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→MCQ 511 Mark
The value of $(1+\text{i})(1+\text{i}^2)(1+\text{i}^3)(1+\text{i}^4)$ is
Answer$(1+\text{i})(1+\text{i}^2)(1+\text{i}^3)(1+\text{i}^4)$
$=(1+\text{i})(1-1)(1-\text{i})(1+1) \ \big(\because\text{i}^2=-1, \beta=-\text{i and} \ \text{i}^4=1\big)$
$=(1+\text{i})(0)(1-\text{i})(2)$
$=0$
View full question & answer→MCQ 521 Mark
The values of $k$ for which the quadratic equation $k x^2+1=k x+3 x-11 x^2$ has real and equal roots are:
- A
$-11, -3$
- B
$5, 7$
- ✓
$5, -7$
- D
AnswerCorrect option: C. $5, -7$
The given equation is $k x^2+1=k x+3 x-11 x^2$ which can be written as.
$ k x^2+11 x^2-k x-3 x+1= $
$ \Rightarrow(k+11) x^2-(k+3) x+1=0$
For equal and real roots, the discriminant of $(k+11) x^2-(k+3) x+1=0$.
$ \therefore(k+3)^2-4(k+11)=0 $
$ \Rightarrow k^2+2 k-35=0 $
$\Rightarrow(k-5)(k+7)=0 $
$ \Rightarrow k=5,-7$
Hence, the equation has real and equal roots when $k=5,-7$.
View full question & answer→MCQ 531 Mark
If $\text{z}=\frac{-2}{1+\sqrt{3}},$ then the value of $\text{arg (z)}$ is:
- A
$\pi$
- B
$\frac{\pi}{3}$
- ✓
$\frac{2\pi}{3}$
- D
$\frac{\pi}{4}$
AnswerCorrect option: C. $\frac{2\pi}{3}$
$\text{z}=\frac{-2}{1+\sqrt{3}}$
Rationalising $z$, we get,
$\text{z}=\frac{-2}{1+\text{i}\sqrt{3}}\times\frac{1-\text{i}\sqrt{3}}{1-\text{i}\sqrt{3}}$
$\Rightarrow\text{z}=\frac{-2+\text{i}2\sqrt{3}}{1+3}$
$\Rightarrow\text{z}=\frac{-1+\text{i}\sqrt{3}}{2}$
$\Rightarrow\text{z}=\frac{-1}{2}+\frac{\text{i}\sqrt{3}}{2}$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=\sqrt{3}$
$\Rightarrow\alpha=\frac{\pi}{3}$
Since, $z$ lies in the second quadrant.
Therefore, $\text{arg(z)}=\pi-\frac{\pi}{3}$
$=\frac{2\pi}{3}$
View full question & answer→MCQ 541 Mark
$6i$ is point on $...........?$
- A
$x -$ axis
- ✓
$y -$ axis
- C
$z -$ axis
- D
AnswerCorrect option: B. $y -$ axis
Since real part of complex number is zero.
So, it is plotted on imaginary axis
i.e. $y -$ axis.
$6i$ is point on $y -$ axis.
View full question & answer→MCQ 551 Mark
If the complex number $\text{z}=\text{x}+\text{iy}$ satisfies the condition $|\text{z}+1|=1,$ then $z$ lies on:
AnswerCorrect option: B. circle with centre $(-1, 0)$ and radius $1$
$|\text{z}+1|=1$
$\Rightarrow|\text{z+1|}^2=1^2$
$\Rightarrow(\text{z}+1)\overline{(\text{z}+1)}=1$
$\Rightarrow(\text{z}+1)(\overline{z}+1)=1$
$\Rightarrow\text{z}\overline{\text{z}}+\text{z}+\overline{\text{z}}+1=1$
$\Rightarrow\text{z}\overline{\text{z}}+\text{z}+\overline{\text{z}}=0$
Since, $\text{z}=\text{x}+\text{iy}$
$\therefore\text{z}\overline{\text{z}}+\text{z}+\overline{\text{z}}=0$
$\Rightarrow(\text{x}+\text{iy})(\text{x}-\text{iy})+\text{x}+\text{iy}+\text{x}- \text{iy}=0$
$\Rightarrow\text{x}^2+\text{y}^2+\text{2x}=0$
$\Rightarrow(\text{x}+1)^2+(\text{y}-0)^2=1^2$
which is the equation of a circle with the center $(-1,\ 0)$ and radius $1.$
View full question & answer→MCQ 561 Mark
Choose the correct answer. 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
AnswerCorrect 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, $|\text{f(z)}|=\sqrt{(1)+\Big(\frac{1}{2}\Big)^2}$
$=\sqrt{1+\frac{1}{4}}$
$=\frac{\sqrt{5}}{2}$
$=\frac{|\text{z}|}{2}$
View full question & answer→MCQ 571 Mark
Let $S$ denotes the set of all real values of the parameter $'a\ '$ for which every solution of the inequality $\log\frac{1}{2} \text{x}^2\geq \log\frac{1}{2} (\text{x} + 2)$ is the solution of the inequality $49\text{x}^2-4\text{a}^4\leq0$ What is the value of $S?$
- A
$(-\infty,-\sqrt{7})\cup(\sqrt{7},\infty)$
- ✓
$(-\infty,-\sqrt{7}\Big]\cup\Big[\sqrt{7},\infty)$
- C
$(-\sqrt{7, }\sqrt{7})$
- D
$\Big[-\sqrt{7, }\sqrt{7}\Big]$
AnswerCorrect option: B. $(-\infty,-\sqrt{7}\Big]\cup\Big[\sqrt{7},\infty)$
We have, $\log\frac{1}{2} \text{x}^2\geq \log\frac{1}{2} (\text{x} + 2)$
$\Rightarrow\text{x}^2\leq\text{x}+2$
$\Rightarrow-1\leq\text{x}\leq2$
And, $49\text{x}^2-4\text{a}^4\leq0$
$\text{ i.}\text{e.}\text{x}^2\leq\frac{4\text{a}^4}{49}$
$\Rightarrow\frac{-2\text{a}^2}{7}\leq\text{a}\leq\frac{2\text{a}^2}{7}$
From the above equations,
$\frac{-2\text{a}^2}{7}\leq-1$ and $2\leq\frac{2\text{a}^2}{7}$
$\text{i.}\text{e.} a^2 \in\frac{7}{2} $ and $\text{a}^2\geq7$
$\Rightarrow\text{a}\in(-\infty,-\sqrt{7}\Big]\cup\Big[\sqrt{7},\infty)$
So, $\text{S}=(-\infty,-\sqrt{7}\Big]\cup\Big[\sqrt{7},\infty)$
View full question & answer→MCQ 581 Mark
If $\text{f(z)}=\frac{7-\text{z}}{1-\text{z}^2},$ where $\text{z}=1+2\text{i},$ then $|\text{f(z)}|$ is:
- ✓
$\frac{|\text{z}|}{2}$
- B
$|\text{z}|$
- C
$2|\text{z}|$
- D
AnswerCorrect option: A. $\frac{|\text{z}|}{2}$
$\text{f(z)}=\frac{7-\text{z}}{1-\text{z}^2}$
$=\frac{7-(1+2\text{i})}{1-(1+2\text{i})^2}$
$=\frac{7-1+2\text{i}}{1-(1^2+2^2\text{i}^2+4\text{i})}$
$=\frac{6-2\text{i}}{1-1+4-4\text{i}}$
$=\frac{6-2\text{i}}{4-4\text{i}}$
$=\frac{6-2\text{i}}{4-4\text{i}}\times\frac{4+4\text{i}}{4+4\text{i}}$
$=\frac{24+24\text{i}-8\text{i}-8\text{i}^2}{4^2-4^2\text{i}^2}$
$=\frac{24+16\text{i}+8}{16+16}$
$=\frac{32+16\text{i}}{32}$
$=1+\frac{1}{2}\text{i}$
since $\text{z}=1+2\text{i},$
$\because|\text{z}|=\sqrt{(1)^2+(2)^2}$
$=\sqrt{1+4}$
$=\sqrt{5}$
$\therefore|\text{f}\text{(z)|}=\sqrt{(1)^1+(\frac{1}{2})^2}$
$=\sqrt{1+\frac{1}{4}}$
$=\frac{\sqrt5}{2}$
$=\frac{|\text{z}|}{2}$
View full question & answer→MCQ 591 Mark
$a + i b > c + id$ can be explained only when:
- A
$b = 0, c = 0$
- ✓
$b = 0, d = 0$
- C
$a = 0, c = 0$
- D
$a = 0, d = 0$
AnswerCorrect option: B. $b = 0, d = 0$
View full question & answer→MCQ 601 Mark
If $(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})=\text{a}+\text{ib},$ then $2.5.10.17...(1+\text{n}^2)=$
- A
$\text{a}-\text{ib}$
- B
$\text{a}^2-\text{b}^2$
- ✓
$\text{a}^2+\text{b}^2$
- D
AnswerCorrect option: C. $\text{a}^2+\text{b}^2$
$(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})=\text{a}+\text{ib}$
Taking modulus on both the sides, we get,
$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|=\text{a}+\text{ib}$
$|(1+\text{i})(1+2\text{i})(1+3\text{i})...(1+\text{ni})|$ can be wriiten as $|(1+\text{i})||(1+2\text{i})||(1+3\text{i})|...|(1+\text{ni})|$
$\therefore\sqrt{1^2+1^2}\times\sqrt{1^2+2^2}\times\sqrt{1^2+3^2}...\times\sqrt{1^2+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$
$\therefore\sqrt{2}\times\sqrt{5}\times\sqrt{10}...\times\sqrt{1^2+\text{n}^2}=\sqrt{\text{a}^2+\text{b}^2}$
Squaring on both the sides, we get:
$2\times5\times10...\times(1+\text{n}^2)=\sqrt{\text{a}^2+\text{b}^2}$
View full question & answer→MCQ 611 Mark
Choose the correct answer.
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
$-1 + 2i$
AnswerCorrect 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 $=\text{z}\cdot\text{e}^{-\big(\frac{\pi}{2}\big)}$
$=(2-\text{i})\text{e}^{-\big(\frac{\pi}{2}\big)}$
$=(2-\text{i})\Big[\cos\Big(\frac{-\pi}{2}\Big)+\text{i}\sin\Big(\frac{-\pi}{2}\Big)\Big]$
$=(2-\text{i})(0-\text{i})$
$=-1-2\text{i}$
View full question & answer→MCQ 621 Mark
If $\text{a}\cos\theta+\text{b}\sin\theta=\text{c}$ have roots $\alpha$ and $\beta$.Then, what will be the value of $\sin\alpha + \sin\beta$
AnswerCorrect option: A. $\frac{2\text{bc}}{(\text{a}^2+\text{b}^2)}$
Given, $\text{a}\cos\theta+\text{b}\sin\theta=\text{c}$
So, this implies $\text{a}\cos\theta=\text{c}-\text{b}\sin\theta$
Now squaring both the sides we get,
$(\text{a}\cos\theta)^2=(\text{c}-\text{b}\sin\theta)^2$
$\text{a}^2\cos^2\theta=\text{c}^2+\text{b}^2\sin^2\theta-2\text{b c}\sin\theta$
$\text{a}^2(1-\sin^2\theta)=\text{c}^2+\text{b}^2\sin^2\theta-2\text{b c}\sin\theta$
$\text{a}^2-\text{a}^2\sin^2\theta=\text{c}^2+\text{b}^2\sin^2\theta-2\text{b c}\sin\theta$
Now rearranging the elements,
$\text{a}^2+\text{b}^2\sin^2\theta-2\text{b c}\sin\theta+(\text{c}^2-\text{a}^2)=\theta$
So, as sum of the roots are in the form $\frac{-\text{b}}{\text{a}}$ if there is a quadratic equation $\text{a}\text{x}^2=\text{bx}+\text{c}=0$
Now, we can conclude that
$\sin\alpha+\sin\beta=\frac{2\text{bc}}{(\text{a}^2+\text{b}^2)}$
View full question & answer→MCQ 631 Mark
The value of $p$ and $q (\text{P}\neq0,\ \text{q}\neq0)$ for which $p, q$ are the roots of the equation $x^2 + px + q = 0$ are:
- ✓
$p = 1, q = −2$
- B
$p = −1, q = −2$
- C
$p = −1, q = 2$
- D
$p = 1, q = 2$
AnswerCorrect option: A. $p = 1, q = −2$
It is given that, $p$ and $q (\text{P}\neq0,\ \text{q}\neq0)$are the roots of the equation $x^2 + px + q = 0$
$\therefore$ Sum of roots $= p + q = −p$
$\Rightarrow 2p + q = 0 ...(1)$
Product of roots $= pq = q$
$\Rightarrow q (p − 1) = 0$
$\Rightarrow p = 1, q = 0$
Now, substituting $p = 1$ in $(1),$ we get,
$2 + q = 0$
$\Rightarrow q = −2$
View full question & answer→MCQ 641 Mark
Value of $i ($iota$)$ is:
- A
$-1$
- B
$1$
- ✓
$(-1)^\frac{1}{2}$
- D
$(-1)^\frac{1}{4}$
AnswerCorrect option: C. $(-1)^\frac{1}{2}$
Iota is used to denote complex number.
