Complex Numbers and Quadratic Equations Test 12

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Complex Numbers and Quadratic Equations Test 12

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$(x - 1) \displaystyle \left ( x + \frac{1}{2} - \frac{\sqrt{3}}{2} i \right )\left ( x + \frac{1}{2} + \frac{\sqrt 3}{2} i \right )$$ is

A
$$x^3 + x^2 + x 1$$

B
$$x^3 -1$$

C
$$x^3 + 1$$

D
$$x^3 - x^2 + x + 1$$

Solution

$$\left( x-1 \right) \left( x+\dfrac { 1 }{ 2 } -\dfrac { \sqrt { 3 } }{ 2 } i \right) \left( x+\dfrac { 1 }{ 2 } +\dfrac { \sqrt { 3 } }{ 2 } i \right) $$
$$=\left( x-1 \right) \left[ \left( x+\dfrac { 1 }{ 2 } \right) -\left( \dfrac { \sqrt { 3 } }{ 2 } i \right) \right] \left[ \left( x+\dfrac { 1 }{ 2 } \right) +\left( \dfrac { \sqrt { 3 } }{ 2 } i \right) \right] $$
$$=\left( x-1 \right) \left[ \left( x+\dfrac { 1 }{ 2 } \right) ^{ 2 }-\left( \dfrac { \sqrt { 3 } }{ 2 } i \right) ^{ 2 } \right] $$
$$=\left( x-1 \right) \left[ { x }^{ 2 }+x+\dfrac { 1 }{ 4 } +\dfrac { 3 }{ 4 } \right] $$
$$=\left( x-1 \right) \left[ { x }^{ 2 }+x+1 \right] $$
$$=x^3+x^2+x-x^2-x-1$$
$$=x^3-1$$
Hence, the answer is $$x^3-1.$$
Question 2
1 / -0

The number of real solutions of the equation $$\displaystyle{\left(\frac{5}{7}\right)^2}=-x^2+2x-3$$ is equal to

A
2

B
zero

C
1

D
none of these

Solution

$$\displaystyle { \left( \frac { 5 }{ 7 }\right)}^{ 2 }=-{ x }^{ 2 }+2x-3$$
$$\displaystyle \Rightarrow { x }^{ 2 }-2x+3-\frac {25 }{ 49 } =0$$
$$\Rightarrow { 49x }^{ 2 }-98x+122=0$$
Hence, $$D={ \left( 98 \right)}^{ 2 }-4\left( 122 \right) \left( 49 \right) <0$$
So, roots are imaginary.
Question 3
1 / -0

Number of integer values of k for which the quadratic equation $$\displaystyle 2x^{2}+kx-4=0$$ will have two rational solutions is

A
$$1$$

B
$$2$$

C
$$4$$

D
$$5$$

Solution

Dividing the equation by the coefficient of $$x^{2}$$ i.e. 2 we get

$$x^{2}+\dfrac{kx}{2}-2=0$$

$$\left(x+\dfrac{k}{4}\right)^{2}-\dfrac{k^{2}}{16}-2=0$$

$$\left(x+\dfrac{k}{4}\right)^{2}=\dfrac{k^{2}+32}{16}$$

Hence for rational roots, $$\dfrac{k^{2}+32}{16}$$ has to be a perfect square.

We get a perfect square at $$k=\pm2$$ for ($$\dfrac{k^{2}+32}{16}$$) i.e. $$\dfrac{36}{16}$$ which becomes $$\dfrac{6}{4}$$ upon removing the square

We get a perfect square at $$k=\pm7$$ for ($$\dfrac{k^{2}+32}{16}$$) i.e. $$\dfrac{81}{16}$$ which becomes $$\dfrac{9}{4}$$ upon removing the square

Hence the number of integral values of k is $$4 (2, -2, 7,-7)$$
Question 4
1 / -0

If $$z_1$$ and $$z_2$$ are any two complex numbers, then $$\displaystyle \frac{z_2 + z_1}{||z_2| - |z_1||}$$ is

