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 $\text{a}^4+\text{b}^4=(\text{a}^2+\text{b}^2)^2-2\text{a}^2\text{b}^2$
$(1+\text{i})^4+(1-\text{i})^4$
$=\Big((1+\text{i})^2+(1-\text{i})^2\Big)^2-2(1+\text{i})^2(1-\text{i})^2$
$=(1+\text{i}^2+2\text{i}+1+\text{i}^2-2\text{i})^2-2(1+\text{i}^2+2\text{i})(1+\text{i}^2-2\text{i})$
$=(1-1+2\text{i}+1-1-2\text{i})^2-2(1-1+2\text{i})(1-1-2\text{i})$
$=(0)-2(2\text{i})(-2\text{i}) \ (\because\text{i}^2=-1)$
$=8\text{i}^2$
$=-8$
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