MCQ Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 5

Find the modulus and amplitude of

$-\sqrt 3-i$

0%
$|z|=2; amp(z)=-\dfrac {5\pi}{6}$
0%
$|z|=4; amp(z)=\dfrac {5\pi}{6}$
0%
$|z|=4; amp(z)=-\dfrac {\pi}{6}$
0%
none of these

Explanation

$z=-\sqrt{3}-i$

$|z|=2$

$=2(\dfrac{-\sqrt{3}-i}{2})$

$=2e^{i(\frac{\pi}{6}-\pi)}$

$=2e^{i\frac{-5\pi}{6}}$

$=|z|.e^{iarg(z)}$

Hence

$|z|=2$ and

$arg(z)=\dfrac{-5\pi}{6}$ .

Find the value of

$x^3 + 7x^2 -x + 16$ , where

$x = 1 + 2i$

0%
$-17 + 24i$
0%
$24 + 17i$
0%
$17 - 24i$
0%
$24 - 17i$

Explanation

$f\left(x\right)={x}^{3}+7{x}^{2}-x+16$

$f(1+2i)=(1+2i)^3+7(1+2i)^2-(1+2i)+16$

$=\left[1+8{i}^{3}+6i\left(1+2i\right)\right]+7\left[1+4{i}^{2}+4i\right]-1-2i+16$

$=\left[1-8i+6i-12\right]+7\left[1-4+4i\right]-2i+15$

$=\left[-11-2i\right]+7\left[4i-3\right]-2i+15$

$=-11-2i+28i-21-2i+15$

$=\left(-32+15\right)+24i$

$=-17+24i$

The amplitude of

$\displaystyle sin \frac{\pi}{5} + i \left ( 1 - cos \frac{\pi}{5} \right )$

0%
$\displaystyle \frac{\pi}{5}$
0%
$\displaystyle \frac{2\pi}{5}$
0%
$\displaystyle \frac{\pi}{10}$
0%
$\displaystyle \frac{\pi}{15}$

Explanation

Given,

$\Rightarrow\displaystyle \sin\frac { \pi }{ 5 } +i\left( 1-\cos\frac { \pi }{ 5 } \right)$

$=2\sin\dfrac { \pi }{ 10 } \cos\dfrac { \pi }{ 10 } +i2\sin^{ 2 }\dfrac { \pi }{ 10 }$

$[\because \sin2\theta=2\sin\theta \cos\theta\quad 2\sin^2\theta=1-\cos\theta]$

$\displaystyle =2\sin\frac { \pi }{ 10 } \left( \cos\frac { \pi }{ 10 } +i\sin\frac { \pi }{ 10 } \right)$

Then

$Amplitude=\tan\theta=\dfrac yx$

$\displaystyle \tan\theta =\frac { \sin\frac { \pi }{ 10 } }{ \cos\frac { \pi }{ 10 } } =\tan\frac { \pi }{ 10 }$

$\Rightarrow \theta =\dfrac { \pi }{ 10 }$

$z_1, z_2, \varepsilon C$ are such that

$|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2$ then

$\displaystyle \frac{z_1}{z_2}$ is

0%
zero
0%
purely real
0%
purely imaginary
0%
complex

Explanation

$\because |z_1 + z_2|^2 = |z_1|^2 + |z_2|^2$

$\Rightarrow |z_1|^2 + |z_2|^2 + 2 R (z_1 \bar z_2) = |z_1|^2 + |z_2|^2$

$\Rightarrow R(z_1 \bar z_2) = 0 \Rightarrow z_1 \bar z_2$ is imaginary

Now

$\displaystyle \frac{z_1}{z_2} = \frac{z_1 \bar z_2}{z_2 \bar z_2} = \frac{\text{imaginary}}{|z_2|^2} =$ imaginary

$(1 + i)^8 + (1 -i)^8 =$

0%
$16$
0%
$-16$
0%
$32$
0%
$-32$

Explanation

${ \left( 1+i \right) }^{ 8 }+{ \left( 1-i \right) }^{ 8 }$

$\Rightarrow { \left( 1+i \right) }^{ 8 }={ \left[ { { \left( 1+i \right) }^{ 2 } } \right] }^{ 4 }={ \left[ { 1+2i+{ i }^{ 2 } } \right] }^{ 4 }={ \left[ { 1+2i-1 } \right] }^{ 4 }$

$={ \left[ { { \left( 2i \right) }^{ 2 } } \right] }^{ 2 }$

$={ \left( -4 \right) }^{ 2 }$

$=16$

Similarely

${ \left( 1-i \right) }^{ 8 }=16$

$\therefore { \left( 1+i \right) }^{ 8 }+{ \left( 1-i \right) }^{ 8 }$

$=16+16$

$=32.$

Hence, the answer is

$32.$

Modulus of

$\displaystyle \frac{cos \theta- isin\theta }{sin\theta - icos\theta} is$

0%
0
0%
2 $\theta$
0%
$\pi - 2\theta$
0%
none of these

Explanation

$\displaystyle |\frac{\cos \theta- i\sin\theta }{\sin\theta - i\cos\theta}|$

$=\displaystyle \frac{|\cos \theta- i\sin\theta| }{|\sin\theta - i\cos\theta|}$

$= \displaystyle \frac{\sin^2 \theta +\cos^2 \theta}{\sin^2 \theta +\cos^2 \theta}=1$

$\therefore \displaystyle |\frac{\cos \theta- i\sin\theta }{\sin\theta - i\cos\theta}|=1$
Hence, option D.

