Complex Numbers and Quadratic Equations Test 36

Result

Complex Numbers and Quadratic Equations Test 36

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

Express the following complex numbers in the standard from $$ a+ib$$ :
$$ \dfrac{5+\sqrt{2}i}{1-\sqrt{2}i}$$

A
$$ 1-2\sqrt{2}i$$

B
$$ 1+\sqrt{2}i$$

C
$$ 1+2\sqrt{2}i$$

D
$$ 1-\sqrt{2}i$$

Solution

$$ \dfrac{5+\sqrt{2}i}{1-\sqrt{2}i}$$

Rationalizing the denominator, we get

$$ = \dfrac{5+\sqrt{2}i}{1-\sqrt{2}i} \times \dfrac{1+\sqrt{2}i}{1+\sqrt{2}i}$$

$$ = \dfrac{5+2i^2+(5+1)\sqrt{2}i}{1+2}$$

$$ = \dfrac{3+6\sqrt{2}i}{1+2}$$......................$$(\because i^2=-1)$$

$$ = 1+2\sqrt{2}i$$
Question 2
1 / -0

If $$a$$ and $$b$$ are the nonzero distinct roots of $$x^2 + ax + b =0$$, then the minimum vlue of $$x^2 + ax+b$$ is

A
$$\dfrac{2}{3}$$

B
$$\dfrac{9}{4}$$

C
$$\dfrac{-9}{4}$$

D
$$\dfrac{-2}{3}$$

Solution

Given, $$x^2 + ax + b = 0$$
For distinct now zero roots
$$D > 0$$
$$\Rightarrow a^2 - 4b > 0$$
Now, $$x^2 + ax + b$$
$$= \left(x + \dfrac{a}{2}\right)^2 + \left(b-\dfrac{a^2}{4}\right)$$

$$= \left(x + \dfrac{a}{2}\right)^2 - \left(a^2-\dfrac{4b}{4}\right)$$
We know, sum of roots $$a + b = -a$$

$$\Rightarrow 2a + b = 0$$

Product of roots $$a\times b = b$$
$$\Rightarrow b(a-1) = 0$$
$$\Rightarrow a = 1, b \neq 0$$
From Equation (i),
$$2a+b = 0$$$
$$2(1) + b = 0$$
$$b = -2$$
Now, $$\left( x + \dfrac{A}{2}\right)^2 - \left(\dfrac{1^2 + 4 \times 2}{4}\right)$$

$$= \left(x+\dfrac{a}{2}\right)^2 - \dfrac{9}{4}$$

$$\therefore$$ Maximum value $$= -\dfrac{9}{4}$$
Question 3
1 / -0

The real part of $$(i - \sqrt{3})^{13}$$ is

A
$$2^{-10}\sqrt3$$

B
$$-2^{12}\sqrt3$$

C
$$2^{-12}\sqrt3$$

D
$$-2^{-12}\sqrt3$$

E
$$-2^{10}\sqrt3$$

Solution

$$w^2=\dfrac{-1-i\sqrt3}{2}$$
$$\Rightarrow iw^2=\dfrac{-i+\sqrt3}{2}$$
$$\Rightarrow -2iw^2=i-\sqrt3$$
$$\Rightarrow (-2iw^2)^{13}=(i-\sqrt3)^{13}$$
$$\Rightarrow -2^{13}i^{13}w^{26}=(i-\sqrt3)^{13}$$
$$\Rightarrow -2^{13}iw^2=(i-\sqrt3)^{13}$$
$$\Rightarrow -2^{13}\left({\dfrac{-i+\sqrt3}{2}}\right)=(i-\sqrt3)^{13}$$
$$\Rightarrow -2^{12}(-i+\sqrt3)=(i-\sqrt3)^{13}$$
$$\therefore (i-\sqrt3)^{13}=-2^{12}\sqrt3+i2^{12}$$
Question 4
1 / -0

Mark against the correct answer in each of the following .
$$i^{326}=$$?

A
$$1$$

B
$$-1$$

C
$$i$$

D
$$-i$$

Solution

$$i^{326}\\=(i^4)^{81}\times i^2 =(1)^{81}\times (-1)\\=1\times (-1)=-1$$
Question 5
1 / -0

Mark against the correct answer in each of the following .
$$i^{91}=$$?

