Complex Numbers and Quadratic Equations Test 21

Result

Complex Numbers and Quadratic Equations Test 21

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

For the equations $$x^2 + bx + c = 0$$ and $$2x^2 + (b + 1)x + c + 1 = 0$$ select the correct alternative

A
both the equations can have integral roots

B
both the equations cant have integral roots simultaneously

C
none of the equations can have integral roots

D
nothing can be said

Solution

$$x^2 + bx + c = 0$$
Applying dharacharya's formula
$$x = \dfrac{-b \pm \sqrt{b^2 - 4c}}{2}$$
$$2x^2 + (b + 1) x + c+ 1 = 0$$
$$x = \dfrac{-b- 1 \pm \sqrt{(b + 1)^2 - 8 (c + 1)}}{4}$$
if $$b^2 - 4c = 0 $$ for integral roots
then $$(b + 1)^2 - 8 (c + 1) \neq 0$$
that mean's only one equation can have integral root.
Question 2
1 / -0

Consider the equation $$x^3 - nx + 1 =0$$, $$n\ \epsilon \ N$$ , $$n \geq 3$$. Then

A
Equation has atleast one rational root

B
Equation has exactly one rational root.

C
Equation has all real roots belonging to $$(0, 1)$$

D
Equation has no rational root.

Solution
Question 3
1 / -0

The equation $$x^2 + nx + m = 0,$$ where $$n, m \in I$$ can not have:

A
integral roots

B
non-integral rational roots

C
irrational roots

D
complex roots

Solution
Question 4
1 / -0

If $$x=1+i$$ is a root of the equation $$x^3-ix+1-i=0$$, then the other real root is

A
$$0$$

B
$$1$$

C
$$-1$$

D
None of the above.

Solution

Given equation is
$${ x }^{ 3 }-ix+1-i=0$$
$${ x }^{ 3 }+1-i(x+1)=0$$
$$(x+1)({ x }^{ 2 }-x+1-i)=0$$
$$\therefore $$ Real root of given equation is $$x=-1$$
Question 5
1 / -0

If the argument of a complex number is $$\cfrac { \pi }{ 2 } $$, then the number is:

A
Purely imaginary

B
Purely real

C
$$0$$

D
Neither real nor imaginary

Solution

Let the complex number be $$z$$.
Hence
$$z=|z|e^{i\arg(z)}$$
$$=|z|e^{i\tfrac{\pi}{2}}$$
$$=|z|\left(\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}\right)$$
$$=i|z|$$
Hence the complex number is purely imaginary.
Question 6
1 / -0

Total number of integral values of $$a$$ so that $$x^2-(a+1)x+a-1=0$$ has integral roots is equal to

A
$$1$$

B
$$2$$

C
$$4$$

D
None of the above.

Solution

For integer roots $$b^{ 2 }-4ac$$ should be a perfect square.
$$x^{ 2 }-(a+1)x+(a-1)=0\\ \Longrightarrow b^{ 2 }-4ac={ (a+1) }^{ 2 }-4(a-1)\\ \quad \quad \quad \quad \quad =a^{ 2 }-2a+5\\ or\quad p^{ 2 }=a^{ 2 }-2a+5\\ \therefore \quad for\quad a=1, a^{ 2 }-2a+5$$
is a perfect square.
$$\therefore$$ for $$a=1,a^{ 2 }-2a+5=4$$,
Only one integral value of $$a$$.
Question 7
1 / -0

If $$\overline { z } $$ lies in the third quadrant then $$z$$ lies in the

A
First quadrant

B
Second quadrant

C
Third quadrant

D
Fourth quadrant

Solution

Fact: $$\overline z$$ is image of $$z$$ in $$x$$-axis
So if $$\overline z$$ lies in third quadrant then $$z$$ will lie in second quadrant
Question 8
1 / -0

If $$a,b,c,d$$ are four consecutive terms of an increasing A.P, then the roots of the equation
$$(x-a)(x-c)+2(x-b)(x-d)=0$$ are

