Complex Numbers and Quadratic Equations Test 13

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Complex Numbers and Quadratic Equations Test 13

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Weekly Quiz Competition

Question 1
1 / -0

Let $$\displaystyle \left| Z_{r}-r\right| \leq r,\forall r=1,2,3,.......n.$$ Then $$\displaystyle \left| \sum_{r=1}^{n} Z_{r} \right| $$ is less than

A
$$n$$

B
$$2n$$

C
$$n(n+1)$$

D
$$\displaystyle \frac{n(n+1)}{2}$$

Solution

$$|Z_r - r| \le r$$

$$\therefore Z_r \le 2r$$

or $$|Z_r| \le 2r$$

also $$\sum_{r=1}^n |Z_r| \le 2 \sum_{r=1}^n r$$

also $$|\sum_{r=1}^n Z_r| \le \sum_{r=1}^n |Z_r|$$ $$\{$$ extending triangle inequality $$\}$$

$$\therefore |\sum_{r=1}^n Z_r| \le 2\times \dfrac{n(n+1)}{2}$$

$$\therefore |\sum_{r=1}^n Z_r| \le {n(n+1)}$$

$$\therefore$$ option C is correct.
Question 2
1 / -0

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 :

A
$$1 \;or\; 2 $$

B
$$1$$

C
$$2$$

D
None

Solution

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) $$2$$
Question 3
1 / -0

Number of complex numbers $$z$$ satisfying $$\left| 2z \right| =\left| 2z-1 \right| =\left| 2z+1 \right| $$ is equal to-

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

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 $$0$$
Question 4
1 / -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 :

A
$$10$$

B
$$7$$

C
$$6$$

D
$$12$$

Solution

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
Question 5
1 / -0

If 0 $$\leq$$ argz $$\leq \displaystyle \frac{\pi}{4}$$, then the least value of $$\sqrt 2 |2z - 4|$$ is

A
6

B
1

C
4

D
2

Solution

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

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 $$C$$
Question 6
1 / -0

If $$\displaystyle z= (i)^{(i)^{(i)}}$$ where $$ i = \sqrt{-1}$$, then $$z$$ is equal to

A
$$-i$$

B
$$-1$$

C
$$1$$

D
$$i$$

Solution

$$Z=(i)^{(i)^{(i)}}$$
$$i=e^{i\pi/2}$$
$$i^i=e^{i.(i\pi/2)}=e^{-\pi/2}$$
$$i^{i^i}=e^{i(-\pi/2)}\\ \implies \cos(\pi/2)-i\sin(\pi/2)\\ \implies 0-i=-i$$
Question 7
1 / -0

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

A
$$\displaystyle \pm 8$$

B
$$\displaystyle \pm \frac{8}{3}$$

C
$$\displaystyle \pm \frac{3}{8}$$

D
$$0$$

Solution

$$\displaystyle 16x^{2}+6kx+4=0$$ has equal roots.
Therefore, $$D = 0$$
$$\displaystyle \Rightarrow (6k)^{2}-4(16)(4)=0$$
$$\displaystyle \Rightarrow 36k^{2}-256=0$$
$$\displaystyle \Rightarrow k^{2}=\frac{256}{36}=\frac{64}{9}$$
$$\displaystyle \Rightarrow k=\pm \frac{8}{3}$$
Question 8
1 / -0

The roots of the equations $$\displaystyle x^{2}-x-3=0$$ are

A
Imaginary

B
Rational

C
Irrational

D
None of these

Solution

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.
Question 9
1 / -0

The quadratic equation $$ax^{2}+bx+c=0$$ will always have imaginary roots if:

A
$$ a<-1,0< c <1,b<0 $$

B
$$a<-1,-1< c < 0,0 < b < 1$$

C
$$a<-1,c<0,b>1 $$

D
$$ a <-1,c < -1,1< b < 2 $$

Solution

The quadratic equation $$ax^{2}+bx+c=0$$ will have imaginary roots if discriminant is negative
So, $$D$$ must be negative,
$$\Rightarrow b^2 - 4ac < 0 $$
$$\Rightarrow b^2 < 4ac $$
Now we need to examine each option, only D will satisfy.
Question 10
1 / -0

Which of the following equation has two equal real toots ?

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

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

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

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

Solution

(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

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Complex Numbers and Quadratic Equations Test 13

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Question 1

$$n$$

$$2n$$

$$n(n+1)$$

$$\displaystyle \frac{n(n+1)}{2}$$

Solution

Question 2

$$1 \;or\; 2 $$

$$1$$

$$2$$

None

Solution

Question 3

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 4

$$10$$

$$7$$

$$6$$

$$12$$

Solution

Question 5

6

1

4

2

Solution

Question 6

$$-i$$

$$-1$$

$$1$$

$$i$$

Solution

Question 7

$$\displaystyle \pm 8$$

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

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

$$0$$

Solution

Question 8

Imaginary

Rational

Irrational

None of these

Solution

Question 9

$$ a<-1,0< c <1,b<0 $$

$$a<-1,-1< c < 0,0 < b < 1$$

$$a<-1,c<0,b>1 $$

$$ a <-1,c < -1,1< b < 2 $$

Solution

Question 10

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

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

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

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

Solution

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