The number of real solutions of the equation

$\displaystyle{\left(\frac{5}{7}\right)^2}=-x^2+2x-3$ is equal to

0%
2
0%
zero
0%
1
0%
none of these

Explanation

$\displaystyle { \left( \frac { 5 }{ 7 }\right)}^{ 2 }=-{ x }^{ 2 }+2x-3$

$\displaystyle \Rightarrow { x }^{ 2 }-2x+3-\frac {25 }{ 49 } =0$

$\Rightarrow { 49x }^{ 2 }-98x+122=0$
Hence,

$D={ \left( 98 \right)}^{ 2 }-4\left( 122 \right) \left( 49 \right) <0$
So, roots are imaginary.

Number of integer values of k for which the quadratic equation

$\displaystyle 2x^{2}+kx-4=0$ will have two rational solutions is

0%
$1$
0%
$2$
0%
$4$
0%
$5$

Explanation

Dividing the equation by the coefficient of

$x^{2}$ i.e. 2 we get

$x^{2}+\dfrac{kx}{2}-2=0$

$\left(x+\dfrac{k}{4}\right)^{2}-\dfrac{k^{2}}{16}-2=0$

$\left(x+\dfrac{k}{4}\right)^{2}=\dfrac{k^{2}+32}{16}$

Hence for rational roots,

$\dfrac{k^{2}+32}{16}$ has to be a perfect square.

We get a perfect square at

$k=\pm2$ for (

$\dfrac{k^{2}+32}{16}$ ) i.e.

$\dfrac{36}{16}$ which becomes

$\dfrac{6}{4}$ upon removing the square

We get a perfect square at

$k=\pm7$ for (

$\dfrac{k^{2}+32}{16}$ ) i.e.

$\dfrac{81}{16}$ which becomes

$\dfrac{9}{4}$ upon removing the square

Hence the number of integral values of k is

$4 (2, -2, 7,-7)$

Find the value of:

$i^2 + i^4 + i^6$ +..... upto

$(2n +1)$ terms.

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

Explanation

$i^2$ +

$i^4$ +

$i^6$ + ...upto

$(2n + 1 )$ terms

$=(-1 +1)+ (-1 +1)+( -1 +$ .....upto

$(2n +1))$ terms

$\Rightarrow$ (odd terms)

$= 0+ 0 + ....-1$

$= -1$

Hence, option D is correct.

$i^n + i^{n + 1} + i^{n + 2}+ i^{n + 3} (n \in N)$ is equal to

0%
$4$
0%
$1$
0%
$0$
0%
$2$

Explanation

$\textbf{Step-1: Expanding the terms}$

${{i}^{n}}+{{i}^{n+1}}+{{i}^{n+2}}+{{i}^{n+3}}$

$= {{i}^{n}}+{{i}^{n}}.i+{{i}^{n}}.{{i}^{2}}+{{i}^{n}}.{{i}^{3}}$

$\textbf{Step-2: Taking } \mathbf{{i}^{n}} \textbf{ common}$

$= {{i}^{n}}(1+i+{{i}^{2}}+{{i}^{3}})$

$= {{i}^{n}}(1+i-1-i)$ $\mathbf{[\because {{i}^{2}}=-1,{{i}^{3}}=-i]}$

$= {{i}^{n}}(0)=0$

$\textbf{Hence the correct option is (C) 0}$

The argument of the complex number

$z = sin \alpha + i (1 - cos \alpha)$ is

0%
$2 sin \displaystyle \frac{\alpha}{2}$
0%
$\displaystyle \frac{\alpha}{2}$
0%
$\alpha$
0%
none of these

Explanation

$z = sin \alpha + i (1 - cos \alpha)$

$\Rightarrow arg (z) = tan^{-1} \displaystyle \left ( \frac{1 - cos \alpha}{sin \alpha} \right ) = tan^{-1} \left (\frac{\displaystyle 2 sin^2 \frac{\alpha}{2}}{2 sin \displaystyle \frac{\alpha}{2} cos \frac{\alpha}{2}} \right )$

$= tan^{-1} \displaystyle \left ( tan \frac{\alpha}{2} \right ) = \frac{\alpha}{2}$

Solve:

$\displaystyle \left|(1 + i)\frac{(2+i)}{(3 + i)}\right|$

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

Explanation

Given,

$\displaystyle |(1 + i)\frac{(2+i)}{(3 + i)}|$ =

$\displaystyle |(1 + i)|\frac{|(2+i)|}{|(3 + i)|}$

Since,

$|Z_{1}Z_{2}|=|Z_{1}||Z_{2}|$ and

$\displaystyle|\frac{Z_{1}}{Z_{2}}|=\frac{|Z_{1}|}{|Z_{2}|}$

$=\displaystyle \sqrt{2}\cdot \frac{\sqrt{5}}{\sqrt{10}}=1$

$\therefore \displaystyle |(1 + i)\frac{(2+i)}{(3 + i)}| = 1$

Hence, option C.

