MCQ Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 6

If quadratic equation

$\displaystyle 6x^{2}-7x+2=0$ has one root

$\displaystyle \frac{1}{2}$ then the second root will be--

0%
$\displaystyle -\frac{1}{2}$
0%
$\displaystyle \frac{2}{3}$
0%
$\displaystyle \frac{3}{2}$
0%
1

Explanation

Let the second root of the quadratic equation

$\displaystyle 6x^{2} - 7x + 2 = 0$ be

$\displaystyle \alpha$

Then as per question--

$\displaystyle \because$ Sum of the roots =

$\displaystyle \alpha +\frac{1}{2}=\frac{7}{6}$

$\displaystyle \Rightarrow$

$\displaystyle \alpha =\frac{7}{6}-\frac{1}{2}=\frac{7-3}{6}=\frac{2}{3}$

$\displaystyle \Rightarrow$ Product of the roots =

$\displaystyle \alpha \times \frac{1}{2}=\frac{2}{6}$

$\displaystyle \Rightarrow$

$\displaystyle \alpha =\frac{1}{3}\times 2=\frac{2}{3}$

$\displaystyle \therefore$ Second root =

$\displaystyle \frac{2}{3}$

The sum of the roots of

$\displaystyle \frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}$ is zero. The product of the roots is

0%
$\displaystyle -\frac{1}{2}(a^{2}+b^{2})$
0%
0
0%
$\displaystyle \frac{1}{2}(a+b)$
0%
$\displaystyle 2(a^{2}+b^{2})$

Explanation

Consider the given equation.

$\dfrac{1}{x+a}+\dfrac{1}{x+b}=\dfrac{1}{c}$

$\dfrac{x+a+x+b}{(x+a)(x+b)}=\dfrac{1}{c}$

$x^2+(a+b-2c)x+[ab-(a+b)c]=0$

$..........(1)$

Let the roots of the given equation be

$\alpha$ and

$\beta$ .

Since,

$\alpha+\beta=0$

$a+b-2c=0$

$2c=a+b$

And

$\alpha\beta=ab-(a+b)c$

$\alpha\beta=ab-(a+b)\dfrac{(a+b)}{2}$

$\alpha\beta=-\dfrac{(a^2+b^2)}{2}$

Hence, this is the answer.

Find the value of P for which the following equation has equal roots

$\displaystyle { px }^{ 2 }-8x+2p=0$

0%
$\displaystyle \pm 2\sqrt { 2 }$
0%
$\displaystyle \pm \sqrt { 2 }$
0%
$\displaystyle \pm \sqrt { 3 }$
0%
$\displaystyle \pm 2\sqrt { 3 }$

Explanation

$\displaystyle { px }^{ 2 }-8x+2p=0$
here,

$\displaystyle a=p,b=-8,c=2p$
For equal roots

$\displaystyle { b }^{ 2 }-4ac=0$
or,

$\displaystyle { \left( -8 \right) }^{ 2 }-4\left( p \right) \left( 2p \right) =0$
or,

$\displaystyle 64-8{ p }^{ 2 }=0$
or,

$\displaystyle -8{ p }^{ 2 }=-64$
or,

$\displaystyle 8{ p }^{ 2 }=64$
or,

$\displaystyle { p }^{ 2 }=\frac { 64 }{ 8 } =8$
or,

$\displaystyle p=\pm \sqrt { 8 } =\pm 2\sqrt { 2 }$

The roots of the equation

${2x^{2 }+ 3x + 2} = 0$ are

0%
Real, rational, and equal
0%
Real,rational and unequal
0%
Real, irrational, and unequal
0%
non real (imaginary)

Explanation

Given quadratic is

$2x^2+3x+2=0$
Comparing it with standard from of quadratic equation

$ax^2+bx+c=0$ , we get

$a=2, b=3, c=2$
Thus discriminant of the given quadratic is

$D=b^2-4ac=9-16=-7<0$
Hence roots of given quadratic are imaginary.
Thus choice

$D$ is correct

Find the product. Write the answer in standard form.

$i\left( 6-2i \right) \left( 7-5i \right)$

0%
$52+16i$
0%
$10{i}^{3}+44{i}^{2}+42i$
0%
$-44-32i$
0%
$44+32i$

Explanation

Given

$z=i(6-2i)(7-5i)$

$z=i[6(7-5i)-2i(7-5i)]$

$=i[42-30i-14i+10i^2]$

$=i[32-44i]$ ............[since,

$i^2=-1$ ]

$=-44i^2+32i$

$z=44+32i$

Which of the following statements has the truth value

$'F'$ ?

0%
A quadratic equation has always a real root
0%
The number of ways of seating $2$ persons in two chairs out of $n$ persons in $P(n,2)$
0%
The cube roots of unity are in GP
0%
None of the above

Explanation

The quadratic equation have real roots if and only if

$D=b^2-4ac\geq0$ .

So, the first statement has truth value F.

The imaginary number

$i$ is defined such that

$i^2=-1$ . What is the value of

$(1 - i \sqrt {5}) ( 1 + i\sqrt {5})$ ?

