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

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

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

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

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

Solution

Question 2

$$1,-2$$

$$0,4$$

$$-1,3$$

$$0,3$$

Question 3

no real roots

two distinct real roots

two equal real roots

more than two real roots

Solution

Question 4

$$2$$

$$3$$

$$-5$$

$$5$$

Solution

Question 5

Real and distinct

Equal

Imaginary

Real

Solution

Question 6

$$2b = ac$$

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

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

$$b = ac$$

Solution

Question 7

zero

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

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

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

Solution

Question 8

$$9, -5$$

$$-9, 5$$

$$9, 5$$

$$-9, -5$$

Solution

Question 9

$$k> 9$$

$$k=6,7$$ only

$$k<9$$

$$k=9$$

Solution

Question 10

$$1-c$$

$$c-1$$

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

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

Solution

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