The value of $i ($iota$)$ is $\sqrt{-1}\text{ i.e.}(-1)^\frac{1}{2}$
View full question & answer→MCQ 651 Mark
Find mirror image of point representing $x + i y$ on real axis:
- A
$(x, y)$
- B
$(-x, -y)$
- C
$(-x, y)$
- ✓
$(x, -y)$
AnswerCorrect option: D. $(x, -y)$
Mirror image of point $(x, y)$ on real axis is $(x, -y).$
Since real axis is acting as mirror $x-$coordinate remains same whereas $y-$coordinate gets inverted.
So, $(x, -y)$ is mirror image of $(x, y)$ on real axis.
View full question & answer→MCQ 661 Mark
If $\frac{1-\text{ix}}{1+\text{ix}}=\text{a}+\text{ib},$ then $\text{a}^2+\text{b}^2=$
Answer$\frac{1-\text{ix}}{1+\text{ix}}=\text{a}+\text{ib}$
Taking modulus on both the sides, we get:
$\Big|\frac{1-\text{ix}}{1+\text{ix}}\Big|=\big|\text{a}+\text{ib}\big|$
$\Rightarrow\frac{\sqrt{1^2+\text{x}^2}}{\sqrt{1^2+\text{x}^2}}=\sqrt{\text{a}^2+\text{b}^2}$
$\Rightarrow\sqrt{\text{a}^2+\text{b}^2}=1$
Squaring both the sides, we get:
$\text{a}^2+\text{b}^2=1$
View full question & answer→MCQ 671 Mark
The set of all values of $m$ for which both the roots of the equation $x^2− (m + 1) x + m + 4 = 0$ are real and negative, is:
AnswerCorrect option: C. $[-4, -3]$
The roots of the quadratic equation $x^2− (m + 1) x + m + 4 = 0$ will be real, if its discriminant is greater than or equal to zero.
$\therefore (m + 1)^2 - 4 (m + 4) > 0$
$\Rightarrow (m - 5) (m + 3) > 0$
$\Rightarrow m < -3$ or $m > 5 ...(1)$
It is also given that, the roots of $x^2− (m + 1) x + m + 4 = 0$ are negative.
So, the sum of the roots will be negative.
$\therefore$ Sum of the roots $< 0$
$\Rightarrow m + 1 < 0 ....(2)$
and product of zeros $> 0$
$\Rightarrow m + 4 > 0 ...(3)$
From $(1), (2)$ and $(3),$ we get,
$[−4, −3]$
View full question & answer→MCQ 681 Mark
In $z = 4 + i,$ what is the real part?
AnswerIn $z = a + b i\, a$ is real part and $b$ is imaginary part.
So, in $4 + i,$ real part is $4.$
View full question & answer→MCQ 691 Mark
According to De Moivre’s theorem what is the value of $\text{z}^\frac{1}{\text{n}}$
- A
$\text{r}^\frac{1}{\text{n}}\big[\cos2\text{kn}+\theta)+\text{i}\sin(2\text{kn}+\theta)\big]$
- B
$\text{r}^\frac{1}{\text{n}}\bigg[\frac{\cos2\text{kn}+\theta)}{\text{n}}-\frac{\text{i}\sin(2\text{kn}+\theta)}{\text{n}}\bigg]$
- ✓
$\text{r}^\frac{1}{\text{n}}\bigg[\frac{\cos2\text{kn}+\theta)}{\text{n}}+\frac{\text{i}\sin(2\text{kn}+\theta)}{\text{n}}\bigg]$
- D
$\text{r}^\frac{1}{\text{n}}\big[\cos2\text{kn}+\theta)-\text{i}\sin(2\text{kn}+\theta)\big]$
AnswerCorrect option: C. $\text{r}^\frac{1}{\text{n}}\bigg[\frac{\cos2\text{kn}+\theta)}{\text{n}}+\frac{\text{i}\sin(2\text{kn}+\theta)}{\text{n}}\bigg]$
If $n$ is any integer, then $(\cos\theta+\text{i}\sin\theta)^\text{n}=\cos(\text{n}\theta)+\text{i}\sin(\text{n}\theta)$
Writing the binomial expansion of $(\cos\theta+\text{i}\sin\theta)^\text{n}$ and equating real parts of $\cos(\text{n}\theta)$ and the imaginary part to $\sin(\text{n}\theta)$ we ge
$\cos\text{n}\theta=\cos^\text{n}\theta-\text{n}\text{c}_2\cos^\text{n}-2\theta\sin^2\theta+\text{nc}_4\cos^\text{n-4}\theta\sin^4\theta+\dots\\\sin(\text{n}\theta)=\text{nc}_1\cos^\text{n-1}\theta\sin\theta-\text{nc}_3\cos^\text{n-3}\theta\sin^3\theta+\dots$
If, $n$ is a rational number, then one of the value of $(\cos\theta+\text{i}\sin\theta)^\text{n}=\cos(\text{n}\theta)+\text{i}\sin(\text{n}\theta)$
If, $\text{n}=\frac{\text{p}}{\text{q}},$
where, $p$ and $q$ are integers $(\text{q}>\theta)$ and $p, q$ have no common factor, then $(\cos\theta+ \text{i}\sinθ)^\text{n}$ has $q$ distinct values one of which is $(\cos\theta+ \text{i}\sinθ)^\text{n}$
If, $\text{z}^\frac{1}{\text{n}}\bigg[\frac{\cos(2\text{k}\pi+\theta)}{\text{n}}\bigg]+\bigg[\frac{\text{i}\sin(2\text{k}\pi+\theta)}{\text{n}}\bigg]$where $k = 0, 1, 2, ……….., n – 1.$
View full question & answer→MCQ 701 Mark
A real number $'a\ '$ is called a good number if the inequality$\frac{(2\text{x}^2 - 2\text{x} -3)}{(\text{x}^2 + \text{x} + 1)}\leq a$ is satisfied for all real $x.$ What is the set of all real numbers?
- A
$\Big(\infty,\frac{10}{3}\Big]$
- B
$\Big(\frac{10}{3 },\infty\Big)$
- ✓
$\Big[\frac{10}{3},\infty\Big)$
- D
$\Big[\frac{-10}{3},\infty\Big)$
AnswerCorrect option: C. $\Big[\frac{10}{3},\infty\Big)$
We have, $\frac{(2\text{x}^2 - 2\text{x} -3)}{(\text{x}^2 + \text{x} + 1)}\leq\text{a}\forall\text{x}\in\text{R}$
$\Rightarrow2\text{x}^2 – 2\text{x} – 3\leq \text{a}(\text{x}^2 + \text{x} + 1)\forall \text{x} \in \text{R}$
$\Rightarrow(2 – \text{a})\text{x}^2 – (2 – \text{a})\text{x} – (3 – \text{a})\forall \text{ x} \in\text{R}$
$2-\text{a}<0$ and $(2 – \text{a})\text{x}^2 – 4(2 – \text{a})(3 – \text{a})\leq 0 \forall \text{ x } \in \text{R}$
So, $\text{a}>2$ and $\text{a}\leq2$ or $\text{a}\geq\frac{10}{3}$
$\Rightarrow\text{a}\geq\frac{10}{3}$
$\therefore\text{a}\in\Big[\frac{10}{3},\infty\Big)$
View full question & answer→MCQ 711 Mark
The sum of two complex numbers $a + ib$ and $c + id$ is a real number if
- A
$a + c = 0$
- ✓
$b + d = 0$
- C
$a + b = 0$
- D
$b + c = 0$
AnswerCorrect option: B. $b + d = 0$
It is given that
$z_1=a+i b$ and
$z_2=c+i d$
Then
$z_1+z_2=(a+c)+i(b+d)$
Now
$\left(z_1+z_2\right)$ is purely real.
Then the imaginary part has to be $0.$
Hence
$b + d = 0.$
View full question & answer→MCQ 721 Mark
Let $z_1$ and $_2$ be two com plex num bers with α and $\beta$ as their principal ar gu ments such that $\alpha+\beta>\pi,$ then principal arg $(z_1z_2)$ is given by:
- A
$\alpha+\beta+\pi$
- ✓
$\alpha+\beta-\pi$
- C
$\alpha+\beta+2\pi$
- D
$\alpha+\beta$
AnswerCorrect option: B. $\alpha+\beta-\pi$
View full question & answer→MCQ 731 Mark
Choose the correct answer. 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. $|\text{z}^2|=|\text{z}|^2$
Let $z=x+y i$
$|z|=|x+y i|$ and $|z|^2=|x+y i|^2$
$\Rightarrow|z|^2=x^2+y^2 \ldots \text { (i) }$
Now, $z^2=x^2+y^2 i^2+2 x y i$
$z^2=x^2-y^2+2 x y i$
$|\text{z}^2|=\sqrt{(\text{x}^2-\text{y}^2)^2+(2\text{xy})^2}$
$=\sqrt{\text{x}^4+\text{y}^4-2\text{x}^2\text{y}^2+4\text{x}^2\text{y}^2}$
$=\sqrt{\text{x}^4+\text{y}^4+2\text{x}^2\text{y}^2}$
$=\sqrt{(\text{x}^2+\text{y}^2)^2}$
So, $|\text{z}|^2=\text{x}^2+\text{y}^2=|\text{z}|^2$
So, $|\text{z}|^2=|\text{z}^2|$
View full question & answer→MCQ 741 Mark
Find mirror image of point representing $x + i y$ on imaginary axis:
- A
$(x, y)$
- B
$(-x, -y)$
- ✓
$(-x, y)$
- D
$(x, -y)$
AnswerCorrect option: C. $(-x, y)$
Mirror image of point $(x, y)$ on imaginary axis is $(-x, y).$
Since imaginary axis is acting as mirror $y-$coordinate remains same whereas $x -$ coordinate gets inverted.
So, $(-x, y)$ is mirror image of $(x, y)$ on imaginary axis.