In the argand diagram, if

$O,P$ and

$Q$ represents respectively the origin and the complex numbers

$z$ and

$z+iz$ then the

$\angle OPQ$ is

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

Explanation

Let

$\displaystyle z=r\left( \cos { \theta } +i\sin { \theta } \right) ,$ then

$\displaystyle z+iz=r\left( \cos { \theta } +i\sin { \theta } \right) +ir\left( \cos { \theta } +i\sin { \theta } \right)$

$\displaystyle=r\left[ \left( \cos { \theta } -\sin { \theta } \right) +i\left( \sin { \theta } +\cos { \theta } \right) \right]$

$\displaystyle=\sqrt { 2 } r\left[ \cos { \left( \theta +\frac { \pi }{ 4 } \right) } +i\sin { \left( \theta =\frac { \pi }{ 4 } \right) } \right]$

$\displaystyle{ \triangle OPQ } ,$

$\displaystyle{ PQ }^{ 2 }={ r }^{ 2 }+{ \left( \sqrt { 2 } r \right) }^{ 2 }-2r\left( \sqrt { 2 } r \right) \cos { \frac { \pi }{ 4 } }$

$\displaystyle={ r }^{ 2 }+2{ r }^{ 2 }-2{ r }^{ 2 }={ r }^{ 2 }$

$\displaystyle\therefore PQ=r,\quad$

$\therefore \angle OPQ=\dfrac { \pi }{ 2 }$

The modulus of (1 + i) (1 + 2i) (1 + 3i) is equal to

0%
$\sqrt10$
0%
$\sqrt 5$
0%
5
0%
10

Explanation

$|(1 + i) (1 + 2i) (1 + 3i)|$ =

$|1+i||1+2i||1+3i|$ (

$\because |Z_{1}Z_{2}|=|Z_{1}||Z_{2}|$ )

$=\sqrt {2}\cdot\sqrt {5}\cdot\sqrt {10}$

$\therefore |(1 + i) (1 + 2i) (1 + 3i)|=10$
Hence, option D.

The number of real roots of the equation

$\displaystyle \frac{A^{2}}{x} +\frac{B^{2}}{x-1}=1$ where A and B are real numbers not equal to zero simultaneously, is :

0%
$1 \;or\; 2$
0%
$1$
0%
$2$
0%
None

Explanation

Given that

$\cfrac { { A }^{ 2 } }{ x } +\cfrac { { B }^{ 2 } }{ x-1 } =1,A,B\epsilon R$

A and B cannot be zero simultaneously

$\left( { A }^{ 2 }+{ B }^{ 2 } \right) x-{ A }^{ 2 }={ x }^{ 2 }-x$

${ x }^{ 2 }-\left( { A }^{ 2 }+{ B }^{ 2 }+1 \right) x+{ A }^{ 2 }=0$

$\triangle ={ \left( { A }^{ 2 }+{ B }^{ 2 }+1 \right) }^{ 2 }-4{ A }^{ 2 }$

$=\left( { A }^{ 2 }+{ B }^{ 2 }+1+2A \right) \left( { A }^{ 2 }+{ B }^{ 2 }+1-2A \right)$

$=\left( { \left( A+1 \right) }^{ 2 }+{ B }^{ 2 } \right) \left( { (A-1) }^{ 2 }+{ B }^{ 2 } \right)$

$\triangle =0$ If

$A=\pm 1$ &

$B=0$

$\therefore \triangle$ is always

$\ge 0$

$\therefore$ Number of real roots

$1$ (or)

$2$

Find the value of

$k$ for which the quadratic equation

$\displaystyle \left ( k-2 \right )x^{2}+2\left ( 2k-3 \right )x+5k-6=0$ has equal roots

0%
1
0%
3
0%
A and B both
0%
none of these

Explanation

Since the above equation has equal roots
Hence

$B^{2}-4AC=0$
Hence

$4(2k-3)^{2}-4(5k-6)(k-2)=0$

$4k^{2}-12k+9-(5k^{2}-16k+12)=0$

$-k^{2}+4k-3=0$
Or

$k^{2}-4k+3=0$

$(k-3)(k-1)=0$
Hence

$k=3$ and

$k=1$ .

