MCQ 11 Mark
If $z$ is a comp lex num ber, then $|3z – 1|= 3|z – 2|$ represents:
AnswerCorrect option: D. a line parallel to $y-$axis
View full question & answer→MCQ 21 Mark
If $z_1 = 2 + 3i$ and $z_2 = 5 + 2i,$ then find sum of two complex numbers:
- A
$4 + 8i$
- B
$3 - i$
- ✓
$7 + 5i$
- D
$7 - 5i$
AnswerCorrect option: C. $7 + 5i$
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 $\text{n}\in\text{N}$
AnswerCorrect 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:
AnswerComparing 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:
AnswerGiven 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
$\Big(\frac{1-5\text{a}}{\text{a}}\Big)<0$
AnswerCorrect 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
$\cos\pi-\text{i}\sin\pi$
- ✓
$\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}$
AnswerCorrect 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 $(x^2 + 2x)^2− (x + 1)^2 − 55 = 0:$
Answer$(x^2 + 2x)^2 - (x + 1)^2 - 55 = 0$
$\Rightarrow (x^2+ 2x + 1 - 1)^2 - (x + 1)^2 - 55 = 0$
$\Rightarrow\Big\{(\text{x}+1)^2-1\Big\}^2-(\text{x}+1)^2-55=0$
$\Rightarrow\Big\{(\text{x}+1)^2\Big\}^2+1-3(\text{x}+1)^2-55=0$
$\Rightarrow\Big\{(\text{x}+1)^2\Big\}^2-3(\text{x}+1)^2-54=0$
Let $p = (x + 1)^2$
$\Rightarrow p^2 - 3p - 54 = 0$
$\Rightarrow p^2- 9p + 6p - 54 = 0$
$\Rightarrow (p + 6) (p - 9) = 0$
$\Rightarrow p = 9$ or $p = -6$
Rejecting $p = −6$
$\Rightarrow (x + 1)^2 = 9$
$\Rightarrow x^2 + 2x - 8 = 0$
$\Rightarrow (x + 4) (x - 2) = 0$
$\Rightarrow x = 2, x = -4$
View full question & answer→MCQ 91 Mark
A quadratic equation $ax^2 + bx + c = 0$ has two distinct real roots, if
- A
$a = 0$
- B
$b^2- 4ac = 0$
- C
$b^2 - 4ac < 0$
- ✓
$b^2- 4ac > 0$
AnswerCorrect option: D. $b^2- 4ac > 0$
If $a = 0,$ it becomes linear equation.
If $b^2 - 4ac = 0$, then there will be real and equal roots.
If $b^2 - 4ac < 0,$ then the roots will be unreal.
Only if $b^2 - 4ac > 0,$ we will get two real distinct roots.
Option $D$ is correct!
View full question & answer→MCQ 101 Mark
What will be the product of $b \times c$ if the equations $x^2 + bx + c = 0$ and $x^2 + 3x + 3 = 0$ have one common root?
AnswerComparing 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
AnswerCorrect option: A. $|z + 3|^2$
Let $z = x + iy.$ Then
$(\text{z}+3)(\bar{\text{z}}+3)=(\text{x}+\text{iy}+3)(\text{x}-\text{iy}+3)$
$=(\text{x}+3)^2-(\text{iy})^2$
$=(\text{x}+3)^2+\text{y}^2$
$=|\text{x}+3+\text{iy}|^2$
$=|\text{z}+3|^2$
View full question & answer→MCQ 121 Mark
For any complex number $z,$ the minimum value of $|z| + |z – 2i|$ is equal to:
View full question & answer→MCQ 131 Mark
What will be the value of $f(x)$ if, $\text{2A, A + B, C}$ are integers and $f(x) = Ax^2 + Bx + C = 0?$
Answer$\text{f}(\text{x}) = A\text{x}^2 + \text{Bx} +\text{C} = 0$
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
$\sqrt{2},\frac{\pi}{4}$
AnswerCorrect option: B. $\sqrt{2},\frac{3\pi}{4}$
$\text{r}=\sqrt{\text{x}^2+\text{y}^2}=\sqrt{(-1)^2+\text{1}^2}=\sqrt{1+1}=\sqrt{2}$
$\text{r}\cos\theta = -1$ 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:
AnswerCorrect 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
$\Big(0,1\Big)$
- ✓
AnswerFor, 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
- B
additive identity element
- C
multiplicative identity element
- ✓
AnswerOn multiplying reciprocal of complex number $\frac{1}{\text{z}}$ to complex number $z,$
we get multiplying inverse one
i.e. $z \times 1 = z$.