A
Non-real complex

B
Real and Equal

C
Integers

D
Real and Distinct

Solution

Let $$a<b<c<d$$ be increasing A.P.
Let $$f\left( x \right) =(x-a)(x-c)+2(x-b)(x-d)\\ f\left( b \right) =(b-a)(b-c)<0$$
and
$$f\left( d \right) =(d-a)(d-c)>0$$
Hence $$f\left( x \right) =0$$ has roots in $$(b,d)$$.
Also $$f\left( a \right) f\left( c \right) <0$$.
So the other root lies in $$(a,c)$$.
Therefore roots are Real and Distinct.
Question 9
1 / -0

$$p + iq = (2 - 3i) (4 + 2i)$$ then $$q$$ is

A
$$14$$

B
$$-14$$

C
$$-8$$

D
$$8$$

Solution

$$p+iq=(2-3i)(4+2i)$$
$$=(8+4i-12i+6)$$
$$=14-8i$$
Hence comparing the real parts give us $$p=14$$ and comparison of the imaginary parts give us $$q=-8$$.
Question 10
1 / -0

If $${ x }^{ 2 }+{ y }^{ 2 }=1$$ then value of $$\dfrac { 1+x+iy }{ 1+x-iy } $$ is

A
$$x-iy$$

B
$$2x$$

C
$$-2iy$$

D
$$x+iy$$

Solution

Since $$x^2+y^2=1$$
So let $$x=\cos\theta$$ and $$y=\sin\theta$$
$$\therefore \dfrac{1+x+iy}{1+x-iy}=\dfrac{(1+\cos\theta)+i\sin\theta}{(1+\cos\theta)-i\sin\theta}$$

$$=\dfrac{2\cos^2\cfrac{\theta}{2}+2i\sin\cfrac{\theta}{2}\cos\cfrac{\theta}{2}}{2\cos^2\cfrac{\theta}{2}-2i\sin\cfrac{\theta}{2}\cos\cfrac{\theta}{2}}$$

$$=\dfrac{\cos\cfrac{\theta}{2}+i\sin\cfrac{\theta}{2}}{\cos\cfrac{\theta}{2}-i\sin\cfrac{\theta}{2}}$$

$$=\dfrac{\cos\cfrac{\theta}{2}+i\sin\cfrac{\theta}{2}}{\cos\cfrac{\theta}{2}-i\sin\cfrac{\theta}{2}}\times$$ $$\dfrac{\cos\cfrac{\theta}{2}+i\sin\cfrac{\theta}{2}}{\cos\cfrac{\theta}{2}+i\sin\cfrac{\theta}{2}}$$

$$=\dfrac{\left(\cos\cfrac{\theta}{2}+i\sin\cfrac{\theta}{2}\right)^2}{\cos^2\cfrac{\theta}{2}-i^2\sin^2\cfrac{\theta}{2}}$$

$$=\dfrac{\cos^2\cfrac{\theta}{2}+i^2\sin^2\cfrac{\theta}{2}+2i\cos\cfrac{\theta}{2}\cos\cfrac{\theta}{2}}{\cos^2\cfrac{\theta}{2}+\sin^2\cfrac{\theta}{2}}$$

$$=\dfrac{\cos^2\cfrac{\theta}{2}-\sin^2\cfrac{\theta}{2}+2i\cos\cfrac{\theta}{2}\cos\cfrac{\theta}{2}}{1}$$

$$=\cos\theta+i\sin\theta=x+iy$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Complex Numbers and Quadratic Equations Test 21

Result

Complex Numbers and Quadratic Equations Test 21

Score

-

Rank

-

SHARING IS CARING

Question 1

both the equations can have integral roots

both the equations cant have integral roots simultaneously

none of the equations can have integral roots

nothing can be said

Solution

Question 2

Equation has atleast one rational root

Equation has exactly one rational root.

Equation has all real roots belonging to $$(0, 1)$$

Equation has no rational root.

Solution

Question 3

integral roots

non-integral rational roots

irrational roots

complex roots

Solution

Question 4

$$0$$

$$1$$

$$-1$$

None of the above.

Solution

Question 5

Purely imaginary

Purely real

$$0$$

Neither real nor imaginary

Solution

Question 6

$$1$$

$$2$$

$$4$$

None of the above.

Solution

Question 7

First quadrant

Second quadrant

Third quadrant

Fourth quadrant

Solution

Question 8

Non-real complex

Real and Equal

Integers

Real and Distinct

Solution

Question 9

$$14$$

$$-14$$

$$-8$$

$$8$$

Solution

Question 10

$$x-iy$$

$$2x$$

$$-2iy$$

$$x+iy$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.