Number Theory Test 42

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Number Theory Test 42

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Question 1
1 / -0

If $$\alpha$$ and $$\beta$$ are the roots of $${ 4x }^{ 2 }-16x+c=0,$$ c>0 such that $$1<\alpha <2<\beta <3$$, then the no.of integer values of c is

A
$$17$$

B
$$14$$

C
$$18$$

D
$$15$$

Solution

Given, $$\alpha$$ and $$\beta$$ are roots of $$4x^2-16x+x=0$$
where $$a=4$$
$$b=-16$$
$$x > 0$$
now, $$\alpha\beta =\dfrac{a}{a}$$
$$\Rightarrow \alpha\beta =\dfrac{c}{4}$$
$$\Rightarrow \beta =\dfrac{c}{4\alpha}$$ ………(i)
we have $$2 < \beta < 3$$
$$\Rightarrow 2 < \dfrac{c}{4\alpha} < 3$$ [from $$(1)$$]
$$\Rightarrow 8 < \dfrac{c}{\alpha} < 12$$
$$\Rightarrow 8\alpha < c < 12\alpha$$ ………(ii)
and $$1 < \alpha < 2$$
$$\Rightarrow \alpha > 1\Rightarrow 8\alpha > 8$$ ……….(iii)
and $$\alpha < 2$$
$$\Rightarrow 12\alpha < 24$$ ………..(iv)
Substituting (iii) and (iv) in (ii) we get
$$8 < 8\alpha < c < 12\alpha < 24$$ $$[\alpha =9, 10, 11, 12, 13, 14, 15, 16, 13, 18, 19, 20, 21, 22, 23]$$
$$\Rightarrow 8 < c < 24$$
Therefore number of integer values of c is $$15$$.
Question 2
1 / -0

$$\sqrt { \left( \log _ { 3 } \tan x \right) }$$ is real for:

A
$$n \pi + \pi / 4 \leq x < n \pi + \pi / 2$$

B
$$n \pi < x < n \pi + \pi / 2$$

C
$$n \pi \pm \pi / 4 \leq x < n \pi \pm \pi / 2$$

D
None of these.
Question 3
1 / -0

Let P$$\left( x \right) ={ x }^{ 3 }-6{ x }^{ 2 }+Bx+C$$ has 1+5i as a zero and B,C real number, then value of (B+C) is

A
-70

B
70

C
24

D
138

Solution

Given $$1+5{i}$$ is the root of $$P(x)=x^3-6{x^2}+B{x}+C$$

$$\implies 1-5{i}$$ is also the root of $$P(x)$$

Let the third root be $$h$$

$$\implies $$sum of roots is $$6$$

$$\implies 1+5{i}+1-5{i}+h=6\implies h=4$$

$$B=(1+5{i}+1-5{i})(4)+(1+5{i})(1-5 i)=8+26=34$$

$$C=-(1+5{i})(1-5{i})(4)=-104$$

$$B+C=34-104=-70$$
Question 4
1 / -0

If $$\left| {z - 3 + 2i} \right|\, \le 4$$ then the difference between the greatest value and the least value of $$\left| z \right|$$ is :

A
$$2\sqrt {13} $$

B
$$8$$

C
$$4 + \sqrt {13} $$

D
$$\sqrt {13} $$

Solution

As we know that $$\left| z - {z}_{0} \right| = c$$ represents equation of circle.
Here,
$${z}_{0} = 3 - 2i$$
$$c = 4$$

Maximum value of $$z = c + \left| z \right|$$
Minimum value of $$z = c - \left| z \right|$$

For $$\left| z - 3 + 2i \right| \le 4$$
Maximum value $$= 4 + \sqrt{{3}^{2} + {\left( -2 \right)}^{2}} = 4 + \sqrt{13}$$
Miniimum value $$= 4 - \sqrt{{3}^{2} + {\left( -2 \right)}^{2}} = 4 - \sqrt{13}$$

Difference between maximum and minimum value
$$= \left( 4 + \sqrt{13} \right) - \left( 4 - \sqrt{13} \right) \\ = 4 + \sqrt{13} - 4 + \sqrt{13} \\ = 2 \sqrt{13}$$

Hence, this is the answer.
Question 5
1 / -0

Let $$\left| z _ { i } \right| = i , i = 1,2,3,4$$ and $$\left| 16 z _ { 1 } z _ { 2 } z _ { 3 } + 9 z _ { 1 } z _ { 2 } z _ { 4 } + 4 z _ { 1 } z _ { 3 } z _ { 4 } + z _ { 2 } z _ { 3 } z _ { 4 } \right| = 48 ,$$ then the value of $$\left| \dfrac { 1 }{ \overline { z } _{ { 1 } } } +\dfrac { 4 }{ \overline { z } _{ { 2 } } } +\dfrac { 9 }{ \overline { z } _{ { 3 } } } +\dfrac { 16 }{ \overline { z } _{ { 4 } } } \right| .$$

