Number Theory Test 47

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Number Theory Test 47

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following is a pair of twin-prime number ?

A
$$ 19 , 21 $$

B
$$ 43 , 47 $$

C
$$ 59 , 61 $$

D
$$ 73 , 79 $$

Solution

The even numbers between $$ 68 $$ and $$ 90 $$ is $$ 70 , 72 , 74 , 76 , 78 , 80 , 82 , 84 , 86 , 88 , = 10 $$ numbers
Question 2
1 / -0

If $$ \displaystyle z_{0}=\frac{1-i}{2}$$, then $$ \displaystyle \left (1+z_{0} \right )\left (1+z_{0}^{{2}^{1}} \right )\left (1+z_{0}^{{2}^{2}} \right ).......... \left (1+z_{0}^{2^n} \right )$$ must be

A
$$(1-i)(1+\dfrac{1}{2^{2^n}})$$ for $$n>1$$

B
$$(1-i)(1-\dfrac{1}{2^{2^n}})$$ for $$n>1$$

C
$$\dfrac{1+i}{2}$$ for $$n>1$$

D
$$(1-i)(1-\dfrac{1}{2^{2^{n+1}}})$$ for $$n>1$$

Solution

Let $$P=(1+z_0)(1+{z_0}^{2^1})+(1+{z_0}^{2^2})....(1+{z_0}^{2^n})$$

$$\Rightarrow (1-z_0)P=(1-z_0)(1+z_0)(1+{z_0}^{2^1})+(1+{z_0}^{2^2})....(1+{z_0}^{2^n})$$

$$\Rightarrow (1-z_0)P=(1-{z_0}^{2^2})(1+{z_0}^{2^2}).......(1+{z_0}^{2^n})$$

$$\Rightarrow (1-z_0)P=1-{z_0}^{2^{n+1}}$$

$$\Rightarrow P=\dfrac{1-{z_0}^{2^{n+1}}}{1-{z_0}}$$

$$\Rightarrow P=\dfrac{1-({z_0}^2)^{2n}}{1-z_0}$$

$$n>1$$
Since, $$z_0=\dfrac{1-i}{2}$$

$${z_0}^2=\dfrac{1-1-2i}{4}$$

$$\Rightarrow {z_0}^2=\dfrac{-i}{2}$$

$$\therefore \ P=\dfrac{1-\left(\dfrac{-i}{2}\right)^{2^n}}{1-z_0}$$

$$\Rightarrow P=\dfrac{1-\dfrac{1}{2^{2^n}}}{\dfrac{1+i}{2}}=\dfrac{\left(1-\dfrac{1}{2^{2^n}}\right)2(1-i)}{1+1}$$

$$\Rightarrow P=(1-i)\left(1-\dfrac{1}{2^{2^n}}\right)$$
Question 3
1 / -0

The number of solutions of $$log _{\frac{1}{5}}log_{\frac{1}{2}}(\left | z \right |^{2}+4\left | z \right |+3)< 0$$ is/are?

A
$$0$$

B
$$2$$

C
$$4$$

D
infinite

Solution

The glven inequality becomes $$log_{\frac{1}{2}}|\mathrm{z}|^{2}+4|\mathrm{z}|+3>1$$, since $$\displaystyle \frac{1}{5}<1$$
$$\displaystyle \Rightarrow|\mathrm{z}|^{2}+4|\mathrm{z}|+3<\frac{1}{2}$$
$$\Rightarrow(|\mathrm{z}|+2)^{2}$$ + 1 $$<\displaystyle \frac{1}{2}$$
$$(|\displaystyle \mathrm{z}|+2)^{2}<\frac{-1}{2}$$
This is not possible. Hence, there is no solution.
Question 4
1 / -0

Let $$z$$ be a complex number and $$c$$ be a real number $$\geq $$ 1 such that z + $$c\left | z+1 \right |+i=0 ,$$ then $$c$$ belongs to

A
$$[2, 3]$$

B
$$(3, 4)$$

C
$$[1,\sqrt{2}]$$

D
None of these

Solution

Since $$c$$ $$\left | z+1 \right |$$ is real
$$z + i$$ is real
let $$z-i = x $$, where $$x$$ is real
$$z+i=-c$$ $$\left | z+1 \right |$$ (given)
$$x^{2}=c^{2}\left \{ x+1 \right \}^{2}+1^{2})$$
$$(c^{2}-1)x^{2}+2c^{2}x+2c^{2}=0$$
As $$x$$ is real $$4c^{4}-8c^{2}(c^{2}-1)\geq 0$$
$$4c^{2}-(c^{2}-2)\leq 0$$
$$c^{2}\leq 2$$
$$c\leq \sqrt{2}$$
But $$c\geq1$$
$$\Rightarrow c\, \, \epsilon [1,\sqrt{2}]$$
Question 5
1 / -0

If $$x = 2 + 5i($$where $$1 i = \sqrt{-1})$$ and $$2(\displaystyle \frac{1}{1! 9!} + \frac{1}{3! 7!}) + \frac{1}{5! 5!} = \frac{2^{a}}{b!}$$ then $$ x^{3}-5x^{2}+33x-10 = $$

