Complex Numbers and Quadratic Equations Test 20

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Complex Numbers and Quadratic Equations Test 20

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Weekly Quiz Competition

Question 1
1 / -0

If $$a,b \in R$$ then the equation $$x^2-abx-a^2=0$$ has

A
One positive and one negative root

B
Both positive root

C
Both negative root

D
Non-real root

Solution

Roots of the given equation are :$$\dfrac {ab\pm \sqrt{b^{2}+4}}{2}$$

Roots =$$a\dfrac {(b\pm\sqrt {b^{2}+4) }}{2}$$

Given $$a, b\in R$$

Here $$\sqrt{b^{2}}=b<\sqrt {b^{2}+4}$$

$$\therefore $$one root is positive and one root is negative
Question 2
1 / -0

If $$b$$ and $$c$$ are odd integers, then the equation $$x^2+bx+c=0$$ has:

A
two integers roots, one even and one odd

B
no integers roots

C
two odd roots

D
none of these

Solution
Question 3
1 / -0

If $$b$$ and $$c$$ are odd integers, then the equation $$x^2+bx+c=0$$ has

A
Two odd roots

B
Two integer roots, one odd and one even

C
No integer roots

D
None of the above

Solution

$${ x }^{ 2 }+bx+c=0$$
b,c are odd integers
roots, $$x=-b\pm \sqrt { { b }^{ 2 }-4c } $$
In order to have integer roots i.e., assume roots are integers
$${ b }^{ 2 }-4c={ m }^{ 2 },m\epsilon z$$
$${ b }^{ 2 }-{ m }^{ 2 }=4c$$
if $${ b }^{ 2 }-{ m }^{ 2 }$$ is divisible by $$2$$. Then both b,m are even or both m,b are odd
$$\therefore m$$ is odd ($$\because b$$ is odd)
Let $$b=2k+1,m=2n+1,k,n\epsilon z$$
$$4\left( { k }^{ 2 }-{ n }^{ 2 } \right) +4\left( k-n \right) =4c$$
$$c=\left( k-n \right) \left( k+n+1 \right) $$
if $$k-n$$ is odd, then $$k+n+1$$ is even.
(Or) if $$k-n$$ is even then $$k+n+1$$ is odd
$$\therefore $$C will be even
But C cannot be even as it is odd (Given)
$$\therefore $$ roots of $${ x }^{ 2 }+bx+c$$ are not integers.
Question 4
1 / -0

If Arg $$(z + i)\, -$$ Arg $$(z - i)$$ $$= \dfrac{\pi}{2}$$, then $$z$$ lies on a ..........

A
Circle

B
Line

C
Coordinate axes

D
None of these

Solution

Putting z = x+iy,

$${tan}^{-1}\dfrac{y+1}{x}$$ - $${tan}^{-1}\dfrac{y-1}{x}$$ = $$\pi$$/2

$$\Rightarrow$$ 1 + ($$\dfrac{y+1}{x})$$($$\dfrac{y-1}{x}$$) = 0

$$\Rightarrow$$ $$x^2 + y^2$$ = 1

It is a circle of center coinciding with origin and radius 1 units.

Hence, option A is correct.
Question 5
1 / -0

Let $$a,b,c \in R$$ such that $$4a+2b+c=0$$ and $$ab>0$$.
Then the equation $$ax^2+bx+c=0$$ has

A
Complex root

B
Exactly one root

C
Real roots

D
None of the above

Solution

$$ab>0\\ a,b,c\in R\\ f(x)={ ax }^{ 2 }+bx+c\\ f(2)=4a+2b+c=0\\ \therefore x=2\quad is\quad a\quad root\quad of\quad f(x)\\ \therefore \quad f(x)\quad has\quad real\quad roots$$
Question 6
1 / -0

If the equation $$x^3 - 3ax^2 + 3bx- c = 0$$ has positive and distinct roots, then

