Complex Numbers and Quadratic Equations Test 37

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Complex Numbers and Quadratic Equations Test 37

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Weekly Quiz Competition

Question 1
1 / -0

Mark against the correct answer in each of the following .
$$i^{273}=$$?

A
$$i$$

B
$$-i$$

C
$$1$$

D
$$-1$$

Solution

$$i^{273}=(i^4)^{68}\times i=(1)^{68}\times i=(1\times i)=i$$
Question 2
1 / -0

$$arg \left(\dfrac{2+6\sqrt{3}i}{5+\sqrt{3}i}\right)=?$$

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

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

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

D
$$\pi$$

Solution

$$\dfrac{(2+6\sqrt{3}i)}{(5+\sqrt{3}i)}\\=\dfrac{(2+6\sqrt{3}i)}{(5+\sqrt{3}i)}\times \dfrac{(5-\sqrt{3}i)}{(5-\sqrt{3}i)}\\=\dfrac{(2+6\sqrt{3}i)(5-\sqrt{3}i)}{(5-\sqrt{3}i)(5-\sqrt{3}i)}$$
$$=\dfrac{(28+28\sqrt{3}i)}{28}\\=\dfrac{28(1+\sqrt{3}i)}{28}=\dfrac{28(1+\sqrt{3}i)}{28}=(1+\sqrt{3}i)=2\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)$$
$$\therefore arg\left(\dfrac{2+6\sqrt{3}i}{5+\sqrt{3}i}\right)=\dfrac{\pi}{3}$$
Question 3
1 / -0

Compare List I with List II and choose the correct answer using codes given below:
List I (Complex number) List II (Its modulus)
$$(4-3i)$$ $$10$$
$$(8+6i)$$ $$\dfrac{1}{5}$$
$$\dfrac{1}{(3+4i)}$$ $$1$$
$$\dfrac{(3-4i)}{(3+4i)}$$ $$5$$

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

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

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

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

Solution
Question 4
1 / -0

Let $$z, w$$ be complex numbers such that $$\bar z + i\bar w =0$$ and arg $$zw = \pi$$. then $$arg \ z$$ equals

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

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

C
$$\dfrac{3\pi}{4}$$

D
$$\dfrac{5\pi}{4}$$

Solution

$$\overline{z} + i\overline{w} = 0$$

$$\Rightarrow z - iw = 0$$

$$\Rightarrow z = iw$$

$$arg \space zw = \pi$$

$$\Rightarrow arg\space z + arg \space w = \pi$$

$$\Rightarrow arg \space z + arg \space \dfrac{z}{i} = \pi$$ [Using (1)]

$$\Rightarrow arg\space z + arg\space z - arg \space i = \pi$$

$$\Rightarrow 2 arg\space z - \dfrac{\pi}{2} = \pi$$

$$\Rightarrow 2 arg \space z = \dfrac{3\pi}{2}$$

$$\Rightarrow arg \space z = \dfrac{3\pi}{4}$$
Question 5
1 / -0

The principal argument of the complex number
$$[(1 + i)^5 (1 + \sqrt{3}i)^2] / [-2i(-\sqrt{3} +i)]$$ is

A
$$\frac{19\pi}{12}$$

B
$$-\frac{17\pi}{12}$$

C
$$-\frac{5\pi}{12}$$

D
$$\frac{5\pi}{12}$$

Solution
Question 6
1 / -0

Let $$a, b $$ and $$ c $$ be real numbers such that $$ 4 a+2 b+c=0 $$ and $$ a b>0 . $$ Then the equation $$ a x^{2}+b x+c=0 $$ has

A
complex root

B
exactly one root

C
real root

D
none of these

Solution

$$ a x^{2}+b x+c=0 $$
$$a, b $$ and $$ c $$ be real numbers and $$4a+2b+c=0$$
Substituting $$x=2$$ in the quadratic equation
we have $$4a+2b+c$$ which is zero according to the given condition
So $$one$$ $$root$$ $$is$$ $$x=2$$ $$which$$ $$is$$ $$real$$
$$So \space other \space root \space is \space also \space real$$
Hence the equation has $$ two$$ $$real$$ $$roots$$.
Question 7
1 / -0

If $$ b_{1} b_{2}=2\left(c_{1}+c_{2}\right), $$ then at least one of the equations $$ x^{2}+b_{1} x $$ $$ +c_{1}=0 $$ and $$ x^{2}+b_{2} x+c_{2}=0 $$ has