View full question & answer→MCQ 751 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 761 Mark
If $x^2+p x+1=0$ and $(a-b) x^2+(b-c) x+(c-a)=0$ have both roots common, then what is the form of $a, b, c?$
- A
$a, b, c$ are in $A.P$
- ✓
$b, a, c$ are in $A.P$
- C
$b, a, c$ are in $G.P$
- D
$b, a, c$ are in $H.P$
AnswerCorrect option: B. $b, a, c$ are in $A.P$
Given, $(a-b) x^2+(b-c) x+(c-a)=0$ and $x^2+p x+1=0$
So, $\frac{1}{(\text{a}-\text{b})}=\frac{p}{(\text{b}-\text{c})}=\frac{1}{(\text{c}-\text{a})}$
Equating the above equation, we get,
$(b – c) = p(a – b)$ and
$(b – c) = p(c – a)$
So, $p(a – b) = p(c – a)$
$\Rightarrow a – b = c – a$
So, $2a = b + c$ which means that $b, a, c$ are in $A.P.$
View full question & answer→MCQ 771 Mark
A real value of $x$ satisfies the equation $\frac{3-4\text{ix}}{3+4\text{ix}}=\text{a}-\text{ib}\Big(\text{a},\text{b}\in\text{R}\Big),$ if $\text{a}^2+\text{b}^2=$
Answer$\text{a}-\text{ib}=\frac{3-4\text{ix}}{3+4\text{ix}}$
$=\frac{3-4\text{ix}}{3+4\text{ix}}\times\frac{3-4\text{ix}}{3-4\text{ix}}$
$=\frac{9+16\text{x}^2\text{i}^2-24\text{xi}}{9-16\text{x}^2\text{i}^2}$
$=\frac{(9-16\text{x}^2)-\text{i}(24\text{x})}{9+16\text{x}^2}$
$\Rightarrow\ |\text{a}-\text{ib}|^2=\Bigg|\frac{(9-16\text{x}^2)-\text{i}(24\text{x})}{9+16\text{x}^2}\Bigg|^2$
$\Rightarrow\text{a}^2+\text{b}^2=\frac{(9-16\text{x}^2)^2+(24\text{x})^2}{(9+16\text{x}^2)^2}$
$=\frac{81+256\text{x}^4-288\text{x}^2+576\text{x}^2}{(9+16\text{x}^2)^2}$
$=\frac{81+256\text{x}^4+288\text{x}^2}{(9+16\text{x}^2)^2}$
$=\frac{(9+16\text{x}^2)^2}{(9+16\text{x}^2)^2}$
$=1$
View full question & answer→MCQ 781 Mark
If $p$ and $q$ are the roots of the equation $x^2+p x+q=0$ then, what are the values of $p$ and $q?$
- ✓
$p = 1, q = -2$
- B
$p = 0, q = 1$
- C
$p = -2, q = 0$
- D
$p = -2, q = 1$
AnswerCorrect option: A. $p = 1, q = -2$
Since, $p$ and $q$ are the roots of the equation $x^2+p x+q=0$
So, $p + q = -p$ and $pq = q$
So, $pq = q$
And, $q = 0$ or $p = 1$
If, $q = 0$ then, $p = 0$ and if $p = 1$ then $q = -2.$
View full question & answer→MCQ 791 Mark
If, $x^4+4 x^3+6 a x^2+6 b x+c$ is divisible by $x^3+3 x^2+9 x+3$. Then, what is the value of $a + b + c ?$
AnswerHere, $f(x) =\text{x}^4 + 4\text{x}^3 + 6\text{ax}^2 + 6\text{bx}+\text{c}$
so, let its roots be, $\alpha,\beta,\gamma,\delta$ and
$g(x) =\text{x}^3 + 3\text{x}^2 +9\text{x} + 3$
so, let its roots be, $\alpha,\beta,\gamma$
So, from here we can conclude,
$\alpha+\beta+\gamma+\delta=-4$ and $\alpha+\beta+\gamma+\delta=-3$
Thus, $\delta=-1$
This means, $(\text{x} + 1)(\text{x}^3 + 3\text{x}^2 + 9\text{x} + 3)$
On solving this equation in simpler form we get,
$\text{x}^4 + 4\text{x}^3 + 12\text{x}^2 + 12\text{x} + 3$
$\Rightarrow6\text{a}=12$
$\Rightarrow\text{a}=2$
$\Rightarrow6\text{b}=12$
$\Rightarrow\text{b}=2$
$\Rightarrow\text{c}=3$
$\Rightarrow\text{a}+\text{b}+\text{c}=7$
View full question & answer→MCQ 801 Mark
Which is the largest negative integer which satisfies $\frac{(\text{x}^2 – 1)}{(\text{x} – 2)(\text{x} – 3)}$ ?
Answer$\frac{(\text{x}^2 – 1)}{(\text{x} – 2)(\text{x} – 3)} > 0$
So, $x = -1, 1, 2, 3$
Thus, $\text{x}\in(-\infty, -1)\cup(1,\text{ 2})\cup(3,\infty)$
Therefore, the largest negative integer is $-2.$
View full question & answer→MCQ 811 Mark
If $\alpha$ and $\beta$ are imaginary cube roots of unity, then the value of $\alpha^4+\beta^{28}+\frac{1}{\alpha\beta}$ is:
View full question & answer→MCQ 821 Mark
The number of real solutions of $\left|2 x-x^2-3\right|=1$ is:
Answer$i.$Given equation: $\left|2 x-x^2-3\right|=1$
$2 x-x^2-3=1$
$\Rightarrow 2 x-x^2-4=0$
$\Rightarrow x^2-2 x+4=0$
$\Rightarrow(x-2)^2=0$
$\Rightarrow x=2,2$
$ii. -2 x+x^2+3=1$
$\Rightarrow x^2-2 x+2=0$
$\Rightarrow x^2-2 x+1+1=0$
$\Rightarrow(x-1+i)(x-1-i)=0$
$\Rightarrow x=1-i, 1+i$
Hence, the real solutions are $2,2 .$
View full question & answer→MCQ 831 Mark
In $z = 4 + i,$ what is imaginary part?
AnswerIn $z = a + b i, a$ is real part and $b$ is imaginary part.
So, in $4 + i,$ imaginary part is $1.$
View full question & answer→MCQ 841 Mark
Roots of a quadratic equation are real when discriminant is $............$
- A
- B
- C
- ✓
greater than or equal to zero
AnswerCorrect option: D. greater than or equal to zero
For a quadratic equation, $a x^2+b x+c=0$, discriminant is $b^2-4 a c$.
Roots are $\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$ For real roots,
radical must be non$-$negative
i.e. discriminant should be greater than or equal to zero.
View full question & answer→MCQ 851 Mark
If $\text{z}=\frac{1}{(1-\text{i})(2+3\text{i})},$ then $|\text{z}|=$
- A
$1$
- ✓
$\frac{1}{\sqrt{26}}$
- C
$\frac{5}{\sqrt{26}}$
- D
AnswerCorrect option: B. $\frac{1}{\sqrt{26}}$
Let $\text{z}=\frac{1}{(1-\text{i})(2+3\text{i})}$
$\Rightarrow\text{z}=\frac{1}{2+\text{i}-3\text{i}^2}$
$\Rightarrow\text{z}=\frac{1}{2+\text{i}+3}$
$\Rightarrow\text{z}=\frac{1}{5+\text{i}}\times\frac{5-\text{i}}{5-\text{i}}$
$\Rightarrow\text{z}=\frac{5-\text{i}}{25-\text{i}^2}$
$\Rightarrow\text{z}=\frac{5-\text{i}}{25+1}$
$\Rightarrow\text{z}=\frac{5-\text{i}}{26}$
$\Rightarrow\text{z}=\frac{5}{26}-\frac{\text{i}}{26}$
$\Rightarrow|\text{z}|=\sqrt{\frac{25}{676}+\frac{1}{676}}$
$\Rightarrow\text{z}=\frac{1}{\sqrt{26}}$
View full question & answer→MCQ 861 Mark
If the equations $\text{x}^2+2\text{x}+3\lambda=0$ and $2\text{x}^2+3\text{x}+5\lambda=0$ have a non$-$zero common roots, then $\lambda=$
AnswerLet a be the common roots of the equations $\text{x}^2+2\text{x}+3\lambda=0$ and $2\text{x}^2+3\text{x}+5\lambda=0$
Therefore
$\alpha^2+2\text{a}+3\lambda=0\ ...(1)$
$2\alpha^2+3\alpha+5\lambda=0\ ...(2)$
Solving $(1)$ and $(2)$ by cross multiplication, we get
$\frac{\alpha^2}{10\lambda-9\lambda}=\frac{\alpha}{6\lambda-5\lambda}=\frac{1}{3-4}$
$\Rightarrow\text{a}^2=-\lambda,\alpha=-\lambda$
$\Rightarrow-\lambda=\lambda^2$
$\Rightarrow\lambda=-1$
View full question & answer→MCQ 871 Mark
If $n$ is a positive integer, then $\Big(\frac{1+\text{i}}{1-\text{i}}\Big)4\text{n}+1$ is equal to:
View full question & answer→MCQ 881 Mark
If $\theta$ is the amplitude of $\frac{\text{a}+\text{ib}}{\text{a}-\text{ib}},$ then $\tan\theta=$
- A
$\frac{2\text{a}}{\text{a}^2+\text{b}^2}$
- ✓
$\frac{2\text{ab}}{\text{a}^2-\text{b}^2}$
- C
$\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}$
- D
AnswerCorrect option: B. $\frac{2\text{ab}}{\text{a}^2-\text{b}^2}$
$\text{z}=\frac{\text{a}+\text{ib}}{\text{a}-\text{ib}}\times\frac{\text{a}+\text{ib}}{\text{a}+\text{ib}}$
$\Rightarrow\text{z}=\frac{\text{a}^2+\text{i}^2\text{b}^2+2\text{abi}}{\text{a}^2-\text{i}^2\text{b}^2}$
$\Rightarrow\text{z}=\frac{\text{a}^2-\text{b}^2+2\text{abi}}{\text{a}^2+\text{b}^2}$
$\Rightarrow\text{z}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2}+\text{i}\frac{2\text{a}\text{b}}{\text{a}^2+\text{b}^2}$
$\Rightarrow\text{Re(z)}=\frac{\text{a}^2-\text{b}^2}{\text{a}^2+\text{b}^2},\text{Im(z)}\frac{2\text{a}\text{b}}{\text{a}^2+\text{b}^2}$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=\frac{2\text{ab}}{\text{a}^2-\text{b}^2}$
$\alpha=\tan^{-1}\Big(\frac{2\text{ab}}{\text{a}^2-\text{b}^2}\Big)$
Since, $z$ lies in the first quadrant .
Therefore,
$\text{arg(z)}=\alpha=\tan^{-1}\Big(\frac{2\text{ab}}{\text{a}^2-\text{b}^2}\Big)$
$\tan\theta=\frac{2\text{ab}}{\text{a}^2-\text{b}^2}$
View full question & answer→MCQ 891 Mark
If $\text{z}=\cos\frac{\pi}{4}+\text{i}\sin\frac{\pi}{6},$ then
- A
$|\text{z}|=1,\text{arg(z)}=\frac{\pi}{4}$
- B
$|\text{z}|=1,\text{arg(z)}=\frac{\pi}{6}$
- C
$|\text{z}|=\frac{\sqrt{3}}{2},\text{arg(z)}=\frac{5\pi}{24}$
- ✓
$|\text{z}|=\frac{\sqrt{3}}{2},\text{arg(z)}=\tan^{-1}\frac{1}{\sqrt{2}}$
AnswerCorrect option: D. $|\text{z}|=\frac{\sqrt{3}}{2},\text{arg(z)}=\tan^{-1}\frac{1}{\sqrt{2}}$
$\text{z}=\cos\frac{\pi}{4}+\text{i}\sin\frac{\pi}{6}$
$\Rightarrow\text{z}=\frac{1}{\sqrt{2}}+\frac{1}{2}\text{i}$
$\Rightarrow|\text{z}|=\sqrt{\Big(\frac{1}{\sqrt{2}}\Big)^2+\frac{1}{4}}$
$\Rightarrow|\text{z}|=\sqrt{\frac{1}{2}+\frac{1}{4}}$
$\Rightarrow|\text{z}|=\sqrt{\frac{3}{4}}$
$\Rightarrow|\text{z}|=\frac{\sqrt{3}}{2}$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=\frac{1}{\sqrt{2}}$
$\Rightarrow\alpha=\tan^{-1}\Big(\frac{1}{\sqrt{2}}\Big)$
Since, the point z lies in the first quadrant.
Therefore, $|\text{z}|=\frac{\sqrt{3}}{2},\text{arg(z)}=\tan^{-1}\frac{1}{\sqrt{2}}$
View full question & answer→MCQ 901 Mark
The value of $\frac{(\text{i}^5+\text{i}^6+\text{i}^7+\text{i}^8+\text{i}^9)}{(1+\text{i})}$ is:
AnswerCorrect option: A. $\frac{1}{2}(1+\text{i})$
$\frac{(\text{i}^5+\text{i}^6+\text{i}^7+\text{i}^8+\text{i}^9)}{(1+\text{i})}$
$=\frac{\text{i}-1-\text{i}+1+\text{i}}{1+\text{i}} \ [$ As, $\text{i}^5=\text{i},\text{i}^6=-1,\text{i}^7=-\text{i},\text{i}^8=1,\text{i}^9=\text{i}]$
$=\frac{\text{i}}{\text{i}+1}$
$=\frac{\text{i}}{\text{i}+1}\times\frac{\text{i}-1}{\text{i}-1}$
$=\frac{\text{i}(\text{i}-1)}{\text{i}^2-1}$
$=\frac{\text{i}^2-\text{i}}{-2}$
$=\frac{1}{2}(1+\text{i})$
View full question & answer→MCQ 911 Mark
If $\alpha,\beta$ are the roots of the equation $a x^2+b x+c=0$, then $\frac{1}{\text{a}\alpha+\text{b}}+\frac{1}{\text{a}\beta+\text{b}}$
- A
$\frac{\text{c}}{\text{ab}}$
- B
$\frac{\text{a}}{\text{bc}}$
- ✓
$\frac{\text{b}}{\text{ac}}$
- D
AnswerCorrect option: C. $\frac{\text{b}}{\text{ac}}$
Given equation: $a x^2+b x+c=0$
Also, $\alpha$ and $\beta$ are the roots of the given equation.
