2}\Big|$
View full question & answer→
MCQ 331 Mark
If, $(\text{a}+1)\text{x}^2+2(\text{a}+1)\text{x}+(\text{a}-2)=0$ then, for what parameter of ‘a’ the given equation have real and distinct roots?
  • A
    $(-\infty, \infty)$
  • ✓
    $(-1,\infty)$
  • C
    $\big[-1,\infty)$
  • D
    $(-1,1)$
Answer
Correct option: B.
$(-1,\infty)$
For, real and distinct roots, $\text{D}>0$
Where, $\text{D}=\text{b}^2-\text{4ac}$
In the equation, $(\text{a}+1)\text{x}^2+2(\text{a}+1)\text{x}+(\text{a}-2)=0$
$\text{D}=\Big[2(\text{a}+1)\big]^2-4(\text{a}+1)(\text{a}-2)$
$=4\text{a}^2+4+\text{8a}-4{(\text{a}^2}-2\text{a}+\text{a}-2)$
$= 4\text{a}^2 + 4 + 8\text{a} – 4\text{a}^2 + 4\text{a} + 8 > 0$
$\Rightarrow12\text{a}+12>0$
$\Rightarrow12\text{a}>-12$
$\Rightarrow\text{a}>-1$
$\therefore\text{a}\in(-1,\infty)$
View full question & answer→
MCQ 341 Mark
Solve – $\text{x}^2+\text{x}-2=0$.
  • A
    $\frac{-1\pm\text{i}\sqrt{7}}{2}$
  • ✓
    $\frac{1\pm\text{i}\sqrt{7}}{2}$
  • C
    $\frac{1\pm\sqrt{7}}{2}$
  • D
    $\frac{1\pm\sqrt{7}}{2}$
Answer
Correct option: B.
$\frac{1\pm\text{i}\sqrt{7}}{2}$
  1. $\frac{1\pm\text{i}\sqrt{7}}{2}$
Solution:
$\text{x}^2+\text{x}-2=0$
$\Rightarrow\text{x}^2-\text{x}+2=0$
$\text{D} =(-1)^2-4\times1\times2=1-8=-7\leq0$
Since $\text{D}\leq0,$ imaginary roots are there.
$\Rightarrow\text{x}=\frac{1\pm\sqrt{\text{D}}}{2.1}=\frac{1\pm\text{i}\sqrt{\text{D}}}{2.1}$
View full question & answer→
MCQ 351 Mark
If $\text{x}+\text{iy}=(1+\text{i})(1+2\text{i})(1+3\text{i}),$ then $\text{x}^2+\text{y}^2=$
  • A
    0
  • B
    1
  • ✓
    100
  • D
    none of these
Answer
Correct option: C.
100
$\therefore\text{x}+\text{iy}=(1+\text{i})(1+2\text{i})(1+3\text{i})$
Taking modulus on both the sides:
$|\text{x}+\text{iy}|=|(1+\text{i})(1+2\text{i})(1+3\text{i})|$
$\Rightarrow|\text{x}+\text{iy}|=|(1+\text{i})|\times|(1+2\text{i})|\times|(1+3\text{i})|$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{1^2+1^2}\sqrt{1^2+2^2}\sqrt{1^2+3^2}$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{2}\sqrt{5}\sqrt{10}$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=\sqrt{100}$
Squaring both the sides,
$\Rightarrow\text{x}^2+\text{y}^2=100$
View full question & answer→
MCQ 361 Mark
Solve $\text{x}^2 + 1 = 0$.
  • A
    x = 1, -1
  • ✓
    x = i, -i
  • C
    x = -1
  • D
    x = i
Answer
Correct option: B.
x = i, -i
  1. x = i, -i
Solution:
$\text{x}^2 + 1 = 0$
$\Rightarrow\text{x}^2=-1\Rightarrow\text{x}$
$=\pm\sqrt{-1}=\pm\text{i}$
View full question & answer→
MCQ 371 Mark
Convert -1 - i into polar form:
  • A
    $\sqrt{2},\frac{5\pi}{4}$
  • B
    $\sqrt{2},\frac{3\pi}{4}$
  • ✓
    $\sqrt{2},\frac{-3\pi}{4}$
  • D
    $\sqrt{2},\frac{\pi}{4}$
Answer
Correct option: C.