Number of complex numbers

$z$ satisfying

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

Explanation

Let the complex number

$z$ be

$x+iy$

Given that

$|2z|=|2z-1|$

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

$\Rightarrow 3{y}^{2}+4x=1$

Given that

$|2z|=|2z+1|$

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

$\Rightarrow 3{y}^{2}-4x=1$

By solving above two equation , we get

$x=0 , y=\pm\frac{1}{\sqrt3}$

But this point doesnt satisfy the equation

$|2z-1|=|2z+1|$

Therefore the number of points which satisfies the equation

$|2z|=|2z+1|=|2z-1|$ is

$0$

If both a and b belong to the set

$\displaystyle \left\{ 1,2,3,4 \right\}$ then the number of equations of the form

$ax^{2}+bx+1=0$ having real roots is :

0%
$10$
0%
$7$
0%
$6$
0%
$12$

Explanation

Real roots

$\Rightarrow b^{2}\geq4a$

1)let b=1

$\Rightarrow 1\geq 4a$

$\rightarrow$ no choice for "a"

2)let b=2

$\Rightarrow 4\geq4a$

$\Rightarrow a=1$

$\rightarrow$ one choice

3)let b=3

$\Rightarrow 9\geq4a$

$\Rightarrow$ a=1,2

$\rightarrow$ 2 choices

4)let b=4

$\Rightarrow 16\geq4a$

$\Rightarrow a=1,2,3$

$\rightarrow$

$3$ choices

$\therefore$ total

$7$ choices

If 0

$\leq$ argz

$\leq \displaystyle \frac{\pi}{4}$ , then the least value of

$\sqrt 2 |2z - 4|$ is

0%
6
0%
1
0%
4
0%
2

Explanation

$\sqrt2|2z-4|=2\sqrt2|z-2|$

Given that

$0\le argz\le \dfrac{\pi}{4}$

The least value of

$|z-2|$ is

$\sqrt2$ and occurs at

$argz=\dfrac{\pi}{4}$

Therefore the least value of

$\sqrt2|2z-4| = 2 \sqrt2 \times \sqrt2=4$

Therefore the correct option is

$C$

$\displaystyle z= (i)^{(i)^{(i)}}$ where

$i = \sqrt{-1}$ , then

$z$ is equal to

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

Explanation

$Z=(i)^{(i)^{(i)}}$

$i=e^{i\pi/2}$

$i^i=e^{i.(i\pi/2)}=e^{-\pi/2}$

$i^{i^i}=e^{i(-\pi/2)}\\ \implies \cos(\pi/2)-i\sin(\pi/2)\\ \implies 0-i=-i$

The roots of the equations

$\displaystyle x^{2}-x-3=0$ are

0%
Imaginary
0%
Rational
0%
Irrational
0%
None of these

Explanation

Give equation is:

$\displaystyle x^{2}-x-3=0$

$\Rightarrow a=1,b=-1,c=-3$

Discriminant

$\displaystyle D = b^{2}-4ac$

$=(-1)^{2}-4\times 1\times (-3)$

$= 1 + 12 = 13$

As the value of

$D$ is not a perfect square

$\therefore$ the roots of the above equation are irrational.

If the roots of the equation

$\displaystyle \left ( a^{2}+b^{2} \right )x^{2}-2b\left ( a+c \right )x+\left ( b^{2}+c^{2} \right )=0$ are equal then

0%
$2b = ac$
0%
$\displaystyle b^{2}=ac$
0%
$\displaystyle b=\frac{2ac}{a+c}$
0%
b = ac

Explanation

Since the roots are equal

$B^{2}-4AC=0$
Hence

$4b^{2}(a+c)^{2}-4(b^{2}+c^{2})(a^{2}+b^{2})=0$

$4b^{2}a^{2}+4b^{2}c^{2}+8b^{2}ac-4b^{2}a^{2}-4b^{4}-4c^{2}a^{2}-4c^{2}b^{2}=0$

$8b^{2}ac-4b^{4}-4c^{2}a^{2}=0$
Or

$(b^{2})^{2}-2b^{2}ac+(ac)^{2}=0$

$b^{2}=\dfrac{2ac\pm\sqrt{4ac^{2}-4ac^{2}}}{2}$

$b^{2}=ac$

The value (s) of

$k$ for which the quadratic equation

$\displaystyle kx^{2}-kx+1=0$ has equal roots is

0%
$k = 0$ only
0%
$k = 4$ only
0%
$k = 0, 4$
0%
$k = -4$

Explanation

For equal roots

$B^{2}-4AC=0$
Hence

$k^{2}-4k=0$

$k(k-4)=0$

$k=0$ or

$k=4$
Now if

$K=0$ , the above equation will be

$1=0$ which is not true.
Hence

$k=4$ is the answer.

Let

$z = x + iy$ & amp

$(e^{z^2})$ = amp

$(e^{(z+i)})$ . If

$y = (x)$ is a function, then

$y(3)$ is equal to

0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{1}{3}$
0%
$\displaystyle \frac{1}{4}$
0%
$\displaystyle \frac{1}{5}$

Explanation

$(e^{z^2} ) = e^{(x^2 - y^2 )+2ixy}=e^{x^2-y^2}.e^{ixy}$

$\Rightarrow$ amp

$(e^{z^2}) = 2xy$
And amp

$(e^{(z+i)}) =$ amp

$(e^{x+(y+1)i})=$ amp

$(e^x.e^{i(y+1)})= (y + 1)$

$\therefore 2xy = y + 1 \Rightarrow \displaystyle y = \frac{1}{(2x - 1)}$

$\therefore y(3) = \dfrac{1}{5}$

The quadratic equations

$\displaystyle x^{2}-5x+3=0$ has

0%
no real roots
0%
two distinct real roots
0%
two equal real roots
0%
more than two real roots

Explanation

The nature of roots of the equation can be given by,

$D=b^{ 2 }4ac=0,$

For equation

$x^{ 2 }-5x+3=0$ ,

$D=(-5)^{ 2 }-4(3)(1)$

$D=13>0$ ,

It has two distinct real roots.