A
$$a^2 > b$$

B
$$ab > c$$

C
$$a^3 > c$$

D
$$a^3 > b^2 > c$$

Solution
Question 7
1 / -0

The number of real roots of the equation $$|x| ^2-3|x|+2=0$$ is

A
$$2$$

B
$$1$$

C
$$4$$

D
$$3$$

Solution

$${ |x| }^{ 2 }-3|x|+2=0\\ let\quad |x|\quad be\quad t,\quad t\ge 0\\ { t }^{ 2 }-3t+2=0\\ (t-2)(t-1)=0\\ \therefore t=2\ or\ 1\\ where\quad t=1\\ |x|=1\\ \therefore x=+1,-1\\ when\quad t=2\\ |x|=2\\ \therefore x=+2,-2\\ so\quad 4\quad real\quad roots$$
Question 8
1 / -0

If $$c > 0 $$ and $$b > c$$ then $$x ^2+bx-c=0$$ will have:

A
exactly one root between $$0$$ and $$1$$

B
both root between $$0$$ and $$1$$

C
no root between $$0$$ and $$1$$

D
none of these
Question 9
1 / -0

Let $$a > 0, b > 0,c >0 $$ then both roots of the equation $$ ax^2 + bx + c = 0$$

A
are real and negative

B
have negative real parts

C
have positive real parts

D
none of these

Solution

Since, $$a,\,b,\,c>0$$ and $$ax^2+bx+c=0$$ then,
$$\Rightarrow$$ $$x=\dfrac{-b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$$

$$Case\,1:$$ When $$b^2-4ac>0$$
$$\Rightarrow$$ $$x=\dfrac{-b}{2a}-\dfrac{\sqrt{b^2-4ac}}{2a}$$ and $$\dfrac{-b}{2a}+\dfrac{\sqrt{b^2-4ac}}{2a}$$ both roots are negative.

$$Case\,2:$$ Ehen $$b^2-4ac=0$$
$$\Rightarrow$$ $$x=\dfrac{-b}{2a}$$ i.e., both roots are equal and negative.

$$Case\,3:$$ When $$b^2-4ac<0$$
$$\Rightarrow$$ $$x=\dfrac{-b}{2a}\pm i\dfrac{\sqrt{4ac-b^2}}{2a}$$ have negative real part.
$$\therefore$$ From above discission both roots have negative real parts.
Question 10
1 / -0

If $$arg(z) < 0$$, then $$arg(-z)-arg(z)=$$

A
$$\pi$$

B
$$-\pi$$

C
$$\dfrac{\pi}{2}$$

D
$$-\dfrac{\pi}{2}$$

Solution

Let $$Z=re^{i\theta_1}$$
$$-Z=-re^{i\theta_1}$$
$$\implies -a\cos\theta_1-ib\sin\theta_1$$
$$\implies -a\cos(\pi+\theta_1)-ib\sin(\pi+\theta_1)$$
$$\implies re^{i(\pi+\theta_1)}$$
$$arg(-Z)-arg(Z)$$
$$\implies \pi+\theta_1-\theta_1\\ \implies \pi$$

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Complex Numbers and Quadratic Equations Test 20

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Question 1

One positive and one negative root

Both positive root

Both negative root

Non-real root

Solution

Question 2

two integers roots, one even and one odd

no integers roots

two odd roots

none of these

Solution

Question 3

Two odd roots

Two integer roots, one odd and one even

No integer roots

None of the above

Solution

Question 4

Circle

Line

Coordinate axes

None of these

Solution

Question 5

Complex root

Exactly one root

Real roots

None of the above

Solution

Question 6

$$a^2 > b$$

$$ab > c$$

$$a^3 > c$$

$$a^3 > b^2 > c$$

Solution

Question 7

$$2$$

$$1$$

$$4$$

$$3$$

Solution

Question 8

exactly one root between $$0$$ and $$1$$

both root between $$0$$ and $$1$$

no root between $$0$$ and $$1$$

none of these

Question 9

are real and negative

have negative real parts

have positive real parts

none of these

Solution

Question 10

$$\pi$$

$$-\pi$$

$$\dfrac{\pi}{2}$$

$$-\dfrac{\pi}{2}$$

Solution

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