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CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 1 - MCQExams.com
CBSE
Class 11 Engineering Maths
Complex Numbers And Quadratic Equations
Quiz 1
Express $$\dfrac{1}{(1 - cos \theta + 2 i sin \theta)}$$ in the form $$x + iy$$
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$$\left(\displaystyle \frac{1}{5 + 3 cos \theta}\right) + \left(\displaystyle \frac{2 cot \theta/2}{5 + 3 cos \theta}\right)i$$
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$$\left(\displaystyle \frac{1}{5 - 3 cos \theta}\right) + \left(\displaystyle \frac{-2 cot \theta/2}{5 - 3 cos \theta}\right)i$$
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$$\left(\displaystyle \frac{1}{5 + 3 cos \theta}\right) + \left(\displaystyle \frac{-2 cot \theta/2}{5 + 3 cos \theta}\right)i$$
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$$\left(\displaystyle \frac{1}{5 - 3 cos \theta}\right) + \left(\displaystyle \frac{2 cot \theta/2}{5 - 3 cos \theta}\right)i$$
Explanation
To Express $$ \dfrac{1}{(1-cos\theta +2isin\theta )}$$ in the form $$x+iy$$
We have , $$ \dfrac{1}{(1-cos\theta +2isin\theta )}$$
$$ = \dfrac{1}{(1-cos\theta) +2isin\theta } \times \dfrac{(1-cos\theta) - 2isin\theta }{(1-cos\theta ) - 2isin\theta }$$
$$ = \dfrac{(1-cos\theta -2isin\theta )}{(1-cos\theta )^{2}+(2sin\theta )^{2}}$$
$$ = \dfrac{(1-cos\theta -2isin\theta )}{1 + cos^{2}\theta -2cos\theta +4sin^{2}\theta }$$
$$ = \dfrac{(1-cos\theta )}{(1-cos\theta )( 3cos\theta + 5)} - \dfrac{2isin\theta }{(1-cos\theta )( 3cos\theta + 5)} $$
$$ = \dfrac{1}{( 3cos\theta + 5)} - \dfrac{2isin\dfrac{\theta }{2}cos\dfrac{\theta }{2} }{2sin^{2}\dfrac{\theta }{2}( 3cos\theta + 5)}$$
$$ = \dfrac{1}{( 3cos\theta + 5)} - \dfrac{2cot\dfrac{\theta }{2} }{( 3cos\theta + 5)}i$$
$$ = \dfrac{1}{( 5 + 3cos\theta )} + \dfrac{(-2cot{\theta }/{2} )}{( 5 + 3cos\theta )}i$$
Hence , Option C
If $$z = x + iy$$ and $$\omega = \dfrac{(1 -iz)}{(z-i)}$$, then $$\left|\omega\right| = 1$$ implies that in the complex plane
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z lies on the imaginary axis
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z lies on the real axis
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z lies on the unit circle
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none of these
Explanation
Given $$w=\dfrac { 1-iz }{ z-i } $$ and
$$\left| w \right| =1$$
$$\Rightarrow \left| \dfrac { 1-iz }{ z-i } \right| =1$$ ...(1)
Substitute $$z=x+iy$$ in equation (1)
$$\Rightarrow \left| \dfrac { 1-i\left( x+iy \right) }{ \left( x+iy \right) -i } \right| =1$$
$$\Rightarrow \left| 1+y-ix \right| =\left| x+i\left( y-1 \right) \right| \\ \Rightarrow { \left( 1+y \right) }^{ 2 }+{ x }^{ 2 }={ x }^{ 2 }+{ \left( y-1 \right) }^{ 2 }\\ \Rightarrow y=0$$
Therefore z lies on the real axis.
