Complex Numbers and Quadratic Equations Test 19

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Complex Numbers and Quadratic Equations Test 19

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Weekly Quiz Competition

Question 1
1 / -0

If $$2,2$$ are the roots of $$x^2+px+b=0$$ and $$1,4$$ are the roots of $$x^2+ax+q=0$$, then roots of the equation $$x^2+ax+b=0$$ are

A
real and distinct

B
imaginary

C
real and equal

D
irrational

Solution

First equation :

-p=4$$\Rightarrow $$p=-4 and b=4(applying sum and product of roots)

Second equation :

$$a=-5$$ and $$q=4$$

Third equation :

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

$$\Rightarrow (x-4)(x-1)=0$$

$$\therefore $$x=4,1

$$\Rightarrow $$real and distinct roots
Question 2
1 / -0

If $$z = \dfrac {(\sqrt {3} + i)^{3} (3i + 4)^{2}}{(8 + 6i)^{2}}$$, then $$|z|$$ is equal to

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

Uisng the identity $$|a+ib|=\sqrt{a^2+b^2}$$
Since $$\bigg|\dfrac{z_1z_2}{z_3} \bigg|=\dfrac{|z_1||z_2|}{|z_3|}$$
Therefore
$$|z| = \bigg|\dfrac {(\sqrt {3} + i)^{3} (3i + 4)^{2}}{(8 + 6i)^{2}}\bigg|$$
$$\quad =\dfrac{|(\sqrt 3+i)^3||(3i+4)^2|}{|(8+6i)^2|}$$
$$\quad =\dfrac{|\sqrt 3+i|^3|3i+4|^2}{|8+6i|^2}$$
$$\quad =\dfrac{2^3\cdot 5^2}{10^2}=2$$
Question 3
1 / -0

The value of $$(a+2i)(b-i)$$ is

A
$$a+b-i$$

B
$$ab+2$$

C
$$ab+(2b-a)i+2$$

D
$$ab-2$$

E
$$ab+(2b-a)i-2$$

Solution

The value of $$(a+2i)(b-i) $$
$$=a(b-i)+2i(b-i)$$
$$= ab-ai+2bi-2{i}^{2} $$
$$= ab+(2b-a)i+2$$
Question 4
1 / -0

If $$l,m$$ are real and $$l\neq m$$, the roots of the equation $$(l-m)x^2-5(l+m)x-2(l-m)=0$$ are-

A
Real and Equal

B
Complex

C
Real and Unequal

D
None of the above

Solution

Let we check the descriminant

Descriminant =$$25(l+m)^{2}+4(l-m)^{2}$$ which is always >0

$$\therefore $$it has real and unequal roots
Question 5
1 / -0

If $$a<c<b$$ then the roots of the equation $$(a-b) ^2x^2+2(a+b-2c)x+1=0$$ are-

A
Imaginary

B
Real

C
One real and one Imaginary

D
Equal and Imaginary

Solution

$$a<c<b$$
$${ \left( a-b \right) }^{ 2 }{ x }^{ 2 }+2\left( a+b-2c \right) x+1=0$$
$$\triangle =4{ \left( a+b-2c \right) }^{ 2 }-4{ \left( a-b \right) }^{ 2 }$$
$$=4{ \left( \left( a-c \right) +\left( b-c \right) \right) }^{ 2 }-4{ \left( \left( a-c \right) -\left( b-c \right) \right) }^{ 2 }$$
$$=4\left( 4\left( a-c \right) \left( b-c \right) \right) $$
$$=16\left( a-c \right) \left( b-c \right) $$
$$a-c<0$$ & $$b-c>0\quad \left( \because a<c<b \right) $$
$$\therefore \triangle <0$$
$$\therefore $$Roots of given equation are imaginary.
Question 6
1 / -0

Total number of integral values of $$a$$ such that $$x^2+ax+a+1=0$$ has integral roots is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$for\quad a\quad equation\quad { ax }^{ 2 }+bx+c=0\quad to\quad have\quad integer\quad roots\quad conditions\quad are\quad \\ a=1,\quad b\in I,\quad c\in I\quad and\quad { b }^{ 2 }-4ac\quad is\quad a\quad perfect\quad square\\ { x }^{ 2 }+ax+a+1=0\\ a\in I\quad \therefore a+1\in I\\ { a }^{ 2 }-4(a+1)is\quad perfect\quad square\\ \therefore { a }^{ 2 }-4(a+1)\quad is\quad perfect\quad square\\ \therefore { a }^{ 2 }-4a-4={ p }^{ 2 }\\ { a }^{ 2 }-4a-4-{ p }^{ 2 }=0\\ { (a-2) }^{ 2 }-8-{ p }^{ 2 }=0\\ { (a-2) }^{ 2 }={ p }^{ 2 }+8\\ \therefore { p }^{ 2 }=1\\ { (a-2) }^{ 2 }=9\\ { (a-2) }^{ 2 }={ 3 }^{ 2 }\\ \therefore a-2=3\quad or\quad a-2=-3\\ a=5\quad or\quad a=-1\\ \therefore 2\quad values\quad of\quad a$$
Question 7
1 / -0

