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Jee Main Advanced Math Chapter Wise Mock Test
Quiz 1
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2
3
4
5
6
7
8
9
10
11
12
13
Q.1
if |z|=1 and $\omega = \frac {z-1}{z+1}$ ( where $z \ne -1$ Then $Re( \omega )$ is
0%
0
0%
$- \frac {1}{|z+1|^2}$
0%
$|\frac {z}{z+1}|. \frac {1}{|z+1\^2}$
0%
$ \frac {\sqrt 2}{|z+1|^2}$
Explanation
Solution:
Write z = e
iθ
. Then ω = (e
iθ
-1)/(e
iθ
+1). Multiplying top and bottom by e
-iθ/2
:
ω = (e
iθ/2
-e
-iθ/2
)/(e
iθ/2
+e
-iθ/2
) = (2i sin(θ/2))/(2cos(θ/2)) = i·tan(θ/2), which is purely imaginary, so Re(ω) = 0.
Q.2
Let z = x + iy be the complex number where x and y are integers. Then the area of the rectangle whose vertices are the roots of the equation $\bar{z} z^3 + z (\bar{z})^3 =350$ is
0%
48
0%
32
0%
40
0%
80
Explanation
Solution:
z̄z³+z(z̄)³ = zz̄(z²+z̄²) = |z|²·2Re(z²). For z=x+iy, Re(z²)=x²-y² and |z|²=x²+y², so the equation becomes 2(x²+y²)(x²-y²) = 350, i.e. x⁴-y⁴ = 175.
Factoring x⁴-y⁴ = (x²-y²)(x²+y²) = 175 and testing integer factor pairs of 175, the pair (x²-y², x²+y²) = (7, 25) gives x²=16, y²=9, i.e. x=4, y=3 -- the only pair that lands on perfect squares.
The four roots (±4, ±3) form a rectangle 8 units wide and 6 units tall, with area 8×6 = 48.
Q.3
if $ z= \left ( \frac {\sqrt 3 + i}{2} \right )^5 + \left ( \frac {\sqrt 3 - i}{2} \right )^5$ , then
0%
Re(z) =0
0%
Im (z) =0
0%
Re(z), Im(z) > 0
0%
Re(z) > 0, Im(z) < 0
Explanation
Solution:
(√3+i)/2 = cos30°+isin30° and (√3-i)/2 = cos30°-isin30° -- these are e
iπ/6
and e
-iπ/6
.
z = e
i5π/6
+ e
-i5π/6
= 2cos(5π/6), which is purely real (the two imaginary parts cancel), so Im(z) = 0.
Q.4
if |z| =1 and $ z \ne 1$ , then all the values of $\frac {z}{1-z^2} $ lie on
0%
x axis
0%
y axis
0%
$|z| = \sqrt 2$
0%
A line not passing through the origin
Explanation
Solution:
Write z = e
iθ
. Then 1-z² = 1-e
2iθ
= -e
iθ
(e
iθ
-e
-iθ
) = -e
iθ
·2i sinθ.
So z/(1-z²) = e
iθ
÷ (-2i sinθ · e
iθ
) = 1/(-2i sinθ) = i/(2sinθ) -- purely imaginary for every valid θ, so every such point lies on the y (imaginary) axis.
Q.5
If $ z_1$ ,$z_2$ ,$z_3$ are complex numbers such that $| z_1| = | z_2|=| z_3| = | \frac {1}{z_1} + \frac {1}{z_2} +\frac {1}{z_3}|=1$,then $|z_1 + z_2 + z_3|$ is
0%
1
0%
< 1
0%
> 1
0%
3
Explanation
Solution:
Since |z
1
|=|z
2
|=|z
3
|=1, each satisfies 1/z = z̄ (the reciprocal of a unit-modulus number is its own conjugate).
So |1/z
1
+1/z
2
+1/z
3
| = |z̄
1
+z̄
2
+z̄
3
| = |conjugate of (z
1
+z
2
+z
3
)| = |z
1
+z
2
+z
3
|, since a number and its conjugate always have the same modulus.
The problem states this quantity is 1, so |z
1
+z
2
+z
3
| = 1 directly.
Q.6
The area of the triangle on the complex plane formed by the complex numbers z, – iz and z + iz is
0%
$|z|^2$
0%
$| \bar{z}|^2$
0%
$\frac {|z|^2}{2}$
0%
$\frac {|z|^2}{4}$
Explanation
Solution:
Testing with z=1 (the formula's answer scales with |z|², so any convenient z will reveal the constant): the three points are z=(1,0), -iz=(0,-1), and z+iz=(1,1).
Shoelace formula: Area = ½|x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)| = ½|1(-1-1)+0(1-0)+1(0-(-1))| = ½|-2+0+1| = ½.
With |z|²=1 here, this matches the formula |z|²/2.