Ans: B
If $$a, b$$ and $$c$$ are real numbers then the roots of the equation $$(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0$$ are always
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Real
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Imaginary
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Positive
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Negative
Explanation
Given equation is $$(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0$$
$$\Rightarrow 3x^{2} - 2(b + a + c) x + ab + bc + ca = 0$$
Now, here $$A = 3, B = -2(a + b + c)$$
$$C = ab + bc + ca$$
Therefore, $$ D = \sqrt {B^{2} - 4AC}$$
$$= \sqrt {(-2(a + b + c))^{2} - 4(3)(ab + bc + ca)}$$
$$= \sqrt {4(a + b +c)^{2} - 12(ab + bc + ca)}$$
$$= 2\sqrt {a^{2} + b^{2} + c^{2} - ab - bc - ca}$$
$$= 2\sqrt {\dfrac {1}{2}\left \{(a - b)^{2} - (b - c)^{2} + (c - a)^{2} \right \}} \geq 0$$
This is always $$\geq 0$$, we have real roots for the equation $$(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0$$
If $$(x+iy)(2-3i)=4+i$$ then (x, y) =
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$$\left ( 1,\dfrac{1}{13} \right )$$
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$$\left ( -\dfrac{5}{13},\dfrac{14}{13} \right )$$
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$$\left ( \dfrac{5}{13},\dfrac{14}{13} \right )$$
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$$\left ( -\dfrac{5}{13},-\dfrac{14}{13} \right )$$
Explanation
$$(x+iy)(2-3i)=4+i$$
$$2x-(3x)i+(2y)i-3yi^{2}=4+i$$
$$\underbrace{2x+3y}_{Real }+\underbrace{(2y-3x)}_{Imaginary }i=4+i$$
Comparing the real & imaginary parts,
$$2x+3y=4$$--------------------------(1)
$$2y-3x=1$$----------------------------(2)
Solving eq(1) & eq(2),
$$4x+6y=8$$
$$-9x+6y=3$$
$$13x=5\Rightarrow x=\dfrac{5}{13}$$
$$y=\dfrac{14}{13}$$
$$\therefore (x,y)=\left (\dfrac{5}{13},\dfrac{14}{13} \right )$$
If $$z =3+5i$$, then $$z^3+z+198=$$
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$$3 - 15i$$
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$$-3 - 15i$$
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$$-3 + 15i$$
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$$3 + 15i$$
Explanation
$$z=3+5i$$
$$z^{3}=(3+5i)^{3}$$
$$=3^{3}+3.3^{2}(5i)+3.3(5i)^{2}+(5i)^{3}$$
$$=27-125i+135i-225$$
$$=-225+27+(135-125)i$$
$$=-198+10i$$
$$\therefore z^{3}+z+198$$
$$=-198+10i+3+5i+198$$
$$=3+15i$$
If $$z=2-3i$$ then $$z^2-4z+13=$$
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$$0$$
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$$1$$
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$$2$$
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$$3$$
Explanation
$$z=2-3i$$
$$z^{2}=2^{2}-3^{2}-12i$$
$$=-5-12i$$
$$\therefore z^{2}-4z+13$$
$$=(-5-12i)-4(2-3i)+13$$
$$=-5-12i-8+12i +13$$
$$=-13+13$$
$$=0$$
The complex number $$\displaystyle \frac{1+2i}{1-i}$$ lies in the quadrant :
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I
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II
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III
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IV
Explanation
Let $$z =\dfrac{1+2i}{1-i}$$
$$\Rightarrow z =\dfrac{(1+2i)}{1-i}\times \dfrac{1+i}{1+i}$$
$$= \dfrac{1+2i+i+2i^2}{1-i^2}$$
$$=\dfrac{1+3i-2}{1+1}$$ ............ $$[\because i^2 = -1]$$
$$\Rightarrow z =\dfrac{-1+3i}{2}$$
$$ =-\dfrac{1}{2}+\dfrac{3}{2}i $$
$$=x+iy$$
Clearly $$x<0$$ and $$y>0$$
Hence $$z$$ lies in $$\text{II}$$ quadrant
$$\sqrt{-3}\sqrt{-75}=$$
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$$15$$
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$$15i$$
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$$-15$$
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$$-15i$$
Explanation
$$\sqrt{-3}\times\sqrt{-75}=\sqrt{3\times(-1)}\sqrt{75\times(-1)}$$
$$=\sqrt{3}i\times\sqrt{75}i$$
$$=\sqrt{225}i^{2}$$
$$= - 15$$
The sum of two complex numbers $$a + ib$$ and $$c +id$$ is a real number if
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$$a + c = 0$$
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$$b + d = 0$$
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$$a + b= 0$$
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$$b + c = 0$$
Explanation
It is given that
$$z_{1}=a+ib$$ and
$$z_{2}=c+id$$
Then
$$z_{1}+z_{2}=(a+c)+i(b+d)$$
Now
$$(z_{1}+z_{2})$$ is purely real.