If $$a < c < b$$ then the roots of the equation $$(a-b)x^2+2(a+b-2c)x+1=0$$ are:

A
imaginary

B
real

C
one real and one imaginary

D
equal and imaginary

Solution
Question 8
1 / -0

The roots of the equation $$(b+c)x^2-(a+b+c)x+a=0$$ $$(a,b,c\ \epsilon \ Q, b+c \neq a)$$ are

A
irrational and different

B
rational and different

C
imaginary and different

D
real and equal

Solution

$$(b+c){ x }^{ 2 }-(a+b+c)x+a=0\\ a,b,c\in q\\ b+c\neq a\\ D={ (a+b+c) }^{ 2 }-4(a)(b+c)\\ ={ a }^{ 2 }+{ b }^{ 2 }+c^{ 2 }-2ab-2ac+2bc\\ ={ (a-b-c) }^{ 2 }\\ as\quad b+c\neq a\\ \therefore D>0$$ always
$$\therefore$$ roots are rational and distinct as $$D$$ is a perfect square and greater than $$0$$
Question 9
1 / -0

Let $$f(x)=x^3+3x^2+9x+6\sin x$$ then the roots of the equation
$$\dfrac 1{x-f(1)}+\dfrac 2{x-f(2)}+\dfrac 3{x-f(3)}=0$$ has

A
No real roots

B
One real root

C
Two real roots

D
More than 2 real roots

Solution

Given $$f\left( x \right) ={ x }^{ 3 }+3{ x }^{ 2 }+9x+6\sin { x } $$
$$f\left( 0 \right) =0$$
$$f'\left( x \right) =3{ x }^{ 2 }+6x+9+6\cos { x } $$
$$=3{ \left( x+1 \right) }^{ 2 }+6\left( 1+\cos { x } \right) $$
$$f'\left( x \right) >0$$
$$\therefore f\left( x \right) $$ is monotonically increasing graph
$$\therefore f\left( 3 \right) >f\left( 2 \right) >f\left( 1 \right) >f\left( 0 \right) $$
$$\therefore f\left( 3 \right) >f\left( 2 \right) >f\left( 1 \right) >0$$ ($$\because $$ if $${ x }_{ 1 }>{ x }_{ 2 }\Rightarrow f\left( { x }_{ 1 } \right) >f\left( { x }_{ 2 } \right) $$)
Let $$f\left( 1 \right) =a;f\left( 2 \right) =9;f\left( 3 \right) =c$$
$$g\left( x \right) =\cfrac { 1 }{ x-f\left( 1 \right) } +\cfrac { 2 }{ x-f\left( 2 \right) } +\cfrac { 3 }{ x-f\left( 3 \right) } =0$$
$$\left( x-b \right) \left( x-c \right) +2\left( x-a \right) \left( x-c \right) +3\left( x-b \right) \left( x-a \right) =0$$
$$\left( x\neq a,b,c \right) $$
$$3{ x }^{ 2 }-\left( b+c+2a+2c+3b+3a \right) x+\left( bc+2ac+3ab \right) =0$$
$$3{ x }^{ 2 }-\left( 5a+4b+3c \right) x+\left( 3ab+2ac+bc \right) =0$$
$$\triangle ={ \left( 5a+4b+3c \right) }^{ 2 }-4\left( 3 \right) \left( 3ab+bc+2ac \right) $$
$$=25{ a }^{ 2 }+16{ b }^{ 2 }+a{ c }^{ 2 }+4ab+6ac+12bc$$
$$\therefore \triangle >0\quad \left( \because a>b>c>0 \right) $$
$$\therefore g\left( x \right) $$ has two distinct real roots.
Question 10
1 / -0

If $$l,m,n $$ are real and $$l \neq m$$, the roots of the equation $$(l-m)x^2-5(l+m)x-2(l-m)=0$$ are-

A
complex

B
real and equal

C
real and unequal

D
none of these

Solution

$$l,m,n\in R\\ l\neq m\\ (l-m){ x }^{ 2 }-5(l+m)x-2(l-m)=0\\ D={ (-5(l+m)) }^{ 2 }+8{ (l-m) }^{ 2 }\\$$
As $${ (l+m) }^{ 2 }$$ and $${ (l-m) }^{ 2 }$$ both greater than zero as $$l\neq m$$ so $$D$$ greater than zero so roots are real and distinct.

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Complex Numbers and Quadratic Equations Test 19

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Question 1

real and distinct

imaginary

real and equal

irrational

Solution

Question 2

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 3

$$a+b-i$$

$$ab+2$$

$$ab+(2b-a)i+2$$

$$ab-2$$

$$ab+(2b-a)i-2$$

Solution

Question 4

Real and Equal

Complex

Real and Unequal

None of the above

Solution

Question 5

Imaginary

Real

One real and one Imaginary

Equal and Imaginary

Solution

Question 6

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 7

imaginary

real

one real and one imaginary

equal and imaginary

Solution

Question 8

irrational and different

rational and different

imaginary and different

real and equal

Solution

Question 9

No real roots

One real root

Two real roots

More than 2 real roots

Solution

Question 10

complex

real and equal

real and unequal

none of these

Solution

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