air force x group - amar ujala
TRANSCRIPT
Q2. If x − iy = √𝑎−𝑖𝑏
𝑐−𝑖𝑑 then find the value of (x²+y²)²
;fn x − iy = √𝑎−𝑖𝑏
𝑐−𝑖𝑑 rks (x²+y²)² dk eku gksxk\
(a) 𝑎2+𝑏2
𝑐2+𝑑²
(b) 𝑎2−𝑏2
𝑐2−𝑑²
(c) 𝑐2+𝑑²
𝑎2+𝑏²
(d) 𝑐2−𝑑²
𝑎2+𝑏²
Q3. Find the modulus of 1+𝑖
1−𝑖−
1−𝑖
1+𝑖
1+𝑖
1−𝑖−
1−𝑖
1+𝑖 dk ekikad D;k gS
(a) 1
(b) 2
(c) -1
(d) -2
Q5. If and are imaginary cube roots of unity then
4 +
4 +1
is
;fn o bdkbZ ds lfEeJ ewy gS rks 4 +
4 +1
dk eku gSA
(a) 3
(b) 0
(c) 1
(d) 2
Q6. If 𝑧 =3+𝑖
3−𝑖 The find amplitude of z is
;fn 𝑧 =3+𝑖
3−𝑖 rks z dk dks.kkad D;k gksxk
(a) 𝜋
2
(b) 𝜋
3
(c) 𝜋
6
(d) 𝜋
4
Q7. If 1+2𝑖
2+𝑖= 𝑟 (𝑐𝑜𝑠 + 𝑖 𝑠𝑖𝑛) then
(a) 𝑟 = 1, = tan−1 3
4
(b) 𝑟 = 5 = tan−1 4
3
(c) 𝑟 = 1, = tan−1 4
3
(d) N.O.T
Q8. If z=x+iy satisfie amp (z-1)=amp (z+3i) then the
value of (x-1) : y is equal to
;fn z=x+iy amp (z-1)=amp (z+3i) dks lUrq’V djrk gS rks (x-1)
: y dk eku gksxk
(a) 2 : 1
(b) 1 : 3
(c) – 1: 3
(d) 2 : 3
Q9. If is an imaginary cube root of uniy then
(1 + 𝜔 − 𝜔2)7 equals
;fn bdkbZ dk ?kuewy gS rks (1 + 𝜔 − 𝜔2)7 dk eku D;k gksxkA
(a) 128
(b) -128
(c) 128²
(d) -128²
Q10. If is a cube root of unity then find (1-) (1-²)
(1+4) (1+8)
;fn bdkbZ dk ?kuewy gS rks (1-) (1-²) (1+4) (1+8) dk eku
gSA
(a) 0
(b) 1
(c) 2
(d) 3
Q16. If 1, , ² be cube roots of unity, then (1-+²)
(1-²+4) (1-4+8)… to 2n factors is equal to:
(a) 2n
(b) 22n
(c) 0
(d) 1
Q18. If x=a+b, y=a+b² and z= a²+b, then
(x3+y3+z3) is equal to:
(a) 3(a3+b3)
(b) 3abc
(c) a3+b3+c3
(d) 0
Q19. If 1, , ² be cube roots of unity, then the
value of (2+5+2²)6 is
(a) 576
(b) 625
(c) 729
(d) None
Q20. If 1, , ² be cube roots of unity, then the (2-)
(2-²) (2-10)(2-11) is (a) 49
(b) 36
(c) 56
(d) None