COMPLEX NUMBERS
Page 13.1 Ex – 13.1
Q1.
Answer :
i i457 =i4×114+1 =i4114×i =i ∵i4=1ii i528 =i4×132 =i4132 =1 ∵i4=1iii 1i58=1i4×14+2 =1i414×i2 =1i2 ∵i4=1 =-1 ∵i2=-1iv i37+1i67=i4×9+1+1i4×16+3 =i49×i+1i416×i3 =i-1i ∵i3=-i =i-1i×ii =i-ii2 =i–i ∵i2=-1 =2i v i41+1i2579 =i4×10+1+1i4×64+19 =i410×i+1i464×i9 =i+1i9 ∵ i4=1
=i+ii29 =i-i9 ∵i2=-1=0
vi i77+i70+i87+i4143=i4×19+1+i4×17+2+i4×21+3+i4×103+23 =i419×i+i417×i2+i421×i3+i4103×i2 =i-1-i-13 ∵i4=1,i3=-i and i2=-1 =-23 =-8vii i30+i40+i60=i4×7+2+i4×10+i4×15 =i47×i2+i410+i415 =-1+1+1 ∵ i4=1,i2=-1 =1 viii i49+i68+i89+i110=i4×12+1+i4×17+i4×22+1+i4×27+2 =i412×i+i417+i422×i+i427×i2 =i+1+i-1 ∵ i4=1,i2=-1 =2i
Q2.
Answer :
1+i10+i20+i30=1+i4×2+2+i4×5+i4×7+2=1+i42×i2+i45+i47×i2=1+i2+1+i2 ∵i4=1=1-1+1-1 ∵ i2=-1=0This is a real number.Hence proved.
Q3.
Answer :
i i49+i68+i89+i110=i4×12+1+i4×17+i4×22+1+i4×27+2=i412×i+i417+i422×i+i427×i2=i+1+i+i2 ∵i4=1=2i+1-1 ∵i2=-1 =2iii i30+i80+i120=i4×7+2+i4×20+i4×30=i47×i2+i420+i430=i2+1+1 ∵i4=1=-1+2 ∵i2=-1 =1iii i+i2+i3+i4=i-1-i+1 ∵i2=-1, i3=-i and i4=1=0 iv i5+i10+i15=i4×1+1+i4×2+2+i4×3+3=i41×i+i42×i2+i43×i3=i+i2+i3 ∵i4=1=i-1-i ∵i2=-1, i3=-i =-1v i592+i590+i588+i586+i584i582+i580+i578+i576+i574=i4×148+i4×147+2+i4×147+i4×146+2+i4×146i4×145+2+i4×145+i4×144+2+i4×144+i4×143+2=i4148+i4147×i2+i4146+i4146×i2+i4146i4145×i2+i4145+i4144×i2+i4144+i4143×i2=1+i2+1+i2+1i2+1+i2+1+i2 ∵i4=1=1-1+1-1+1-1+1-1+1-1 ∵i2=-1=-1
(vi) 1+i2+i4+i6+i8+…+i20∵i2=-1,i4=1,i6=-1,i8=1,…..i20=1∴ 1+i2+i4+i6+i8+…+i20=1+-1+1+-1+1+-1+…+1+-1+1=5×1+-1+1 As, there are 11 terms=5×0+1=1
Page 13.30 Ex – 13.2
Q1.