0%
$\sqrt5$
0%
$5$
0%
$6$
0%
$\sqrt6$

Explanation

Imaginary number is

$i=\sqrt {-1}$

Value of

$(1-i\sqrt{5})(1+i\sqrt{5})$

It is in the form of

$(a+b)(a-b)=a^2-b^2$

So,

$(1-i\sqrt{5})(1+i\sqrt{5})={(1)}^2-{(i\sqrt {5})}^2$

$\Rightarrow 1-(i^2\times 5)$

$\Rightarrow 1-(-1\times 5)$

$\Rightarrow 6$

$(1-i\sqrt{5})(1+i\sqrt{5})=6$

The product of

$(3-2i)$ and

$\left(\dfrac { 5 }{ 2 } -4i\right)$ , if

$i=\sqrt { -1 }$ , is:

0%
$-\dfrac { 1 }{ 2 } -17i$
0%
$14+\dfrac{9}{2}i$
0%
$2-8i-14{i}^{2}$
0%
$i\left(8+\dfrac{9}{2}\right)$

Explanation

$(3 - 2i) \times \left(\dfrac{5}{2} - 4i\right)$

$= 3\left(\dfrac{5}{2} - 4i\right) - 2i\left(\dfrac{5}{2} - 4i\right)$

$= \dfrac{15}{2} - 12i - 5i - 8$

$= \dfrac{-1}{2} - 17i$

The argument of

$\dfrac {1 + i\sqrt {3}}{\sqrt {3} + i}$ is

0%
$\dfrac {\pi}{3}$
0%
$\dfrac {\pi}{4}$
0%
$\dfrac {2\pi}{3}$
0%
$\dfrac {\pi}{6}$

Explanation

Let

$z = \dfrac {1 + i\sqrt {3}}{\sqrt {3} + i}$

$= \dfrac {1 + i\sqrt {3}}{(\sqrt {3} + i)} \times \dfrac {(\sqrt {3} - i)}{(\sqrt {3} - i)}$

$= \dfrac {\sqrt {3} - i + 3i + \sqrt {3}}{3 + 1} = \dfrac {\sqrt {3} + i}{2}$

$\arg(z) = \tan^{-1}\left (\dfrac {1}{\sqrt {3}}\right )$

$= \dfrac {\pi}{6}$

The principal amplitude of

$\displaystyle { \left( \sin { { 40 }^{ \circ }+i\cos { { 40 }^{ \circ } } } \right) }^{ 5 }$ is

0%
$\displaystyle { 70 }^{ \circ }$
0%
$\displaystyle { -110 }^{ \circ }$
0%
$\displaystyle { 110 }^{ \circ }$
0%
$\displaystyle { -70 }^{ \circ }$

Explanation

$\displaystyle { \left( \sin { { 40 }^{ \circ } } +i\cos { { 40 }^{ \circ } } \right) }^{ 5 }$

$\displaystyle ={ i }^{ 5 }{ \left( \cos { { 40 }^{ \circ } } -i\sin { { 40 }^{ \circ } } \right) }^{ 5 }$

$\displaystyle =i\left( \cos { { 200 }^{ \circ } } -i\sin { { 200 }^{ \circ } } \right)$

$\displaystyle =i\left[ \cos { \left( { 180 }^{ \circ }+{ 20 }^{ \circ } \right) } -i\sin { \left( { 180 }^{ \circ }+{ 20 }^{ \circ } \right) } \right]$

$\displaystyle =i\left( -\cos { { 20 }^{ \circ } } +i\sin { { 20 }^{ \circ } } \right)$

$\displaystyle =-i\cos { { 20 }^{ \circ } } -\sin { { 20 }^{ \circ } }$

$\displaystyle =\cos { \left( -{ 11 }0^{ \circ } \right) + } i\sin { \left( -{ 11 }0^{ \circ } \right) }$

$\displaystyle \therefore$ Principal amplitude =

$\displaystyle -{ 110 }^{ \circ }$ .

Let

${ X }_{ n }=\left\{ z=x+iy:{ \left| z \right| }^{ 2 } \le \dfrac { 1 }{ n } \right\}$ for all integers

$n\ge 1$ . Then,

$\displaystyle\bigcap _{ n=1 }^{ \infty }{ { X }_{ n } }$ is

0%
A singleton set
0%
Not a finite set
0%
An empty set
0%
A finite set with more than one element

Explanation

Given,

${ X }_{ n }=\left\{ z=x+iy:{ \left| z \right| }^{ 2 }\le \dfrac { 1 }{ n } \right\}$

$=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le \dfrac { 1 }{ n } \right\}$

$\therefore { X }_{ 1 }=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le 1 \right\}$

${ X }_{ 2 }=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le \dfrac { 1 }{ 2 } \right\}$

${ X }_{ 3 }=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le \dfrac { 1 }{ 3 } \right\}$
.............................................
.............................................
.............................................

${ X }_{ \infty }=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le 0 \right\}$

$\therefore \bigcap _{ n=1 }^{ \infty }{ { X }_{ n } } ={ X }_{ 1 }\cap { X }_{ 2 }\cap { X }_{ 3 }\cap \dots \cap { X }_{ \infty }$

$=\left\{ { x }^{ 2 }+{ y }^{ 2 }=0 \right\}$
Hence,

$\bigcap _{ n=1 }^{ \infty }{ { X }_{ n } }$ is a singleton set.