A
$$a + b$$

B
$$b - a$$

C
$$a-b$$

D
$$-a-b$$

E
$$(a - b)(a + b)$$

Solution

$$ \displaystyle \frac{10! 2^{a}}{b!} = 2[^{10}C_{1} + ^{10}C_{3}] + ^{10} C_{5}$$

$$ \displaystyle \frac{10! 2^{a}}{b!} = ^{10}C_{1} + ^{10}C_{3} + ^{10}C_{5} + ^{10}C_7 + ^{10}C_{9} = 2^{9}$$

$$ \therefore a = 9, b = 10$$

$$ x = 2 + 5i$$
$$ (x - 2)^{2} = - 25$$
$$x^{2} - 4x + 29 = 0$$
$$ \therefore x^{3} - 5x^{2} + 33x - 10 = (x - 1)(x^{2} - 4x + 29) + 19$$
$$\Rightarrow 19 = a + b$$
Question 6
1 / -0

lf $$\displaystyle \log_{\tan 30^{\circ}}\left(\frac{2|Z|^{2}+2|Z|-3}{|z|+1}\right) <-2$$ then

A
$$|\displaystyle \mathrm{z}|<\frac{3}{2}$$

B
$$|z|>\displaystyle \frac{3}{2}$$

C
$$|z|>2$$

D
$$|z|<2$$

Solution

we have,
$$\Rightarrow log_{(\dfrac{1}{\sqrt{3}})}(\dfrac{2|z|^{2}+2|z|-3}{|z|+1})<-2$$

$$\Rightarrow \dfrac{2|z|^{2}+2|z|-3}{|z|+1}>(\dfrac{1}{\sqrt{3}})^{-2}=(\sqrt{3})^{2}=3$$ [as$$\dfrac{1}{\sqrt{3}}<1$$ so equality sign changes]

$$\Rightarrow 2|z|^{2}+2|z|-3>3|z|+3$$

$$\Rightarrow 2|z|^{2}-|z|-6>0$$

$$\Rightarrow 2|z|^{2}-4|z|+3|z|-6>0$$

$$\Rightarrow 2|z| (|z|-2)+3 (|z|-2)>0$$

$$\Rightarrow (2|z|+3) (|z|-2)>0$$

$$\Rightarrow |z|>2$$
Question 7
1 / -0

Given $$z$$ is a complex number with modulus $$1$$. Then the equation in $$a$$, $$\left(\dfrac{1+ia}{1-ia}\right )^4=z$$ has

A
all roots real and distinct.

B
two real and two imaginary.

C
three roots real and one imaginary.

D
one root real and three imaginary.

Solution

$$\left (\dfrac{1+ia}{1-ia}\right )^4 = z$$

$$\left (\dfrac{-i(i-a)}{-i(i+a)}\right )^4 = z$$

$$\left (\dfrac{a-i}{a+i}\right )^4= z$$

$$\left |\dfrac{a-i}{a+i}\right |^4= |z| =1$$

$$|a-i| ^4 = |a+i|^4$$

$$|a-i| = |a+i|$$

$$\therefore a$$ lies on the perpendicular bisector of $$i$$ and $$-i$$.
$$a $$ lies on real axis. Hence the roots all are real and distinct.
Question 8
1 / -0

Which of the following is not a composite

number?

A
$$2 \times 3 \times 5\times 13\times 17 + 13$$

B
$$7\times 6\times 5\times 4\times 3\times 2\times 1 + 5$$

C
$$17\times 41\times 43\times 61 + 43$$

D
$$2 \times 3\times 43 + 13$$
Question 9
1 / -0

Let $$x_1, x_2,$$ are the roots of quadratic equation $$x^2 + ax + b = 0$$, Where $$ a, b$$ are complex numbers and $$y_1, y_2$$ are the roots of the quadratic equation $$y^2 + \left|a\right| y + \left|b\right| =0$$. If $$\left|x_1\right| = \left|x_2\right| = 1$$, then

A
$$\left|y_1\right| = \left|y_2\right|= 1$$.

B
$$\left|y_1\right| = \left|y_2\right|\neq 1$$.

C
$$\left|y_1\right| \neq 1, \left|y_2\right|= 1$$.

D
$$\left|y_1\right| =1, \left|y_2\right|\neq 1$$.

