Complex Numbers and Quadratic Equations Test 14

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Complex Numbers and Quadratic Equations Test 14

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Weekly Quiz Competition

Question 1
1 / -0

Find the value of
$arg\left ( \left ( 1+i \right )^{i} \right )$

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

B
$\displaystyle \dfrac{1}{2}ln\left ( 2 \right )$

C
$\displaystyle \dfrac{1}{2}ln\left (\dfrac{1}{ 2} \right )$

D
$\displaystyle ln\left ( 2 \right )$

Solution

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 $K=e^{-(8n+1)\dfrac{\pi}{4}}$ ]
Question 2
1 / -0

Find the value of $k$ for which the equation $(k+1){x}^{2}-2(k-1)x+1=0$ has real and equal roots

A
$1,-2$

B
$0,4$

C
$-1,3$

D
$0,3$
Question 3
1 / -0

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

A
no real roots

B
two distinct real roots

C
two equal real roots

D
more than two real roots

Solution

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.
Question 4
1 / -0

If the roots of the equation $\displaystyle x^{2}+px-6=0$ are $6$ and $-1$ then the value of $p$ is

A
$2$

B
$3$

C
$-5$

D
$5$

Solution

GIven equation is :
$x^{2}+px-6=0$ ...... $(i)$
$6$ and $-1$ are the roots of $(i)$
So, they satisfies $(i)$
If $6$ 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 $(i)$
$\implies (x-6)(x+1)=x^{2}+px-6$
$\implies x^{2}+x-6x-6=x^{2}+px-6$
$\implies x^{2}-5x-6=x^{2}+px-6$
$\implies p=-5$
Question 5
1 / -0

The roots of the equation, where $\displaystyle a\: \: \epsilon \: \: R$ are
$\displaystyle x^{2}+ax-4=0$

A
Real and distinct

B
Equal

C
Imaginary

D
Real

Solution

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.
Question 6
1 / -0

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 =

A
$2b = ac$

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

C
$\displaystyle b=\frac{2ac}{a+c}$

D
$b = ac$

Solution

$\displaystyle (a^{2}+b^{2})x^{2}-2b(a+c)x+(b^{2}+c^{2})=0$
Roots are real and equal $\therefore$ $D = 0$
$D=b^2-4ac=0$
$\displaystyle \Rightarrow [-2b(a+c)]^{2}-4(a^{2}+b^{2})(b^{2}+c^{2})=0$
$\displaystyle \Rightarrow b^{2}(a^{2}+c^{2}+2ac)-(a^{2}b^{2}+a^{2}c^{2}+b^{4}+c^{2}c^{2})=0$
$\displaystyle \Rightarrow b^{2}a^{2}+b^{2}c^{2}+2acb^{2}-a^{2}b^{2}-a^{2}c^{2}-b^{4}-b^{2}c^{2}=0$
$\displaystyle \Rightarrow 2acb^{2}-a^{2}c^{2}-2acb^{2}=0$
$\displaystyle \Rightarrow (b^{2}-ac)^{2}=0$
$\displaystyle \Rightarrow b^{2}=ac$
Question 7
1 / -0

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

A
zero

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

C
$\displaystyle -\frac{1}{2}$

D
either $\displaystyle \frac{1}{2}$ or $\displaystyle -\frac{1}{2}$

Solution

The equation $x^{ 2 }-2(k+1)x+k^{ 2 }=0$ will have equal roots when $D=0$ ,
$D=b^{ 2 }-4ac,$
$\Longrightarrow 2(k+1)^{ 2 }-4(1)(k)^{ 2 }=(k+1)^{ 2 }-k^{ 2 }$
$\therefore \Longrightarrow 2k+1=0\\ k=-1/2$
Question 8
1 / -0

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

A
$9, -5$

B
$-9, 5$

C
$9, 5$

D
$-9, -5$

Solution

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 $k-9=0 \ or\ k=9$
Or $k-5=0 \ or\ k=5$
Then values are $9,5$
Question 9
1 / -0

Find the value of $k$ for which the equation ${x}^{2}-6x+k=0$ has distinct roots.

A
$k> 9$

B
$k=6,7$ only

C
$k<9$

D
$k=9$

Solution

Equation ${x}^{2}-6x+k=0$ has
discriminant $D={b}^{2}-4ac$
$=36-4k$
For real roots $D>0$
$\Rightarrow$ $36-4k>0$
$\Rightarrow$ $k<9$
Question 10
1 / -0

If ${z}_{1},{z}_{2}$ are two complex numbers and $c>0$ such that ${ \left| { z }_{ 1 }+{ z }_{ 2 } \right| }^{ 2 }\le \left( 1+c \right) { \left| { z }_{ 1 } \right| }^{ 2 }+k{ \left| { z }_{ 2 } \right| }^{ 2 },$ then $k=$