Then the imaginary part has to be $$0$$.
Hence
$$b+d=0$$.
The locus of complex number z such that z is purely real and real part is equal to - 2 is
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Negative y-axis
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Negative x-axis
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The point (-2, 0)
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The point (2, 0)
Explanation
$$z=x+iy$$
$$(x,y)$$
$$z$$ is purely real and the real part equals $$-2$$
$$\therefore y=0$$ & $$x=-2$$
$$z=-2$$
Hence, this would be represented by the point (-2,0) on the Argand Plane.
$$\dfrac{1}{i-1}+\dfrac{1}{i+1}$$ is
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positive rational number
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purely imaginary
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positive Integer
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negative integer
Explanation
Let $$Z= \dfrac{1}{i-1}+\dfrac{1}{i+1}$$
$$=\dfrac{i+1+i-1}{(i-1)(i+1)}$$
$$=\dfrac{i+i}{(i^2-1^2)}$$
$$=\dfrac{2i}{-2}$$
$$ \therefore Z=-i$$
The argument of every complex number is
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Double valued
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Single valued
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Many valued
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Triple valued
Explanation
$$z=x+iy$$
amplitude $$= tan^{-1}\frac{y}{x}$$
$$\Rightarrow $$ amplitude $$=\theta \pm 2k\pi $$ where $$\theta \epsilon \left [ -\pi ,\pi \right ]$$ $$\forall k\epsilon R$$
since $$k\epsilon R$$
$$\Rightarrow $$ Amplitude of any complex number is many valued.
The sum of two complex numbers $$a + ib$$ and $$c+ id$$ is purely imaginary if
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$$a + c = 0$$
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$$a + d = 0$$
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$$b + d = 0$$
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$$b + c = 0$$
Explanation
It is given that
$$z_{1}=a+ib$$ and
$$z_{2}=c+id$$
$$z_{1}+z_{2}=(a+c)+i(b+d)$$
$$z_{1}+z_{2}$$ is purely imaginary. (Given)
Then the real part has to be $$0$$.
Hence
$$a+c=0$$.
For $$a < 0$$, arg $$(ia) = $$
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$$\dfrac{\pi }{2}$$
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$$-\dfrac{\pi }{2}$$
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$$\pi $$
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$$-\pi $$
Explanation
Let $$z= ia $$
$$ a <0 $$ . Hence $$z$$ must lie on the negative imaginary axis ,
Hence, $$arg(z) = -\dfrac{\pi}{2}$$
The principal value of the argument of $$-\sqrt{3}+i$$ is :
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$$\dfrac{\pi }{6}$$
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$$\dfrac{3\pi }{6}$$
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$$\dfrac{5\pi }{6}$$
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$$\dfrac{7\pi }{6}$$
Explanation
Given, $$z=-\sqrt 3+i$$
We have $$| z| = \sqrt{\sqrt{3}^2 +1^2} =2 $$
Therefore, $$ \cos \theta = \dfrac{-\sqrt{3} }{2} $$
and $$\sin \theta = \dfrac{1}{2} $$
Hence,
$$\theta =\dfrac{5\pi}{6}$$
Amplitude of $$\dfrac{1+i}{1-i}$$ is :
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$$0$$
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$$\pi $$
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$$\dfrac{\pi }{2}$$
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$$-\pi $$
Explanation
$$\dfrac{1+i}{1-i}$$
$$=\dfrac{(1+i)(1+i)}{(1-i)(1+i)}$$
$$=\dfrac{(1+i)^{2}}{1^{2}-i^{2}}$$
$$=\dfrac{1-1+2i}{2}$$
$$=i$$
therefore amplitude $$=\dfrac{\pi }{2}$$
Which of the following equations has two distinct real roots ?