The modulus of

$\dfrac { 1-i }{ 3+i } +\dfrac { 4i }{ 5 }$ is

0%
$\sqrt { 5 }$ unit
0%
$\dfrac { \sqrt { 11 } }{ 5 }$ unit
0%
$\dfrac { \sqrt { 5 } }{ 5 }$ unit
0%
$\dfrac { \sqrt { 12 } }{ 5 }$ unit

Explanation

Let

$z=\dfrac { 1-i }{ 3+i } +\dfrac { 4i }{ 5 }$

$=\dfrac { 5-5i+12i-4 }{ 5\left( 3+i \right) } =\dfrac { 1+7i }{ 5\left( 3+i \right) }$

$=\dfrac { \left( 1+7i \right) \left( 3-i \right) }{ 5\left( 9+1 \right) }$

$=\dfrac { 10+20i }{ 50 } =\dfrac { 1+2i }{ 5 }$

$\therefore \left| z \right| =\sqrt { { \left( \dfrac { 1 }{ 5 } \right) }^{ 2 }+{ \left( \dfrac { 2 }{ 5 } \right) }^{ 2 } } =\dfrac { 1 }{ 5 } \sqrt { 1+4 } =\dfrac { \sqrt { 5 } }{ 5 }$

Simplify

$(2+8i)(1-4i)-(3-2i)(6+4i)$

(Note

$:i=\sqrt{-1}$ )

0%
$8$
0%
$26$
0%
$34$
0%
$50$

Explanation

The value of

$\left ( 2+8i \right )\left ( 1-4i \right )-\left ( 3-2i \right )\left ( 6+4i \right )$

$=$

$\left ( 2+8i-8i-32i^{2} \right )-\left ( 18-12i+12i-8i^{2} \right )$

$=$

$2-32i^{2}-18+8i^{2}$

Put the given value of

$i=\sqrt{-1}$ , we get

$=$

$2-32(\sqrt{-1})^{2}-18+8(\sqrt{-1})^{2}$

$=$

$2-32(-1)-18+8(-1)$

$=$

$2+32-18-8$

$=$

$34-26=8$

$\displaystyle \left|\dfrac{\sqrt{3}+i}{(1+i)(1+\sqrt{3}i)}\right|=$

0%
$1$
0%
$\sqrt{2}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{\sqrt{2}}$

Explanation

The value of

$\displaystyle \left |\frac{\sqrt{3} + i}{(1 + i)(1 + \sqrt{3}i)} \right |$

$= \dfrac{|\sqrt{3} + i|}{|1 + i||1 + \sqrt{3}i|} = \dfrac{\sqrt{3 + 1}}{\sqrt{1 + 1} \times \sqrt{1 + 3}}$

$= \dfrac{2}{2\sqrt{2}}$

$= \dfrac{1}{\sqrt{2}}$

What is the approximate magnitude of

$8 + 4i$ ?

0%
4.15
0%
8.94
0%
12.00
0%
18.64
0%
32.00

Explanation

Magnitude of

$8+4i$ is

$=\sqrt { { 8 }^{ 2 }+{ 4 }^{ 2 } }$

$=\sqrt { 64+16 }$

$=\sqrt { 80 }$

$=8.94$

Find the value of

$k$ for which the number lies between the roots of the equation

${ k }^{ 2 }+(1-2k)x+({ x }^{ 2 }-k-2)=0$ .

0%
$3< k< 4$
0%
$3< k< 5$
0%
$2< k< 6$
0%
$2< k< 5$

Given that

$4$ is a root of the quadratic equation

${x}^{2}-5x+q=0$ . Find the value of

$q$ and the other root.

0%
$4$ and $1$ respectively
0%
$1$ and $4$ respectively
0%
$4$ and $3$ respectively
0%
$3$ and $1$ respectively

The real part of

${ \left( 1-\cos { \theta } +i\sin { \theta } \right) }^{ -1 }$ is

0%
$\cfrac{1}{2}$
0%
$\cfrac { 1 }{ 1+\cos { \theta } }$
0%
$\tan { \cfrac { \theta }{ 2 } }$
0%
$\cot { \cfrac { \theta }{ 2 } }$

Explanation

Given that:

${ \left( 1-\cos { \theta } +i\sin { \theta } \right) }^{ -1 }=\dfrac{1}{1-\cos\theta+i\sin\theta}$

$=\dfrac{1}{2\sin^2\frac{\theta}{2}+i\cdot 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}=\dfrac{1}{2\sin\frac{\theta}{2}}\cdot \dfrac{1}{\sin\frac{\theta}{2}+i\cos\frac{\theta}{2}}$

$=\dfrac{1}{2\sin\frac{\theta}{2}}\cdot \dfrac{1}{\sin\frac{\theta}{2}+i\cos\frac{\theta}{2}}\cdot \dfrac{\sin\frac{\theta}{2}-i\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}-i\cos\frac{\theta}{2}}$ , rationalize denominator