Then, sum of the roots $=\alpha+\beta=-\frac{\text{b}}{\text{a}}$
Product of the roots $=\alpha\beta=\frac{\text{c}}{\text{a}}$
$\therefore\frac{1}{\text{a}\alpha+\text{b}}+\frac{1}{\text{a}\beta+\text{b}}=\frac{\text{a}\beta+\text{b}+\text{a}\alpha+\text{b}}{(\text{a}\alpha+\text{b})(\text{a}\beta+\text{b})}$
$=\frac{\text{a}(\alpha+\beta)+2\text{b}}{\text{a}^2\alpha\beta+\text{ab}\alpha+\text{ab}\beta+\text{b}^2}$
$=\frac{\text{a}(\alpha+\beta)+2\text{b}}{\text{a}^2\alpha\beta+\text{ab}(\alpha+\beta)+\text{b}^2}$
$=\frac{\text{a}\big(-\frac{\text{b}}{\text{a}}\big)+2\text{b}}{\text{a}^2\big(\frac{\text{c}}{\text{a}}\big)+\text{ab}\big(-\frac{\text{b}}{\text{a}}\big)+\text{b}^2}$
$=\frac{\text{b}}{\text{ac}}$
View full question & answer→MCQ 921 Mark
The value of $\frac{\text{i}^{592}+\text{i}^{590}+\text{i}^{588}+\text{i}^{586}+\text{i}^{584}}{\text{i}^{582}+\text{i}^{580}+\text{i}^{578}+\text{i}^{576}+\text{i}^{574}}-1$ is
Answer$\frac{\text{i}^{592}+\text{i}^{590}+\text{i}^{588}+\text{i}^{586}+\text{i}^{584}}{\text{i}^{582}+\text{i}^{580}+\text{i}^{578}+\text{i}^{576}+\text{i}^{574}}-1$
$=\frac{\text{i}^{4\times148}+\text{i}^{4\times147+2}+\text{i}^{4\times147}+\text{i}^{4\times146+2}+\text{i}^{4\times146}}{\text{i}^{4\times145+2}+\text{i}^{4\times145}+\text{i}^{4\times144+2}+\text{i}^{4\times144}+\text{i}^{4\times143+2}}-1$ $[\because\text{i}^4=1$ and $\text{i}^2=-1]$
$=\frac{1+\text{i}^2+1+\text{i}^2+1}{\text{i}^2+1+\text{i}^2+1+\text{i}^2}-1$
$=\frac{1}{-1}-1$
$=-2$
View full question & answer→MCQ 931 Mark
If $x^2+a x+b=0$ and $x^2+b x+a=0$ have exactly $1$ common root then what is the value of $(a + b)$
AnswerSubtracting the equation $x^2+a x+b=0$ to $x^2+b x+a=0$ by solving the equation simultaneously, we get,
$(a – b)x + (b – a) = 0$
So, $(a – b) x = (a – b)$
Therefore, $x = 1$
Now, putting the value of $x = 3$ in any one of the equation, we get,
$1 + a + b = 0$
Therefore, $a + b = -1.$
View full question & answer→MCQ 941 Mark
Solve $2\text{x}^2+\sqrt{2\text{x}+2}=0$
- ✓
$\frac{-1\pm\text{i}\sqrt{7}}{2\sqrt{2}}$
- B
$\frac{1\pm\text{i}\sqrt{7}}{2\sqrt{2}}$
- C
$\frac{1\pm\sqrt{7}}{2\sqrt{2}}$
- D
$\frac{-1\pm\sqrt{7}}{2\sqrt{2}}$
AnswerCorrect option: A. $\frac{-1\pm\text{i}\sqrt{7}}{2\sqrt{2}}$
$2\text{x}^2+\sqrt{2\text{x}+2}=0$
$\Rightarrow\text{D}=(\sqrt{2})^2-4.2.2=2-16=-14$
Since $\text{D}\leq0,$ imaginary roots are there.
$\Rightarrow\text{x}=\frac{-\sqrt{2\pm}\sqrt{\text{D}}}{2.2}$
$=\frac{-\sqrt{2\pm\text{i}}\sqrt{\text{D}}}{2.2}$
$=\frac{-1\pm\text{i}\sqrt{7}}{2\sqrt{2}}$
View full question & answer→MCQ 951 Mark
If $\left|z_1\right|=4,\left|z_2\right|=3$, then what is the value of $\left|z_1+z_2+3+4 i\right|$
- A
Less than $2$
- B
Less than $5$
- C
Less than $7$
- ✓
Less than $12$
AnswerCorrect option: D. Less than $12$
As, we know $|\text{Z}_2+\text{Z}_2+.....+\text{Z}_\text{n}|\leq|\text{Z}_1|+|\text{Z}_2+......+|\text{Z}_\text{n}|$
So, $ |\text{Z}_1 + \text{Z}_2 + 3 + 4\text{i}|\leq|\text{Z}_1| + |\text{Z}_2| + |3 + 4\text{i}|$
Now, putting the given values in the equation, we get,
$\Rightarrow|\text{Z}_1 + \text{Z}_2 + 3 + 4\text{i}|\leq4+3+\sqrt{9+16}$
$\Rightarrow|\text{Z}_1 + \text{Z}_2 + 3 + 4\text{i}|\leq4+3+5$
$\Rightarrow|\text{Z}_1 + \text{Z}_2 + 3 + 4\text{i}|\leq12$
View full question & answer→MCQ 961 Mark
The least positive integer $n$ such that $\Big(\frac{2\text{i}}{1+\text{i}}\Big)^\text{n}$ is a positive integer, is:
AnswerLet $\text{z}=\Big(\frac{2\text{i}}{1+\text{i}}\Big)$
$\Rightarrow\text{z}=\frac{2\text{i}}{1+\text{i}}\times\frac{1-\text{i}}{1-\text{i}}$
$\Rightarrow\text{z}=\frac{2\text{i}(1-\text{i})}{1-\text{i}^2}$
$\Rightarrow\text{z}=\frac{2\text{i}(1-\text{i})}{1+1} \ [\because\text{i}^2=-1]$
$\Rightarrow\text{z}=\frac{2\text{i}(1-\text{i})}{2}$
$\Rightarrow\text{z}=\text{i}-\text{i}^2$
$\Rightarrow\text{z}=\text{i}+1$
Now, $\text{z}^\text{n}=(1+\text{i})^\text{n}$
For $\text{n}=2,$
$\text{z}^2=(1+\text{i})^2$
$=1+\text{i}^2+2\text{i}$
$=1-1+2\text{i}$
$=2\text{i} \ ...(1)$
Since this is not a positive integer,
For $\text{n}=4,$
$\text{z}^4=(1+\text{i})^4$
$=\big[(1+\text{i})^2\big]^2$
$=(2\text{i})^2 [$Using $(1)]$
$=(4\text{i})^2$
$=-4 \ ...(2)$
This is a negative integer.
For $\text{n}=8,$
$\text{z}^8=(1+\text{i})^8$
$=\big[(1+\text{i})^4\big]^2$
$=(-4)^2 [$Using $(2)]$
$=16$
This is a positive integer.
Thus, $\text{z}=\Big(\frac{2\text{i}}{1+\text{i}}\Big)^\text{n}$ is positive for $\text{n}=8.$
Therefore, $8$ is the least positive integer such that $\Big(\frac{2\text{i}}{1+\text{i}}\Big)^\text{n}$ is a positive integer.
View full question & answer→MCQ 971 Mark
The amplitude of $\frac{1}{\text{i}}$ is equal to:
- A
$0$
- B
$\frac{\pi}{2}$
- ✓
$-\frac{\pi}{2}$
- D
$\pi$
AnswerCorrect option: C. $-\frac{\pi}{2}$
Let $\text{z}=\frac{1}{\text{i}}$
$\text{z}=\frac{1}{\text{i}}\times\frac{\text{i}}{\text{i}}$
$\text{z}=\frac{\text{i}}{\text{i}^2}$
$\text{z}=-\text{i}$
Since, $z (0, -1)$ lies on the negative imaginary axis .
Therefore, $\text{arg(z)}=\frac{-\pi}{2}$
View full question & answer→MCQ 981 Mark
The common roots of the equations $x^{12}-1=0$ and $x^4+x^2+1=0$ are:
AnswerCorrect option: C. $\pm\text{w,}\pm\text{w}^2$
View full question & answer→MCQ 991 Mark
If $\text{z}=\text{a}+\text{ib}$ lies in third quadrant, then $\frac{\bar{\text{z}}}{\text{z}}$ also lies in third quadrant if:
- A
$\text{a}>\text{b}>0$
- B
$\text{a}<\text{b}<0$
- ✓
$\text{b}<\text{a}<0$
- D
$\text{b}>\text{a}>0$
AnswerCorrect option: C. $\text{b}<\text{a}<0$
Since, $\text{z}=\text{a}+\text{ib}$ lies in third quadrant.
$\Rightarrow\text{a}<0$ and $\text{b}<0 \ ...(1)$
Now,
$\frac{\bar{\text{z}}}{\text{z}}=\frac{\overline{\text{a}+\text{ib}}}{\text{a}+\text{ib}}$
$=\frac{\text{a}-\text{ib}}{\text{a}+\text{ib}}$
$=\frac{\text{a}-\text{ib}}{\text{a}+\text{ib}}\times\frac{\text{a}-\text{ib}}{\text{a}-\text{ib}}$
$=\frac{\text{a}^2+\text{i}^2\text{b}^2-2\text{abi}}{\text{a}^2-\text{i}^2\text{b}^2}$
$=\frac{\text{a}^2-\text{b}^2-2\text{abi}}{\text{a}^2+\text{b}^2}$
Since, $\frac{\bar{\text{z}}}{\text{z}}$ also lies in third quadrant.
$\Rightarrow\text{a}^2-\text{b}^2<0$
$\Rightarrow(\text{a}-\text{b})(\text{a}+\text{b})<0$
$\Rightarrow\text{a}-\text{b}>0$ and $\text{a}+\text{b}<0$
$\Rightarrow\text{a}>\text{b} \ ...(2)$
From $(1)$ and $(2),$
$\text{b}<\text{a}<0$
View full question & answer→MCQ 1001 Mark
The equation of the smallest degree with real coefficients having $1 + i$ as one of the roots is:
- A
$x^2+x+1=0$
- ✓
$x^2-2 x+2=0$
- C
$x^2+2 x+2=0$
- D
$x^2+2 x-2=0$
AnswerCorrect option: B. $x^2-2 x+2=0$
We know that, imaginary roots of a quadratic equation occur in conjugate pair.
It is given that, $1 + i$ is one of the roots.
So, the other root will be $1−i1 - i.$
Thus, the quadratic equation having roots $1 + i$ and $1 - i$ is,
$x^2-(1+i+1-i) x+(1+i)(1-i)=0$
$\Rightarrow x^2-2 x+2=0$
View full question & answer→MCQ 1011 Mark
If $\alpha,\beta$ are the roots of the equation $x^2+p x+1=0; \gamma,\delta$ the roots of the equation $x^2+q x+1=0$, then $(\alpha-\gamma)(\alpha+\delta)(\beta-\delta)(\beta+\delta)=$
- ✓
$q^2-p^2$
- B
$p^2-q^2$
- C
$p^2+q^2$
- D
AnswerCorrect option: A. $q^2-p^2$
Given: $\alpha$ and $\beta$ are the roots of the equation $x^2+p x+1=0$.
