Free Objective Test 02 Practice Test - 11th and 12th
Question 1
1+7i(2−i)2=
√2 [cos3π4+isin3π4]
√2 [cosπ4+isinπ4]
[cos3π4+isin3π4]
None of these
SOLUTION
Solution : A
1+7i(2−i)2=(1+7i)(3+4i)(3−4i)(3+4i)=−25+25i25= -1+i
Let z=x+iy = -1+i
∴ rcosθ =-1 And rsinθ =1 ∴ θ=3π4 and r=√2
Thus 1+7i(2−i)2=√2 [cos3π4+isin3π4]
Alter: ∣∣∣1+7i(2−i)2∣∣∣=∣∣1+7i3−4i∣∣=√2
And arg (1+7i(2−i)2)=tan−17−tan−1(−43)
=tan−17+tan−1(43)=3π4
∴ 1+7i(2−i)2=√2 [cos3π4+isin3π4]
Question 2
If z = 3+5i, then z3 + ¯z + 198 =
-3-5i
-3+5i
3+5i
3-5i
SOLUTION
Solution : C
z = 3+5i , ¯z = 3-5i
⇒ z3=(3+5i)3=33+(5i)3+3.3.5i(3+5i)
= -198+10i
Hence, z3+¯z + 198 = 10i-198+3-5i+198 = 3+5i .
Question 3
The least positive integer n which will reduce (i−1i+1)nto a real number , is
2
3
4
5
SOLUTION
Solution : A
(i−1i+1×i−1i−1)n=(−2i−2)n=in
Hence, to make the real number the least positive integar is 2.
Question 4
If a=2√2i then which of the following is correct?
a=1+i
a=√2×(1+i)
1
None of these
SOLUTION
Solution : A
We have a =√2i=√2(cosπ2+isinπ2)12=√2(eiπ2)12=√2eiπ4=√2(1√2+i√2i)=1+i
Trick:Check with options.
Question 5
The value of i1+3+5+......+(2n+1) is
i if n is even, - i if n is odd
1 if n is even, - 1 if n is odd
1 if n is odd, i if n is even
i if n is even, - 1 if n is odd
SOLUTION
Solution : C
Let z = i[1+3+5+......+(2n+1)]
Clearly series is A.P with common difference = 2
∵ Tn = 2n-1 And Tn+1 = 2n+1
So, number of terms in A.P. = n+1
Now, sn+1=n+12 [2.1+(n+1-1)2]
⇒ Sn+1=n+12[2+2n]=(n+1)2 i.e., i(n+1)2
Now put n = 1,2,3,4,5,........
n = 1, z = i4 = 1, n = 2, z = i5 = -1 ,
n = 3, z = i8 = 1, n = 4, z = i10 = -1 ,
n = 5, z = i12 = 1, .........
Question 6
If z is a complex number ¯¯¯¯¯¯¯¯z−1(¯z) = , then
1
-1
0
None of these
SOLUTION
Solution : A
Let z = x + iy, ¯z = x -iy and z−1=1x+iy
⇒ ¯¯¯¯¯¯¯¯z−1=x+iyx2+y2 ; ∴ ¯¯¯¯¯¯¯¯z−1¯z=x+iyx2+y2(x−iy) = 1
Question 7
The complex numbers sinx+icos2x and cosx-isin2x are conjugate to each other for
x = nπ
x = (n+12)π
x=0
No value of x
SOLUTION
Solution : D
sinx+icos2x and cosx-isin2xare conjugate to each other if sinx=cosx and cos2x=sin2x
Or tan x = 1 ⇒ x = π4,5π4,9π4, .......... ...............(i)
And tan 2x = 1 ⇒ 2x = π4,5π4,9π4, ..........
or x = π8,5π8,9π8, .......... ...............(ii)
There exists no value of x common in (i) and (ii). Therefore there is no value of x for which the given complex numbers are conjugate.