$=\dfrac{1}{2\sin\frac{\theta}{2}}(\sin\frac{\theta}{2}-i\cos\frac{\theta}{2})$

$=\dfrac{1}{2}-\dfrac{i}{2}\cot\frac{\theta}{2}$

The real part is

$\dfrac{1}{2}$

$z = \dfrac {(\sqrt {3} + i)^{3} (3i + 4)^{2}}{(8 + 6i)^{2}}$ , then

$|z|$ is equal to

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

Uisng the identity

$|a+ib|=\sqrt{a^2+b^2}$
Since

$\bigg|\dfrac{z_1z_2}{z_3} \bigg|=\dfrac{|z_1||z_2|}{|z_3|}$

Therefore

$|z| = \bigg|\dfrac {(\sqrt {3} + i)^{3} (3i + 4)^{2}}{(8 + 6i)^{2}}\bigg|$

$\quad =\dfrac{|(\sqrt 3+i)^3||(3i+4)^2|}{|(8+6i)^2|}$

$\quad =\dfrac{|\sqrt 3+i|^3|3i+4|^2}{|8+6i|^2}$

$\quad =\dfrac{2^3\cdot 5^2}{10^2}=2$

If the roots of the equations

$ax^{2} + 2bx + c = 0$ and

$bx^{2} - 2\sqrt {ac} x + b = 0$ are simultaneously real, then

0%
$b^{2} = 4ac$
0%
$b^{2} = ac$
0%
$2b^{2} = 9ac$
0%
none

Explanation

for eq

$^{n}$ to be have equal roots

$D\geq 0$

$\Rightarrow (2b)^{2}-4ac\geq 0\Rightarrow \boxed{b^{2}> ac}$

$(2\sqrt{ac})^{2}-4(b)(b)\geq 0\Rightarrow \boxed{ac\geq b^{2}}$

for roots to be simultaneously equal

$\boxed{b^{2} = ac}$

1125089_516963_ans_157df317651d434d846a7298ee32b4ce.jpg

If the roots of the equation

${ x }^{ 2 }+2(3a+5)x+2(9{ a }^{ 2 }+25)=0$ are real, then find

$a$ .

0%
$\cfrac{5}{3}$
0%
$\cfrac{7}{3}$
0%
$\cfrac{2}{3}$
0%
None of these

Explanation

As roots of equation are real

$\Rightarrow b^{2}-4ac \geq 0$

$\Rightarrow [2(3a+5)]^{2}-4\times 1\times (18a^{2}+50) \geq 0$

$\Rightarrow 4(9a^{2}+25+30a)-72a^{2}-200 \geq 0$

$\Rightarrow 36a^{2}+100+120a-72a^{2}-200 \geq 0$

$\Rightarrow 36a^{2}+100-120a \leq 0$

$\Rightarrow (6a-10)^{2} \leq 0 \Rightarrow a\leq \dfrac{5}{3}$

For what values of 'k', the equation

$x^{2} + 2(k - 4) x + 2k = 0$ has equal roots?

0%
$8, 2$
0%
$6, 4$
0%
$12, 2$
0%
$10, 4$

Explanation

As eq

$^{n}$ has equal roots

$\Rightarrow D = 0$

$\Rightarrow \left [ 2(k-4) \right ]^{2}-4\times 2k\times 1 = 0$

$\Rightarrow k^{2}-8k+16-2k = 0$

$\Rightarrow k^{2}-10k+16 = 0$

$\Rightarrow (k-2)(k-8) = 0$

$\Rightarrow \boxed{k = 8,2}$

1106569_517065_ans_8e6698c94f9744dd9ba0bff5fa9ef24f.jpg

What is the smallest integral value of

$k$ such that

$2x (kx - 4) - x^{2} + 6 = 0$ has no real roots?

0%
$-1$
0%
$2$
0%
$3$
0%
$4$

Explanation

$2x(kx-4)-x^2+6=0$

$\therefore 2kx^2-8x-x^2+6=0$

$\therefore (2k-1)x^2-8x+6=0$

The condition for a quadratic equation to have no real roots is

$Discriminant<0$

$\therefore (-8)^2-4(2k-1)(6)<0$

$\therefore 64-48k+24<0$

$\therefore k>\dfrac{88}{48}$

$\therefore k>\dfrac{11}{6}$

So, the smallest integer value of k is

$2$

The answer is option (B)

Perform the indicated operations:

$(5+3i)(3-2i)$

0%
$21-2i$
0%
$19-3i$
0%
$11-2i$
0%
$21-i$

Explanation

The value of

$(5+3i)(3-2i)$

$=5(3-2i)+3i(3-2i)$

$= 15-10i+9i-6{i}^{2}$

$= 21-i$

The roots of the equation

$(b+c)x^2-(a+b+c)x+a=0$

$(a,b,c\ \epsilon \ Q, b+c \neq a)$ are

0%
irrational and different
0%
rational and different
0%
imaginary and different
0%
real and equal

Explanation

$(b+c){ x }^{ 2 }-(a+b+c)x+a=0\\ a,b,c\in q\\ b+c\neq a\\ D={ (a+b+c) }^{ 2 }-4(a)(b+c)\\ ={ a }^{ 2 }+{ b }^{ 2 }+c^{ 2 }-2ab-2ac+2bc\\ ={ (a-b-c) }^{ 2 }\\ as\quad b+c\neq a\\ \therefore D>0$ always

$\therefore$ roots are rational and distinct as

$D$ is a perfect square and greater than

$0$

The number of real roots of

$\left (x+\dfrac{1}{x}\right)^2-4=0$ is

0%
$0$
0%
$2$
0%
$4$
0%
none of these

Explanation

${ (x+\cfrac { 1 }{ x } ) }^{ 2 }-4=0\\ { (x+\cfrac { 1 }{ x } ) }^{ 2 }-{ 2 }^{ 2 }=0\\ { (x+\cfrac { 1 }{ x } -2) }(x+\cfrac { 1 }{ x } +2)=0\\ \therefore x+\cfrac { 1 }{ x } -2=0\\ { x }^{ 2 }-2x+1=0\\ { (x-1) }^{ 2 }=0\\ x=1\\ or\quad \\ x+\cfrac { 1 }{ x } +2=0\\ { x }^{ 2 }+2x+1=0\\ { (x+1) }^{ 2 }=0\\ \therefore x=-1\\ \therefore 2\quad real\quad roots$

If the argument of a complex number is

$\cfrac { \pi }{ 2 }$ , then the number is:

0%
Purely imaginary
0%
Purely real
0%
$0$
0%
Neither real nor imaginary

Explanation

Let the complex number be

$z$ .
Hence

$z=|z|e^{i\arg(z)}$

$=|z|e^{i\tfrac{\pi}{2}}$

$=|z|\left(\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}\right)$

$=i|z|$
Hence the complex number is purely imaginary.