$\sqrt{2},\frac{-3\pi}{4}$
  1. $\sqrt{2},\frac{-3\pi}{4}$
Solution:
$\text{r}=\sqrt{\text{x}^2+\text{y}^2}=\sqrt{(-1)^2+\text{1}^2}=\sqrt{1+1}=\sqrt{2}$
$\text{r}\cos\theta=-1$ and $\text{r}\sin\theta=-1\Rightarrow\theta$ is in $3^{rd}$ quadrant since sin and cos both negative.
$\tan\theta=1\Rightarrow\theta=\frac{-3\pi}{4}$
View full question & answer→
MCQ 381 Mark
If $\text{i}^2=-1,$ then the sum $\text{i}+\text{i}^2+\text{i}^3+...$ upto 1000 terms is equal to:
  • A
    1
  • B
    -1
  • C
    i
  • ✓
    0
Answer
Correct option: D.
0
$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4...\text{i}^{1000}$
$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4 \ [\because\text{i}^2=-1,\text{i}^3=-\text{i} \ \text{and} \ \text{i}^4=1]$
$=\text{i}-1-\text{i}+1$
$=0$
Similarly, the sum of the next four terms of the series will be equal to 0. This is because the powers of i follow a cyclicity of 4. Hence, the sum of all terms, till 1000, will be zero.
$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4...\text{i}^{1000}=0$
View full question & answer→
MCQ 391 Mark
If, $\alpha$ and $\beta$ are the roots of the equation $2 \mathrm{x}^2-3 \mathrm{x}-6=0$, then what is the equation whose roots are $\mathrm{a}^2+2$ and $\beta^2+2$
 
  • A
    $4 \mathrm{x}^2+49 \mathrm{x}+118=0$
  • ✓
    $4 x^2-49 x+118=0$
  • C
    $4 x^2-49 x-118=0$
  • D
    $x^2-49 x+118=0$
Answer
Correct option: B.
$4 x^2-49 x+118=0$
  1. $4 x^2-49 x+118=0$
Solution:
Let, $y=x^2+2$
Then, $2 x^2-3 x-6=0$
So, $(3 x)^2=\left(2 x^2-6\right)^2$
$[2(y-2)-6]^2=9(y-2)$
$=4 x^2-49 x+118=0$
View full question & answer→
MCQ 401 Mark
( x + 3 ) + i ( y - 2 ) = 5 + i2, find the values of x and y.
  • A
    x = 8 and y = 4
  • ✓
    x = 2 and y = 4
  • C
    x = 2 and y = 0
  • D
    x = 8 and y = 0
Answer
Correct option: B.
x = 2 and y = 4
If two complex numbers are equal, then corresponding parts are equal i.e. real parts of both are equal and imaginary parts of both are equal.
x + 3 = 5 and y - 2 = 2
x = 5 - 3 and y = 2 + 2
x = 2 and y = 4.
View full question & answer→
MCQ 411 Mark
The number of solutions of $x^2+|x-1|=1$ is:
  • A
    0
  • B
    1
  • ✓
    2
  • D
    3
Answer
Correct option: C.
2
  1. 2
Solution:
$x^2+|x-1|=x^2+x-, x>1$
$=x^2-x+1, x<1$
i. $x^2+x-1=1$
$\Rightarrow x^2+x-2=0$
$\Rightarrow x^2+2 x-x-2=0$
$\Rightarrow x(x+2)-1(x+2)=0$
$\Rightarrow x+2=0 \text { or } x-1=0$
$\Rightarrow x=-2 \text { or } x=1$
Since -2 does not satisfy the condition $x \geq 1$
ii. $x^2-x+1=1$
$\Rightarrow x^2-x=0$
$\Rightarrow x(x-1)=0$
$\Rightarrow x=0 \text { or }(x-1)=0$
$\Rightarrow x=0, x=1$
$\mathrm{x}=1$ does not satisfy the condition $\mathrm{x}<1$
So, there are two solutions.