Find the value of

$arg\left ( \left ( 1+i \right )^{i} \right )$

0%
$\displaystyle \dfrac{1}{4}ln\left ( 2 \right )$
0%
$\displaystyle \dfrac{1}{2}ln\left ( 2 \right )$
0%
$\displaystyle \dfrac{1}{2}ln\left (\dfrac{1}{ 2} \right )$
0%
$\displaystyle ln\left ( 2 \right )$

Explanation

Write

$(1+i)$ as

$\sqrt 2 e^{[\dfrac{i\pi}{4}]}$

Therefore

$(1+i)^i=\sqrt 2^{i}e^{-\pi/4}=[e^{-(8n+1)\tfrac{\pi}{4}}]\times e^{i\tfrac{ln2}{2}}$

So,

$arg(1+i)^i=arg[e^{-(8n+1)\tfrac{\pi}{4}}]\times e^{i\tfrac{ln2}{2}}=arg[ke^{i\tfrac{ln2}{2}}]=\dfrac{ln2}{2}$ .........[where

$K=e^{-(8n+1)\dfrac{\pi}{4}}$ ]

Find the value of

$k$ for which the equation

$(k+1){x}^{2}-2(k-1)x+1=0$ has real and equal roots

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

If the roots of the equation

$\displaystyle x^{2}+px-6=0$ are

$6$ and

$-1$ then the value of

$p$ is

0%
$2$
0%
$3$
0%
$-5$
0%
$5$

Explanation

GIven equation is :

$x^{2}+px-6=0$ ......

$(i)$

$6$ and

$-1$ are the roots of

$(i)$

So, they satisfies

$(i)$

$6$ is the root of

$(i)$

$\implies (6)^{2}+p(6)-6=0\Rightarrow 36+6p-6=0\Rightarrow 6p=-36+6\Rightarrow 6p=-30\Rightarrow p=-5$

If the root is

$-1$ then

$(-1)^{2}+p(-1)-6=0\Rightarrow 1-p-6=0\Rightarrow- p=6-1\Rightarrow p=-5$

$\therefore p=-5$

Alternate Method

$6$ and

$-1$ are the roots of

$(i)$

$\implies (x-6)(x+1)=x^{2}+px-6$

$\implies x^{2}+x-6x-6=x^{2}+px-6$

$\implies x^{2}-5x-6=x^{2}+px-6$

$\implies p=-5$

The roots of the equation, where

$\displaystyle a\: \: \epsilon \: \: R$ are

$\displaystyle x^{2}+ax-4=0$

0%
Real and distinct
0%
Equal
0%
Imaginary
0%
Real

Explanation

The equation can be written as

$x^{ 2 }+ax-4=0$ ,

$b^{ 2 }-4a$ will determine the nature of roots.

$\Longrightarrow b^{ 2 }-4ac=a^{ 2 }-4(1)(-4)$

$\Longrightarrow b^{ 2 }-4ac=a^{ 2 }+16>0$

$\therefore$ roots are real and distinct.

If the roots of the equation

$\displaystyle (a^{2}+b^{2})x^{2}-2b(a+c)x+(b^{2}+c^{2})=0$ are equal then =

0%
$2b = ac$
0%
$\displaystyle b^{2}=ac$
0%
$\displaystyle b=\frac{2ac}{a+c}$
0%
$b = ac$

Explanation

$\displaystyle (a^{2}+b^{2})x^{2}-2b(a+c)x+(b^{2}+c^{2})=0$
Roots are real and equal

$\therefore$

$D = 0$

$D=b^2-4ac=0$

$\displaystyle \Rightarrow [-2b(a+c)]^{2}-4(a^{2}+b^{2})(b^{2}+c^{2})=0$

$\displaystyle \Rightarrow b^{2}(a^{2}+c^{2}+2ac)-(a^{2}b^{2}+a^{2}c^{2}+b^{4}+c^{2}c^{2})=0$

$\displaystyle \Rightarrow b^{2}a^{2}+b^{2}c^{2}+2acb^{2}-a^{2}b^{2}-a^{2}c^{2}-b^{4}-b^{2}c^{2}=0$

$\displaystyle \Rightarrow 2acb^{2}-a^{2}c^{2}-2acb^{2}=0$

$\displaystyle \Rightarrow (b^{2}-ac)^{2}=0$

$\displaystyle \Rightarrow b^{2}=ac$

If the equation

$\displaystyle x^{2}-2kx-2x+k^{2}=0$ has equal roots the value of k must be

0%
zero
0%
either zero or $\displaystyle -\frac{1}{2}$
0%
$\displaystyle -\frac{1}{2}$
0%
either $\displaystyle \frac{1}{2}$ or $\displaystyle -\frac{1}{2}$

Explanation

The equation

$x^{ 2 }-2(k+1)x+k^{ 2 }=0$ will have equal roots when

$D=0$ ,

$D=b^{ 2 }-4ac,$

$\Longrightarrow 2(k+1)^{ 2 }-4(1)(k)^{ 2 }=(k+1)^{ 2 }-k^{ 2 }$

$\therefore \Longrightarrow 2k+1=0\\ k=-1/2$

For what value of k will

$\displaystyle x^{2}-\left ( 3k-1 \right )x+2k^{2}+2k=11$ have equal roots?