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
$-\frac{3}{4}$
AnswerCorrect option: B. $-\frac{3}{7}$
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
$210^\circ$
- ✓
$240^\circ$
AnswerCorrect 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 $\text{arg(z) – arg(w)} =\frac{\pi}{2}$ then $\overline{\text{z}} w$ is equal to:
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
$0, -1$
AnswerCorrect option: B. $6, 0$
View full question & answer→MCQ 221 Mark
If $\text{a}=\cos\theta+\text{i}\sin\theta,$ then $\frac{1+\text{a}}{1-\text{a}}=$
AnswerCorrect 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 $x + y = 1$
AnswerCorrect option: B. The $x-$axis
$\Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|=1$
$\Rightarrow \Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|^2=1^2$
$\Rightarrow\Big(\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big)\overline{\Big(\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big)}=1$
$\Rightarrow\Big(\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big)\Big(\frac{-\text{i}+\overline{\text{z}}}{-\text{i}-\overline{\text{z}}}\Big)=1$
$\Rightarrow\Big(\frac{-\text{i}^2-\text{zi}+\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}}{-\text{i}^2+\text{z}\text{i}-\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}}\Big)=1$
$\Rightarrow-\text{i}^2-\text{zi}+\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}=-\text{i}^2+\text{zi}-\overline{\text{z}}\text{i}+\text{z}\overline{\text{z}}$
$\Rightarrow-\text{zi}+\overline{\text{z}}\text{i}=\text{zi}-\overline{\text{z}}\text{i}$
$\Rightarrow\overline{\text{z}}\text{i}+\overline{\text{z}}\text{i}=\text{zi}-\text{zi}$
$\Rightarrow2\overline{\text{z}}\text{i}=2\text{zi}$
$\Rightarrow\overline{\text{z}}=\text{z}$
$ \Rightarrow\text{z}$ is purely real.
View full question & answer→MCQ 241 Mark
If $z = 2 - 3i$ then $z^2 - 4z + 13 =$
Answer$z = 2 - 3i$
$z^2 = 2^2 - 3^2 - 12i$
$= -5 - 12i$
$\therefore z^2 - 4z + 13$
$= (-5 - 12i) - 4(2 - 3i) + 13$
$= -5 - 12i - 8 + 12i + 13$
$= -13 + 13$
$= 0$
View full question & answer→MCQ 251 Mark
If one root of the equation $x^2+ px + 12 = 0,$ is $4,$ while the equation $x^2+ px + q = 0$ has equal roots, the value of $q$ is:
- ✓
$\frac{49}{4}$
- B
$\frac{4}{49}$
- C
$4$
- D
AnswerCorrect option: A. $\frac{49}{4}$
It is given that, $4$ is the root of the equation $x^2+ px + 12 = 0.$
$\therefore 16 + 4p + 12 = 0$
$\Rightarrow p = -7$
It is also given that, the equation $x^2 + px + q = 0$ has equal roots.
So, the discriminant of:
$x^2 + px + q = 0$ will be zero.