Given :

$u = 1+i \sqrt{3}$ and

$v = \sqrt{3} + i$

Calculate $\dfrac{u^3 }{ v^4}$

0%
$(1/4) - i \sqrt{1/4}$
0%
$(3/4) - i \sqrt{3}/4$
0%
$(1/4) - i \sqrt{3}/4$
0%
none of these

Explanation

$u=1+i\sqrt 3 \implies \sqrt{1+3}e^{i(\tan^{-1}\sqrt 3)}\\ \implies 2e^{i\pi/3}$

$v=i+\sqrt 3 \implies \sqrt{1+3}e^{i(\tan^{-1} 1/\sqrt 3)}\\ \implies 2e^{i\pi/6}$

$\cfrac{u^3}{v^4}=\cfrac{2^3e^{i\pi}}{2^4e^{2i\pi/3}}=(1/2)e^{-i\pi/3}$

$\implies \cfrac{1}{2} [\cos(\pi/3)-i \sin(\pi/3)] \implies \cfrac{1}{2}(\cfrac{1}{2}-\cfrac{i\sqrt 3}{2})$

$\implies (\cfrac{1}{4}-\cfrac{i\sqrt 3}{4})$

$\overline { z }$ lies in the third quadrant then

$z$ lies in the

0%
First quadrant
0%
Second quadrant
0%
Third quadrant
0%
Fourth quadrant

Explanation

Fact:

$\overline z$ is image of

$z$ in

$x$ -axis

So if

$\overline z$ lies in third quadrant then

$z$ will lie in second quadrant

$l,m,n$ are real and

$l \neq m$ , the roots of the equation

$(l-m)x^2-5(l+m)x-2(l-m)=0$ are-

0%
complex
0%
real and equal
0%
real and unequal
0%
none of these

Explanation

$l,m,n\in R\\ l\neq m\\ (l-m){ x }^{ 2 }-5(l+m)x-2(l-m)=0\\ D={ (-5(l+m)) }^{ 2 }+8{ (l-m) }^{ 2 }\\$

${ (l+m) }^{ 2 }$ and

${ (l-m) }^{ 2 }$ both greater than zero as

$l\neq m$ so

$D$ greater than zero so roots are real and distinct.

$arg(z) < 0$ , then

$arg(-z)-arg(z)=$

0%
$\pi$
0%
$-\pi$
0%
$\dfrac{\pi}{2}$
0%
$-\dfrac{\pi}{2}$

Explanation

Let

$Z=re^{i\theta_1}$

$-Z=-re^{i\theta_1}$

$\implies -a\cos\theta_1-ib\sin\theta_1$

$\implies -a\cos(\pi+\theta_1)-ib\sin(\pi+\theta_1)$

$\implies re^{i(\pi+\theta_1)}$

$arg(-Z)-arg(Z)$

$\implies \pi+\theta_1-\theta_1\\ \implies \pi$

If Arg

$(z + i)\, -$ Arg

$(z - i)$

$= \dfrac{\pi}{2}$ , then

$z$ lies on a ..........

0%
Circle
0%
Line
0%
Coordinate axes
0%
None of these

Explanation

Putting z = x+iy,

${tan}^{-1}\dfrac{y+1}{x}$ -

${tan}^{-1}\dfrac{y-1}{x}$ =

$\pi$ /2

$\Rightarrow$ 1 + (

$\dfrac{y+1}{x})$ (

$\dfrac{y-1}{x}$ ) = 0

$\Rightarrow$

$x^2 + y^2$ = 1

It is a circle of center coinciding with origin and radius 1 units.

Hence, option A is correct.

Determine the nature of the roots of the equation:

$x^2-8x+12=0$

0%
Real and unequal
0%
Real and equal
0%
Imaginary
0%
None of these

Explanation

$\Rightarrow$

$x^2-8x+12=0$

$\Rightarrow$

$x^2-6x-2x+12=0$

$\Rightarrow$

$x(x-6)-2(x-6)=0$

$\Rightarrow$

$(x-6)(x-2)=0$

$\therefore$

$x=6$ and

$x=2$

$\therefore$ The nature of the roots of the equation

$x^2-8x+12=0$ are

$real\,\,and\,\,unequal$

The complex number

$z$ satisfying the equation

$|z - i| = |z + 1| = 1$ is

0%
$0$
0%
$1 + i$
0%
$-1 + i$
0%
$1 - i$

Explanation

Given :

$|z-i|=|z+1|=1$

Let

$z=x+iy$

$|x+iy-i|=1$

$\implies x^{2}+(y-1)^{2}=1$

$\implies x^{2}+y^{2}-2y+1=1$

$x^{2}+y^{2}-2y=0$ ....

$(i)$

Also,

$|z+1|=1$

$\implies |x+iy+1|=1$

$\implies (x+1)^{2}+y^{2}=1$

$\implies x^{2}+y^{2}+2x=0$ .....