0%
$9, -5$
0%
$-9, 5$
0%
$9, 5$
0%
$-9, -5$

Explanation

For this equation to have equal roots

$b^{2}=4ac$ Here

$a=1 ,b=-(3k-1)$ and

$c=2k^{2}+2k-11$

Then

$\left ( -(3k-1) \right )^{2}=4\left ( 1(2k^{2}+2k-11) \right )$

$\Rightarrow 9k^{2}-8k+1=8k^{2}+8k-44$

$\Rightarrow 9k^{2}-8k^{2}-6k-8k+1+44=0$

$\Rightarrow k^{2}-14k+45=0$

$\Rightarrow k^{2}-9k-5k+45=0$

$\Rightarrow k(k-9)-5(k-9)=0$

$\Rightarrow (k-9)(k-5)=0$

Then

$k-9=0 \ or\ k=9$

$k-5=0 \ or\ k=5$

Then values are

$9,5$

If the equation

$\displaystyle 16x^{2}+6kx+4=0$ has equal roots, then the value of

$k$ is

0%
$\displaystyle \pm 8$
0%
$\displaystyle \pm \frac{8}{3}$
0%
$\displaystyle \pm \frac{3}{8}$
0%
$0$

Explanation

$\displaystyle 16x^{2}+6kx+4=0$ has equal roots.

Therefore,

$D = 0$

$\displaystyle \Rightarrow (6k)^{2}-4(16)(4)=0$

$\displaystyle \Rightarrow 36k^{2}-256=0$

$\displaystyle \Rightarrow k^{2}=\frac{256}{36}=\frac{64}{9}$

$\displaystyle \Rightarrow k=\pm \frac{8}{3}$

Which of the following equation has two equal real toots ?

0%
$\displaystyle 3x^{2}+14x-5=0$
0%
$\displaystyle 4x^{2}+2x-1=0$
0%
$\displaystyle 9x^{2}-6x+1=0$
0%
$\displaystyle x^{2}-5x+4=0$

Explanation

(A)

$3x^{2}+14x-5=0$

Then

$b^{2}-4ac=0$ is if roots are equal

$b^{2}-4ac\Rightarrow (14)^{2}-4\times 3\times (-5)=196+60=256> 0$

Then roots are not equal

(B)

$4x^{2}+2x-1=0$

Then

$b^{2}-4ac=0$ is if roots are equal

$b^{2}-4ac\Rightarrow (2)^{2}-4\times 4\times (-1)=4+16=20> 0$

Then roots are not equal

(C)

$9x^{2}-6x+1=0$

Then

$b^{2}-4ac=0$ is if roots are equal

$b^{2}-4ac\Rightarrow (-6)^{2}-4\times 9\times (1)=36-36=0$

Then roots are equal

(D)

$x^{2}-5x+4=0$

Then

$b^{2}-4ac=0$ is if roots are equal

$b^{2}-4ac\Rightarrow (-5)^{2}-4\times 1\times (4)=25-16=9> 0$

Then roots are not equal

Then option (C)

$9x^{2}-6x+1=0$ have equal roots

For what values of

$k$ will the quadratic equation :

$\displaystyle { 2x }^{ 2 }-kx+1=0$ have real and equal roots?

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

Explanation

Given equation is:

$\displaystyle { 2x }^{ 2 }-kx+1=0$
Hence,

$\displaystyle a=2,b=-k.c=1$

$\displaystyle D={ b }^{ 2 }-4ac={ \left( -k \right) }^{ 2 }-4\times 2\times 1-{ k }^{ 2 }-8$
We know that a quadratic equation has real and equal roots, if

$\displaystyle D=0\Leftarrow { k }^{ 2 }-8=0\Rightarrow { k }^{ 2 }=8$

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

The roots of

$2{x}^{2}-6x+3=0$ are

0%
rational
0%
equal
0%
irrational
0%
imaginary

Explanation

$2x^2-6x+3=0$

$x=\dfrac{6\pm\sqrt{36-24}}{2}$

$x=\dfrac{6\pm\sqrt{12}}{2}$

Given that

$kx(x-2)+6=0$ has real and equal roots, the root is

0%
$2$
0%
$-1$
0%
$1$
0%
$\cfrac{1}{2}$

Explanation

Given equation is:

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

$kx^{2}-2kx+6=0$
Then

$a=k ,b=-2k ,c=6$
Then

$b^{2}-4ac=0$
So

$(-2k)^{2}-4k(6)=0$

$4k^{2}-24k=0$

$4k(k-6)=0$

$k=0$ and

$k=6$
Put value of

$k$
Then

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

$6x^{2}-6x-6x+6=0$

$6x(x-1)-6(x-1)=0$

$\therefore x=1$

Find the value of 'm' so that the equation

$\displaystyle { 9x }^{ 2 }-8mx-9=0$ has one root as the negative of the other.