Question 8
If a=cosα+isinα, b=cosβ+isinβ,
c=cosγ+isinγ and bc+ca+ab=1,Then
cos(β-γ)+cos(γ-α)+cos(α-β) is equal to
SOLUTION
Solution : D
bc=cosβ+isinβcosγ+isinγ×cosγ−isinγcosγ−isinγ
⇒bc=cos(β−γ)+isin(β−γ).....(i)
Similarly, ca=cos(γ−α)+isin(γ−α).......(ii)
And ab=cos(α−β)+isin(α−β).......(ii)
Equating real and imaginary parts,
cos(β−γ)+cos(γ−α)+cos(α−β)=1
Question 9
If cosα+cosβ+cosγ=0=sinα+sinβ+sinγ then
cos2α+cos2β+cos2γ equals
2cos(α+β+γ)
cos2(α+β+γ)
0
1
SOLUTION
Solution : C
cosα+cosβ+cosγ=0 ......(i)
sinα+sinβ+sinγ=0 .......(ii)
Let a=cosα+isinα, b=cosβ+isinβ
c=cosγ+isinγ
⇒ a+b+c=0 .......(iii)
1a+1b+1c
=[cosα+isinα]−1+[cosβ+isinβ]−1+[cosγ+isinγ]−1
=cosα-isinα+cosβ-isinβ+cosγ-isinγ
⇒ ab+bc+ca=0 .....(iv) [by (i) and (ii)]
Squaring both sides of equations (iii),
We get a2+b2+c2+2ab+2bc+2ca=0
or a2+b2+c2 [by (iv)]
[cosα+isinα]2+[cosβ+isinβ]2+[cosγ+isinγ]2=0
⇒ (cos2α+cos2β+cos2γ) + i(sin2α+sin2β+sin2γ)
Equating the real part to zero, we have
⇒cos2α+cos2β+cos2γ=0
Question 10
If sinα+sinβ+sinγ=0= cosα+cosβ+cosγ,
value of sin2α+sin2β+sin2γ
23
32
12
1
SOLUTION
Solution : B
cosα+cosβ+cosγ = 0............(i)
sinα+sinβ+sinγ =0...........(ii)
Let a=cosα+isinα, b=cosβ+isinβ
c=cosγ+isinγ
⇒ a+b+c=0 [By (i) and (ii)]....(iii)
1a+1b+1c
=cosα-isinα+cosβ-isinβ+cosγ-isinγ
⇒ ab+bc+ca=0 ......(iv) [by (i) and (ii)]
Squaring both sides of equation (iii),
we get a2+b2+c2+2ab+2bc+2ca=0
or a2+b2+c2=0 [by (iv)]
[cosα+isinα]2 + [cosβ+isinβ]2 + [cosγ+isinγ]2 = 0
⇒ (cos2α+cos2β+cos2γ) + i(sin2α+sin2β+sin2γ)=0
Separation of real and imaginary part,
⇒ cos2α+cos2β+cos2γ=0 .........(v)
And sin2α+sin2β+sin2γ =0 .........(iv)
⇒ 1-2sin2α+1-2sin2β+1-2sin2γ=0 [By eq.(v)]
Therefore, sin2α+sin2β+sin2γ= 32
Question 11
The maximum distance from the origin of coordinates to the point z satisfying the equation ∣∣z+1z∣∣=a is
12(2√(a2+1)+a)
12(2√(a2+2)+a)
12(2√(a2+4)+a)
None of these
SOLUTION
Solution : C
let z=r (cosθ+isinθ)
Then ∣∣z+1z∣∣=a ⇒ ∣∣z+1z∣∣2=a2
⇒ r2+1r2+2cosθ = a2 ⋯ ⋯(i)
Differentiating w.r.t θ we get
2rdrdθ-2r3drdθ-4sin2θ
Putting drdθ=0, we get θ=0,π2
r is maximum for θ = 0, π2, therefore from (i)
r2+1r2−2=a2⇒r−1r=a⇒r=a+2√a2+42
Question 12
√3+i=(a+ib)(c+id),thentan−1ba+tan−1dc
has the value
2nπ+π3, nϵI
nπ+π6, nϵI
nπ-π3, nϵI
2nπ-π6, nϵI
SOLUTION
Solution : B
√3+i=(a+ib)(c+id)
∴ac−bd=√3 and ad+bc=1
Now tan−1(ba)+tan−1(dc)
= tan−1(ab+dc1−ba.dc)= tan−1(bc+adac−bd)= tan−1(1√3)
=nπ+π6, nϵI
Question 13
The number of non-zero integral solutions of the equation |1−i|x=2x is
zero
1
2
None of these
SOLUTION
Solution : A
Since1-i = √2 [cosπ4−isinπ4], |1-i|
∴ |1−i|x=2x ⇒ (√2)x=2x ⇒ 2x/2 =2x
⇒ x2=x then x=0.
Therefore, the number of non-zero integral solutions is nil or Zero.
Question 14
If c+ic−i = a+ib, where a,b,c are real, then a2+b2 =
1
-1
c2
−c2
SOLUTION
Solution : A
c+ic−i = a+ib .........(i)
∴ c−ic+i = a-ib .........(ii)
Multiplying (i) and (ii), we get
c2+1c2+1=a2+b2⇒a2+b2=1.
Question 15
The square root of 3 - 4i is
±(2+i).
±(2-i).
±(1-2i).
±(1+2i).
SOLUTION
Solution : B
Let √3−4i=x+iy ⇒ 3-4i= x2-y2+2ixy
⇒ x2-y2=3, 2xy=-4 ..........(i)
⇒ (x2+y2)2=(x2−y2)2+4x2y2=(3)2+(−4)2=25
⇒ (x2+y2)2=5 ............(ii)
From equation (i) and (ii) ⇒ x2=4 ⇒x=±2,
y2=1 ⇒y=±1.Hence the square root of (3-4i)
is ±(2-i).