View full question & answer→
MCQ 421 Mark
If z $=\text{i}-\frac{\text{i}\sqrt{3}}{1},+\sqrt{3},$ then arg(z) is equal to:
  • A
    $\frac{\pi}{3}$
  • B
    $\frac{2\pi}{3}$
  • ✓
    $\frac{-2\pi}{3}$
  • D
    $\text{None of these}$
Answer
Correct option: C.
$\frac{-2\pi}{3}$
  1. $\frac{-2\pi}{3}$
View full question & answer→
MCQ 431 Mark
$\big(\sqrt{-2}\big)\big(\sqrt{-3}\big)$ is equal to:
  • A
    $\sqrt{6}$
  • ✓
    $-\sqrt{6}$
  • C
    $\text{i}\sqrt{6}$
  • D
    none of these.
Answer
Correct option: B.
$-\sqrt{6}$
$\sqrt{-2}\times\sqrt{-3}$
$=\sqrt{2}\times\sqrt{3}\times\sqrt{-1}\times\sqrt{-1}$
$=\sqrt{6}\times\text{i}\times\text{i}$
$=\sqrt{6}\times\text{i}^2$
$=-\sqrt{6} \ [\because\text{i}^2=-1]$
View full question & answer→
MCQ 441 Mark
Choose the correct answer.$\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|$ 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.
$\arg(\text{z}_1)=\arg(\text{z}_2)$
  1. $\text{arg}(\text{z}_1)=\text{arg}(\text{z}_2)$
Solution:
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, $\text{arg}(\text{z}_1)=\text{arg}(\text{z}_2)$
View full question & answer→
MCQ 451 Mark
If $\frac{3+2\text{i}\sin\theta}{1-2\text{i}\sin\theta}$ is a real number and $0 < \theta < 2\pi,$ then $\theta=$
  • ✓
    $\pi$
  • B
    $\frac{\pi}{2}$
  • C
    $\frac{\pi}{3}$
  • D
    $\frac{\pi}{6}$
Answer
Correct option: A.
$\pi$
Given:
$\frac{3+2\text{i}\sin\theta}{1-2\text{i}\sin\theta}$ is a real number
On rationalising, we get,
$\frac{3+2\text{i}\sin\theta}{1-2\text{i}\sin\theta}\times\frac{{1+2\text{i}\sin\theta}}{{1+2\text{i}\sin\theta}}$
$=\frac{(3+2\text{i}\sin\theta)({1+2\text{i}\sin\theta})}{(1)^2-(2\text{i}\sin\theta)^2}$
$=\frac{3+2\text{i}\sin\theta+{6\text{i}\sin\theta}+4\text{i}^2\sin^2\theta}{1+4\sin^2\theta}$
$=\frac{3-4\text{i}\sin^2\theta+{8\text{i}\sin\theta}}{1+4\sin^2\theta} \ [\because\text{i}^2=-1]$
$=\frac{3-4\text{i}\sin^2\theta}{1+4\sin^2\theta}+\text{i}\frac{8\sin\theta}{1+4\sin^2\theta}$
For the above term to be real, the imaginary part has to be zero.
$\therefore\frac{8\sin\theta}{1+4\sin^2\theta}=0$
$\Rightarrow8\sin\theta=0$
For this to be zero,
$\sin\theta=0$
$\Rightarrow\theta=0,\pi,2\pi,3\pi...$
But $0<\theta<2\pi$
Hence, $\theta=\pi$
View full question & answer→
MCQ 461 Mark
The square root of (- 15 – 8i) is:
  • ✓
    $\pm(1-4\text{i})$
  • B
    $\pm(1+4\text{i})$
  • C
    $\pm(-2+4\text{i})$
  • D
    $\text{None of these}$
Answer
Correct option: A.