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$$2x^2-3\sqrt 2 x+\dfrac 94=0$$
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$$x^{2}+x-5=0$$
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$$x^{2}+3x+2\sqrt{2}=0$$
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$$5x^{2}-3x+1=0$$
Explanation
An equation is said to have two distinct and real roots if the discriminant $$b^2-4ac>0$$
Case (i): For equation: $$2x^2-3\sqrt 2 x+\dfrac 94=0$$.
Here $$a=2, b=-3\sqrt 2, c=\dfrac 94$$
The discrimant is $$(-3\sqrt 2)^2-4(2)\left(\dfrac 94\right)=18-18=0$$
Hence this equation has equal real roots
Case (ii): For equation: $$x^2+x-5=0$$.
Here $$a=1, b=1, c=-5$$
The discrimant is $$1^2-4(1)(-5)=1+20=21>0$$
Hence this equation has two distinct real roots
Case (iii): For equation: $$x^2+3x+2\sqrt 2=0$$.
Here $$a=1, b=3, c=2\sqrt 2$$
The discrimant is $$3^2-4(1)(2\sqrt2)=9-8\sqrt 2<0$$
Hence this equation has no real roots
Case (iv): For equation: $$5x^2-3x+1=0$$.
Here $$a=5, b=-3, c=1$$
The discrimant is $$(-3)^2-4(5)(1)=9-20<0$$
Hence this equation has no real roots
A quadratic equation $$ax^2 + bx+c=0$$ has two distinct real roots, if
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$$a=0$$
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$$b^2-4ac = 0$$
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$$b^2-4ac < 0$$
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$$b^2-4ac > 0$$
Explanation
If $$a=0$$, it becomes linear equation.
If $${ b }^{ 2 }-4ac=0$$, then there will be real and equal roots.
If $${ b }^{ 2 }-4ac<0$$, then the roots will be unreal.
Only if $${ b }^{ 2 }-4ac>0$$, we will get two real distinct roots.
Option D is correct!
For $$a > 0$$, arg $$(ia) =$$
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$$\dfrac{\pi }{2}$$
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$$-\dfrac{\pi }{2}$$
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$$\pi $$
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$$-\pi $$
Explanation
$$z= 0 + ia$$
$$\therefore arg(z) = arg (ia) = \tan^{-1} \cfrac a0$$
= $$\cfrac {\pi}{2}$$ ...(since a is greater than or equal to zero)
The modulus of $$\sqrt{2}i-\sqrt{-2}i$$ is:
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2
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$$\sqrt{2}$$
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0
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$$2\sqrt{2}$$
Explanation
Let $$z=\sqrt{2}i-\sqrt{-2}i$$
$$\Rightarrow z=\sqrt{2}i-\sqrt{2}i^2$$
$$\Rightarrow z=\sqrt{2}+\sqrt{2}i$$
Now, $$|z|=\sqrt{2+2}=2$$
The roots of the equation $$3x^{2} - 4x + 3 = 0$$ are :
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real and unequal
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real and equal
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imaginary
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none of these
Explanation
Given equation is $$3x^2-4x+3=0$$
To find, the nature of the roots of the equation
An equation is said to have
(i) two distinct and real roots if the discriminant $$b^2-4ac>0$$
(ii) equal real roots if $$b^2-4ac=0$$
(iii) no real roots or imaginary roots if $$b^2-4ac<0$$
In the given equation $$a=3, b=-4, c=3$$
Hence the discriminant is $$(-4)^2-4(3)(3)=16-36=-20<0$$
Therefore the roots of the given equation are imaginary in nature.
For $$a<0$$, arg $$a=$$
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$$\dfrac{\pi }{2}$$
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$$\dfrac{-\pi }{2}$$
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$$\pi $$
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$$-\pi $$
Explanation
It is given that $$a<0$$
Thus, 'a' is purely real.
Hence $$arg(a)$$ will be $$0$$ or, $$\pi$$
But it is given that $$a<0$$
Hence $$cos(arg(a))=-1$$
Or
$$arg(a)=\pi$$.