Answer :
i 1+i 1+2i=1+2i+i+2i2=1+3i-2 ∵i2=-1 =-1+3iii 3+2i-2+i=3+2i-2+i×-2-i-2-i=-6-3i-4i-2i24-i2 ∵i2=-1=-6-7i+24+1=-4-7i5=-45-75iiii12+i2=14+i2+4i ∵i2=-1=13+4i=13+4i×3-4i3-4i=3-4i9-16i2=3-4i9+16=325-425i
iv1-i1+i=1-i1+i×1-i1-i=1+i2-2i1-i2 ∵ i2=-1=-2i2=-iv2+i32+3i=4+i2+4i2+i2+3i ∵ i2=-1=8+2i2+8i+4i+i3+4i22+3i =2+11i2+3i×2-3i2-3i=4-6i+22i-33i24-9i2=37+16i4+9=3713+1613ivi1+i1+3i1-i=1+i1+3i1-i×1+i1+i=1+3i1+i2+2i1-i2 ∵ i2=-1=1+3i2i2=i1+3i=i+3i2=-3+ivii 2+3i4+5i=2+3i4+5i×4-5i4-5i=8-10i+12i-15i216-25i2 ∵ i2=-1=23+2i16+25=2341+241i
(viii) 1-i31-i3=1+i2-2i1-i1-i3 ∵ i2=-1=-2i1-i1-i3×1+i31+i3=-2i1+i3-i-i41-i6=-2i1-i-i-11-i2=-2i-2i2=-2+0i
(ix) 1+2i-3=11+2i3=11+8i3+6i+12i2=11-8i+6i-12 ∵i2=-1 & i3=-i=1-2i-11=1-2i-11×-2i+11-2i+11=-2i+114i2-121=-2i+11-4-121=-2i+11-125=-11125+2i125
(x) 3-4i4-2i1+i=3-4i4+2i-2i2 ∵i2=-1=3-4i6+2i=3-4i6+2i×6-2i6-2i=18-6i-24i+8i236-4i2=18-30i-836+4 =10-30i40=14-34i
(xi) 11-4i-21+i3-4i5+i=1+i-2+8i1-4i1+i3-4i5+i=-1+9i1-3i+4i23-4i5+i=-1+9i5-3i3-4i5+i ∵i2=-1=-3+4i+27i-36i225+5i-15i-3i2=33+31i28-10i=33+31i28-10i×28+10i28+10i=924+330i+868i+310i2784-100i2=614+1198i884=307442+599442i
(xii) 5+2i1-2i=5+2i1-2i×1+2i1+2i=5+52i+2i+2i21-2i2=5+62i-21+2 ∵i2=-1=3+62i3=1+22i
Q2.
Answer :
i x+iy2-3i=4+i2x-3ix+2iy-3i2y=4+i2x+3y+i-3x+2y=4+iComparing both the sides: 2x+3y=4 ….(1) -3x+2y=1 ….(2)Multiplying equation (1) by 3 and equation (2) by 2: 6x+9y=12 …(3)-6x+4y=2 …(4)Adding equations (3) and (4):13y=14y=1413Substituting the value of y in equation (1): 2x+3×1413=4⇒2x=4-4213⇒2x=1013⇒x=513∴ x=513 and y=1413
(ii) 3x-2iy2+i2=10 1+i⇒ 3x-2iy4+i2+4i=101+i⇒ 3x-2iy3+4i=101+i⇒9x+12xi-6iy-8i2y=10+10i⇒9x+8y+i12x-6y=10+10iComparing both the sides:9x+8y=10 ….(1)12x-6y=10or, 6x-3y=5 …(2)Multiplying equation (1) by 3 and equation (2) by 8,27x+24y=30 ….(3) 48x-24y=40 ….(4)Adding equations (3) and (4): 75x=70∴ x=1415Substituting the value of x in equation (1): 9×1415+8y=10⇒12615+8y=10⇒8y=10-12615⇒8y=2415⇒y=15
(iii) 1+ix-2i3+i+2-3iy+i3-i=i⇒1+i3-ix-2i3-i+2-3i3+iy+i3+i3+i3-i=i⇒3x-ix+3ix-i2x-6i+2i2+6y+2iy-9iy-3i2y+3i+i29-i2=i⇒4x+2ix-3i+9y-7iy-310=i⇒4x+9y-3+i2x-3-7y=10iComparing both the sides: 4x+9y-3=0⇒4x+9y=3 ….(1) 2x-3-7y=10⇒2x-7y=13 …(2)Multiplying equation (2) by 2: 4x-14y=26 …(3) Subtracting equation (3) from (1): 4x+9y= 3 4x-14y=26 – + – 23y=-23∴ y=-1Substituting the value of y in equation (1): 4x-9=3⇒4x=12⇒x=3∴ x=3 and y=-1
(iv) 1+ix+iy=2-5i⇒x+iy+ix+i2y=2-5i⇒x+iy+ix-y=2-5i⇒x-y+iy+x=2-5iComparing both the sides,x-y=2 …(1) x+y=-5 …(2)Adding equations (1) and (2), 2x=-3⇒x=-32Substituting the value of x in equation (1),-32-y=2⇒y=-32-2⇒y=-72∴x=-32 and y=-72
Page 13.31 Ex – 13.2
Q3.
Answer :
i Let z=4-5i ∴ z¯=4+5i z=a+ib, so z¯=a-ibii Let z=13+5i =13+5i×3-5i3-5i =3-5i9-25i2 =3-5i9+25 =3-5i34∴ z¯=3+5i34iii Let z=11+i =11+i×1-i1-i =1-i1-i2 =1-i2⇒ z¯=1+i2iv Let z=3-i22+i =9-6i-12+i =8-6i2+i×2-i2-i =16-8i-12i+6i24-i2 =10-20i5 =2-4i∴ z =2+4iv Let z=1+i2+i3+i =2+i+2i+i23+i =1+3i3+i =1+3i3+i×3-i3-i =3-i+9i-3i29-i2 =6+8i10 =3+4i5⇒ z¯=3-4i5vi Let z=3-2i2+3i1+2i2-i =6+9i-4i-6i22-i+4i-2i2 =6+6+5i2+2+3i =12+5i4+3i×4-3i4-3i =48-36i+20i-15i216-9i2 =63-16i25∴ z=63+16i25
Q4.