A
$$1$$

B
$$-1$$

C
$$i$$

D
$$-i$$

Solution

$$i^{91}\\=(1^4)^{22}\times i^3 \\=1\times i^2 \times i\\=1\times (-1)\times i\\=-i$$
Question 6
1 / -0

$$(1-\sqrt{-1})(1+\sqrt{-1})(5-\sqrt{-7})(5+\sqrt{-7})=?$$

A
$$(25+7i)$$

B
$$(32+5i)$$

C
$$(29-3i)$$

D
$$none\ of\ these$$

Solution

Given expression
$$=(1-i)(1+i)(5-\sqrt 7 i)(5+\sqrt 7i)$$
$$=(1-i^2)(25-7i^2)\\=(1+1)(25+7)\\=(2\times 32)=64$$.
Question 7
1 / -0

$$arg (-1+i\sqrt{3})=?$$

A
$$\dfrac{\pi}{3}$$

B
$$\dfrac{2\pi}{3}$$

C
$$\pi$$

D
$$none\ of\ these$$

Solution

$$(-1+i\sqrt{3})\\ =\dfrac{-1}{2}+i\dfrac{\sqrt 3}{2} \\=2\left(\cos \dfrac{2\pi}{3}+i\sin \dfrac{2\pi}{3}\right)\\ \Rightarrow arg (-1+i\sqrt{3})=\dfrac{2\pi}{3}$$
Question 8
1 / -0

For any complex numbers $$z_{1}$$ and $$z_{2}$$ compare List I with with List II and choose the correct answer, using codes given below:
List I List II
$$arg (z_{1},z_{2})$$ $$\dfrac{\pi}{2}$$
$$arg \left(\dfrac{z_{1}}{z_{2}}\right)$$ $$arg (z_{1}-arg (z_{2})$$
$$arg (z)+arg (\bar{z})$$ $$arg (z_{1})+arg (z_{2})$$
$$arg (i)$$ $$2\pi$$

A
$$(i)-(q), (ii)-(r), (iii)-(s), (iv)-(p)$$

B
$$(i)-(r), (ii)-(q), (iii)-(p), (iv)-(s)$$

C
$$(i)-(r), (ii)-(q), (iii)-(s), (iv)-(p)$$

D
$$none\ of\ these$$

Solution
Question 9
1 / -0

$$(2-3i)(-3+4i)=?$$

A
$$(6+17i)$$

B
$$(6-17i)$$

C
$$(-6+17i)$$

D
$$none\ of\ these$$

Solution

$$(2-3i)(-3+4i)\\=(-6+8i+9i-12i^2)\\=(-6+17i+12)\\=(6+17i)$$.
Question 10
1 / -0

$$arg (1+i)=?$$

A
$$\pi$$

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

C
$$\dfrac{\pi}{3}$$

D
$$\dfrac{\pi}{4}$$

Solution

$$(1+i)\\=\sqrt 2 \left(\dfrac {1}{\sqrt2}+\dfrac{i}{\sqrt2}\right)\\=\sqrt{2}\left(\cos \dfrac{\pi}{4}+i\sin \dfrac{\pi}{4}\right)\\ \Rightarrow arg (1+i)=\dfrac{\pi}{4}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

List I	List II
$$arg (z_{1},z_{2})$$	$$\dfrac{\pi}{2}$$
$$arg \left(\dfrac{z_{1}}{z_{2}}\right)$$	$$arg (z_{1}-arg (z_{2})$$
$$arg (z)+arg (\bar{z})$$	$$arg (z_{1})+arg (z_{2})$$
$$arg (i)$$	$$2\pi$$

Complex Numbers and Quadratic Equations Test 36

Result

Complex Numbers and Quadratic Equations Test 36

Score

-

Rank

-

SHARING IS CARING

Question 1

$$ 1-2\sqrt{2}i$$

$$ 1+\sqrt{2}i$$

$$ 1+2\sqrt{2}i$$

$$ 1-\sqrt{2}i$$

Solution

Question 2

$$\dfrac{2}{3}$$

$$\dfrac{9}{4}$$

$$\dfrac{-9}{4}$$

$$\dfrac{-2}{3}$$

Solution

Question 3

$$2^{-10}\sqrt3$$

$$-2^{12}\sqrt3$$

$$2^{-12}\sqrt3$$

$$-2^{-12}\sqrt3$$

$$-2^{10}\sqrt3$$

Solution

Question 4

$$1$$

$$-1$$

$$i$$

$$-i$$

Solution

Question 5

$$1$$

$$-1$$

$$i$$

$$-i$$

Solution

Question 6

$$(25+7i)$$

$$(32+5i)$$

$$(29-3i)$$

$$none\ of\ these$$

Solution

Question 7

$$\dfrac{\pi}{3}$$

$$\dfrac{2\pi}{3}$$

$$\pi$$

$$none\ of\ these$$

Solution

Question 8

$$(i)-(q), (ii)-(r), (iii)-(s), (iv)-(p)$$

$$(i)-(r), (ii)-(q), (iii)-(p), (iv)-(s)$$

$$(i)-(r), (ii)-(q), (iii)-(s), (iv)-(p)$$

$$none\ of\ these$$

Solution

Question 9

$$(6+17i)$$

$$(6-17i)$$

$$(-6+17i)$$

$$none\ of\ these$$

Solution

Question 10

$$\pi$$

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

$$\dfrac{\pi}{3}$$

$$\dfrac{\pi}{4}$$

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.