A
$$\leq 1$$

B
$$\geq 1$$

C
$$\geq -1$$

D
none of these

Solution
Question 5
1 / -0

Find the value of $$k$$ for which the quadratic equation $$\displaystyle \left ( k-2 \right )x^{2}+2\left ( 2k-3 \right )x+5k-6=0$$ has equal roots

A
1

B
3

C
A and B both

D
none of these

Solution

Since the above equation has equal roots
Hence
$$B^{2}-4AC=0$$
Hence
$$4(2k-3)^{2}-4(5k-6)(k-2)=0$$
$$4k^{2}-12k+9-(5k^{2}-16k+12)=0$$
$$-k^{2}+4k-3=0$$
Or
$$k^{2}-4k+3=0$$
$$(k-3)(k-1)=0$$
Hence
$$k=3$$ and $$k=1$$.
Question 6
1 / -0

The locus represented by $$|z -1|=|z + i|$$ is

A
a circle of radius $$1$$ unit

B
an ellipse with foci at $$(1, 0)$$ and $$(0, 1)$$

C
a straight line through the origin

D
a circle on the line joining $$(1, 0)$$ and $$(0, 1)$$ as diameter

Solution

Given,
$$ |z−1|=|z+i|$$
To find the locus of the equation.
Solution,
Given that,
$$|z−1|=|z+i|$$
Say $$x=x+iy$$
$$ \left| x+iy-1 \right| =\left| x+iy+1 \right| $$
$$ \left| (x-1)+iy \right| =\left| x+i(y+1) \right| $$
$$\Rightarrow \sqrt{(x-1)^2+y^2}=\sqrt{x^2+(y+1)^2}$$
Squaring both sides,
$$x^2+1-2x+y^2=x^2+y^2+1+2y$$
$$\Rightarrow 1-2x=2y+1$$
$$\therefore \ x+y=0$$ [line passes through origin]
Hence, C is the right answer
Question 7
1 / -0

If the roots of the equation $$\displaystyle \left ( a^{2}+b^{2} \right )x^{2}-2b\left ( a+c \right )x+\left ( b^{2}+c^{2} \right )=0 $$ are equal then

A
$$2b = ac$$

B
$$\displaystyle b^{2}=ac $$

C
$$\displaystyle b=\frac{2ac}{a+c} $$

D
b = ac

Solution

Since the roots are equal
$$B^{2}-4AC=0$$
Hence
$$4b^{2}(a+c)^{2}-4(b^{2}+c^{2})(a^{2}+b^{2})=0$$
$$4b^{2}a^{2}+4b^{2}c^{2}+8b^{2}ac-4b^{2}a^{2}-4b^{4}-4c^{2}a^{2}-4c^{2}b^{2}=0$$
$$8b^{2}ac-4b^{4}-4c^{2}a^{2}=0$$
Or
$$(b^{2})^{2}-2b^{2}ac+(ac)^{2}=0$$
$$b^{2}=\dfrac{2ac\pm\sqrt{4ac^{2}-4ac^{2}}}{2}$$
$$b^{2}=ac$$
Question 8
1 / -0

In the argand diagram, if $$O,P$$ and $$Q$$ represents respectively the origin and the complex numbers $$z$$ and $$z+iz$$ then the $$\angle OPQ$$ is