Also, $\gamma$ and $\delta\gamma$ and $\delta$ are the roots of the equation $x^2+q x+1=0$
Then, the sum and the product of the roots of the given equation are as follows:
$\alpha+\beta=-\frac{\text{p}}{1}=-\text{p}$
$\alpha\beta=\frac{1}{1}=1$
$\gamma+\delta=-\frac{\text{q}}{1}=-\text{P}$
$\gamma\delta=\frac{1}{1}=1$
Moreover, $(\gamma-\delta)^2=\gamma^2+\delta^2+2\gamma\delta$
$\Rightarrow\gamma^2+\delta^2=\text{q}^2-2$
$\therefore(\alpha-\gamma)(\alpha-\delta)(\beta-\gamma)(\beta+\delta)=(\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)$
$=(\alpha\beta-\alpha\gamma-\beta\gamma+\gamma^2)(\alpha\beta+\alpha\gamma+\beta\delta+\delta^2)$
$=[\alpha\beta-\gamma(\alpha+\beta)+\gamma^2][\alpha\beta+\delta(\alpha+\beta)+\delta^2])$
$=(1-\gamma(-\text{p})+\gamma^2)(1+\delta(-\text{p}+\delta^2))$
$=(1+\gamma\text{p}+\gamma^2)(1-\delta\text{p}+\delta^2)$
$=(1+\gamma\text{p}+\gamma^2)(1-\delta\text{p}+\delta^2)$
$=1-\text{p}\delta+\delta^2+\text{p}\gamma-\text{p}^2\gamma\delta+\text{p}\gamma\delta^2+\gamma^2-\text{p}\delta\text{y}^2+\gamma^2\delta^2$
$=1-\text{p}\delta+\text{p}\gamma+\delta^2-\text{p}^2\gamma\delta+\text{p}\gamma\delta^2+\gamma^2=\text{p}\delta\gamma^2+\gamma^2\delta^2$
$=1-\text{p}(\delta-\gamma)-\text{p}^2\gamma\delta+\text{p}\gamma\delta(\delta-\gamma)+(\gamma^2+\delta)+1$
$=1-\text{p}^2+(\delta-\gamma)\text{p}(\delta-\gamma)+(\gamma^2+\delta^2)+1$
$=-\text{P}^22+(\delta-\gamma)\text{p}(1-1)+\text{q}^2$
$=\text{q}^2-\text{p}^2$
View full question & answer→MCQ 1021 Mark
Which axis is known as imaginary axis in argand plane?
- A
$x -$ axis
- ✓
$y -$ axis
- C
$z -$ axis
- D
AnswerCorrect option: B. $y -$ axis
The plane having a complex number assigned to each of its point is called the
complex plane or the Argand plane.
When $(x + y i)$ is plotted in argand plane, $y -$ axis is imaginary axis.
View full question & answer→MCQ 1031 Mark
If $\text{z}=\Big(\frac{1+2\text{i}}{1-(1-\text{i})^2}\Big),$ then $\text{arg(z)}$ equal:
- ✓
$0$
- B
$\frac{\pi}{2}$
- C
$\pi$
- D
Answer$\text{z}=\frac{1+2\text{i}}{1-(1-\text{i})^2}$
$\Rightarrow\text{z}=\frac{1+2\text{i}}{1-(1+\text{i}^2-2\text{i})}$
$\Rightarrow\text{z}=\frac{1+2\text{i}}{1-(1-1-2\text{i})}$
$\Rightarrow\text{z}=\frac{1+2\text{i}}{1+2\text{i}}$
$\Rightarrow\text{z}=1$
Since point $(1,0)$ lies on the positive direction of real axis, we have:
$\text{arg(z)} = 0$
View full question & answer→MCQ 1041 Mark
The value of $a$ such that $x^2-11 x+a=0$ and $x^2-14 x+2 a=0$ may have a common root is:
AnswerLet $\alpha$ be the common roots of the equations
$x^2-11 x+a=0$ and $x^2-14 x+2 a=0$
Therefore,
$\alpha^2-11\alpha+\alpha=0\ ...(1)$
$\alpha^2-14\alpha+2\alpha=0\ ...(2)$
Solving $(1)$ and $(2)$ by cross multiplication, we get,
$\frac{\alpha^2}{-22\alpha+14\alpha}=\frac{\alpha}{\alpha-2\alpha}=\frac{1}{-14+11}$
$\Rightarrow\alpha^2=\frac{-22\alpha+14\alpha}{-14+11},\alpha=\frac{\alpha-2\alpha}{-14+11}$
$\Rightarrow\alpha^2=\frac{-8\alpha}{-3}=\frac{8\alpha}{3},\alpha=\frac{-\alpha}{-3}=\frac{\alpha}{3}$
$\Rightarrow\Big(\frac{\alpha}{3}\Big)^2=\frac{8\alpha}{3}$
$\Rightarrow\alpha^2=24\alpha$
$\Rightarrow\alpha^2-24\alpha=0$
$\Rightarrow\alpha(\alpha-24)=0$
$\Rightarrow\alpha=0$ or $\alpha=24$
View full question & answer→MCQ 1051 Mark
For the equation $|x|^2+|x|-6=0$, the sum of the real roots is:
Answer$\text { Let } P=|x|$
$\Rightarrow p^2+p-6=0$
$\Rightarrow p^2+3 p-2 p-6=0$
$\Rightarrow(p+3)(p-2)=0$
$\Rightarrow p=-3,2$
Also, $|x|=p$
$\Rightarrow|x|=2$, or $|x|=-3$
Modules can not be negative,
$\therefore|x|=2$
$\Rightarrow x= \pm 2$
$\Rightarrow x=2$ or $-2$
Sum of the roots of $x$ is $0.$
View full question & answer→MCQ 1061 Mark
If $P$ and $Q$ are conjugate complex numbers then their points on argand plane are mirror image on $............?$
- ✓
$x -$ axis
- B
$y -$ axis
- C
$z - $axis
- D
AnswerCorrect option: A. $x -$ axis
Conjugate complex numbers means their real part is same and imaginary part is inverted
i.e.same $x$ part and opposite imaginary part.
So, they are mirror image on real axis
i.e. $x -$ axis.
View full question & answer→MCQ 1071 Mark
If $\text{z}=\frac{1}{1-\cos\theta-\text{i}\sin\theta},$ then $\text{Re(z)}=$
AnswerCorrect option: B. $\frac{1}{2}$
$\text{z}=\frac{1}{1-\cos\theta-\text{i}\sin\theta}$
$\text{z}=\frac{1}{1-\cos\theta-\text{i}\sin\theta}\times\frac{1-\cos\theta+\text{i}\sin\theta}{1-\cos\theta+\text{i}\sin\theta}$
$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{(1-\cos\theta)^2-(\text{i}\sin\theta)^2}$
$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{1+\cos^2\theta-2\cos\theta+\text{i}\sin^2\theta}$
$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{1+1-2\cos\theta}$
$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{2(1-\cos\theta)}$
$\Rightarrow\text{Re(z)}=\frac{(1-\cos\theta)}{2(1-\cos\theta)}=\frac{1}{2}$
View full question & answer→MCQ 1081 Mark
$2 + i 0$ is point on $...........?$
- ✓
$x -$ axis
- B
$y -$ axis
- C
$z -$ axis
- D
AnswerCorrect 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.
View full question & answer→MCQ 1091 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 $zw$ is equal to:
View full question & answer→MCQ 1101 Mark
If $\text{x}+1\text{ x}=2\cos\frac{\pi}{10},$ then $\text{x}^5+\frac{1}{\text{x}^5}$ is equal to:
View full question & answer→MCQ 1111 Mark
Determine the nature of roots of the equation $\text{x}^2+2\text{x}\sqrt{3}+3=0.$
- A
- B
Non$-$real and distinct
- ✓
- D
Non$-$real and equal
AnswerThe nature of the roots can be determined from the discriminant $b^2-4 a c$
$\therefore \text{b}^2-4\text{ac}=(2\sqrt{3})2−(4×1×3)$
$\Rightarrow b^2-4 a c=12-12$
$\Rightarrow b^2-4 a c=0$
$\therefore b^2-4 a c=0$
There are two real and equal roots.
View full question & answer→MCQ 1121 Mark
$-z$ is, $............$ for complex number $z.$
- ✓
- B
additive identity element
- C
multiplicative identity element
- D
AnswerOn adding negative of complex number $( -z )$ to complex number $z,$
we get additive identity element zero
i.e. $z + ( -z ) = 0.$
View full question & answer→MCQ 1131 Mark
Choose the correct answer.
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
$y > x > 0$
AnswerCorrect 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 $\text{x, y}<0$
$\Rightarrow\text{y}<\text{x}<0$
View full question & answer→MCQ 1141 Mark
The locus of complex number $z$ such that $z$ is purely real and real part is equal to $- 2$ is:
- A
Negative $y-$axis
- B
Negative $x-$axis
- ✓
The point $(-2, 0)$
- D
The point $(2, 0)$
AnswerCorrect option: C. The point $(-2, 0)$
$z = x + iy$
$(x, y)$
$z$ is purely real and the real part equals $-2$
$\therefore y = 0\ \&\ x = -2$
$z = -2$
Hence, this would be represented by the point $(-2, 0)$ on the Argand Plane.
View full question & answer→MCQ 1151 Mark
What will be the sum of the real roots of the equation $x^2+5|x|+6=0$?
- A
Equal to $5$
- B
Equal to $10$
- C
Equal to $-5$
- ✓
AnswerSince, $x^2, 5|x|$ and 6 are positive.
So, $x^2+5|x|+6=0$ does not have any real root.
Therefore, sum does not exist.
View full question & answer→MCQ 1161 Mark
$0 + 0 i$ is $...........$ for complex number $z.$
- A
- ✓
additive identity element.
- C
multiplicative identity element.
- D
AnswerCorrect option: B. additive identity element.
On adding zero $(0 + 0 i)$ to a complex number, we get same complex number
so $0 + 0 i$ is additive identity
element for complex number $z.$
i.e. $z + 0 = z.$
View full question & answer→MCQ 1171 Mark
Choose the correct answer. Which of the following is correct for any two complex numbers $z_1$ and $z_2$?
- ✓
$|\text{z}_1\text{z}_2|=|\text{z}_1||\text{z}_2|$
- B
$\arg(\text{z}_1\text{z}_2)=\arg(\text{z}_1)\cdot\arg(\text{z}_2)$
- C
$|\text{z}_1+\text{z}_2|=|\text{z}_1|+|\text{z}_2|$
- D
$|\text{z}_1+\text{z}_2|\geq|\text{z}_1|-|\text{z}_2|$
AnswerCorrect option: A. $|\text{z}_1\text{z}_2|=|\text{z}_1||\text{z}_2|$
Let $\text{z}_1=\text{r}_1(\cos\theta_1+\text{i}\sin\theta_2)$
$\therefore\ |\text{z}_1|=\text{r}_1$
and $\text{z}_2=\text{r}_2(\cos\theta_1+\text{i}\sin\theta_2)$
$\therefore\ |\text{z}_2|=\text{r}_2$
$\text{z}_1\text{z}_2=\text{r}_1(\cos\theta_1+\text{i}\sin\theta_1)\cdot\text{r}_2(\cos\theta_2+\text{i}\sin\theta_2)$
$=\text{r}_1\text{r}_2(\cos\theta_1+\text{i}\sin\theta_1)\cdot(\cos\theta_2+\text{i}\sin\theta_2)$
$=\text{r}_1\text{r}_2(\cos\theta_1\cos\theta_2+\text{i}\sin\theta_2\cos\theta_1+\text{i}\sin\theta_1\cos\theta_2+\text{i}_2\sin\theta_1\sin\theta_2)$
$=\text{r}_1\text{r}_2\big[(\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2)+\text{i}(\sin\theta_1\cos\theta_2+\cos\theta_1+\sin\theta_2)\big]$
$=\text{r}_1\text{r}_2\big[\cos(\theta_1+\theta_2)+\text{i}\sin(\theta_1+\theta_2)\big]$
$\therefore\ |\text{z}_1\text{z}_2|=|\text{z}_1||\text{z}_2|$
View full question & answer→MCQ 1181 Mark
The argument of every complex number is:
Answer$z = x + iy$
amplitude $=\tan^{-1}\frac{\text{y}}{\text{x}}$
$\Rightarrow$ amplitude $=\theta\pm2\text{k}\pi$
where $\theta\in[-\pi,\pi]\ \forall\text{k}\in\text{R}$
since $\text{k}\in\text{R}$
$\Rightarrow$ Amplitude of any complex number is many valued.
