Complex Numbers and Quadratic Equations Test 38

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Complex Numbers and Quadratic Equations Test 38

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Weekly Quiz Competition

Question 1
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 2
1 / -0

If $$ \alpha, \beta $$ are the roots of equation $$(k + 1) x^{2} - (20k + 14)x + 91k + 40 = 0 ; (\alpha < \beta)$$, $$k > 0$$, then
The nature of the roots of this equation is

A
imaginary

B
real and distinct

C
equal real roots

D
None of these

Solution

Given equation is

$$(k + 1) x^{2} - (20k + 14)x + 91k + 40 = 0$$

$$D=(20k+14)^{2}-4(k+1)(91k+40)=discriminant$$

$$\Rightarrow D=36(k^{2}+k+1)\ge0$$..........$$(\because k^2 \ and \ k \ are \ always \ positive \ for \ all \ k>0)$$

$$\because D>0$$

$$\therefore \ the \ equation \ has \ real \ roots$$

Hence, option $$B$$ is correct.
Question 3
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 4
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 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

If $$b_1b_2=2(c_1+c_2)$$, then at least one of the equations $$x^2+b_1x+c_1=0$$ and $$x^2+b_2x+c_2=0$$ has

A
imaginary roots

B
real roots

C
purely imaginary roots

D
none of these

Solution

Suppose the equations $$x^2+b_1x+c_1=0$$ & $$x^2+b_2x+c_2=0$$ have real roots.
Then $${ b_{ 1 } }^{ 2 }\ge 4c_{ 1 }$$ ...(1)
and $${ b_{ 2 } }^{ 2 }\ge 4c_{ 2 }$$ ...(2)
given that $$b_{ 1 }b_{ 2 }=2(c_{ 1 }+c_{ 2 })$$
On squaring $${ b_{ 1 } }^{ 2 }{ b_{ 2 } }^{ 2 }=4\left( { c_{ 1 } }^{ 2 }+{ c_{ 2 } }^{ 2 }+2c_{ 1 }c_{ 2 } \right) =4\left[ { \left( c_{ 1 }-c_{ 2 } \right) }^{ 2 }+4c_{ 1 }c_{ 2 } \right] $$
$$\Rightarrow { b_{ 1 } }^{ 2 }{ b_{ 2 } }^{ 2 }-16c_{ 1 }c_{ 2 }=4{ \left( c_{ 1 }-c_{ 2 } \right) }^{ 2 }\ge 0$$
Multiplying (1) & (2), we get
$${ b_{ 1 } }^{ 2 }{ b_{ 2 } }^{ 2 }\ge 16c_{ 1 }c_{ 2 }$$
Therefore, at least one equation have real roots.

Ans: B
Question 7
1 / -0

If $$x=9^{\frac {1}{3}} 9^{\frac {1}{9}} 9^{\frac {1}{27}} .....\infty, y=4^{\frac {1}{3}} 4^{\frac {-1}{9}} 4^{\frac {1}{27}}....\infty,$$ and $$z=\sum_{r=1}^{\infty} (1+i)^{-r}$$, then $$arg (x+yz)$$ is equal to

A
0

B
$$\tan^{-1}(\frac {\sqrt 2}{3})$$

C
$$-\tan^{-1}(\frac {\sqrt 2}{3})$$

D
$$- \tan^{-1}(\frac {2}{\sqrt 3})$$

Solution

$$x=9^{\dfrac{1}{3}+\dfrac{1}{3^{2}}+\dfrac{1}{3^{3}}...\infty}$$
$$x=9^{\dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}}$$
$$x=9^{\dfrac{1}{2}}$$
$$x=3$$
$$y=4^{\dfrac{1}{3}-\dfrac{1}{3^{2}}+\dfrac{1}{3^{3}}...\infty}$$
$$y=4^{\dfrac{\dfrac{1}{3}}{1+\dfrac{1}{3}}}$$
$$y=4^{\dfrac{1}{4}}$$
$$y=2^{\dfrac{1}{2}}$$
$$y=\sqrt{2}$$
$$z=\dfrac{1}{(1+i)^{1}}+\dfrac{1}{(1+i)^{2}}+\dfrac{1}{(1+i)^{3}}...\infty$$
Since it is a G.P with a common ratio of $$\dfrac{1}{(1+i)}$$ we get the sum as
$$=\dfrac{\dfrac{1}{1+i}}{1-\dfrac{1}{1+i}}$$
$$=\dfrac{1}{i}$$
$$=-i$$.
Hence $$x+yz =3-\sqrt{2}i$$
$$\tan\theta=\dfrac{-\sqrt{2}}{3}$$
$$\theta=\tan^{-1}(\dfrac{-\sqrt{2}}{3})$$
$$=-\tan^{-1}(\dfrac{\sqrt{2}}{3})$$
Question 8
1 / -0