$\therefore p^2 - 4q = 0$
$\Rightarrow 4q = (-7)^2 = 49$
$\Rightarrow\text{q}=\frac{49}{4}$
View full question & answer→MCQ 261 Mark
Which one is the complete set of values of $x$ satisfying log $\text{x}^2(\text{x+1})>0$
AnswerCorrect option: D. $(1,\text{ 0})\cup(1,\infty)$
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
AnswerCorrect 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 $\log_3 (x^2+ 4x + 12) = 2$ are:
- A
$2, −4$
- B
$1, −3$
- C
$−1, 3$
- ✓
$−1, −3$
AnswerCorrect option: D. $−1, −3$
The given equation is $\log_3(x^2 + 4x + 12) = 2$
$\Rightarrow x^2 + 4x + 12 = 3^2= 9$
$\Rightarrow x^2 + 4x + 3 = 0$
$\Rightarrow (x + 1) (x + 3) = 0$
$\Rightarrow x = -1, -3$
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
$|\text{z}|^2\geq|\text{z}|^2$
AnswerCorrect 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
$b + c = 0$
AnswerCorrect option: A. $a + c = 0$
It is given that
$z_1=a+i b$ 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
$a + c = 0.$
View full question & answer→MCQ 311 Mark
The complete set of values of $k$, for which the quadratic equation $x^2− kx + k + 2 = 0$ has equal roots, consists of:
- A
$2+\sqrt{2}$
- ✓
$2\pm\sqrt{12}$
- C
$2-\sqrt{12}$
- D
$-2-\sqrt{12}$
AnswerCorrect option: B. $2\pm\sqrt{12}$
Since the equation has real roots.
$\Rightarrow D = 0$
$\Rightarrow b^2 - 4ac = 0$
$\Rightarrow K^2 - 4 (1)(K + 2) = 0$
$\Rightarrow K^2 - 4K - 8 = 0$
$\Rightarrow\text{K}=\frac{4\pm\sqrt{16-4(1)(-8)}}{2(1)}$
$\Rightarrow\text{K}=\frac{4\pm2\sqrt{12}}{2}$
$\Rightarrow\text{K}=2\pm\sqrt{12}$
View full question & answer→MCQ 321 Mark
If $\text{z}=1-\cos\theta+\text{i}\sin\theta,$ then $|\text{z}|=$
AnswerCorrect 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
$(-1,1)$
AnswerCorrect 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 $x^2 + x – 2 = 0$.
AnswerCorrect option: B. $\frac{1\pm\text{i}\sqrt{7}}{2}$
$\text{x}^2+\text{x}-2=0$
$\Rightarrow\text{x}^2-\text{x}+2=0$
$\text{D} =(-1)^2-4\times1\times2=1-8=-7\leq0$
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 $\text{x}^2+\text{y}^2=$
Answer$\therefore\text{x}+\text{iy}=(1+\text{i})(1+2\text{i})(1+3\text{i})$
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 $x^2 + 1 = 0$.
- A
$x = 1, -1$
- ✓
$x = i, -i$
- C
$x = -1$
- D
$x = i$
AnswerCorrect option: B. $x = i, -i$
$\text{x}^2 + 1 = 0$
$\Rightarrow\text{x}^2=-1$
$\Rightarrow\text{x}$
$=\pm\sqrt{-1}=\pm\text{i}$
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
$\sqrt{2},\frac{\pi}{4}$
AnswerCorrect option: C. $\sqrt{2},\frac{-3\pi}{4}$
$\text{r}=\sqrt{\text{x}^2+\text{y}^2}$
$=\sqrt{(-1)^2+\text{1}^2}$
$=\sqrt{1+1}$
$=\sqrt{2}$
$\text{r}\cos\theta=-1$ 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:
Answer$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4...\text{i}^{1000}$
$\text{i}+\text{i}^2+\text{i}^3+\text{i}^4$
$[\because\text{i}^2=-1,\text{i}^3=-\text{i}$ 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 $2x^2 – 3x – 6 = 0$, then what is the equation whose roots are $\text{a}^2+2$ and $\beta^2+2$
- A
$4x^2 + 49x + 118 = 0$
- ✓
$4x^2 – 49x + 118 = 0$
- C
$4x^2 – 49x – 118 = 0$
- D
$x^2 – 49x + 118 = 0$
AnswerCorrect option: B. $4x^2 – 49x + 118 = 0$
Let, $y=x^2+2$
Then, $2 x^2-3 x-6=0 $
