Free Objective Test 01 Practice Test - 11th and 12th
Question 1
For positive integers n1,n2 the value of the expression (1+i)n1+(1+i3)n1+(1+i5)n2 + (1+i7)n2 where i=2√−1 is a real number if and only if
n1=n2+1
n1=n2-1
n1=n2
n1>0,n2>0
SOLUTION
Solution : D
Using i3=−i,i5=i and i7=−i, we can write the given expression as
(1+i)n1+(1+i3)n1+(1+i5)n2+(1+i7)n2 where i=2√1=2[n1C0+n1C2(i)2+n1C4(i)4+n1C6(i)6+⋯⋯]+2[n2C0+n2C2(i)2+n2C4(i)4+n2C6(i)6+⋯⋯]=2[n1C0−n1C2+n1C4−n1C6+⋯⋯]+2[n2C0−n2C2+n2C4−n2C6+⋯⋯]
This is a real number irrespective of values of n1 and n2.
Question 2
The conjugate of (2+i)23+i , in the form of a+ib, is
132+i(152)
1310+i(−152)
1310+i(−910)
1310+i(910)
SOLUTION
Solution : C
z = (2+i)23+i=3+4i3+i×3−i3−i=1310+i910
Conjugate = 1310−i910.
Question 3
If (1+i)(1+2i)(1+3i)......(1+ni) = a+ib, then 2.5.10.....(1+n2) is equal to
SOLUTION
Solution : B
we have(1+i)(1+2i)(1+3i)......(1+ni) = a+ib .......(i)
⇒ (1+i)(1+2i)(1+3i)......(1+ni) = a-ib ........(ii)
Multiplying (i) and (ii), we get 2.5.10.....(1+n2) = a2+b2
Question 4
If x=-5+√−4, then the value of the expression x4+9x3+35x2-x+4 is
160
-160
60
-60
SOLUTION
Solution : B
x+5 =4i ⇒ x2+10x+25=-16
Now, x4+9x3+35x2-x+4
=(x2+10x+41)(x2-x+4)-160=-160
Question 5
1−i1+i is equal to
cosπ2+isinπ2
cosπ2-isinπ2
sinπ2+icosπ2
None of these
SOLUTION
Solution : B
1−i1+i=(1−i)(1−i)(1+i)(1−i)=1+(i)2−2i1+1=-i
Which can be written as cosπ2-isinπ2
Question 6
The values of x and y for which the numbers 3+ix2y and x2 +y+4i are conjugate complex are
(-2,-1) or (2,-1)
(1,-2) or (-2,1)
(1,2) or (-1,-2)
None of these
SOLUTION
Solution : A
According to condition, 3-ix2 y = x2 + y + 4i
⇒ x2 + y = 3 And x2y = -4 ⇒ x = ±2, y = -1
⇒ (x,y) = (2,-1) or (-2,-1)
Question 7
For the complex number z, which of the following is true?
z+¯z is real and z¯z is imaginary
z+¯z is imaginary and z¯z is real
Both z+¯z and z¯z are real.
Both z+¯z and z¯z are imaginary numbers
SOLUTION
Solution : C
Let z=x+iy
Here, z+¯z=(x+iy)+(x−iy)=2x (Real)And z¯z=(x+iy)(x−iy)=x2+y2 (Real).
Question 8
The number of solutions of the equation z2 + ¯z = 0 is
1
2
3
4
SOLUTION
Solution : D
Let z = x+iy, so that ¯z = x - iy, therefore
z2+¯z=0⇔(x2−y2+x)+i(2xy−y) = 0
Equating real and imaginary parts , we get
x2−y2+x = 0 .......(i)
And 2xy - y = 0 ⇒ y = 0 or x = 12
if y = 0 , then (i) gives x2 + x = 0 ⇒ x = 0 or
x = -1
If x = 12,
Then x2−y2+x=0⇒y2=14+12=34⇒y=±√32
Hence, there are four solutions in all.
Question 9
If z is a complex number, then z.¯z = 0 if and only if
z=0
Re(z)=0
Im(z)=0
None of these
SOLUTION
Solution : A
Let z = x+iy, ¯z = x-iy
∴ z¯z = 0 ⇒ (x+iy)(x-iy) = 0 ⇒ x2+y2 = 0
It is possible onle when x and y oth simultaneously zero
i.e, z = 0 +0i = 0
Question 10
If z is a complex number such that z2 = (¯z)2,then
z is purely real
z is purely imaginary
Either z is purely real or purely imaginary
None of these
SOLUTION
Solution : C
Let z = x+iy, then its coonjucate ¯z = x-iy
Given that z2=(¯z)2
⇒ x2 - y2 + 2ixy = x2 - y2 - 2ixy ⇒ 4ixy = 0
if x ≠ 0 then y = 0 and if y ≠ 0 then x = 0
Question 11
If -1+√−3=reiθ, then θ is equal to
π3
-π3
2π3
-2π3
SOLUTION
Solution : C
Here -1+√−3=reiθ ⇒ -1+i√3=reiθ
=rcosθ=-1 And rsinθ=√3
Hence tanθ=-√3 ⇒ tanθ=tan2π3.Hence θ=2π3
Question 12
If z=reiθ,then |eiz|=
ersinθ
e−rsinθ
e−rcosθ
ercosθ
SOLUTION
Solution : B
if z=reiθ=r(cosθ + isinθ)
⇒ iz=ir(cosθ + isinθ)=-rcosθ + irsinθ
or eiz = e−rcosθ+irsinθ=e−sinθericosθ
or |eiz|=|e−rsinθ||eircosθ|
=e−rsinθ[cos2(rcosθ)+sin2(rcosθ)]2=e−rsinθ
Question 13
Real part of eeiθ is
ecosθ [cos(sinθ)]
ecosθ [cos(cosθ)]
esinθ [cos(sinθ)]
esinθ [sin(sinθ)]
SOLUTION
Solution : A
eeiθ=ecosθ+isinθ=ecosθeisinθ=ecosθ[cos(sinθ)+isin(sinθ)]∴Real part ofeeiθ is ecosθ[cos(sinθ)]
Question 14
If z1,z2 and z3 are complex numbers such that
|z1|=|z2|=|z3|=∣∣1z1+1z2+1z3∣∣=1,
then |z1+z2+z3|
Equal to 1
Less than 1
Greater than 3
Equal to 3
SOLUTION
Solution : A
1=∣∣1z1+1z2+1z3∣∣=∣∣∣z1∗¯z1z1+z2∗¯z2z2+z3∗¯z3z3∣∣∣
(hence, |z1|2=1=z1¯¯¯¯¯z1, etc)
⇒ |z1+z2+z3|=|¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯z1+z2+z3|= |z1+z2+z3|
(hence,|¯¯¯¯¯z1|=|z1|)
Question 15
Find the complex number z satisfying the equations ∣∣z−12z−8i∣∣=53, ∣∣z−4z−8∣∣=1
6
6±8i
6+8i, 6+17i
None of these
SOLUTION
Solution : C
We have ∣∣z−12z−8i∣∣=53, ∣∣z−4z−8∣∣=1
Let z=x+iy, then
∣∣z−12z−8i∣∣=53 ⇒ 3|z-12| = 5|z-8i|
⇒ 9(x−12)2+9y2 = 25x2+25(y−8)2 ⋯⋯(i)
∣∣z−4z−8∣∣=1 ⇒ |z-4| = |z-8|
⇒ (x−4)2+y2 = y2+(x−8)2 ⇒ x=6
Putting x=6 in (i), we get y2-25y+136=0
y=17, 8
Hence z=6+17i or z=6+8i
Trick: Check it with options.