$(ii)$

Equating

$(i)$ and

$(ii)$ we get

$-y=x$

For

$y=1, x=-1$

$\therefore z=-1+i$

Also,

$y=0\implies x=0$

$\therefore z=0$

$\left( \dfrac{1 + i}{1 - i} \right)^m = 1$ , then the least positive integral value of m is

0%
1
0%
4
0%
2
0%
3

Explanation

Consider

$(\dfrac{1+i}{1-i})$

Rationalize the denominator we get,

$\Rightarrow (\dfrac{1+i}{1-i})=\dfrac{(1+i)(1+i)}{(1-i)(1+1)}=\dfrac{(1+i)^2}{(1-i)(1+i)}$

Here we see that denominator is in the form of

$(a+b)(a-b)=a^2-b^2$

$\Rightarrow \dfrac{1+i}{1-i}=\dfrac{(1+i)^2}{1^2-i^2}$

$\Rightarrow \dfrac{1+i}{1-i}=\dfrac{(1+i)^2}{1-i^2}$

we know that

$(a+b)^2=a^2+2ab+b^2$

$\Rightarrow \dfrac{1+i}{1-i}=\dfrac{1^2+(2\times 1\times i)+i^2}{1-i^2}$

$\Rightarrow \dfrac{1+i}{1-i}=\dfrac{1^2+2i+i^2}{1-i^2}$

We know that

$i^2=-1$ , substituting this we get,

$\Rightarrow \dfrac{1+i}{1-i}=\dfrac{1+2i+(-1)}{1-(-1)}$

$\Rightarrow \dfrac{1+i}{1-i}=\dfrac{1+2i-1}{1+1}$

$\Rightarrow \dfrac{1+i}{1-i}=\dfrac{2i}{2}$

$\Rightarrow \dfrac{1+i}{1-i}=i$

Given that

$\Rightarrow (\dfrac{1+i}{1-i})^m=1$

$\Rightarrow (\dfrac{1+i}{1-i})^m=i^m$

So,

$i^m=1$

We know that

$i^2=-1$

Squaring on both sides we get,

$i^4=(-1)^2$

$i^4=1$

Therefore the smallest value of

$m$ for which

$(\dfrac{1+i}{1-i})^m=1$ is

$m=4$

The roots of the equation

${ x }^{ 2 }-8x+16=0$

0%
Are imaginary
0%
Are distinct and real
0%
Are equal and real
0%
Cannot be ascertained

Explanation

Given equation is

$x^2-8x+16=0$

Discriminant is

${(-8)}^2-4\times 1\times 16=64-64=0$

Since the discriminant is zero, the quadratic equation has equal and real roots.

$p, q$ are odd integers, then the roots of the equation

$2px^{2} + (2p + q) x + q = 0$ are

0%
Rational
0%
Irrational
0%
Non-real
0%
Equal

Explanation

Given equation is

$2px^{2} + (2p + q) x + q = 0$ ........

$(i)$

Now,

$D = (2p + q)^{2} - 8pq$

$=4p^{2}+4pq+q^{2}-8pq$

$=4p^{2}-4pq+q^{2}$

$= (2p - q)^{2}\rightarrow$ always a perfect square.

$\therefore$ Roots are given by

$x=\dfrac{-(2p+q)\pm \sqrt{(2p+q)^{2}-8pq}}{4p}$

$=\dfrac{-2p-q\pm (2p-q)}{4p}$

$x=\dfrac{-2p-q+2p-q}{4p}=-\dfrac{q}{2p}$ which is rational as

$p, q$ are odd integers and

$2$ is even

$x=\dfrac{-2p-q-2p+q}{4p}=-\dfrac{4p}{4p}=-1$ which is also rational

Thus, the roots of equation

$(i)$ are rational.

The quadratic equation

${ x }^{ 2 }+bx+4=0$ will have real roots if

0%
$b\le -4$ only
0%
$b\ge 4$ only
0%
$-4 < b < 4$
0%
$b\le -4,b\ge 4$

Explanation

In any quadratic equation, real roots exist if discriminant is greater than or equal to zero.

So for

$x^2+bx+4=0$

Discriminant is

$={ b }^{ 2 }-4\times 1\times 4$

So,

$b^{2}-16\geq 0$

$\Rightarrow b\ge 4,b\le -4$

The principal argument of the complex number

$z=\cfrac { 1+\sin { \cfrac { \pi }{ 3 } } +i\cos { \cfrac { \pi }{ 3 } } }{ 1+\sin { \cfrac { \pi }{ 3 } } -i\cos { \cfrac { \pi }{ 3 } } }$ is?

0%
$\cfrac { \pi }{ 3 }$
0%
$\cfrac { \pi }{ 6 }$
0%
$\cfrac { 2\pi }{ 3 }$
0%
$\cfrac { \pi }{ 2 }$
0%
$\cfrac { \pi }{ 4 }$

Explanation

$arg(z)=arg(Numerator)-arg(Denominator)$

$=\tan ^{ -1 }{ \left| \cfrac { \cos { \cfrac { \pi }{ 3 } } }{ 1+\sin { \cfrac { \pi }{ 3 } } } \right| } +\tan ^{ -1 }{ \left| \cfrac { \cos { \cfrac { \pi }{ 3 } } }{ 1+\sin { \cfrac { \pi }{ 3 } } } \right| }$

$=2\tan ^{ -1 }{ \left| \cfrac { \cos { \cfrac { \pi }{ 3 } } }{ 1+\sin { \cfrac { \pi }{ 3 } } } \right| }$

$=2\tan ^{ -1 }{ (2-\sqrt { 3 } } )=2\times {15}^{o}=\cfrac { \pi }{ 6 }$

$iz^{3} + z^{2} - z + i = 0$ , then

$|z|$ is equal to

0%
$0$
0%
$1$
0%
$2$
0%
None of these

Explanation

Given :

$iz^{3} + z^{2} - z + i = 0$
Dividing both sides by

$i$ and using

$\dfrac {1}{i} = -i$ .
We have,

$z^{3} - iz^{2} + iz + 1 = 0$

$\Rightarrow z^{2}(z - i) + i(z - i) = 0$ ......

