Number Theory Test 15

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Number Theory Test 15

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Weekly Quiz Competition

Question 1
1 / -0

$$\left| \dfrac { (3+i)(2-i) }{ 1+i } \right|=$$

A
$$\sqrt{5}$$

B
$$5\sqrt{2}$$

C
$$\sqrt{10}$$

D
$$5$$

Solution

$$= | \dfrac{(3+i)(2-i)}{1+i}|$$

$$=|\dfrac{6-3i+2i+1}{1+i}|$$

$$=|\dfrac{7-i}{1+i}|$$

$$=|\dfrac{(7-i)(1-i)}{2}|$$

$$=|\dfrac{7-7i-i-1}{2}|$$

$$=|\dfrac{6-8i}{2}|$$

$$=|3-4i|$$

$$=\sqrt{3^{2}+4^{2}}$$

$$=\sqrt{25}$$
$$=5$$
Question 2
1 / -0

If $$(a_1+ib_1)(a_2+ib_2)...(a_n+ib_n)=A+iB$$ then $$(a_1^2+b_1^2)(a_2^2+b_2^2)...(a_n^2+b_n^2)=$$

A
$$A^2-B^2$$

B
$$A^2+B^2$$

C
$$A-B$$

D
$$A+B$$

Solution

$$ (a_1+ib_1) (a_2+ib_2) ...(a_n+i b_n)=A+iB$$
Taking modulus & then squaring,
$$\left | (a_{1}+ib_{1})(a_{2}+ib_{2})-(a_{n}+ib_{n}) \right |^{2}=\left | A+iB \right |^{2}$$

$$\Rightarrow |a_{1}+ib_{1}|^{2}|a_{2}+ib_{2}|^{2}...|a_{n}+ib_{n}|^{2}=A^{2}+B^{2}$$
$$\Rightarrow (a_{1}^{2}+b_{1}^{2})(a^{2}_{2}+b^{2}_{2})...a^{2}_{n}+b^{2}_{n}=A^{2}+B^{2}$$
Question 3
1 / -0

The simplified value of $$\displaystyle \frac{1-i}{1+i}$$ is:

A
$$i$$

B
$$-i$$

C
$$1$$

D
$$-2i$$

Solution

Let $$Z= \dfrac{1-i}{1+i}$$

$$=\dfrac{(1-i)(1-i)}{(1+i)(1-i)}$$

  $$=\dfrac{(1-i)^{2}}{1^{2}-(i)^{2}} \quad \dots (a^2-b^2=(a+b)(a-b))$$

  $$=\dfrac{(1-i)^{2}}{1+1} \quad \dots (i^2=-1)$$

  $$=\dfrac{1-1-2i}{2}$$

  $$=\dfrac{-2i}{2}$$

$$\therefore \dfrac{1-i}{1+i}$$$$=-i$$
Question 4
1 / -0

If $$\left ( 5+3i \right )(x+iy)=3-4i$$ then $$34x = $$

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$(5+3i)(x+iy)=3-4i$$
L.H.S. $$=(5x-3y)+(5y+3x)i=3-4i$$
$$5x-3y=3$$
$$5y+3x=-4$$
Solving the equations simultaneously,
$$x=\dfrac{3}{34} ; y=\dfrac{-29}{34}$$
$$\therefore 34x=34\times \dfrac{3}{34}$$
$$=3$$
Question 5
1 / -0

The value of $$1+(1+i)+(1+i)^2+(1+i)^3=$$

A
$$0$$

B
$$5i$$

C
$$4i$$

D
$$3i$$

Solution

The given series is a G.P with a common ratio of $$(1+i)$$.
Hence
$$1+(1+i)+(1+i)^{2}+(1+i)^{3}$$

$$=\dfrac{1-(1+i)^{4}}{1-(1+i)}$$

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

$$=\sqrt{2}(e^{i\frac{\pi}{4}})$$
Hence
$$\dfrac{1-(1+i)^{4}}{-i}$$

$$=\dfrac{1-(\sqrt{2}(e^{i\frac{\pi}{4}}))^{4}}{-i}$$

$$=\dfrac{1-4(e^{i\pi})}{-i}$$

$$=\dfrac{1-4(-1)}{-i}$$

$$=\dfrac{5}{-i}$$

$$=5i$$
Question 6
1 / -0

If $$z=-1+3i$$ then $$z^2+2z+10=$$

A
$$0$$

B
$$1$$

C
$$–1$$

D
$$2$$

Solution

$$z^{2}+2z+10$$

$$=(z+1)^{2}+10-1$$

$$=(z+1)^{2}+9$$

$$=(-1+3i+1)^{2}+9$$

$$=(3i)^{2}+9$$

$$=-9+9$$

$$=0$$

Hence, $$z^{2}+2z+10=0$$ where $$z=-1+3i$$.
Question 7
1 / -0

If $$\alpha $$ and $$\beta $$ are real then $$\left| \dfrac { \alpha +i\beta }{ \beta +i\alpha } \right|=$$

