Complex Numbers and Quadratic Equations Test 18

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Complex Numbers and Quadratic Equations Test 18

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$25 - 4i$$

B
$$25 - 20i$$

C
$$21$$

D
$$29$$

E
$$0$$

Solution

Since $$i^2=-1$$ and $$(a+b)(a-b)=a^2-b^2$$
$$\Rightarrow (5+2i)(5-2i)$$ $$=(5)^2-(2i)^2$$
$$=25-4i^2$$
$$=25-4(-1)$$
$$=25+4$$
$$=29$$
Question 2
1 / -0

What is the smallest integral value of $$k$$ such that $$2x (kx - 4) - x^{2} + 6 = 0$$ has no real roots?

A
$$-1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$2x(kx-4)-x^2+6=0$$
$$\therefore 2kx^2-8x-x^2+6=0$$
$$\therefore (2k-1)x^2-8x+6=0$$

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)
Question 3
1 / -0

What can you say about the graph of $$y = x^{2} - 12x + 40$$?

A
Intersects $$X$$-axis at $$1$$ point

B
Intersects $$X$$-axis at $$2$$ points

C
Intersects $$X$$-axis at $$3$$ points

D
Does not intersect $$X$$-axis at any point

Solution

To find the intersection point of a curve with the $$X$$-axis, put $$y=0$$.

$$\therefore x^2-12x+40=0$$

Calculating the discriminant of the equation,

$$\Delta = 12^2 - 4(40) = 144-160 = -16 < 0$$

Since, $$\Delta < 0$$, the equation has no real roots.

Hence, the curve does not intersect the $$X$$-axis at any point.
Question 4
1 / -0

For what value of 'p' are the roots of $$x^{2} - 5x = 2(3 + x) - 4p$$ real and equal?

A
$$\dfrac {23}{9}$$

B
$$\dfrac {35}{17}$$

C
$$\dfrac {53}{12}$$

D
$$\dfrac {73}{16}$$

Solution

$${ x }^{ 2 }-5x=2(3+x)-4p\\ $$
for roots to be real and equal
$$D=0$$
$${ x }^{ 2 }\\ -5x-2x-6+4p=0$$
$${ x }^{ 2 }\\ -7x+4p-6=0$$
$$D=0$$
$$49-4(4p-6)=0$$
$$49-16p+24=0$$
$$16p=73$$
$$p=\dfrac { 73 }{ 16 } $$
Question 5
1 / -0

Express $$\dfrac {(-5-i)(-7+8i)}{(2-4i)}$$ in the form of a complex number $$a+bi$$.

A
$$\dfrac{109}{10}-\dfrac{53}{10}i$$

B
$$-\dfrac{109}{10}-\dfrac{53}{10}i$$

C
$$\dfrac{109}{10}+\dfrac{53}{10}i$$

D
$$-\dfrac{109}{10}+\dfrac{53}{10}i$$

Solution

We need to express $$\dfrac {(-5-i)(-7+8i)}{(2-4i)}$$ in the form of $$a+bi$$
On rationalising the given expression as follows:
$$\dfrac {(-5-i)(-7+8i)(2+4i)}{(2-4i)(2+4i)}$$
$$=\dfrac {(35-40i+7i-8i^2)(2+4i)}{(4-16i^2)}$$
$$=\dfrac {(43-33i)(2+4i)}{(4+16)}$$
$$=\dfrac {(86+172i-66i-132i^2)}{20}$$
$$=\dfrac {(218+106i)}{20}$$
$$=\dfrac {109}{10}+\dfrac {53}{10}i$$
Question 6
1 / -0

If $$a, b, c$$ are real, what is the nature of the roots of the equation $$(a - x)(b - x) = c^{2}$$?