$(\because i^{2} = -1)$

$\Rightarrow (z - i)(z^{2} + i) = 0$

$\therefore z = i$ or

$z^{2} = -i$

$\therefore |z| = |i| = 1$
and

$|z|^{2}=|z^{2}| = |-i| = 1$

$\therefore |z| =1$ .

Let

$z$ =

$\cos\theta + i \sin\theta$ . Then the value of

$\sum\limits_{m=1}^15Im( z^{2m-1})$ at

$\theta = 2^0$ is

0%
$\dfrac{1}{sin2^0}$
0%
$\dfrac{1}{3sin2^0}$
0%
$\dfrac{1}{2sin2^0}$
0%
$\dfrac{1}{4sin2^0}$

Explanation

using the sum of series of sine

$\sum_{m=1}^{n}sin(a+k.d)=\dfrac{sin\dfrac{n\times d}{2}}{sin\dfrac{d}{2}}\times sin(\dfrac{2a+(n-1).d}{2})$

Given,

$\sum_{m=1}^{15}Im(z^{2m-1})\rightarrow$

$sin\theta+sin3\theta+sin5\theta+.......sin29\theta=$

$\rightarrow\dfrac{sin\dfrac{15\times 2\theta}{2}}{sin\dfrac{2\theta}{2}}\times sin(\dfrac{2\theta+(14\times2\theta)}{2})$

$\rightarrow\dfrac{sin^215\theta}{sin\theta}$

put

$\theta = 2^{\circ}$

$\rightarrow\dfrac{sin^230^{\circ}}{sin2^{\circ}}$

$\rightarrow\dfrac{1}{4.sin2^{\circ}}$

$z$ is a complex number such that

$z + |z| = 8 + 12i$ , then the value of

$|z^{2}|$ is

0%
$228$
0%
$144$
0%
$121$
0%
$169$
0%
$189$

Explanation

Let

$z = x + iy$
Then, we have

$z + |z| = 8 + 12i$

$\Rightarrow(x + iy) + |x + iy| = 8 + 12l$

$\Rightarrow (x + \sqrt {x^{2} + y^{2}}) + iy = 8 + 12i$
On comparing the real and imaginary part, we get

$y = 12$
and

$x + \sqrt {x^{2} + y^{2}} = 8$

$\Rightarrow \sqrt {x^{2} + 144} = 8 - x$
On squaring both sides, we get

$x^{2} + 144 = 64 + x^{2} - 16x$

$\Rightarrow 16x = -80$

$\Rightarrow x = -5$

$\therefore z = x + iy = -5 + i\cdot 12$
Then,

$|z| = \sqrt {25 + 144} = \sqrt {169} = 13$

$\Rightarrow |z|^{2} = 169$

$\Rightarrow |z^{2}| = 169 (\because |z^{n}| = |z|^{n})$ .

$z=1+i$ , then the argument of

${ z }^{ 2 }{ e }^{ z-i }$ is

0%
$\dfrac { \pi }{ 2 }$
0%
$\dfrac { \pi }{ 6 }$
0%
$\dfrac { \pi }{ 4 }$
0%
$\dfrac { \pi }{ 3 }$
0%
$0$

Explanation

Let

$w={ z }^{ 2 }{ e }^{ z-i }$
Put

$z=1+i$

$\Rightarrow w={ \left( 1+i \right) }^{ 2 }{ e }^{ 1+i-i }=\left( 1+2i+{ i }^{ 2 } \right) e$

$=\left( 1+2i-1 \right) e=2ei$
Now, argument of

$w=arg\left( w \right)$

$=\tan ^{ -1 }{ \left( \dfrac { 2e }{ 0 } \right) } =\tan ^{ -1 }{ \infty } =\dfrac { \pi }{ 2 }$

$z + \sqrt {2}|z + 1| + i = 0$ and

$z = x + iy$ , then

0%
$x = -2$
0%
$x = 2$
0%
$y = -2$
0%
$y = 1$

Explanation

On putting

$z=x+iy$ , We get

=>

$x+iy+\sqrt{2(x+1)^2+2y^2} +i=0$

=>

$x+\sqrt{2(x+1)^2+2y^2} +i(y+1)=0$

=> On comparing real part and imaginary part, We get

=>

$x+\sqrt{2(x+1)^2+2y^2} )=0$ and

$y+1=0$

So,

$y=-1$

and ,

$\sqrt{2(x+1)^2+2y^2} )=-x$

=>

${2(x+1)^2+2} =x^2$

=>

$x=-2$

The principal value of

$arg(z)$ lies in the interval:

0%
$\left[ 0,\cfrac { \pi }{ 2 } \right]$
0%
$(-\pi ,\pi ]$
0%
$\left[ 0,\pi \right]$
0%
$(-\pi ,0]$

Explanation

For a complex number $z=re^{\phi i}$ the principal value of the square root is: $pv\: \sqrt{z}=\sqrt{r}e^{i\phi/2}$ with argument $-\pi < 0 \leq \pi$

Thus the principal value of $arg(z)$ lies in the interval: $(-\pi,\pi]$ .