$ \text { 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
$( x + 3 ) + i ( y - 2 ) = 5 + i^2,$ 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 $y = 0$
AnswerCorrect 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:
Answer$ x^2+|x-1|=x^2+x-, x>1 $
$ =x^2-x+1, x<1 $
$x^2 + x − 1 = 1$
$\Rightarrow x^2+ x - 2 = 0$
$\Rightarrow x^2 + 2x - x - 2 = 0$
$\Rightarrow x(x + 2) -1 (x + 2) = 0$
$\Rightarrow x + 2 = 0$ or $x - 1 = 0$
$\Rightarrow x = -2$ or $x = 1$
Since $−2$ does not satisfy the condition $x \geq 1$
$x^2 - x + 1 = 1$
$\Rightarrow x^2 - x = 0$
$\Rightarrow x (x - 1) = 0$
$\Rightarrow x = 0$ or $(x - 1) = 0$
$\Rightarrow x = 0, x = 1$
$x = 1$ does not satisfy the condition $x < 1$
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 $\text{arg(z)}$ is equal to:
- A
$\frac{\pi}{3}$
- B
$\frac{2\pi}{3}$
- ✓
$\frac{-2\pi}{3}$
- D
$\text{None of these}$
AnswerCorrect option: C. $\frac{-2\pi}{3}$
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
AnswerCorrect 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. $|z_1 + z_2| = |z_1| + |z_2|$ 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
$|\text{z}_1|=|\text{z}_2|$
AnswerCorrect option: C. $\arg(\text{z}_1)=\arg(\text{z}_2)$
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, $\text{arg}(\text{z}_1)=\text{arg}(\text{z}_2)$
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
$\frac{\pi}{6}$
AnswerGiven:
$\frac{3+2\text{i}\sin\theta}{1-2\text{i}\sin\theta}$ 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
$\text{None of these}$
AnswerCorrect option: A. $\pm(1-4\text{i})$
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 $x + y = 1$
AnswerCorrect option: B. The $x-$axis
Given that, $\Big|\frac{\text{i}+\text{z}}{\text{i}-\text{z}}\Big|=1$
Let $\text{z}=\text{x}+\text{yi}$
$\therefore\ \Big|\frac{\text{i}+\text{x}+\text{yi}}{\text{i}-\text{x}-\text{yi}}\Big|=1$
$\Rightarrow\ \bigg|\frac{\text{x}-(\text{y}+1)\text{i}}{-\text{x}-(\text{y}-1)\text{i}}\bigg|=1$
$\Rightarrow|\text{x}+(\text{y}+1)\text{i}|=|-\text{x}-(\text{y}-1)\text{i}|$
$\Rightarrow\sqrt{\text{x}^2+(\text{y}+1)^2}=\sqrt{\text{x}^2+(\text{y}-1)^2}$
$\Rightarrow\ \text{x}^2+(\text{y}^2+1)^2=\text{x}^2+(\text{y}-1)^2$
$\Rightarrow(\text{y}+1)^2=(\text{y}-1)^2$
$\Rightarrow\text{y}^2+2\text{y}+1=\text{y}^2-2\text{y}+1$
$\Rightarrow2\text{y}=-2\text{y}$
$\Rightarrow2\text{y}+2\text{y}=0$
$\Rightarrow4\text{y}=0$
$\Rightarrow\text{y}=0$
$\Rightarrow\text{x-axis}$
View full question & answer→MCQ 481 Mark
If $\text{arg (z – 1) = arg (z + 3i),}$ then $x – 1 : y$ is equal to:
- A
$3 : 1$
- ✓
$1 : 3$
- C
$3 : 2$
- D
$2 : 3$
AnswerCorrect option: B. $1 : 3$
View full question & answer→MCQ 491 Mark
If the difference of the roots of $x^2 − px + q = 0$ is unity, then:
AnswerCorrect option: B. $p^2 − 4q = 1$
Given equation: $x^2 + px + q = 0$
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 $=\alpha\beta=\frac{\text{Constant term}}{\text{Coefficient of x}^2}=\text{q}$
$\therefore(\alpha+\beta)^2-(\alpha-\beta)^2=4\alpha\beta$
$\Rightarrow p^2 - 1 = 4q$
$\Rightarrow p^2 - 4q =1$.
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
$(\text{-4} \sqrt{\text{3, }}4)$
AnswerCorrect 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→