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

Explanation

Given equation is:

$9x^2-8mx-9=0$

Comparing with standard equation

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

$a=9, b=-8m, c=-9$

Let

$\displaystyle \alpha ,\beta$ be the roots
Then,

$\alpha +\beta =-\dfrac{b}{a}=\dfrac { 8m }{ 9 }$
But,

$\alpha =-\beta$
Thus,

$\alpha +\beta =0$

$\Rightarrow \dfrac{8m}{9}=0$

$\Rightarrow m=0$

The value of

$k$ for which

$4{x}^{2}-4\sqrt {3}x+k=0$ is satisfied by only one real value of

$x$ is

0%
$\sqrt 6$
0%
$-3$
0%
$3$
0%
$\cfrac{1}{\sqrt {3}}$

Explanation

We know that

$(a-b)^{2}=a^{2}-2ab+b^{2}$
Given Equation is

$4x^{2}-4\sqrt{3}x+k$
Co-pairing both equations, we get

$a=2x$ and

$k=3$

$kx^2-2\sqrt 5x +4 = 0$

For what value of

$k$ will the quadratic equation have real and equal roots ?

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

Explanation

Given equation is

$\displaystyle { kx }^{ 2 }-2\sqrt { 5 } +4=0$

$\displaystyle a=k,b=-2\sqrt { 5 } ,c=4$

$\displaystyle \therefore D={ b }^{ 2 }-4ac={ \left( -2\sqrt { 5 } \right) }^{ 2 }-4\times k\times 4=20-16$
We know that a quadratic equation has real and equal roots, if

$D=0$ .

$\displaystyle \Rightarrow 20-16k=0$

$\displaystyle \Rightarrow -16k=-20$

$\displaystyle \Rightarrow k=\frac { 20 }{ 16 } =\frac { 5 }{ 4 }$

For what value of 'k' the equation

$\displaystyle \left( k+3 \right) { x }^{ 2 }-\left( 5-k \right) x+1=0$ has coincident roots ?

0%
$1, 13$
0%
$1, 12$
0%
$3, 13$
0%
None of these

Explanation

For coincident roots,

$\displaystyle D=0$
Now,

$\displaystyle { \left[ -\left( 5-k \right) \right] }^{ 2 }-4\times \left( k+3 \right) \times 1=0$

$\displaystyle \Rightarrow \quad 25+{ k }^{ 2 }-10k-4k-12=0$

$\displaystyle \Rightarrow \quad { k }^{ 2 }-14k+13=0$

$\displaystyle \Rightarrow \quad \left( k-13 \right) \left( k-1 \right) =0$

$\displaystyle \therefore \quad k=13,1$

For what value of

$k$ will the quadratic equation

$\displaystyle { 2x }^{ 2 }+3x+k=0$ have real equal roots ?

0%
$\displaystyle \frac { 3 }{ 8 }$
0%
$\displaystyle \frac { 8 }{ 3 }$
0%
$\displaystyle \frac { 9 }{ 8 }$
0%
$\displaystyle \frac { 8 }{ 9 }$

Explanation

Given equation is

$\displaystyle 2{ x }^{ 2 }+3x+k=0$
Here,

$\displaystyle a=2,b=3,c=k$

$\displaystyle \because D={ b }^{ 2 }-4ac={ \left( 3 \right) }^{ 2 }-4\times 2\times k=9-8k$ .
We know that a quadratic equation has real and equal roots if $$D = 0$$.

$\displaystyle \Rightarrow 9-8k=0\Rightarrow -8k=-9$

$\displaystyle \Rightarrow k=\frac { 9 }{ 8\\ }$

Find the value of

$k$ for which the equation

${x}^{2}+kx+81=0$ and

${x}^{2}-6\sqrt {2}x+k=0$ has both real roots,

$k> 0$

0%
$k=9$
0%
$k=18$
0%
$k=3\sqrt {2}$
0%
No such value

Explanation

For,

${x}^{2}+kx+81=0$ to have real roots

${k}^{2}-324 \ge 0$

$\Rightarrow$

$k\ge 18.......(1)$

For

${x}^{2}-6\sqrt {2}x+k=0$ to have real roots

$72-4k\ge 0$

$\Rightarrow$

$k\le 18..........(2)$

For both equations to have real roots

$k=18$

Find the value of

$k$ for which the equation

${x}^{2}-6x+k=0$ has distinct roots.

0%
$k> 9$
0%
$k=6,7$ only
0%
$k<9$
0%
$k=9$

Explanation

Equation

${x}^{2}-6x+k=0$ has
discriminant

$D={b}^{2}-4ac$

$=36-4k$
For real roots

$D>0$

$\Rightarrow$

$36-4k>0$

$\Rightarrow$

$k<9$

The values of

$b$ for which the equation

$\displaystyle { 2bx }^{ 2 }-40x+25=0$ has equal root is :

0%
$5$
0%
$6$
0%
$7$
0%
$8$

Explanation

Given equation is

$2bx^2-40x+25=0$

$\therefore a=2b, b=-40, c=25$

For equal roots,

$D = 0$

$\therefore b^2-4ac=0$

$\displaystyle \Rightarrow { \left( 40 \right) }^{ 2 }-4\times 25\times 2b=0$

$\displaystyle \Rightarrow 200b=1600$

$\displaystyle \Rightarrow b=8$

For what value of

$k$ will the quadratic equation:

$\displaystyle { kx }^{ 2 }+4x+1=0$ have real and equal roots ?