A
$$1$$

B
$$2$$

C
$$4$$

D
$$8$$

Solution

$$\begin{array}{l} \left| { { z_{ i } } } \right| =i\, \, \, \, i=1,2,3,4 \\ \left| { { z_{ 1 } } } \right| =1 \\ \left| { { z_{ 2 } } } \right| =2 \\ \left| { { z_{ 3 } } } \right| =3 \\ \left| { { z_{ 4 } } } \right| =4 \\ \left| { 16{ z_{ 1 } }{ z_{ 2 } }{ z_{ 3 } }+9{ z_{ 1 } }{ z_{ 2 } }{ z_{ 3 } }+4{ z_{ 1 } }{ z_{ 2 } }{ z_{ 3 } }+{ z_{ 2 } }{ z_{ 3 } }{ z_{ 4 } } } \right| =48 \\ \left| { { z_{ 1 } }{ z_{ 2 } }{ z_{ 3 } }{ z_{ 4 } } } \right| \left| { \overline { { z_{ 4 } } } +\overline { { z_{ 3 } } } +\overline { { z_{ 2 } } } +\overline { { z_{ 1 } } } } \right| =48 \\ 1\times 2\times 3\times 4\times \left| { { z_{ 1 } }+{ z_{ 2 } }+{ z_{ 3 } }+{ z_{ 4 } } } \right| =48 \\ \Rightarrow \left| { { z_{ 1 } }+{ z_{ 2 } }+{ z_{ 3 } }+{ z_{ 4 } } } \right| =2 \\ =\left| { \frac { 1 }{ { { z_{ 1 } } } } +\frac { 4 }{ { { z_{ 2 } } } } +\frac { 9 }{ { { z_{ 3 } } } } +\frac { { 16 } }{ { { z_{ 4 } } } } } \right| .\left| { \frac { { { z_{ 1 } }{ { \overline { z } }_{ 1 } } } }{ { { z_{ 1 } } } } +\frac { { { z_{ 2 } }{ { \overline { z } }_{ 2 } } } }{ { { z_{ 2 } } } } +\frac { { { z_{ 3 } }{ { \overline { z } }_{ 3 } } } }{ { { z_{ 3 } } } } +\frac { { { z_{ 4 } }{ { \overline { z } }_{ 4 } } } }{ { { z_{ 4 } } } } } \right| \\ =\left| { { z_{ 1 } }+{ z_{ 2 } }+{ z_{ 3 } }+{ z_{ 4 } } } \right| =2 \\ Option\, \, B\, \, is\, \, correct\, \, answer. \end{array}$$
Question 6
1 / -0

Find the value of $${x}^{3}+7{x}^{2}-x+16$$, when $$x=1+2i$$

A
$$-11+24i$$

B
$$-17+24i$$

C
$$-17-24i$$

D
$$-1+24i$$

Solution

Given function $$f(x)=x^3+7x^2-x+16$$
We know that $$(a+b)^3=a^2+3ab(a+b)+b^3\\(a+b)^2=a^2+b^2+2ab$$
When $$x=1+2$$
$$f(1+2i)=(1+2i)^2+(1+2i)^2-(1+2i)+16\\=1+8i^3+3\times2i\times(1+2i)+7\times(1+4i+4i^2)-(1+2i)+16\\=1+8i^3+6i+12i^2+7+28i+28i^2-1-2i+16\\=1-8i+6i-12+7+28i-28-1-2+16[\because i^3=-i,i^2=-1]\\=-17+24i$$
Question 7
1 / -0

Let 'z' be a complex number satisfying $$|z-2-i|\le 5,$$ Then |z-14-6i| lies in

A
{8,18}

B
{2,8}

C
{0,2}

D
{3,7}

Solution

Here, $$|z-2-i|\leq 5$$ ………..$$(1)$$
Now, $$|z-14-6i|=|z-2-i-12-5i|=|(z-2-i)-(12+5i)|$$
$$\leq |z-2-i|+|12+5i|$$
$$\leq 5+|12+5i|$$
$$=5+\sqrt{(12)^2+(5)^2}$$
$$=5+\sqrt{144+25}$$
$$=5+\sqrt{169}$$
$$=5+13$$
$$=18$$.
$$\therefore |z-14-6i|\leq 18$$
and $$|z-14-6i|=|z-2-i-12-5i|$$
$$=|-(12+5i)+(z-2-i)|$$
$$\geq ||-(12+5i)|-|z-2-i||$$
$$\geq |12+5i|-5$$ $$[\because |z-2-i|\leq 5$$ $$\Rightarrow -|z-2-i|\geq -5]$$
$$=13-5$$
$$=8$$
$$\therefore |z-14-6i|\geq 8$$
$$\therefore |z-14-6i|$$ lies in $$\{8, 18\}$$.
Question 8
1 / -0

If $$w = \dfrac { z } { z - \dfrac { 1 } { 3 } i }$$ and $$| w | = 1$$ then $$z$$ lies on