$\pm(1-4\text{i})$
  1. $\pm(1-4\text{i})$
View full question & answer→
MCQ 471 Mark
Choose the correct answer. 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
  1. The x-axis
Solution:
Given that, $\Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|=1$
Let $\text{z}=\text{x}+\text{yi}$
$\therefore\ \Big|\frac{\text{i}+\text{x}+\text{yi}}{\text{i}-\text{x}-\text{yi}}\Big|=1$
$\Rightarrow\ \bigg|\frac{\text{x}-(\text{y}+1)\text{i}}{-\text{x}-(\text{y}-1)\text{i}}\bigg|=1$
$\Rightarrow|\text{x}+(\text{y}+1)\text{i}|=|-\text{x}-(\text{y}-1)\text{i}|$
$\Rightarrow\sqrt{\text{x}^2+(\text{y}+1)^2}=\sqrt{\text{x}^2+(\text{y}-1)^2}$
$\Rightarrow\ \text{x}^2+(\text{y}^2+1)^2=\text{x}^2+(\text{y}-1)^2$
$\Rightarrow(\text{y}+1)^2=(\text{y}-1)^2$
$\Rightarrow\text{y}^2+2\text{y}+1=\text{y}^2-2\text{y}+1$
$\Rightarrow2\text{y}=-2\text{y}$
$\Rightarrow2\text{y}+2\text{y}=0$
$\Rightarrow4\text{y}=0$
$\Rightarrow\text{y}=0$
$\Rightarrow\text{x-axis}$
View full question & answer→
MCQ 481 Mark
If arg (z – 1) = arg (z + 3i), then x – 1 : y is equal to:
  • A
    3 : 1
  • ✓
    1 : 3
  • C
    3 : 2
  • D
    2 : 3
Answer
Correct option: B.
1 : 3
  1. 1 : 3
View full question & answer→
MCQ 491 Mark
If the difference of the roots of $x^2-p x+q=0$ is unity, then:
  • A
    $p^2+4 q=1$
  • ✓
    $p^2-4 q=1$
  • C
    $p^2+4 q^2=(1+2 q)^2$
  • D
    $4 p^2+q^2=(1+2 p)^2$
Answer
Correct option: B.
$p^2-4 q=1$
  1. $p^2-4 q=1$
Solution:
Given equation: $x^2-p x+q=0$
Also, $\alpha$ and $\beta$ are the roots of the equation such that $\alpha-\beta=1$
Sum of the roots $=\alpha+\beta=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}=-\Big(\frac{-\text{p}}{1}\Big)=\text{p}$
Product of the roots $=\alpha\beta=\frac{\text{Constant term}}{\text{Coefficient of x}^2}=\text{q}$
$\therefore(\alpha+\beta)^2-(\alpha-\beta)^2=4\alpha\beta$
$\Rightarrow p^2-1=4 q$
$\Rightarrow p^2-4 q=1.$
View full question & answer→
MCQ 501 Mark
Convert $\Big(8, \frac{2\pi}{3}\Big)$ into Argand plane representation:
  • ✓
    $(\text{-4, } 4\sqrt{3})$
  • B
    $(\text{4, } 4\sqrt{3})$
  • C
    $(\text{4} \sqrt{\text{3, }}4)$
  • D
    $(\text{-4} \sqrt{\text{3, }}4)$
Answer
Correct option: A.
$(\text{-4, } 4\sqrt{3})$
To convert polar representation $(\text{r, }\theta)$ into argand plane $(\text{x, }\text{y}),$ substitute $\text{x}=\text{r}\cos\theta$ and $\text{y}=\text{r}\sin\theta$
$\text{x}=8\cos\frac{2\pi}{3}=8\cos\Big(\frac{\pi-\pi}{3}\Big)=8\Big(\frac{-1}{2}\Big)=-4$
$\text{y}=8\cos\frac{2\pi}{3}=8\cos\Big(\frac{\pi-\pi}{3}\Big)=8\Big(\frac{\sqrt{3}}{2}\Big)=4\sqrt{3}$
View full question & answer→
Generate Paper FreeAll COMPLEX NUMBERS AND QUADRATIC EQUATIONS topics
MCQ - MATHS STD 11 Science Questions - Vidyadip