The argument of the complex number $$\sin \frac{6\pi }{5}+i\left ( 1+\cos \frac{6\pi }{5} \right )$$ is

A
$$\displaystyle \frac{6\pi }{5}$$

B
$$\displaystyle\frac{5\pi }{6}$$

C
$$\displaystyle\frac{9\pi }{10}$$

D
$$\displaystyle\frac{2\pi }{5}$$

Solution

$$\sin\dfrac{6\pi}{5}+i(1+\cos\dfrac{6\pi}{5})$$
$$=2\sin\dfrac{3\pi}{5} \cos\dfrac{3\pi}{5} +2i\cos^2\dfrac{3\pi}{5}$$
$$=-2\cos\dfrac{3\pi}{5}(-\sin\dfrac{3\pi}{5}-i\cos\dfrac{3\pi}{5})$$
$$=-2\cos\dfrac{3\pi}{5}(\cos\dfrac{9\pi}{10}+i\sin\dfrac{9\pi}{10})$$
So argument is $$\dfrac{9\pi}{10}$$
Question 9
1 / -0

Find the value of $$x$$ such that $$\displaystyle \frac{(x + \alpha)^2 - (x + \beta)^2}{ \alpha + \beta} = \frac{sin 2 \theta}{sin^2 \theta}$$. when $$\alpha$$ and $$\beta $$ are the roots of $$t^2 - 2t + 2 = 0$$

A
$$x = icot \, \, \, \theta - 1$$

B
$$x = -(icot \, \, \, \theta + 1$$)

C
$$x = icot \, \, \, \theta $$

D
$$x = itan \, \, \, \theta - 1$$

Solution

$$t^{2}-2t+2=0$$
$$(t-1)^{2}-1+2=0$$
$$(t-1)^{2}=-1$$
$$t=1\pm i$$
Hence
$$\alpha+\beta=2$$
$$\alpha-\beta=2i$$
Now let $$n=2$$
We get
$$\dfrac{2x(\alpha-\beta)+\alpha^{2}-\beta^{2}}{2}=\dfrac{sin2\theta}{sin^{2}\theta}$$
Hence
$$\dfrac{4ix+2i-(-2i)}{2}=\dfrac{2sin\theta.cos\theta}{sin^{2}\theta}$$

$$2ix+2i=2cot\theta$$
$$ix=cot\theta-i$$
$$-x=icot\theta+1$$
$$x=-(1+icot\theta)$$
Question 10
1 / -0

The complex numbers $$\sin x + i \cos 2x$$ and $$\cos x - i \sin 2x$$ are conjugate to each other, for

A
$$x = n\pi$$

B
$$x \displaystyle = \left ( n + \frac{1}{2} \right ) \pi$$

C
$$x = 0$$

D
No value of $$x$$

Solution

We have,
$$\sin x = \cos x $$ - (i)
and $$\cos 2x = \sin 2x $$ -(ii)
Hence,
$$ \cos^2x - \sin^2x = 2 \sin x \cos x$$
$$\Rightarrow 0 = 2 \sin^2 x $$
$$\Rightarrow \sin x =0 $$
However, $$\sin x \neq \cos x$$
Hence, there is no value of $$x$$ satisfying the given condition.

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Complex Numbers and Quadratic Equations Test 38

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SHARING IS CARING

Question 1

$$0$$

$$2$$

$$4$$

infinite

Solution

Question 2

imaginary

real and distinct

equal real roots

None of these

Solution

Question 3

$$[2, 3]$$

$$(3, 4)$$

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

None of these

Solution

Question 4

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

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

$$|z|>2$$

$$|z|<2$$

Solution

Question 5

$$a + b$$

$$b - a$$

$$a-b$$

$$-a-b$$

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

Solution

Question 6

imaginary roots

real roots

purely imaginary roots

none of these

Solution

Question 7

0

$$\tan^{-1}(\frac {\sqrt 2}{3})$$

$$-\tan^{-1}(\frac {\sqrt 2}{3})$$

$$- \tan^{-1}(\frac {2}{\sqrt 3})$$

Solution

Question 8

$$\displaystyle \frac{6\pi }{5}$$

$$\displaystyle\frac{5\pi }{6}$$

$$\displaystyle\frac{9\pi }{10}$$

$$\displaystyle\frac{2\pi }{5}$$

Solution

Question 9

$$x = icot \, \, \, \theta - 1$$

$$x = -(icot \, \, \, \theta + 1$$)

$$x = icot \, \, \, \theta $$

$$x = itan \, \, \, \theta - 1$$

Solution

Question 10

$$x = n\pi$$

$$x \displaystyle = \left ( n + \frac{1}{2} \right ) \pi$$

$$x = 0$$

No value of $$x$$

Solution

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