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

Explanation

The given quadratic equation is :

$\displaystyle { kx }^{ 2 }+4x+1=0$
Here,

$\displaystyle a=k,b=4,c=1$
Now,

$\displaystyle D={ b }^{ 2 }-4ac={ \left( 4 \right) }^{ 2 }-4\times k\times 1=16-4k$
We know that a quadratic equation has real and equal roots
If

$\displaystyle D=0$

$\displaystyle \Rightarrow 16-4k=0$

$\displaystyle \Rightarrow -4k=-19$

$\displaystyle \therefore k=\frac { -16 }{ -4 } =4$

$z_1, z_2$ and

$z_3$ are complex numbers such that

$|z_1| = |z_2| = |z_3| = \left | \displaystyle \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \right | = 1,$ then

$|z_1 + z_2 + z_3|$ is

0%
equal to 1
0%
less than 1
0%
greater than 3
0%
equal to 3

Explanation

$|z_1| = |z_2| = |z_3| = 1$ (given)
Now

$|z_1| = 1 \Rightarrow |z_r|^2 = 1$

$\Rightarrow z_1 \bar z_1 = 1$
Similarly

$z_2 \bar z_2 = 1, z_3 \bar z_3 = 1$
Now

$\displaystyle \left | \dfrac{1}{z_1} + \dfrac{1}{z_2} + \dfrac{1}{z_3} \right | = 1$

$\Rightarrow |\bar z_1 + \bar z_2 + \bar z_3| = 1$

$\Rightarrow |\overline{z_1 + z_2 + z_3}| = 1 \Rightarrow |z_1 + z_2 + z_3| = 1$

If k be the ratio of the roots of the equation

$\displaystyle x^{2}-px+q=0$ , the value of

$\displaystyle \frac{k}{1+k^{2}}$ is

0%
$\displaystyle \frac{q^{2}-2p}{p}$
0%
$\displaystyle \frac{q}{p^{2}-2q}$
0%
$\displaystyle \frac{p}{q^{2}-2p}$
0%
$\displaystyle \frac{p}{p^{2}-2q}$

Explanation

Consider the given equation.

$x^2-px+q=0$

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

Let the roots are

$\alpha, \beta$ .

Since,

$k=\dfrac{\alpha}{\beta}$

Since,

$=\dfrac{k}{1+k^2}$

$=\dfrac{\dfrac{\alpha}{\beta}}{1+\dfrac{\alpha^2}{\beta^2}}$

$=\dfrac{\alpha \beta}{\alpha^2+\beta^2}$

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

$=\dfrac{q}{p^2-2q}$

Hence, this is the answer.

For what values of k, will quadratic equation

$\displaystyle { 9x }^{ 2 }+3kx+4=0$ have real and equal roots?

0%
$\displaystyle \pm 4$
0%
$\displaystyle \pm 3$
0%
$\displaystyle \pm 2$
0%
$\displaystyle \pm 1$

Explanation

Given equation is Comparing with

$\displaystyle { ax }^{ 2 }+bx+c=0$

$\displaystyle \Rightarrow a=9,b=3k,c=4$ , we get

$\displaystyle D={ b }^{ 2 }-4ac={ \left( 3k \right) }^{ 2 }-4\left( 9 \right) \left( 4 \right) =9{ k }^{ 2 }-144$
Since the quadratic equation has real and equal roots.
Therefore,

$\displaystyle D=0$

$\Rightarrow \displaystyle { 9k }^{ 2 }-144=0$

$\Rightarrow \displaystyle { k }^{ 2 }=\frac { 144 }{ 9 } =16$

$\Rightarrow \displaystyle k=\pm \sqrt { 16 } =\pm 4$

$\Rightarrow \displaystyle k=4$ or

$-4$

For what values of k will the quadratic equation :

$\displaystyle { 3x }^{ 2 }+kx+3=0$ have real equal roots?

0%
$\displaystyle \pm \sqrt { 3 }$
0%
$\displaystyle \pm \sqrt { 6 }$
0%
$\displaystyle \pm 6$
0%
$\displaystyle \pm 3$

Explanation

Given equation is:

$\displaystyle { 3x }^{ 2 }+kx+3=0$
Here,

$\displaystyle a=3,b=k,c=3$
Now,

$\displaystyle D={ b }^{ 2 }-4ac={ k }^{ 2 }-4\times 3\times 3{ k }^{ 2 }-36$
We know that a quadratic equation has real and equal roots,
If

$\displaystyle D=0\Rightarrow { k }^{ 2 }-36=0\Rightarrow { k }^{ 2 }=36$

$\displaystyle \therefore \quad k=\pm \sqrt { 36 } =\pm 6$

CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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