View full question & answer→MCQ 1191 Mark
If $z = 3 + 5i,$ then $z^3+ z + 198 =$
- A
$3 - 15i$
- B
$-3 - 15i$
- C
$-3 + 15i$
- ✓
$3 + 15i$
AnswerCorrect option: D. $3 + 15i$
$ z=3+5 i $
$ z^3=(3+5 i)^3$
$ =3^3+3.3^2(5 i)+3.3(5 i)^2+(5 i)^3 $
$ =27-125 i+135 i-225 $
$ =-225+27+(135-125) i $
$ =-198+10 i $
$ \therefore z^3+z+198 $
$ =-198+10 i+3+5 i+198 $
$ =3+15 i $
View full question & answer→MCQ 1201 Mark
Which axis is known as real axis in argand plane?
- ✓
$x -$ axis
- B
$y -$ axis
- C
$z -$ axis
- D
AnswerCorrect option: A. $x -$ axis
The plane having a complex number assigned to each of its point is called the
complex plane or the Argand plane. When $(x + y i)$ is plotted in argand plane, $x-$axis is real axis.
View full question & answer→MCQ 1211 Mark
The amplitude of $\frac{1+\text{i}\sqrt{3}}{\sqrt{3}+\text{i}}$ is:
- A
$\frac{\pi}{3}$
- B
$-\frac{\pi}{3}$
- ✓
$\frac{\pi}{6}$
- D
$-\frac{\pi}{6}$
AnswerCorrect option: C. $\frac{\pi}{6}$
Let $\text{z}=\frac{1+\text{i}\sqrt{3}}{\sqrt{3}+\text{i}}$
$\Rightarrow\text{z}=\frac{1+\text{i}\sqrt{3}}{\sqrt{3}+\text{i}}\times\frac{\sqrt{3}-\text{i}}{\sqrt{3}-\text{i}}$
$\Rightarrow\text{z}=\frac{\sqrt{3}+2\text{i}-\sqrt{3}\text{i}^2}{3-\text{i}^2}$
$\Rightarrow\text{z}=\frac{\sqrt{3}+\sqrt{3}+2\text{i}}{4}$
$\Rightarrow\text{z}=\frac{2\sqrt{3}+2\text{i}}{4}$
$\Rightarrow\text{z}=\frac{\sqrt{3}}{2}+\frac{1}{2}\text{i}$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=\frac{1}{\sqrt{3}}$
$\Rightarrow\alpha=\frac{\pi}{6}$
Since, $z$ lies in the first quadrant.
Therefore, $\text{arg(z)}=\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)=\frac{\pi}{6}$
View full question & answer→MCQ 1221 Mark
Choose the correct answer. If $\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^{\text{x}}=1,$ then:
AnswerCorrect option: B. $x = 4n$
$\Rightarrow\Big[\frac{(1+\text{i})(1+\text{i})}{(1-\text{i})(1+\text{i})}\Big]^{\text{x}}=1$
$\Rightarrow\Big[\frac{1+2\text{i}+\text{i}^2}{1-\text{i}^2}\Big]^{\text{x}}=1$
$\Rightarrow\Big[\frac{2\text{i}}{1+1}\Big]^{\text{x}}=1$
$\Rightarrow\text{i}^{\text{x}}=1$
$\Rightarrow\text{x}=4\text{n},\text{n}\in\text{N}$
View full question & answer→MCQ 1231 Mark
If $\text{a}=1+\text{i},$ then $a^2$ equals:
AnswerCorrect option: B. $2\text{i}$
$\text{a}=1+\text{i}$
On squaring both the sides, we get,
$\text{a}^2=(1+\text{i})^2$
$\Rightarrow\text{a}^2=1+\text{i}^2+2\text{i}$
$\Rightarrow\text{a}^2=1-1+2\text{i} \ (\because\text{i}^2=-1)$
$\Rightarrow\text{a}^2=2\text{i}$
View full question & answer→MCQ 1241 Mark
The principal value of the amplitude of $(1 + i)$ is:
- ✓
$\frac{\pi}{4}$
- B
$\frac{\pi}{12}$
- C
$\frac{3\pi}{4}$
- D
$\pi$
AnswerCorrect option: A. $\frac{\pi}{4}$
Let $\text{z}=(1+\text{i})$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=1$
$\Rightarrow\alpha=\frac{\pi}{4}$
Since, $z$ lies in the first quadrant.
Therefore, $\text{arg(z)}=\frac{\pi}{4}$
View full question & answer→MCQ 1251 Mark
What is the value of $x$ if $(\text{a} + 2\text{b} – 3\text{c})\text{x}^2 + (\text{b} + 2\text{c} – 3\text{a})\text{x} + (\text{c} + 2\text{a} – 3\text{b}) = 0$ where $a, b, c$ are in $A.P?$
- A
$\frac{1}{2}$
- ✓
$\frac{1}{4}$
- C
$\frac{2}{3}$
- D
$\frac{3}{4}$
AnswerCorrect option: B. $\frac{1}{4}$
If the coefficients of $ (\text{x}^2 +\text{x} +\text{c}) = 0,$ then
$x$ will always be $= 1$
Therefore, here, $(\text{a} + 2\text{b} – 3\text{c}) + (\text{b} + 2\text{c} – 3\text{a}) + (\text{c} + 2\text{a} – 3\text{b}) = 0$
So, $x = 1.$
As, one of its root is $1$
so, we will calculate the other one.
As, $a, b, c$ are in $A.P$
so, $\text{b}=\Big(\frac{\text{a}+\text{c}}{2}\Big)$
Thus, product of the roots $\alpha\beta =\frac{(\text{c} + 2\text{a} – 3\text{b})}{(\text{a} + 2\text{b} – 3\text{c})}$
As, a root say $\alpha=1$ then,
$\beta=\frac{\Big(\frac{\text{c}+2\text{a}-3(\text{a}+\text{c})}{2}\Big)}{\Big(\frac{\text{a}+2(\text{a}+\text{c})}{2-3\text{c}}\Big)}$
We get the value of $\beta=\frac{1}{4}$
View full question & answer→MCQ 1261 Mark
In polar representation of a complex number $\Big(\text{r},\frac{\pi}{2}\Big)$lies on $........?$
- A
$x -$ axis
- ✓
$y -$ axis
- C
$z -$ axis
- D
AnswerCorrect option: B. $y -$ axis
To convert polar representation $(\text{r, }\theta)$ into argand plane $(\text{x, y}),$ substitute $\text{x}=\text{r}\cos\theta$ and $\text{y}=\text{r}\sin\theta.$
$\text{x}=\text{r}\cos\frac{\pi}{2}=0$ and $\text{y}=\text{r}\cos\frac{\pi}{2}=\text{r}$ Argand plane representation is $(0, r).$
Since real part is zero,
so it lies on imaginary axis
i.e. $y -$ axis.
View full question & answer→MCQ 1271 Mark
Choose the correct answer. 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 $
AnswerCorrect option: D. $ 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 $
View full question & answer→MCQ 1281 Mark
If $x$ is real and $\text{K}=\frac{\text{x}^2-\text{x}+1}{\text{x}^2+\text{x}+1},$ then
AnswerCorrect option: A. $\text{K}\in\Big[\frac{1}{3,3}\Big]$
$\text{K}=\frac{\text{x}^2-\text{x}+1}{\text{x}^2+\text{x}+1}$
$\Rightarrow kx^2+ kx + K = x^2 - x + 1$
$\Rightarrow (k - 1)x^2 + (k + 1) x + k - 1 = 0$
For real values of $x,$ the discriminant of $(k - 1)x^2 + (k + 1) x + k - 1 = 0$ should be greater than or equal to zero.
$\therefore$ If $\text{k}\neq1$
$(k + 1)^2 - 4(k - 1) (k - 1) > 0$
$\Rightarrow(\text{k}+1)^2-\big\{2(\text{k}-1)\big\}^2>0$
$\Rightarrow (3k - 1) (k - 3) < 0$
$\Rightarrow\frac{1}{3}<\text{K}<3$
And if $k = 1,$ then,
$x = 0,$ which is real $...(ii)$
So, from $(i)$ and $(ii)$, we get,
$\text{k}\in\Big[\frac{1}{3},3\Big]$
View full question & answer→MCQ 1291 Mark
Choose the correct answer. The value of $\text{arg(x),}$ when $x < 0$ is:
- A
$0$
- B
$\frac{\pi}{2}$
- ✓
$\pi$
- D
AnswerLet $z = -x + 0i$ 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 $(\text{z})=\pi$
View full question & answer→MCQ 1301 Mark
If $\alpha,\beta$ are roots of the equation $4x^2+ 3x + 7 = 0$, then $\frac{1}{\alpha}+\frac{1}{\beta}$ is equal to:
- A
$\frac{7}{3}$
- B
$\frac{-7}{3}$
- C
$\frac{3}{7}$
- ✓
$\frac{-3}{7}$
AnswerCorrect option: D. $\frac{-3}{7}$
Given equation: $4x^2+ 3x + 7 = 0$
Also, $\alpha$ and $\beta$ are the roots of the equation.
Sum of the roots $=\alpha+\beta=\frac{-\text{Coefficient of x}}{\text{Coefficient of x}^2}=-\frac{3}{4}$
Product of the roots $=\alpha\beta=\frac{\text{Coefficient term}}{\text{Coefficient of x}^2}=\frac{7}{4}$
$\therefore\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{-\frac{3}{4}}{\frac{7}{4}}=-\frac{3}{7}$
View full question & answer→MCQ 1311 Mark
The number of roots of the equation $\frac{(\text{x}+2)(\text{x}-5)}{(\text{x}+3)(\text{x}+6)}=\frac{\text{x}-2}{\text{x}+4}$ is:
Answer$\frac{(\text{x}+2)(\text{x}-5)}{(\text{x}+3)(\text{x}+6)}=\frac{\text{x}-2}{\text{x}+4}$
$ \Rightarrow\left(x^2-3 x-10\right)(x+4)=\left(x^2+3 x-18\right)(x-2) $
$ \Rightarrow x^3+4 x^2-3 x^2-12 x-10 x-40=x^3-2 x^2+3 x^2-6 x-18 x+36 $
$ \Rightarrow x^2-22 x-40=x^2-24 x+36 $
$ \Rightarrow 2 x=76 $
$ \Rightarrow x=38 $
Hence, the equation has only $1$ root.
View full question & answer→MCQ 1321 Mark
If $\alpha\beta$ are the roots of the equation $x^2+ px + q = 0$ then $-\frac{1}{\alpha}+\frac{1}{\beta}$ are the roots of the equation:
- A
$ x^2-p x+q=0 $
- B
$ x^2+p x+q=0 $
- ✓
$ q x^2+p x+1=0 $
- D
$ q x^2-p x+1=0 $
AnswerCorrect option: C. $ q x^2+p x+1=0 $
Given equation: $x^2+ px + q = 0$
Also, $\alpha$ and $\beta$ are the roots of the given equation.
Then, sum of the roots $=\alpha+\beta=-\text{p}$
Product of the roots $=\alpha\beta=\text{q}$
Now, for roots $-\frac{1}{\alpha,}-\frac{1}{\beta},$ we have:
Sum of the roots $= -\frac{1}{\alpha}-\frac{1}{\beta}=-\frac{\alpha+\beta}{\alpha\beta}=-\Big(\frac{-\text{p}}{\text{q}}\Big)=\frac{\text{p}}{\text{q}}$
Product of the roots $= \frac{1}{\alpha\beta}=\frac{1}{\text{q}}$
Hence, the equation involving the roots $-\frac{1}{\alpha},-\frac{1}{\beta}$ is as follows:
$\text{x}^2+(\alpha+\beta)\text{x}+\alpha\beta=0$
$\Rightarrow\text{x}^2-\frac{\text{p}}{\text{q}}\text{x}+\frac{1}{\text{q}}=0$
$\Rightarrow\text{qx}^2-\text{px}+1=0$
View full question & answer→MCQ 1331 Mark
If $(\text{x}+\text{iy})^{\frac{1}{3}}=\text{a}+\text{ib,}$ then $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=$
Answer$(\text{x}+\text{iy)}^{\frac{1}{3}}=\text{a}+\text{ib}$
Cubing on both the sides, we get:
$\text{x}+\text{iy}=(\text{a}+\text{ib})^3$
$\Rightarrow\text{x}+\text{iy}=\text{a}^3+(\text{ib})^3+3\text{a}^2\text{bi}+3\text{a}(\text{bi})^2$
$\Rightarrow\text{x}+\text{iy}=\text{a}^3+\text{i}^3\text{b}^3+3\text{a}^2\text{bi}+3\text{i}^2\text{ab}^2$
$\Rightarrow\text{x}+\text{iy}=\text{a}^3-\text{i}\text{b}^3+3\text{a}^2\text{bi}-3\text{ab}^2 \ (\because\text{i}^2=-1,\text{i}^3=-\text{i})$
$\Rightarrow\text{x}+\text{iy}=\text{a}^3-3\text{a}\text{b}^2+\text{i}(-\text{b}^3+3\text{a}^2\text{b})$
$\therefore\text{x}=\text{a}^3-3\text{a}\text{b}^2$ and $\text{y}=3\text{a}^2\text{b}-\text{b}^3$
or, $\frac{\text{x}}{\text{a}}=\text{a}^2-3\text{b}^2$ and $\frac{\text{y}}{\text{b}}=3\text{a}^2-\text{b}^2$
$\Rightarrow\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=\text{a}^2-3\text{b}^2+3\text{a}^2-\text{b}^2$
$\Rightarrow\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=4\text{a}^2-4\text{b}^2$
View full question & answer→MCQ 1341 Mark
If $\alpha,\beta$ are the roots of the equation $x^2− p(x + 1) − c = 0$, then $(\alpha+1)(\beta+1)=$
AnswerCorrect option: C. $1 - c$
Given equation:
$x^2 − p(x + 1) − c = 0$
or $x^2 − px − p − c = 0$
Also $\alpha $ and $\beta$ are the roots of the equation.