If the square of $$(a + ib)$$ is real, then $$ ab=$$
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$$0$$
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$$1$$
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$$-1$$
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$$2$$
Explanation
$$(a+ib)^{2}=a^{2}-b^{2}+2iab$$ is given to be real
$$\Rightarrow ab=0$$
Find the argument of
$$-1 - i\sqrt{3}$$
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$$\theta= -2\pi/3$$
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$$\theta= 2\pi/3$$
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$$\theta= -4\pi/3$$
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$$\theta= 4\pi/3$$
Explanation
Let
$$ z = -1 -i\sqrt{3}$$
Then $$\alpha = \tan^{-1}\left|b/a\right| = \tan^{-1} \left|\sqrt{3}/1\right| = \pi/3$$
Here, $$z$$ is in the third quadrant.
Therefore, argument is $$\theta= -(\pi - \alpha) = -(\pi - \pi/3) = -2\pi/3$$
Ans: A
The roots of $$x^{2}-x+1=0$$ are:
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Real and equal
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Real and not equal
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Imaginary
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Reciprocals
Explanation
Given equation is $$x^2-x+1=0$$
We know $$D=b^{2}-4ac$$
Here $$a=1, b=-1, c=1$$
Therefore, $$D=(-1)^{2}-4(1)(1)$$
$$=1-4$$
$$=-3< 0$$
Thus roots are imaginary.
Nature of the roots of the quadratic equation $$2x^{2}-2\sqrt{6}x+3=0$$ is:
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Real, irrational, unequal
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Real, rational, equal
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Real, rational, unequal
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Complex
Explanation
(B) $$b^{2}-4ac=(-2\sqrt{6})^{2}-4(2)(3)$$
$$=24-24$$
$$=0$$
$$\therefore $$ Real, rational, equal
Determine the nature of roots of the equation $$x^2 + 2x\sqrt{3}+3=0$$.
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Real and distinct
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Non-real and distinct
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Real and equal
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Non-real and equal
Explanation
The nature of the roots can be determined from the discriminant $$b^2-4ac$$
$$\therefore b^2-4ac=(2\sqrt {3})^2-(4\times 1\times 3)$$
$$\Rightarrow b^2-4ac=12-12$$
$$\Rightarrow b^2-4ac=0$$
$$\because b^2-4ac=0$$
There are two real and equal roots.
Find the value of $$x$$ of the equation $${ \left( 1-i \right) }^{ x }={ 2 }^{ x }$$
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$$1$$
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$$2$$
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$$0$$
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none of these
Explanation
Then $${ \left( 1-i \right) }^{ x }={ 2 }^{ x }$$
$$\Rightarrow |{(1-i)}|^{x} =|2|^{x}$$
$$\Rightarrow { \left( \sqrt {1+1}\right)}^{x}={2}^{x}$$
$$\Rightarrow { \left( \sqrt { 2 } \right) }^{ x }={ 2 }^{ x }$$
$$\Rightarrow \dfrac{x}{2}=x$$
$$\Rightarrow 2x=x\Rightarrow x=0$$
Hence, option C is correct.
If the discriminant of a quadratic equation is negative, then its roots are:
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unequal
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equal
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inverse
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imaginary
Explanation
If the discriminant $$ b^2 - 4ac $$ of a quadratic equation is negative, then its roots are imaginary.
Solve $$\displaystyle \left ( 1-i \right )x+\left ( 1+i \right )y= 1-3i,$$
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$$\displaystyle x= -1, y= 2.$$
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$$\displaystyle x= 2, y= -1.$$
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$$\displaystyle x= 2, y= 1.$$
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$$\displaystyle x= 1, y= 2.$$
Explanation
$$\displaystyle \left ( 1-i \right )x+\left ( 1+i \right )y= 1-3i.$$
Equating real and imaginary parts, we get
$$\displaystyle x+y= 1$$ and $$\displaystyle -x+y= -3.$$
Adding both equations we get $$y=-1$$
Substituting this value of $$y$$ in any of the 2 equations, we get $$x=2$$
$$\therefore \displaystyle x= 2, y= -1.$$
Ans: B
The roots of $$4x^{2}-2x+8=0$$ are:
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Real and equal
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Rational and not equal
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Irrational
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Not real
Explanation
Given equation is $$4x^2-2x+8=0$$
We know value of discriminant $$D=b^{2}-4ac$$
Here $$a=4, b=-2, c=8$$
Therefore, $$D=(-2)^{2}-4(4)(8)$$
$$=4-128$$
$$=-124< 0$$
Hence, roots are not real.