Q.7
The value of arg (x) when x < 0 is:
0%
0
0%
$\frac {\pi}{2}$
0%
$\pi$
0%
none of these
Explanation
Solution:
A negative real number sits on the negative real axis of the complex plane, which corresponds to an angle of π (180°) from the positive real axis.
Q.8
if $\omega$ ( $\ne 1$) is a cube root of the unity and $(1 + \omega )^7 = A + B \omega $, then A and B are respectively the numbers
0%
0,1
0%
1,1
0%
1,0
0%
-1 ,1
Explanation
Solution:
Since ω is a primitive cube root of unity, 1+ω+ω²=0, so 1+ω = -ω².
(1+ω)⁷ = (-ω²)⁷ = -ω¹⁴ = -(ω³)⁴·ω² = -(1)·ω² = -ω² (using ω³=1).
And -ω² = 1+ω (from the same identity 1+ω+ω²=0 rearranged as -ω²=1+ω), so (1+ω)⁷ = 1+ω = A+Bω with A=1, B=1.
Q.9
sinx + i cos 2x and cos x – i sin 2x are conjugate to each other for
0%
$x = n \pi$
0%
x=0
0%
$x = (n + \frac {1}{2}) \frac {\pi}{2}$
0%
No value of x
Explanation
Solution:
For cos x - i sin 2x to be the conjugate of sin x + i cos 2x, the real parts must match (cos x = sin x) and the imaginary parts must be opposite (-sin 2x = -cos 2x, i.e. sin 2x = cos 2x).
cos x = sin x gives x = nπ + π/4. sin 2x = cos 2x gives x = mπ/2 + π/8.
Setting these equal and simplifying leads to 8n+1 = 4m -- an odd number equal to an even number, which is impossible for any integers n, m. So no value of x satisfies both conditions at once.
Q.10
The point $z_1$, $z_2$, $z_3$,$z_4$ in the complex plane are the vertices of a parallelogram taken in order if and only if
0%
$z_1 +z_4 =z_2 + z_3$
0%
$z_1 +z_2=z_3 + z_4$
0%
$z_1 +z_3 =z_2 + z_4$
0%
None of these
Explanation
Solution:
A quadrilateral z
1
z
2
z
3
z
4
(vertices taken in order) is a parallelogram exactly when its diagonals bisect each other, i.e. the midpoint of z
1
z
3
equals the midpoint of z
2
z
4
:
(z
1
+z
3
)/2 = (z
2
+z
4
)/2 ⇒ z
1
+z
3
= z
2
+z
4
.
Q.11
What is the smallest positive integer n, for which $(1 + i)^2n = (1 – i)^2n$?
0%
2
0%
3
0%
4
0%
8
Explanation
Solution:
(1+i)² = 1+2i-1 = 2i, and (1-i)² = 1-2i-1 = -2i.
So (1+i)
2n
= (2i)
n
and (1-i)
2n
= (-2i)
n
. Setting them equal: (2i)
n
= (-2i)
n
⇒ (-1)
n
= 1, which holds only for even n.
The smallest positive even integer is 2.
Q.12
if $z_1 = a+ ib$ and $z_2 =c+ id$ such that $|z_1|= |z_2|=1$ and $Re(z_1 \bar{z_2}) =0$, then the pair of the complex number $W_1=a+ic$ and $W_2 =b+id$ satisfies
0%
$|W_1|=1$
0%
$|W_2|=1$
0%
$Re(W_1 \bar{W_2}) =0$
0%
All the above
Explanation
Solution:
Since (a,b) and (c,d) are unit vectors (|z
1
|=|z
2
|=1) with ac+bd=0 (that's what Re(z
1
z̄
2
)=0 means), they are perpendicular unit vectors, so (c,d) = (-b,a) or (b,-a).
Taking (c,d)=(-b,a): |W
1
|²=a²+c²=a²+b²=1, |W
2
|²=b²+d²=b²+a²=1, and Re(W
1
W̄
2
)=ab+cd=ab+(-b)(a)=0. All three hold (the other sign choice for (c,d) gives the same three results).
Q.13
For all complex numbers $z_1$ ,$z_2$ satisfying $|z_1| =12$ and $|z_2 -3 -4i|=5$, the minimum value of $|z_1 -z_2|$ is
0%
0
0%
2
0%
7
0%
17
Explanation
Solution:
|z
1
|=12 places z
1
on a circle of radius 12 about the origin. |z
2
-3-4i|=5 places z
2
on a circle of radius 5 centred at 3+4i, and |3+4i| = √(3²+4²) = 5 -- that centre is itself 5 units from the origin.
So z
2
's circle reaches from 0 up to 5+5=10 units from the origin -- entirely inside z
1
's circle of radius 12.
The closest the two circles ever get is when z
2
is at its farthest point from the origin (10 units out, in the same direction as z
1
): 12 - 10 = 2.
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