A
Equal

B
Imaginary

C
Real

D
Irrational

Solution

$$\left( a-x \right) \left( b-x \right) ={ c }^{ 2 },\quad a,b,c\epsilon R$$ (real)
$${ x }^{ 2 }-\left( a+b \right) x+ab={ c }^{ 2 }$$
$${ x }^{ 2 }-\left( a+b \right) x+ab-{ c }^{ 2 }=0$$
$$\triangle ={ \left( a+b \right) }^{ 2 }-4ab+4{ c }^{ 2 }$$
$$={ \left( a-b \right) }^{ 2 }+4{ c }^{ 2 }$$
$${ \left( a-b \right) }^{ 2 }\ge 0,4{ c }^{ 2 }\ge 0$$
$$\therefore \triangle \ge 0$$
$$\therefore x$$ is real.
Question 7
1 / -0

If $$(2 - i)\times(a - bi) = 2 + 9i$$, where i is the imaginary unit and a and b are real numbers, then a equals

A
$$3$$

B
$$2$$

C
$$1$$

D
$$0$$

E
$$-1$$

Solution

Given, $$(2-i)(a-bi) = 2a-2bi-ai-b $$
$$= (2a-b)+i(-a-2b)$$
$$ = 2+9i$$
By comparing, we get $$2a-b=2$$ and $$a+2b=-9$$
By solving, we get $$a=-1$$ and $$b=-4$$
Question 8
1 / -0

What is the approximate magnitude of $$8 + 4i$$?

A
4.15

B
8.94

C
12.00

D
18.64

E
32.00

Solution

Magnitude of $$8+4i$$ is
$$=\sqrt { { 8 }^{ 2 }+{ 4 }^{ 2 } } $$
$$=\sqrt { 64+16 } $$
$$=\sqrt { 80 }$$
$$ =8.94$$
Question 9
1 / -0

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

A
$$\cfrac{1}{2}$$

B
$$\cfrac { 1 }{ 1+\cos { \theta } } $$

C
$$\tan { \cfrac { \theta }{ 2 } } $$

D
$$\cot { \cfrac { \theta }{ 2 } } $$

Solution

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 $$\dfrac{1}{2}$$
Question 10
1 / -0

The number of real roots of $$\left (x+\dfrac{1}{x}\right)^2-4=0$$ is

A
$$0$$

B
$$2$$

C
$$4$$

D
none of these

Solution

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

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Complex Numbers and Quadratic Equations Test 18

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Complex Numbers and Quadratic Equations Test 18

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SHARING IS CARING

Question 1

$$25 - 4i$$

$$25 - 20i$$

$$21$$

$$29$$

$$0$$

Solution

Question 2

$$-1$$

$$2$$

$$3$$

$$4$$

Solution

Question 3

Intersects $$X$$-axis at $$1$$ point

Intersects $$X$$-axis at $$2$$ points

Intersects $$X$$-axis at $$3$$ points

Does not intersect $$X$$-axis at any point

Solution

Question 4

$$\dfrac {23}{9}$$

$$\dfrac {35}{17}$$

$$\dfrac {53}{12}$$

$$\dfrac {73}{16}$$

Solution

Question 5

$$\dfrac{109}{10}-\dfrac{53}{10}i$$

$$-\dfrac{109}{10}-\dfrac{53}{10}i$$

$$\dfrac{109}{10}+\dfrac{53}{10}i$$

$$-\dfrac{109}{10}+\dfrac{53}{10}i$$

Solution

Question 6

Equal

Imaginary

Real

Irrational

Solution

Question 7

$$3$$

$$2$$

$$1$$

$$0$$

$$-1$$

Solution

Question 8

4.15

8.94

12.00

18.64

32.00

Solution

Question 9

$$\cfrac{1}{2}$$

$$\cfrac { 1 }{ 1+\cos { \theta } } $$

$$\tan { \cfrac { \theta }{ 2 } } $$

$$\cot { \cfrac { \theta }{ 2 } } $$

Solution

Question 10

$$0$$

$$2$$

$$4$$

none of these

Solution

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