Evaluate :
$$\sqrt{-25} + 3 \sqrt{-4} +2 \sqrt{-9}$$
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$$-17i$$
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$$5i$$
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$$17i$$
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$$6i$$
Explanation
$$\sqrt{-25} + 3 \sqrt{-4} +2 \sqrt{-9} $$
$$=5\sqrt{-1}+6\sqrt{-1}+6\sqrt{-1}$$ we know, $$\sqrt{-1}=i$$
$$= 5i + 6i + 6i = 17i$$
If $$x^{2}-2px+8p-15=0$$ has equal roots, then $$p=$$
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$$3$$ or $$-5$$
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$$3$$ or $$5$$
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$$-3$$ or $$5$$
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$$-3$$ or $$-5$$
Explanation
Given equation is $$x^2-2px+8p-15=0$$
Here $$a=1, b=-2p, c=8p-15$$
We know for equal roots, $$b^{2}-4ac=0$$
Therefore, $$(-2p)^{2}-4(1)(8p-15)=0$$
$$\Rightarrow 4p^{2}-32p+60=0$$
$$\Rightarrow p^{2}-8p+15=0$$
$$\Rightarrow (p-5)(p-3)=0$$
i.e., $$p=5 $$ or $$ 3$$
Determine the values of $$p$$ for which the quadratic equation $$2x^2 + px + 8 = 0$$ has
equal roots.
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$$p=\pm 64$$
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$$p=\pm 8$$
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$$p=\pm 4$$
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$$p=\pm 16$$
Explanation
In $$ax^2+bx+c=0,$$
Discriminant $$D$$ $$=b^2-4ac$$,
D$$=0$$, for the roots to be real and equal
In
$$2x^{2}+px+8=0$$
Thus, we have $$a=2 ,b=p$$ and $$c=8$$
Then $$D$$
$$=b^2-4ac=0$$
$$\therefore p^{2}-4\times 2\times 8=0\Rightarrow p^{2}-64=0$$
$$\Rightarrow p=\pm 8$$.
Find the values of $$k$$ for the following quadratic equation, so that they have two real and equal roots:
$$2x^2 + k x + 3 = 0$$
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$$k = \pm 2\sqrt 3$$
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$$k = \pm 2\sqrt 6$$
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$$k = \pm \sqrt 6$$
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$$k = \pm \sqrt 3$$
Explanation
In $$ax^2+bx+c=0$$
Discriminant $$D=b^2-4ac$$
If $$D=0$$, so they have equal and real roots.
Then,
$$2x^{2}+kx+3=0$$
$$a=2$$, $$b=k$$ and $$c=3$$
$$D=k^{2}-4\times 2\times 3=0$$
$$\Rightarrow k^{2}-24=0$$
$$\Rightarrow k=\pm \sqrt{24}$$
$$\Rightarrow k=\pm 2\sqrt{6}$$
$$\displaystyle \frac{\displaystyle i^{4n + 3} + (-i)^{8n - 3}}{\displaystyle(i)^{12 n- 1} - i^{2 - 16 n}}, n \varepsilon N$$ is equal to
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1 + i
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2i
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-2i
0%
-1 - i
Explanation
Given expression
$$= \displaystyle \frac{\displaystyle i^3 + (-1) (i)^{-3}}{(-i)^{-1} - (i)^2} = \frac{\displaystyle -i - i}{\displaystyle i + 1}$$
$$=\displaystyle \frac{\displaystyle -2i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{\displaystyle -2i - 2}{\displaystyle 1 + 1} = - 1 - i$$
1+$$i^2 + i^4 + i^6 + ........+ i^{2n}$$ is
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Positive
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Negative
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Zero
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Cannot be determined
Explanation
$$1+i^2+i^4+........+i^{2n}$$
$$=1-1+1-1+1-1......$$
Hence it can b positive or negative depending on the value of $$2n$$
Hence It cannot be determined from the given data.