Sum of the roots $=\alpha+\beta=\text{p}$
Product of the roots $=\alpha\beta=-(\text{c}+\text{p})$
Then, $(\alpha+1)(\beta+1)=\alpha\beta+\alpha+\beta+1$
$=-(\text{c}+\text{p})+\text{p}+1$
$=1-\text{c}$
$=-\text{c}-\text{p}+\text{p}+1$
View full question & answer→MCQ 1351 Mark
If $\frac{1+7\text{i}}{(2-\text{i})^2},$ then:
AnswerCorrect option: D. $\text{amp(z)}=\frac{3\pi}{4}$
$\text{z}=\frac{1+7\text{i}}{(2-\text{i})^2}$
$\Rightarrow\text{z}=\frac{1+7\text{i}}{4-1-4\text{i}} \ [\because\text{i}^2=-1]$
$\Rightarrow\text{z}=\frac{1+7\text{i}}{3-4\text{i}}$
$\Rightarrow\text{z}=\frac{1+7\text{i}}{3-4\text{i}}\times\frac{3+4\text{i}}{3+4\text{i}}$
$\Rightarrow\text{z}=\frac{3+4\text{i}+21\text{i}+28\text{i}^2}{9-16\text{i}^2}$
$\Rightarrow\text{z}=\frac{3-28+25\text{i}}{9+16}$
$\Rightarrow\text{z}=\frac{-25+25\text{i}}{25}$
$\Rightarrow\text{z}=-1+\text{i}$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=1$
$\Rightarrow\alpha=\frac{\pi}{4}$
Since, $z$ lies in the second quadrant.
Therefore, $\text{amp(z)}=\pi-\alpha$
$=\pi-\frac{\pi}{4}$
$=\frac{3\pi}{4}$
View full question & answer→MCQ 1361 Mark
If $\text{z}=\frac{1}{(2+3\text{i})^2},$ then $|\text{z}|=$
- ✓
$\frac{1}{13}$
- B
$\frac{1}{5}$
- C
$\frac{1}{12}$
- D
AnswerCorrect option: A. $\frac{1}{13}$
Let $\text{z}=\frac{1}{(2+3\text{i})^2}$
$\Rightarrow\text{z}=\frac{1}{4+9\text{i}^2+12\text{i}}$
$\Rightarrow\text{z}=\frac{1}{4-9+12\text{i}}$
$\Rightarrow\text{z}=\frac{1}{-5+12\text{i}}$
$\Rightarrow\text{z}=\frac{1}{-5+12\text{i}}\times\frac{-5-12\text{i}}{-5-12\text{i}}$
$\Rightarrow\text{z}=\frac{-5-12\text{i}}{25+144}$
$\Rightarrow\text{z}=\frac{-5}{169}-\frac{12\text{i}}{169}$
$\Rightarrow|\text{z}|=\sqrt{\frac{25}{169^2}+\frac{144}{169^2}}$
$\Rightarrow|\text{z}|=\frac{1}{\sqrt{169}}$
$\Rightarrow|\text{z}|=\frac{1}{13}$
View full question & answer→MCQ 1371 Mark
What is the set of values of $p$ for which the roots of the equation $3\text{x}^2 + 2\text{x} + \text{p}(\text{p} – 1) = 0$ are of opposite sign?
- A
$(\infty,0)$
- ✓
$(0,1)$
- C
$(1,\infty)$
- D
$(0,\infty)$
AnswerCorrect option: B. $(0,1)$
Since the roots of the given equation are of opposite sing,
So, products $f$ the roots $< 0$
$\frac{\text{p}(\text{p}-1)}{3<0}$
$\text{p}(\text{p-1})<0$
$0<\text{p}<1$
For real and distinct roots $\frac{1}{2}-\sqrt{\frac{21}{6}}<\text{p}<\frac{1}{2}+\sqrt{\frac{21}{6}}$
View full question & answer→MCQ 1381 Mark
Solve $\sqrt{3\text{x}^2}+\text{x}+\sqrt{3}=0$
- ✓
$\frac{-1\pm\text{i}\sqrt{11}}{6\sqrt{3}}$
- B
$\frac{1\pm\text{i}\sqrt{11}}{6\sqrt{3}}$
- C
$\frac{1\pm\sqrt{11}}{6\sqrt{3}}$
- D
$\frac{-1\pm\sqrt{11}}{6\sqrt{3}}$
AnswerCorrect option: A. $\frac{-1\pm\text{i}\sqrt{11}}{6\sqrt{3}}$
$\sqrt{3\text{x}^2}+\text{x}+\sqrt{3}=0$
$\Rightarrow3\text{x}^2+\sqrt{3\text{x}}+3=0$
$\Rightarrow\text{D}=(\sqrt{3})^2-4.3.3=3-36=-33$
Since $\text{D}\leq0,$ imaginary roots are there.
View full question & answer→MCQ 1391 Mark
If $z$ is a non$-$zero complex number, then $\Big|\frac{|\bar{\text{z}}|^2}{\text{z}\bar{\text{z}}}\Big|$ is equal to:
AnswerCorrect option: A. $\Big|\frac{\bar{\text{z}}}{\text{z}}\Big|$
$\Big|\frac{|\bar{\text{z}}|^2}{\text{z}\bar{\text{z}}}\Big|=\Big|\frac{|\bar{\text{z}}|^2}{|\text{z}|^2}\Big| \ \Big(\because\text{z}\bar{\text{z}}=|\text{z}|^2\Big)$
Let $\text{z}=\text{a}+\text{ib}$
$\Rightarrow|\text{z}|=\sqrt{\text{a}^2+\text{b}^2}$
Let $\bar{\text{z}}=\text{a}-\text{ib}$
$\Rightarrow|\bar{\text{z}}|=\sqrt{\text{a}^2+\text{b}^2}$
$\therefore\Big|\frac{|\bar{\text{z}}|^2}{\text{z}\bar{\text{z}}}\Big|=\Big|\frac{|\bar{\text{z}}|^2}{|\text{z}|^2}\Big|$
$=\Big|\frac{\bar{\text{z}}}{\text{z}}\Big|$
View full question & answer→MCQ 1401 Mark
Which of the following is correct for any two complex numbers $z_1$ and $z_2$?
- ✓
$|\text{z}_1\text{z}_2|=|\text{z}_1||\text{z}_2|$
- B
$\text{arg}(\text{z}_1\text{z}_2)=\text{arg}(\text{z}_1)\text{arg}(\text{z}_2)$
- C
$|\text{z}_1+\text{z}_2|=|\text{z}_1|+|\text{z}_2|$
- D
$|\text{z}_1+\text{z}_2|\geq|\text{z}_1|+|\text{z}_2|$
AnswerCorrect option: A. $|\text{z}_1\text{z}_2|=|\text{z}_1||\text{z}_2|$
since we know that
$|\text{z}_1\text{z}_2|=|\text{z}_1||\text{z}_2|$
$\text{arg}(\text{z}_1\text{z}_2)=\text{arg}(\text{z}_1)+\text{arg}(\text{z}_2)$ and
$|\text{z}_1+\text{z}_2|\le|\text{z}_1|+|\text{z}_2|$
View full question & answer→MCQ 1411 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 imaginary roots?
- ✓
$(-\infty,-1)$
- B
$(-1,\infty)$
- C
$(-1, 1)$
- D
$(-\infty,\infty)$
AnswerCorrect option: A. $(-\infty,-1)$
For, imaginary roots, $\text{D}>0$
Where, $\text{D}=\text{b}^2-4\text{ac}$
In the equation, $(\text{a}+1)\text{x}^2+2(\text{a}+1)\text{x}+(\text{a}-2)=0$
$\text{D} = \Big[2(\text{a}+1)\Big]^2 – 4 (\text{a} + 1)(\text{a} – 2)$
$= 4\text{a}^2 + 4 + 8\text{a} – 4({\text{a}^2 – 2\text{a} + \text{a} – 2})$
$= 4\text{a}^2 + 4 + 8\text{a} – 4{\text{a}^2 – 4\text{a} +8<0}$
$\Rightarrow12\text{a}+12<0$
$\Rightarrow12\text{a}<-12$
$\Rightarrow\text{a}<-1$
$\therefore\text{a}\in(-\infty,-1)$
View full question & answer→MCQ 1421 Mark
If $\frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}=\text{x}+\text{iy},$ then $\text{x}^2+\text{y}^2$ is equal to:
- ✓
$\frac{(\text{a}^2+1)^4}{4\text{a}^2+1}$
- B
$\frac{(\text{a}+1)^2}{4\text{a}^2+1}$
- C
$\frac{(\text{a}^2-1)^2}{(4\text{a}^2-1)^2}$
- D
AnswerCorrect option: A. $\frac{(\text{a}^2+1)^4}{4\text{a}^2+1}$
$\text{x}+\text{iy}=\frac{(\text{a}^2+1)^2}{2\text{a}-\text{i}}$
Taking modulus on both the sides, we get:
$\sqrt{\text{x}^2+\text{y}^2}=\frac{(\text{a}^2+1)^2}{\sqrt{4\text{a}^2+1}}$
Squaring both sides, we get,
$\text{x}^2+\text{y}^2=\frac{(\text{a}^2+1)^4}{{4\text{a}^2+1}}$
View full question & answer→MCQ 1431 Mark
The argument of $\frac{1-\text{i}}{1+\text{i}}$ is:
- ✓
$-\frac{\pi}{2}$
- B
$\frac{\pi}{2}$
- C
$\frac{3\pi}{2}$
- D
$\frac{5\pi}{2}$
AnswerCorrect option: A. $-\frac{\pi}{2}$
Let $\text{z}=\frac{1-\text{i}}{1+\text{i}}$
$\Rightarrow\text{z}=\frac{1-\text{i}}{1+\text{i}}\times\frac{1-\text{i}}{1-\text{i}}$
$\Rightarrow\text{z}=\frac{1+\text{i}^2-2\text{i}}{1-\text{i}^2}$
$\Rightarrow\text{z}=\frac{1-1-2\text{i}}{1+1}$
$\Rightarrow\text{z}=\frac{-2\text{i}}{2}$
$\Rightarrow\text{z}=\text{i}$
Since, $z$ lies on negative direction of imaginary axis.
Therefore, $\text{arg(z)}=\frac{-\pi}{2}$
View full question & answer→MCQ 1441 Mark
$\text{arg} (\bar{\text{z}})$ is equal to:
- A
$\pi-\text{arg}\text{(z)}$
- ✓
$2\pi-\text{arg}\text{(z)}$
- C
$\pi+\text{arg}\text{(z)}$
- D
$2\pi+\text{arg}\text{(z)}$
AnswerCorrect option: B. $2\pi-\text{arg}\text{(z)}$
View full question & answer→MCQ 1451 Mark
Choose the correct answer. 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 $\alpha^2+\beta^2=$
AnswerGiven 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, $\alpha^2+\beta^2=1$
View full question & answer→MCQ 1461 Mark
If the square of $(a + ib)$ is real, then $ab =$
Answer$(a + ib)^2 = a^2 - b^2 + 2iab$ is given to be real
$\Rightarrow ab = 0$
View full question & answer→MCQ 1471 Mark
If $a, b$ are the roots of the equation $x^2+ x + 1 = 0$, then $a^2+ b^2=$
AnswerGiven equation: $x^2+ x + 1 = 0$
Also, $a$ and $b$ are the roots of the given equation.