A
imaginary roots

B
real roots

C
purely imaginary roots

D
none of these

Solution

Let $$ D_{1}$$ and $$ D_{2} $$ be discriminants of $$ x^{2}+b_{1} x+c_{1}=0 $$ and $$ x^{2}+b_{2} x $$ $$ +c_{2}=0, $$ respectively.
Then,
$$D_{1}+D_{2} =b_{1}^{2}-4 c_{1}+b_{2}^{2}-4 c_{2} $$
$$=\left(b_{1}^{2}+b_{2}^{2}\right)-4\left(c_{1}+c_{2}\right) $$
$$=b_{1}^{2}+b_{2}^{2}-2 b_{1} b_{2} \quad\left[\because b_{1} b_{2}=2\left(c_{1}+c_{2}\right)\right]$$
$$=\left(b_{1}-b_{2}\right)^{2} \geq 0$$
$$ \Rightarrow \quad D_{1} \geq 0 $$ or $$ D_{2} \geq 0 $$ or $$ D, $$ and $$ D_{2} $$ both are positive
Hence, at least one of the equations has real roots.
Question 8
1 / -0

The modulus and amplitude of the complex number $$\left[e^{{3}-i\dfrac{\pi}{4}}\right]^{3}$$ are respectively.

A
$$e^{6},\dfrac{-3\pi}{4}$$

B
$$e^{9},\dfrac{\pi}{2}$$

C
$$e^{9},\dfrac{-3\pi}{4}$$

D
$$e^{9},\dfrac{-\pi}{2}$$

Solution

Let $$z=\left[e^{3-i\dfrac{\pi}{4}}\right]^3$$

$$\implies z=\left[e^3\times e^{-i\dfrac{\pi}{4}}\right]^3$$

$$\implies z=[e^3]^3\times \left[e^{-i\dfrac{\pi}{4}}\right]^3$$

$$\implies z=e^9\times e^{-i\dfrac{3\pi}{4}}$$

This is in the form of $$z=|z|e^{i \text{arg} (z)}$$ where $$|z|$$ is the modulus and $$arg(z)$$ is the argument of $$z$$

$$\implies |z|=e^9,\text{arg}(z)=-\dfrac{3\pi}{4}$$

$$\implies $$ modulus is $$e^9$$ and argument is $$-\dfrac{3\pi}{4}$$
Question 9
1 / -0

If roots of quadratic equation $$x^2-kx+4=0$$ then $$k$$ will be

A
$$2$$

B
$$1$$

C
$$4$$

D
$$3$$

Solution

Given equation $$x^2-kx+4=0$$
Comparing it with $$ax^2+bx+c=0$$
We get $$a=1, b=-k$$ and $$c=4$$
$$D=b^2-4ac=(-k)^2-4\times 1\times 4=k^2-16$$
Roots are same
$$D=0$$
$$\Rightarrow k^2-16=0$$
$$\Rightarrow k^2=16$$
$$\Rightarrow k=\pm \sqrt{16}$$
$$\Rightarrow k=\pm 4$$
Hence, option (C) is correct.
Question 10
1 / -0

Nature of roots of quadratic equation $$4x^2-12x-9=0$$ is:

A
Real and same

B
Real and distinct

C
Imaginary and same

D
Imaginary and distinct

Solution

Given equations $$4x^2-12x-9=0$$
where $$a=4, b=-12, c=-9$$
Discriminant $$(D)=b^2-4ac$$
$$=(-12)^2-4\times 4\times (-9)$$
$$=144+144$$
$$=288 > 0$$
Hence, option (B) is correct.

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Re-Attempt Weekly Quiz Competition

List I (Complex number)	List II (Its modulus)
$$(4-3i)$$	$$10$$
$$(8+6i)$$	$$\dfrac{1}{5}$$
$$\dfrac{1}{(3+4i)}$$	$$1$$
$$\dfrac{(3-4i)}{(3+4i)}$$	$$5$$

Complex Numbers and Quadratic Equations Test 37

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Complex Numbers and Quadratic Equations Test 37

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Question 1

$$i$$

$$-i$$

$$1$$

$$-1$$

Solution

Question 2

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

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

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

$$\pi$$

Solution

Question 3

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

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

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

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

Solution

Question 4

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

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

$$\dfrac{3\pi}{4}$$

$$\dfrac{5\pi}{4}$$

Solution

Question 5

$$\frac{19\pi}{12}$$

$$-\frac{17\pi}{12}$$

$$-\frac{5\pi}{12}$$

$$\frac{5\pi}{12}$$

Solution

Question 6

complex root

exactly one root

real root

none of these

Solution

Question 7

imaginary roots

real roots

purely imaginary roots

none of these

Solution

Question 8

$$e^{6},\dfrac{-3\pi}{4}$$

$$e^{9},\dfrac{\pi}{2}$$

$$e^{9},\dfrac{-3\pi}{4}$$

$$e^{9},\dfrac{-\pi}{2}$$

Solution

Question 9

$$2$$

$$1$$

$$4$$

$$3$$

Solution

Question 10

Real and same

Real and distinct

Imaginary and same

Imaginary and distinct

Solution

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