Find the modulus and the principal value of the argument of the number $$1-i$$
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$$\displaystyle \sqrt{2},\pi/4$$
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$$\displaystyle \sqrt{2},-\pi/4$$
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$$\displaystyle \sqrt{2},-\pi/3$$
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$$\displaystyle \sqrt{2},3\pi/4$$
Explanation
Let $$\displaystyle 1=r\cos \theta ,-1=r\sin \theta$$
Squaring and adding these relations, we get $$\displaystyle r^{2}\left ( \cos ^{2}\theta +\sin ^{2}\theta \right )=1^{2}+\left ( -1 \right )^{2}=2$$.
Since, $$cos^2\theta+sin^2\theta=1, \Rightarrow r=\pm\sqrt{2}$$
$$r$$ cannot be negative
Hence. $$r=\sqrt{2}$$
Then $$\displaystyle \cos \theta =\dfrac{1}{\sqrt{2}},\sin \theta =\dfrac{-1}{\sqrt{2}},$$
The value of $$\displaystyle \theta $$ between-$$\displaystyle \pi $$ and $$\displaystyle \pi $$ which satisfies these equations is $$\displaystyle -\left ( \pi /4 \right )$$ Thus $$\displaystyle \left | 1-i \right |=r=\sqrt{2}$$ and $$arg (1-i)=-\displaystyle \left ( \pi /4 \right ).$$
Ans: B
If $$i^2 = - 1$$, then the value of $$\displaystyle \sum^{200}_{n = 1} i^n $$ is
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50
0%
-50
0%
0
0%
100
Explanation
Given, $$i^2=-1$$
$$\displaystyle \sum_{n = 1}^{200} i^n = i + i^2 + i^3 + ......... + i^{200} = \displaystyle \frac{i (1 - i^{200})}{1 - i} $$ (since G. P.)
$$= \displaystyle \frac{\displaystyle i (1 - 1)}{\displaystyle 1 - i} = 0$$
If i = $$\sqrt {-1}, then 1 + i^2 + i^3 -i^6 + i^8 $$ is equal to -
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2- i
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1
0%
-3
0%
-1
Explanation
Given, $$i= \sqrt {-1}$$
$$\therefore 1+i^{ 2 }+i^{ 3 }-i^{ 6 }+i^{ 8 }=1-1-i+1+1$$
$$=2-i$$
Check whether $$2x^2 - 3x + 5 = 0$$ has real roots or no.
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0%
The equation has real roots.
0%
The equation has no real roots.
0%
Data insufficient
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None of these
Explanation
Here the quadratic equation is $$2x^2 - 3x + 5 = 0$$
Comparing it with $$ax^2+bx+c=0$$, we get
$$a=2, b=-3, c=5$$
Therefore, discriminant, $$D=b^2-4ac$$
$$(-3)^2-4\times 2\times 5$$
$$=9-40$$
$$=-31$$
Here, $$D<0$$
Therefore the equation has no real roots.
($$i^{10}+1) (i^9 + 1)(i^8 +1).......(i+1)$$ equal to
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0%
-1
0%
1
0%
0
0%
i
Explanation
$$(i^{10}+1) (i^9 + 1)(i^8 +1).......(i+1)$$
$$=(i^{10}+1) (i^9 + 1)(i^8 +1).......(i^3+1)(i^2+1)(i+1)$$
$$=(i^{10}+1) (i^9 + 1)(i^8 +1).......(i^3+1)(-1+1)(i+1)[\because i^2=-1]$$
$$=(i^{10}+1) (i^9 + 1)(i^8 +1).......(i^3+1)(0)(i+1) =0$$
If $$i^2$$ $$= -1$$, then find the odd one out of the following expressions.
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$$-i^2$$
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$$(-i)^2$$
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$$i^4$$
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$$(-i)^4$$
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$$-i^6$$
Explanation
$$i$$ is an imagiary number which has a value of $$i=\sqrt{-1}$$. Then
A.
$$-i^2=-1*-1=1$$
B.
$${(-i)}^2=i^2=-1$$
C.
$$i^4={(i^2)}^2={(-1)}^2=1$$
D.
$${(-i)}^4={(i^2)}^2={(-1)}^2=1$$
E.
$$-i^6=-{(i^2)}^3=-1\times {(-1)}^3=-1\times (-1)=1$$
All are $$1$$ except option $$B$$ which is $$-1$$.