A
$1-c$

B
$c-1$

C
$1+{c}^{-1}$

D
$1-{c}^{-1}$

Solution

Since Re $\displaystyle \left( { z }_{ 1 }{ \overline { z } }_{ 2 } \right) \le \left| { z }_{ 1 }{ \overline { z } }_{ 2 } \right|$

$\displaystyle \therefore { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+Re\left( { z }_{ 1 }{ \overline { z } }_{ 2 } \right) \le { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+2\left| { z }_{ 1 }{ \overline { z } }_{ 2 } \right|$

$\displaystyle \Rightarrow { \left| { z }_{ 1 }+{ z }_{ 2 } \right| }^{ 2 }\le { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+2\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right|$ ...(1)

Also, Since $A.M.\ge G.M.$

$\displaystyle \therefore \frac { { \left( \sqrt { c } \left| { z }_{ 1 } \right| \right) }^{ 2 }+{ \left( \frac { 1 }{ \sqrt { c } } \left| { z }_{ 2 } \right| \right) }^{ 2 } }{ 2 } \ge { \left\{ c.{ \left| { z }_{ 1 } \right| }^{ 2 }.\frac { 1 }{ c } { \left| { z }_{ 2 } \right| }^{ 2 } \right\} }^{ \frac { 1 }{ 2 } }\left( \because c>0 \right)$

$\displaystyle \Rightarrow c{ \left| { z }_{ 1 } \right| }^{ 2 }+\frac { 1 }{ c } { \left| { z }_{ 1 } \right| }^{ 2 }\ge 2\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right|$

$\displaystyle \therefore { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 1 } \right| }^{ 2 }+2\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \le { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+c{ \left| { z }_{ 1 } \right| }^{ 2 }+\frac { 1 }{ c } { \left| { z }_{ 2 } \right| }^{ 2 }$

$\displaystyle \Rightarrow { \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+2\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \le \left( 1+{ c }^{ -1 } \right) \left( { \left| { z }_{ 2 } \right| }^{ 2 } \right)$ ...(2)

From (1) and (2), we get

$\displaystyle { \left| { z }_{ 1 }+{ z }_{ 2 } \right| }^{ 2 }\le \left( 1+c \right) { \left| { z }_{ 1 } \right| }^{ 2 }+\left( 1+{ c }^{ -1 } \right) { \left| { z }_{ 2 } \right| }^{ 2 }$

$\displaystyle \therefore k=1+{ c }^{ -1 }$

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Complex Numbers and Quadratic Equations Test 14

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Question 1

14ln(2)\displaystyle \dfrac{1}{4}ln\left ( 2 \right )41​ln(2)

12ln(2)\displaystyle \dfrac{1}{2}ln\left ( 2 \right )21​ln(2)

12ln(12)\displaystyle \dfrac{1}{2}ln\left (\dfrac{1}{ 2} \right )21​ln(21​)

ln(2)\displaystyle ln\left ( 2 \right )ln(2)

Solution

Question 2

1,−21,-21,−2

0,40,40,4

−1,3-1,3−1,3

0,30,30,3

Question 3

no real roots

two distinct real roots

two equal real roots

more than two real roots

Solution

Question 4

222

333

−5-5−5

555

Solution

Question 5

Real and distinct

Equal

Imaginary

Real

Solution

Question 6

2b=ac2b = ac2b=ac

b2=ac\displaystyle b^{2}=acb2=ac

b=2aca+c\displaystyle b=\frac{2ac}{a+c}b=a+c2ac​

b=acb = acb=ac

Solution

Question 7

zero

either zero or −12\displaystyle -\frac{1}{2} −21​

−12\displaystyle -\frac{1}{2} −21​

either 12\displaystyle \frac{1}{2} 21​ or −12\displaystyle -\frac{1}{2} −21​

Solution

Question 8

9,−59, -59,−5

−9,5-9, 5−9,5

9,59, 59,5

−9,−5-9, -5−9,−5

Solution

Question 9

k>9k> 9k>9

k=6,7k=6,7k=6,7 only

k<9k<9k<9

k=9k=9k=9

Solution

Question 10

1−c1-c1−c

c−1c-1c−1

1+c−11+{c}^{-1}1+c−1

1−c−11-{c}^{-1}1−c−1

Solution

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

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

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

$\displaystyle ln\left ( 2 \right )$

$1,-2$

$0,4$

$-1,3$

$0,3$

$2$

$3$

$-5$

$5$

$2b = ac$

$\displaystyle b^{2}=ac$

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

$b = ac$

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

$\displaystyle -\frac{1}{2}$

either $\displaystyle \frac{1}{2}$ or $\displaystyle -\frac{1}{2}$

$9, -5$

$-9, 5$

$9, 5$

$-9, -5$

$k> 9$

$k=6,7$ only

$k<9$

$k=9$

$1-c$

$c-1$

$1+{c}^{-1}$

$1-{c}^{-1}$