Sum of the roots $=\text{a}+\text{b}=\frac{-\text{Co-efficient of x}}{\text{C-oefficient of x}^2}=-\frac{1}{1}=-1$
Product of the roots $=\text{ab}=\frac{\text{constant term}}{\text{Coefficient of x}}=\frac{1}{1}=1$
$\therefore(a+b)^2=a^2+b^2+2 a b $
$\Rightarrow(-1)^2=a^2+b^2+2 \times 1 $
$\Rightarrow 1-2=a^2+b^2 $
$\Rightarrow a^2+b^2=-1 $
View full question & answer→MCQ 1481 Mark
The least value of $k$ which makes the roots of the equation $x^2+ 5x + k = A0$ imaginary is:
AnswerThe roots of the quadratic equation $x^2+ 5x + k=0$ will be imaginary if its discriminant is less than zero.
$\therefore25-4\text{0k}<0$
$\Rightarrow\text{k}>\frac{25}{4}$
Thus, the minimum integral value of $k$ for which the roots are imaginary is $7.$
View full question & answer→MCQ 1491 Mark
If $\text{z}_1,\text{z}_2$ and $\text{z}_1,\text{z}_4$ are two pairs of conjugate complex numbers, then $\text{arg}\Big(\frac{\text{z}^1}{\text{z}^4}\Big)+\text{arg}\Big(\frac{\text{z}^2}{\text{z}^3}\Big)$is equal to:
- ✓
$0$
- B
$\frac{\pi}{2}$
- C
$\frac{3\pi}{2}$
- D
$\pi$
View full question & answer→MCQ 1501 Mark
The value of $(1+\text{i})^4+(1-\text{i})^4$ is:
AnswerUsing $\text{a}^4+\text{b}^4=(\text{a}^2+\text{b}^2)^2-2\text{a}^2\text{b}^2$
$(1+\text{i})^4+(1-\text{i})^4$
$=\Big((1+\text{i})^2+(1-\text{i})^2\Big)^2-2(1+\text{i})^2(1-\text{i})^2$
$=(1+\text{i}^2+2\text{i}+1+\text{i}^2-2\text{i})^2-2(1+\text{i}^2+2\text{i})(1+\text{i}^2-2\text{i})$
$=(1-1+2\text{i}+1-1-2\text{i})^2-2(1-1+2\text{i})(1-1-2\text{i})$
$=(0)-2(2\text{i})(-2\text{i}) \ (\because\text{i}^2=-1)$
$=8\text{i}^2$
$=-8$
View full question & answer→MCQ 1511 Mark
Choose the correct answer. Let $x, y \in R,$ then $x + iy$ is a non real complex number if:
- A
$x = 0$
- B
$y = 0$
- C
$x \neq 0$
- ✓
$y \neq 0$
AnswerCorrect option: D. $y \neq 0$
$x + yi$ is a non$-$real complex number if $y ≠ 0.$ If $x, y \in R$
View full question & answer→MCQ 1521 Mark
Determine the values of $p$ for which the quadratic equation $2x^2+ px + 8 = 0$ has equal roots.
- A
$p = ±64$
- ✓
$p = ±8$
- C
$p = ±4$
- D
$p = ±16$
AnswerCorrect option: B. $p = ±8$
In $a x^2+b x+c=0$
Discriminant $D=b^2-4 a c$
$D=0$, for the roots to be real and equal
In $2 x^2+p x+8=0$
Thus, we have $a=2, b=p$ and $c=8$
Then $D=b^2-4 a c=0$
$p^2-4 \times 2 \times 8=0$
$\Rightarrow p^2-64=0$
$\Rightarrow \mathrm{p}= \pm 8$
View full question & answer→MCQ 1531 Mark
If $z_1=2+3 i$ and $z_2=5+2 i$, then find $z_1-z_2$.
- ✓
$-3 + 1i$
- B
$3 - i$
- C
$7 + 5i$
- D
$7 - 5i$
AnswerCorrect option: A. $-3 + 1i$
In subtracting one complex number from other,
difference of corresponding parts of two complex numbers is calculated.
So, $z_1-z_2=(2-5)+(3-2) i=-3+1 {~i}$.
View full question & answer→MCQ 1541 Mark
$1 + 0 i$ is, $...........$ for complex number $z.$
- A
- B
additive identity element
- ✓
multiplicative identity element
- D
AnswerCorrect option: C. multiplicative identity element
On multiplying one $( 1 + 0 i )$ to a complex number, we get same complex number so $1 + 0i$ is multiplicative identity element for complex number $z.$
i.e. $z \times 1 = z.$
View full question & answer→MCQ 1551 Mark
If $\text{x}+\text{iy}=\frac{3+5\text{i}}{7-6\text{i}},$ then $\text{y}=$
- A
$\frac{9}{85}$
- B
$\frac{-9}{85}$
- ✓
$\frac{53}{85}$
- D
AnswerCorrect option: C. $\frac{53}{85}$
$\text{x}+\text{iy}=\frac{3+5\text{i}}{7-6\text{i}}$
$\Rightarrow\text{x}+\text{iy}=\frac{3+5\text{i}}{7-6\text{i}}\times\frac{7+6\text{i}}{7+6\text{i}}$
$\Rightarrow\text{x}+\text{iy}=\frac{21+53\text{i}+30\text{i}^2}{49-36\text{i}^2}$
$\Rightarrow\text{x}+\text{iy}=\frac{21-30+53\text{i}}{49+36}$
$\Rightarrow\text{x}+\text{iy}=\frac{-9}{85}+\text{i}\frac{53}{85}$
On comparing both the sides:
$\text{y}=\frac{53}{85}$
View full question & answer→MCQ 1561 Mark
$\frac{1+2\text{i}+3\text{i}^2}{1-2\text{i}+3\text{i}^2}$ equals:
AnswerLet $\text{z}=\frac{1+2\text{i}+3\text{i}^2}{1-2\text{i}+3\text{i}^2}$
$\Rightarrow\text{z}=\frac{1+2\text{i}-3}{1-2\text{i}-3}$
$\Rightarrow\text{z}=\frac{-2+2\text{i}}{-2-2\text{i}}\times\frac{-2+2\text{i}}{-2+2\text{i}}$
$\Rightarrow\text{z}=\frac{(-2+2\text{i})^2}{(-2)^2-(2\text{i})^2}$
$\Rightarrow\text{z}=\frac{4+4\text{i}^2-8\text{i}}{4+4}$
$\Rightarrow\text{z}=\frac{4-4-8\text{i}}{8}$
$\Rightarrow\text{z}=\frac{-8\text{i}}{8}$
$\Rightarrow\text{z}=-\text{i}$
View full question & answer→MCQ 1571 Mark
What is the number of solution$(s)$ of the equation $|\sqrt{\text{x}-2}|+\sqrt{\text{x} (\sqrt{\text{x}-4})}+2=0$
- ✓
$2$
- B
$4$
- C
- D
Infinitely many solutions
AnswerWe have $|\sqrt{\text{x}-2}|+\sqrt{\text{x}}(\sqrt{\text{x}-4})+2=0$
$|\sqrt{\text{x}} – 2| + \sqrt{(\text{x})^2} – 4\sqrt{\text{x}} + 2 = 0$
$|\sqrt{\text{x}} – 2| + |\sqrt{\text{x}} -2|^2 – 2 = 0$
$|\sqrt{\text{x}} – 2| = -2,\text{ 1}$
Thus, $\sqrt{\text{x}}-2=+1,-1$ or $\text{x}=1,\text{ 9}$
View full question & answer→MCQ 1581 Mark
If $\text{z}=\Big(\frac{1+\text{i}}{1-\text{i}}\Big),$ then $z^4$ equals:
AnswerLet $\text{z}=\frac{1+\text{i}}{1-\text{i}}$
Rationalising the denominator:
$\text{z}=\frac{1+\text{i}}{1-\text{i}}\times\frac{1+\text{i}}{1+\text{i}}$
$\Rightarrow\text{z}=\frac{1+\text{i}^2+2\text{i}}{1-\text{i}^2}$
$\Rightarrow\text{z}=\frac{1-1+2\text{i}}{1+1}$
$\Rightarrow\text{z}=\frac{2\text{i}}{2}$
$\Rightarrow\text{z}=\text{i}$
$\Rightarrow\text{z}^4=\text{i}^4$
Since $\text{i}^2=-1,$ we have:
$\Rightarrow\text{z}^4=\text{i}^2\times\text{i}^2$
$\Rightarrow\text{z}^4=1$
View full question & answer→MCQ 1591 Mark
Choose the correct answer.
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
AnswerCorrect 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
$\frac{3\cos\theta}{1+4\cos^2\theta}=0$
$\Rightarrow3\cos\theta=0$
$\Rightarrow\cos\theta=0$
$\therefore\ \theta=(2\text{n}+1)\frac{\pi}{2},\text{ n}\in\text{N}$
View full question & answer→MCQ 1601 Mark
For what value of $\theta, 1$ lies between the roots of the quadratic equation $3\text{x}^2 – 3\sin\theta \text{x} – 2\cos^2\theta = 0$
- ✓
$2\text{n}\pi+\frac{\pi}{6}<\theta<2\text{n}\pi+\frac{5\pi}{6}$
- B
$2\text{n}\pi+\frac{\pi}{3}<\theta<2\text{n}\pi+\frac{5\pi}{3}$
- C
$2\text{n}\pi+\frac{\pi}{6}<\theta\leq2\text{n}\pi+\frac{5\pi}{6}$
- D
$2\text{n}\pi+\frac{\pi}{3}<\theta\leq2\text{n}\pi+\frac{5\pi}{3}$
AnswerCorrect option: A. $2\text{n}\pi+\frac{\pi}{6}<\theta<2\text{n}\pi+\frac{5\pi}{6}$
Let, $\text{f}(\text{x})=3\text{x}^2-3\sin\theta\text{x}-2\cos^2\theta$
The coefficient of $\text{x}^2>0$
$\text{f}(1)<0$
So, $3-3\sin\theta-2\cos^2\theta<0$
$\Rightarrow2\sin^2\theta-3\sin\theta+1<0$
$\Rightarrow(2\sin\theta-1)(\sin\theta-1)<0$
$\Rightarrow\frac{1}{2}2\text{n}\pi+\frac{\pi}{6}<\theta<2\text{n}\pi+\frac{5\pi}{6}$
View full question & answer→MCQ 1611 Mark
Solve $2x^2+ x + 1 = 0.$
- ✓
$\frac{-1\pm\text{i}\sqrt{7}}{4}$
- B
$\frac{1\pm\text{i}\sqrt{7}}{4}$
- C
$\frac{1\pm\sqrt{7}}{4}$
- D
$\frac{-1\pm\sqrt{7}}{4}$
AnswerCorrect option: A. $\frac{-1\pm\text{i}\sqrt{7}}{4}$
$2\text{x}^2+\text{x}+1=0$
$\text{D}=1^2-4\times2\times1=1-8=-7\leq0$
Since $\text{D}\leq0,$ imaginary roots are there.
$\Rightarrow\text{x}=\frac{-1\pm\sqrt{1^2-42.1}}{2.2}=\frac{-1\pm\text{i}\sqrt{7}}{4}$
View full question & answer→MCQ 1621 Mark
Roots of a quadratic equation are imaginary when discriminant is $..............?$
- A
- B
- ✓
- D
greater than or equal to zero
AnswerFor a quadratic equation, $ax^2 + bx + c = 0$, discriminant is $b^2$-4ac.
Roots are $\frac{-\text{b}\pm\sqrt{b^2 -4\text{ac}}}{2\text{a}}$ for imaginary roots, radical is negative
i.e. discriminant should be less than zero.
View full question & answer→MCQ 1631 Mark
The argument of $\frac{(1-\text{i}\sqrt{3})}{(1+\text{i}\sqrt{3})}$ is:
- ✓
$60^\circ$
- B
$120^\circ$
- C
$210^\circ$
- D
$240^\circ$
AnswerCorrect option: A. $60^\circ$
View full question & answer→