A
Lies betwen 0 and 1

B
= 1

C
>1

D
2

Solution

We know that

$$|\dfrac{z_{1}}{z_{2}}|$$ $$=\dfrac{|z_{1}|}{|z_{2}|}$$
Hence

$$|\dfrac{\alpha+i\beta}{\beta+i\alpha}|$$

$$=\dfrac{|\alpha+i\beta|}{|\beta+i\alpha|}$$

$$=\dfrac{\sqrt{\alpha^{2}+\beta^{2}}}{\sqrt{\beta^{2}+\alpha^{2}}}$$

$$=1$$.
Question 8
1 / -0

If $$(x+iy)(2-3i)=4+i\left ( \dfrac{1}{2} \right )$$ then $$x + y =$$

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

B
$$\dfrac{1}{2}$$

C
$$0$$

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

Solution

$$(x+iy)(2-3i)=4+i\dfrac{1}{2}$$

$$2x+2iy-3ix+3y=4+\dfrac{i}{2}$$

$$2x+3y+i(2y-3x)=4+\dfrac{i}{2}$$
Hence
$$2x+3y=4$$ ................................(i)
And
$$2y-3x=\dfrac{1}{2}$$ .............................(ii)

Solving both the equations give us
$$y=1$$

$$x=\dfrac{1}{2}$$

Hence

$$x+y$$

$$=1+\dfrac{1}{2}$$

$$=\dfrac{3}{2}$$.
Question 9
1 / -0

The minimum value of $$|\mathrm{z}|+|\mathrm{z}-1|+|\mathrm{z}-2|$$ is

A
$$0$$

B
$$1$$

C
$$2$$

D
$$4$$

Solution

Let $$z = x+iy $$
$$f(z) = |z| + |z-1| +|z-2| $$
$$f(x,y) = \sqrt{x^2 +y^2} +\sqrt { (x-1)^2+ y^2} +\sqrt {(x-2)^2 +y^2} $$
For minimum value, $$y= 0 $$
$$f(x) = |x| + |x-1| +|x-2| $$ for all $$x$$
The function can be expressed as follows
$$ f(x) = 3x -3 $$, $$x \geq 2 $$
$$f(x) = x+1, $$ $$1\leq x<2 $$
$$f(x) = 3- x $$, $$0\leq x\leq 1$$
$$f(x) = -3x +3 $$, $$ x <0 $$
Hence, $$f(min) = f(1) = 2 $$
Question 10
1 / -0

If $$m_1$$, $$m_2$$, $$m_3$$ and $$m_4$$ respectively denote the moduli of the complex numbers $$1 + 4i, 3 + i, 1 – i \ and\ 2 – 3i$$ then the correct order among the following is :

A
$$m_1$$<$$m_2$$<$$m_3$$<$$m_4$$

B
$$m_4$$<$$m_3$$<$$m_2$$<$$m_1$$

C
$$m_3$$<$$m_2$$<$$m_4$$<$$m_1$$

D
$$m_3$$<$$m_1$$<$$m_2$$<$$m_4$$

Solution

$$m_{1}=|1+4i|=\sqrt{1+16}=\sqrt{17}$$.
$$m_{2}=|3+i|=\sqrt{9+1}=\sqrt{10}$$
$$m_{3}=|1-i|=\sqrt{1+1}=\sqrt{2}$$
$$m_{4}=|2-3i|=\sqrt{4+9}=\sqrt{13}$$
Hence
$$m_{1}>m_{4}>m_{2}>m_{3}$$

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Re-Attempt Weekly Quiz Competition

Number Theory Test 15

Result

Number Theory Test 15

Score

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Rank

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SHARING IS CARING

Question 1

$$\sqrt{5}$$

$$5\sqrt{2}$$

$$\sqrt{10}$$

$$5$$

Solution

Question 2

$$A^2-B^2$$

$$A^2+B^2$$

$$A-B$$

$$A+B$$

Solution

Question 3

$$i$$

$$-i$$

$$1$$

$$-2i$$

Solution

Question 4

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 5

$$0$$

$$5i$$

$$4i$$

$$3i$$

Solution

Question 6

$$0$$

$$1$$

$$–1$$

$$2$$

Solution

Question 7

Lies betwen 0 and 1

= 1

>1

2

Solution

Question 8

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

$$\dfrac{1}{2}$$

$$0$$

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

Solution

Question 9

$$0$$

$$1$$

$$2$$

$$4$$

Solution

Question 10

$$m_1$$<$$m_2$$<$$m_3$$<$$m_4$$

$$m_4$$<$$m_3$$<$$m_2$$<$$m_1$$

$$m_3$$<$$m_2$$<$$m_4$$<$$m_1$$

$$m_3$$<$$m_1$$<$$m_2$$<$$m_4$$

Solution

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