When $$(3-2i)$$ is subtracted from $$(4 + 7i)$$, then the result is
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0%
$$1 + 5i$$
0%
$$1 + 9i$$
0%
$$7 + 5i$$
0%
$$7 + 9i$$
Explanation
We need to subtract $$(3-2i)$$ from $$(4+7i)$$
$$\therefore 4 + 7i - (3 - 2i)$$
$$=4 + 7i - 3 + 2i $$
$$= 1 + 9i$$
The value of k for which polynomial $$x^{2} - kx + 4$$ has equal zeroes is
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0%
$$4$$
0%
$$2$$
0%
$$-4$$
0%
$$-2$$
Explanation
$$ x^{2}-kx+4 $$ has equal roots $$ \Rightarrow D = 0 $$
$$ \Rightarrow k^{2}-4\times 4\times 1\Rightarrow k^{2} = 16 \Rightarrow \boxed{k = \pm 4} $$
If the discriminant of a quadratic equation is negative, then its roots are
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0%
Unequal
0%
Equal
0%
Inverses
0%
Imaginary
Explanation
As $$ x = \dfrac{-b\pm \sqrt{D}}{2a} $$ if D is $$-$$ve $$\Rightarrow $$ imaginary Roots
If the equation $$(1 + m^{2}) x^{2} + 2mcx + (c^{2} - a^{2}) = 0$$ has equal roots, then $$c^{2} =$$
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0%
$$ a^{2} (1 + m^{2})$$
0%
$$ a (1 + m^{2})$$
0%
$$ a^{4} (1 - m^{2})$$
0%
$$ a^{2} (1 - m^{2})$$
Explanation
As the eq$$^{n}$$ has equal roots $$ \Rightarrow D = 0 $$
$$ \Rightarrow (2mc)^{2}-4(1+m^{2})(c^{2}-a^{2}) = 0 $$
$$ \Rightarrow 4-m^{2}c^{2}-4[c^{2}-a^{2}+m^{2}c^{2}-m^{2}a^{2}] = 0 $$
$$ \Rightarrow c^{2} = m^{2}a^{2}+a^{2}\Rightarrow \boxed{c^{2} = a^{2}(1+m^{2})} $$
Amplitude of $$\displaystyle \frac{1 +\sqrt 3i}{ \sqrt3 + i} $$
is
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0%
$$\displaystyle \frac {\pi}{3}$$
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$$\displaystyle \frac {\pi}{2}$$
0%
0
0%
$$\displaystyle \frac {\pi}{6}$$
Explanation
$$\displaystyle Z= \frac{1 +\sqrt 3i}{ \sqrt3 + i} $$
$$\displaystyle\Rightarrow \left( \frac { 1+\sqrt { 3 } i }{ \sqrt { 3 } +i } \right) \left( \frac { \sqrt { 3 } -i }{ \sqrt { 3 } -i } \right) =\frac { 2\sqrt { 3 } +2i }{ 4 } $$
$$\therefore$$ Amplitude of $$Z$$ is $$\tan ^{ -1 }\left({\displaystyle \frac { 1 }{ \sqrt3 } } \right)=\displaystyle\frac{\pi}{6}$$
Hence, option D.
For $$i=\sqrt{-1}$$, what is the sum $$\left(7+3i\right) + \left(-8+9i\right)$$?
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0%
$$-1+12i$$
0%
$$-1-6i$$
0%
$$15+12i$$
0%
$$15-6i$$
Explanation
Given: $$(7+3i)+(-8+9i)$$
$$= (7+(-8))+(3i+9i)$$
$$=(7-8)+12i$$
$$=-1+12i$$
Option A is correct.
If the equation $$\displaystyle x^{2}-bx+1=0$$ does not possess real roots then
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0%
$$-3 < b < 3$$
0%
$$-2 < b < 2$$
0%
$$b > 2$$
0%
$$b < -2$$
Explanation
Given equation is $$x^{2}-bx+1=0$$
If the equation doesnot possess real roots,then
$$b^2-4ax<0$$
$$=>(-b)^2-4(1)(1)<0$$
$$=>b^2<4$$
$$=>b<\pm 2$$
$$\therefore -2<b<2$$
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