A
a circle

B
an ellipse

C
a parabola

D
a straight line

Solution

$$\begin{array}{l} Given \\ \omega =\frac { { 3z } }{ { 3z-i } } \, \, \, \, \, \, \therefore \left| \omega \right| =\frac { { 3\left| z \right| } }{ { \left| { 3z-i } \right| } } \\ \Rightarrow \left| { 3z-i } \right| =3\left| z \right| \\ \Rightarrow \left| { 3\left( x \right) +i\left( { 3y-1 } \right) } \right| =\left| { 3\left( { x+iy } \right) } \right| \, \, \, \, \, \left( { z=x+iy } \right) \\ \Rightarrow { \left( { 3x } \right) ^{ 2 } }+{ \left( { 3y-1 } \right) ^{ 2 } }=9\left( { { x^{ 2 } }+{ y^{ 2 } } } \right) \\ \Rightarrow 6y-1=0 \\ Which\, is\, straight\, line \\ Hence,\, option\, D\, is\, the\, correct\, answer. \end{array}$$
Question 9
1 / -0

The real part of $$\left[ 1 + \cos \left( \dfrac { \pi } { 5 } \right) + i \sin \left( \dfrac { \pi } { 5 } \right) \right] ^ { - 1 }$$ is

A
$$1$$

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

C
$$\dfrac { 1 } { 2 } \cos \left( \dfrac { \pi } { 10 } \right)$$

D
$$\dfrac { 1 } { 2 } \cos \left( \dfrac { \pi } { 5 } \right)$$

Solution

We have,
$$\begin{array}{l} \dfrac { 1 }{ { 1+\cos \left( { \dfrac { \pi }{ 5 } } \right) +i\sin \left( { \dfrac { \pi }{ 5 } } \right) } } \\ =\dfrac { 1 }{ { 2{ { \cos }^{ 2 } }\dfrac { \pi }{ { 10 } } +i2\sin \left( { \dfrac { \pi }{ { 10 } } } \right) \cos \left( { \dfrac { \pi }{ { 10 } } } \right) } } \\ =\dfrac { 1 }{ { 2\cos \dfrac { \pi }{ { 10 } } \left[ { \cos \dfrac { \pi }{ { 10 } } +i\sin \dfrac { \pi }{ { 10 } } } \right] } } \\ =\dfrac { { \cos \dfrac { \pi }{ { 10 } } -i\sin \dfrac { \pi }{ { 10 } } } }{ { 2\cos \dfrac { \pi }{ { 10 } } } } \\ Thus,\, { { Re } }al\, part=\dfrac { 1 }{ 2 } \\ Hence,\, the\, option\, B\, is\, the\, required\, answer. \end{array}$$

Hence, option $$B$$ is the correct answer.
Question 10
1 / -0

$$\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}=$$

A
$$\dfrac{24}{13}+\dfrac{11}{13}i$$

B
$$\dfrac{24}{13}-\dfrac{11}{13}i$$

C
$$\dfrac{10}{13}+\dfrac{24}{13}i$$

D
$$\dfrac{10}{13}-\dfrac{24}{13}i$$

Solution

$$ \frac{1-2i}{2+1} +\frac{4-1}{(3+2i)} $$

$$ \frac{(1-2i)(3+2i)+(4-i)(2+i)}{(2+i)(3+2i)}$$

$$ \frac{3+2i-6i+4+8+4i-2i+1}{6+4i+3i-2}$$

$$ \frac{3-4i+12+2i+1}{6+7i-2}$$

$$ \frac{-2i+16}{7i+4}$$

$$ \frac{16-2i}{4+7i}\times \frac{(4-7i)}{4-7i}$$

$$ \frac{64-112i-8i-14}{16+49}$$

$$ \frac{50-120i}{65}$$

$$ = \frac{10-24i}{13}$$

$$ = \frac{10}{13}-\frac{24i}{13}$$

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Re-Attempt Weekly Quiz Competition

Number Theory Test 42

Result

Number Theory Test 42

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SHARING IS CARING

Question 1

$$17$$

$$14$$

$$18$$

$$15$$

Solution

Question 2

$$n \pi + \pi / 4 \leq x < n \pi + \pi / 2$$

$$n \pi < x < n \pi + \pi / 2$$

$$n \pi \pm \pi / 4 \leq x < n \pi \pm \pi / 2$$

None of these.

Question 3

-70

70

24

138

Solution

Question 4

$$2\sqrt {13} $$

$$8$$

$$4 + \sqrt {13} $$

$$\sqrt {13} $$

Solution

Question 5

$$1$$

$$2$$

$$4$$

$$8$$

Solution

Question 6

$$-11+24i$$

$$-17+24i$$

$$-17-24i$$

$$-1+24i$$

Solution

Question 7

{8,18}

{2,8}

{0,2}

{3,7}

Solution

Question 8

a circle

an ellipse

a parabola

a straight line

Solution

Question 9

$$1$$

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

$$\dfrac { 1 } { 2 } \cos \left( \dfrac { \pi } { 10 } \right)$$

$$\dfrac { 1 } { 2 } \cos \left( \dfrac { \pi } { 5 } \right)$$

Solution

Question 10

$$\dfrac{24}{13}+\dfrac{11}{13}i$$

$$\dfrac{24}{13}-\dfrac{11}{13}i$$

$$\dfrac{10}{13}+\dfrac{24}{13}i$$

$$\dfrac{10}{13}-\dfrac{24}{13}i$$

Solution

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