Let

$P(e^{i\theta_1})$ ,

$Q(e^{i\theta_2})$ and

$R(e^{i\theta_3})$ be the vertices of a triangle PQR in the Argand Plane. The orthocenter of the triangle PQR is

0%
$2e^(\theta_1+\theta_2+\theta_3)$
0%
$\frac{2}{3}e^({\theta_1+\theta_2+\theta_3})$
0%
$e^{\theta_1}$ + $e^{\theta_2}$ + $e^{\theta_3}$
0%
None of these

Explanation

$P,Q,R$ lie on the unit circle and hence origin is the cirumcenter , The centrid is easily found as

$\dfrac{(p+q+r)}{3}$ ,Using euler line theorem that centroid divides the line joining the line joining orthocenter and circumcenter in the ratio of

$2:1$

Let orthocenter be

$x,y$

$Using \quad section \quad formula;$

$Re(\dfrac{(p+q+r)}{3})=\dfrac{(1.x+2.0)}{3}$

$Im(\dfrac{(p+q+r)}{3})=\dfrac{(1.y+2.0)}{3}$

we get orthocenter as

$p+q+r\rightarrow$

$e^{\theta_1}+$

$e^{\theta_2}+$

$e^{\theta_3}$

$(1 - i\sqrt {3})^{2} (z) (4i) = (1 + i\sqrt {3})$ , then

$Amp\ z$ is

0%
$\pi$
0%
$\dfrac {\pi}{2}$
0%
$\dfrac {\pi}{3}$
0%
$0$

Explanation

$(1 - i\sqrt {3})^{2} (z) (4i) = (1 + i\sqrt {3})$

$\Rightarrow 04iz (1 + i\sqrt {3}) = \dfrac {1 + i\sqrt {3}}{2} \Rightarrow z = \dfrac {0 + i}{8} = \dfrac {i}{8}$

$\therefore$ Amp of

$z$ or argument of

$z = \tan^{-1} \left (\dfrac {1/8}{0}\right )$

$\therefore$ Argument of

$z = \dfrac {\pi}{2}$ .

Which of the given alternatives represent a point in Argand plane, equidistant from roots of the equation

$(z+1)^4= 16z^4$ ?

0%
$(0,0)$
0%
$\left(-\dfrac{1}{3},0\right)$
0%
$\left(\dfrac{1}{3},0\right)$
0%
$\left(0,\dfrac{2}{\sqrt5}\right)$

Explanation

Consider the given equation

$(z+1)^4=16z^4$

$\Rightarrow (z+1)^4=16z^4$

$\Rightarrow z+1=(2^4z^4)^{\frac{1}{4}}$

$\Rightarrow |z+1|=2|z|$

We know that

$z=x+iy$

Therefore

$|x+iy+1|=2|x+iy|$

$\Rightarrow \sqrt{(x+1)^2+y^2}=2\sqrt{x^2+y^2}$

$\Rightarrow (x+1)^2+y^2=4(x^2+y^2)$

$\Rightarrow x^2+2x+1+y^2=4x^2+4y^2$

$\Rightarrow 3x^2+3y^2-2x-1=0$

Divide throughout by 3 we get,

$\Rightarrow x^2+y^2-\dfrac{2}{3}x-\dfrac{1}{3}=0$ , which represents a circle.

We know that for the circle equation of the form

$x^2+y^2+2gx+2hy+c=0$ the center of the circle is given by

$(-g,-h)$

We have

$3x^2+3y^2-2x-1=0$ where

$g=-\dfrac{1}{3}, h=0$ .

Hence the center is

$(\dfrac{1}{3},0)$ which is equidistant from the root of the equation.

Let

$z,\omega$ be complex numbers such that

$\vec{z}+i\vec{\omega}=0$ and

$Arg(z\omega)=\pi$ then

$Arg(z)=$ .

0%
$\displaystyle\frac{\pi}{4}$
0%
$\displaystyle\frac{5\pi}{4}$
0%
$\displaystyle\frac{3\pi}{4}$
0%
$\displaystyle\frac{\pi}{2}$

Explanation

Since

$z+iw=0$

$\Rightarrow z=-iw$

$\Rightarrow w=-iz$

Also

$arg(zw)=\pi$

$\Rightarrow arg(-iz^2)=\pi$

$\Rightarrow arg(-i)+2arg(z)=\pi$

$\Rightarrow \dfrac{-\pi}{2}+2arg(z)=\pi$ .......

$\left[\because arg(-i)=\dfrac{-\pi}{2}\right]$

$\Rightarrow 2arg(z)=\dfrac{3\pi}{2}$

$\Rightarrow arg(z)=\dfrac{3\pi}{4}$

Hence, the answer is

$\dfrac{3\pi}{4}$ .

Study the statements carefully.
Statement I: Both the roots of the equation

$x^2-x+1=0$ are real.
Statement II: The roots of the equation

$ax^2+bx+c=0$ are real if and only if

$b^2-4ac \geq 0$ .
Which of the following options hold?

0%
Both Statement I and Statement II are true
0%
Both Statement I and Statement II are false
0%
Statement I is true and Statement II is false
0%
Statement I is false and Statement II is true

Explanation

As we know condition for the roots to be real is

$b^2-4ac \geq 0$

so as doing for the statement 1 we are getting

$(-1)^2-4*1*1=(-3)$ which is less than

$0$ so the roots will be unequal so statement 1 is false and statement 2 gives the right condition so its true.

Hence option d is correct.

CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 6 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 6 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.