Solution

Now
$$x_{1}.x_{2}=b$$
Hence
$$|x_{1}.x_{2}|$$
$$=|x_{1}|.|x_{2}|$$
$$=1.1$$
$$=1$$
$$=|b|$$ ...(i)
And
$$-a=x_{1}+x_{2}$$
$$|a|=|x_{1}+x_{2}|$$
Now applying triangle inequality, we get
$$|x_{1}+x_{2}|\leq |x_{1}|+|x_{2}|$$
Hence
$$|a|\leq |x_{1}|+|x_{2}|$$
Hence
$$|a|\leq 2$$
Considering
$$|a|=2$$ we get
$$y^{2}+2y+1=0$$
$$(y+1)^{2}=0$$
$$y=-1,-1$$
Hence
$$|y_{1}|=|y_{2}|=1$$.
Question 10
1 / -0

Dividing f(z) by $$z-i$$, we obtain the remainder $$i$$ and dividing it by $$z+i$$, we get the remainder $$1+i$$, then remainder upon the division of f(z) by $$z^2+1$$ is

A
$$\displaystyle \frac {1}{2}(z+1)+i$$

B
$$\displaystyle \frac {1}{2}(iz+1)+i$$

C
$$\displaystyle \frac {1}{2}(iz-1)+i$$

D
$$\displaystyle \frac {1}{2}(z+i)+1$$

Solution

$$\textbf{Step -1: Using remainder theorem to get the following equations.}$$
$$\text{Given, Dividing }f(z) \text{ by } z - i \text{, we get remainder }i$$
  $$\text{Therefore by remainder theorem, }f(i) = i$$ $$\dots \text{(i)}$$
  $$\text{Given, Dividing }f(z) \text{ by } z + i \text{, we get remainder }1 + i$$
  $$\text{Therefore by remainder theorem, }f(-i) = 1+i$$ $$\dots \text{(ii)}$$
  $$\text{Let }R(x) \text{ be the remainder when }f(z)\text{is divided by }z^2 + 1$$
  $$\text{degree of R(z) is 1, hence }R(z) = az+b$$
  $$\text{we can simplify }z^2 + 1 = (z -i) (z+i)$$
  $$\therefore f(z) = Q(z)(z-i)(z+i) + R(z)$$
$$\textbf{Step -2: Solving for a and b.}$$
  $$\text{From (i) and (ii)}$$
  $$f(i) = i = ai + b$$
  $$\Rightarrow b = i - ai$$
  $$f(-i) = 1+i = -ai +b$$
  $$\Rightarrow -ai + i - ai = 1+i$$
  $$\Rightarrow -2ai = 1$$
  $$\Rightarrow a = -\dfrac{1}{2i}$$
    $$= -\dfrac{1}{2i} \times \dfrac{i}{i}$$
         $$= -\dfrac{i}{2(-1)}$$
  $$\therefore a = \dfrac{i}{2}$$
  $$ b = i - ai$$
  $$\Rightarrow b = i - \dfrac{i}{2} (i)$$
  $$\Rightarrow b = i - \dfrac{(-1)}{2}$$
  $$\Rightarrow b = i + \dfrac{1}{2}$$
  $$\therefore R(z) = az+b$$
       $$= \dfrac{i}{2}z + i + \dfrac{1}{2}$$
  $$R(z) = \dfrac{1}{2}(iz + 1) + i$$
$$\textbf{Hence option B is correct}$$

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Re-Attempt Weekly Quiz Competition

Number Theory Test 47

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Number Theory Test 47

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SHARING IS CARING

Question 1

$$ 19 , 21 $$

$$ 43 , 47 $$

$$ 59 , 61 $$

$$ 73 , 79 $$

Solution

Question 2

$$(1-i)(1+\dfrac{1}{2^{2^n}})$$ for $$n>1$$

$$(1-i)(1-\dfrac{1}{2^{2^n}})$$ for $$n>1$$

$$\dfrac{1+i}{2}$$ for $$n>1$$

$$(1-i)(1-\dfrac{1}{2^{2^{n+1}}})$$ for $$n>1$$

Solution

Question 3

$$0$$

$$2$$

$$4$$

infinite

Solution

Question 4

$$[2, 3]$$

$$(3, 4)$$

$$[1,\sqrt{2}]$$

None of these

Solution

Question 5

$$a + b$$

$$b - a$$

$$a-b$$

$$-a-b$$

$$(a - b)(a + b)$$

Solution

Question 6

$$|\displaystyle \mathrm{z}|<\frac{3}{2}$$

$$|z|>\displaystyle \frac{3}{2}$$

$$|z|>2$$

$$|z|<2$$

Solution

Question 7

all roots real and distinct.

two real and two imaginary.

three roots real and one imaginary.

one root real and three imaginary.

Solution

Question 8

$$2 \times 3 \times 5\times 13\times 17 + 13$$

$$7\times 6\times 5\times 4\times 3\times 2\times 1 + 5$$

$$17\times 41\times 43\times 61 + 43$$

$$2 \times 3\times 43 + 13$$

Question 9

$$\left|y_1\right| = \left|y_2\right|= 1$$.

$$\left|y_1\right| = \left|y_2\right|\neq 1$$.

$$\left|y_1\right| \neq 1, \left|y_2\right|= 1$$.

$$\left|y_1\right| =1, \left|y_2\right|\neq 1$$.

Solution

Question 10

$$\displaystyle \frac {1}{2}(z+1)+i$$

$$\displaystyle \frac {1}{2}(iz+1)+i$$

$$\displaystyle \frac {1}{2}(iz-1)+i$$

$$\displaystyle \frac {1}{2}(z+i)+1$$

Solution

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