A
$$\displaystyle\frac {\pi}{4}$$

B
$$\displaystyle\frac {\pi}{3}$$

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

D
$$\displaystyle\frac {2\pi}{3}$$

Solution

Let $$\displaystyle z=r\left( \cos { \theta } +i\sin { \theta } \right) ,$$ then
$$\displaystyle z+iz=r\left( \cos { \theta } +i\sin { \theta } \right) +ir\left( \cos { \theta } +i\sin { \theta } \right) $$
$$\displaystyle=r\left[ \left( \cos { \theta } -\sin { \theta } \right) +i\left( \sin { \theta } +\cos { \theta } \right) \right] $$
$$\displaystyle=\sqrt { 2 } r\left[ \cos { \left( \theta +\frac { \pi }{ 4 } \right) } +i\sin { \left( \theta =\frac { \pi }{ 4 } \right) } \right] $$
In $$\displaystyle{ \triangle OPQ } ,$$
$$\displaystyle{ PQ }^{ 2 }={ r }^{ 2 }+{ \left( \sqrt { 2 } r \right) }^{ 2 }-2r\left( \sqrt { 2 } r \right) \cos { \frac { \pi }{ 4 } } $$
$$\displaystyle={ r }^{ 2 }+2{ r }^{ 2 }-2{ r }^{ 2 }={ r }^{ 2 }$$
$$\displaystyle\therefore PQ=r,\quad$$
$$ \therefore \angle OPQ=\dfrac { \pi }{ 2 } $$
Question 9
1 / -0

The value (s) of $$k$$ for which the quadratic equation $$\displaystyle kx^{2}-kx+1=0$$ has equal roots is

A
$$k = 0$$ only

B
$$k = 4$$ only

C
$$k = 0, 4$$

D
$$k = -4$$

Solution

For equal roots
$$B^{2}-4AC=0$$
Hence
$$k^{2}-4k=0$$
$$k(k-4)=0$$
$$k=0$$ or $$k=4$$
Now if $$K=0$$, the above equation will be $$1=0$$ which is not true.
Hence $$k=4$$ is the answer.
Question 10
1 / -0

Let $$z = x + iy$$ & amp $$(e^{z^2})$$ = amp $$(e^{(z+i)})$$. If $$y = (x)$$ is a function, then $$y(3)$$ is equal to

A
$$\displaystyle \frac{1}{2}$$

B
$$\displaystyle \frac{1}{3}$$

C
$$\displaystyle \frac{1}{4}$$

D
$$\displaystyle \frac{1}{5}$$

Solution

$$(e^{z^2} ) = e^{(x^2 - y^2 )+2ixy}=e^{x^2-y^2}.e^{ixy}$$
$$\Rightarrow$$ amp $$(e^{z^2}) = 2xy$$
And amp$$(e^{(z+i)}) =$$ amp$$(e^{x+(y+1)i})=$$ amp$$(e^x.e^{i(y+1)})= (y + 1)$$
$$\therefore 2xy = y + 1 \Rightarrow \displaystyle y = \frac{1}{(2x - 1)}$$
$$\therefore y(3) = \dfrac{1}{5}$$

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Complex Numbers and Quadratic Equations Test 12

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Question 1

$$x^3 + x^2 + x 1$$

$$x^3 -1$$

$$x^3 + 1$$

$$x^3 - x^2 + x + 1$$

Solution

Question 2

2

zero

1

none of these

Solution

Question 3

$$1$$

$$2$$

$$4$$

$$5$$

Solution

Question 4

$$\leq 1$$

$$\geq 1$$

$$\geq -1$$

none of these

Solution

Question 5

1

3

A and B both

none of these

Solution

Question 6

a circle of radius $$1$$ unit

an ellipse with foci at $$(1, 0)$$ and $$(0, 1)$$

a straight line through the origin

a circle on the line joining $$(1, 0)$$ and $$(0, 1)$$ as diameter

Solution

Question 7

$$2b = ac$$

$$\displaystyle b^{2}=ac $$

$$\displaystyle b=\frac{2ac}{a+c} $$

b = ac

Solution

Question 8

$$\displaystyle\frac {\pi}{4}$$

$$\displaystyle\frac {\pi}{3}$$

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

$$\displaystyle\frac {2\pi}{3}$$

Solution

Question 9

$$k = 0$$ only

$$k = 4$$ only

$$k = 0, 4$$

$$k = -4$$

Solution

Question 10

$$\displaystyle \frac{1}{2}$$

$$\displaystyle \frac{1}{3}$$

$$\displaystyle \frac{1}{4}$$

$$\displaystyle \frac{1}{5}$$

Solution

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