Answer :
i Let z=1-i .Then ,1z=11-i =11-i×1+i1+i =1+i1-i2 =121+i =12+12iii z=1+3i2 =1+3i2+23i =-2+23iThen, 1z=1-2+23i×-2-23i-2-23i =-2-23i4-12i2 =-2-23i16 =-18-38iiii z=4-3iThen, 1z=14-3i×4+3i4+3i =4+3i16-9i2 =4+3i25 =425+325iiv z=5+3iThen, 1z=15+3i×5-3i5-3i =5-3i5-9i2 =5-3i5+9 =5-3i14 =514-314i
Q5.
Answer :
i x=3-5i2⇒x2=3-5i22 =9+25i2-30i4 =-16-30i4⇒x3=-16-30i4×3-5i2 =-48+80i-90i+150i28 =-198-10i8∴ 2×3+2×2-7x+72=2-198-10i8+2-16-30i4-73-5i2+72 =-198-10i-32-60i-42+70i+2884 =16 4 =4ii x=3+2i⇒x2=3+2i2 =9+4i2+12i =5+12i⇒x3=x2×x =5+12i×3+2i =15+10i+36i-24 =-9+46i⇒x4=x22 =5+12i2 =25+144i2+120i =-119+120i⇒x4-4×3+4×2+8x+44=-119+120i-4-9+46i+45+12i +83+2i+44 =-119+120i+36-184i+20+48i+24+16i+44 = 5iii x=-1+2i⇒x2=-1+2i2 =1+2i2-22i =-1-22i⇒x3=-1-22i×-1+2i =1-2i+22i-4i2 =5+2i⇒x4=-1-22i2 =1+8i2+42i =-7+42i⇒x4+4×3+6×2+4x+9=-7+42i+45+2i+6-1-22i+4-1+2i+9 =-7+42i+20+42i-6-122i-4+42i+9 =12iv x=1+i2⇒x2=1+i22 =1+i2+2i2 =2i2 =i⇒x6=x23 =i3 =-i⇒x2=i⇒x4=x22 =i2 =-1Now, x6+x4+x2+1=-i-1+i+1 =0
Q6.
Answer :
Given: z1=2-i, z2=1+i∴ z1+z2+1z1-z2+i=2-i+1+i+12-i-1-i+i =41-i =41-i Also, 1-i=12+i2 ∵a+bi=a2+b2 =2 ∴z1+z2+1z1-z2+i=42
Q7.
Answer :
i z1=2-i, z2=-2+i, z1=2+i∴ z1z2z1=2-i-2+i2+i =-4+2i+2i-i22+i =-3+4i2+i =-3+4i2+i×2-i2-i =-6+3i+8i-4i222-i2 =-2+11i4–1 =-2+11i5Rez1z2z1=-25ii1z1z1=12-i2+i =122-i2 =15Im1z1z1=0 Since no term containing i is present
Q8.
Answer :
1+i1-i-1-i1+i=1+i1+i-1-i1-i1-i1+i=1+i2+2i-1-i2+2i12-i2=4i2 ∵i2=-1=2i∴ 2i=02+22 =2 ∵a+bi=a2+b2⇒1+i1-i-1-i1+i=2
Q9.
Answer :
x+iy=a+iba-ibTaking mod on both the sides:x+iy=a+iba-ib⇒x2+y2=a2+b2a2+b2⇒x2+y2=1⇒x2+y2=1Hence proved.
Q10.
Answer :
1+i1-in=1+i1-i×1+i1+in=1+i2+2i1-i2n=1-1+2i1+1n=2i2n=inFor in to be real, the least positive value of n will be 2. As i2=-1
Q11.
Answer :
1+icosθ1-2icosθ=1+icosθ1-2icosθ×1+2icosθ1+2icosθ=1+2icosθ+icosθ-2cosθ1+4cos2θ=1-2cosθ+i3cosθ1+4cos2θFor it to be purely real, the imaginary part must be zero.3cosθ=0This is true for odd multiples of π2.∴θ=2n+1π2, n∈Z
Q12.
Answer :
1+im1-im-2=1+im1-im×1-i2=1+i1-i×1+i1+im×1+i2-2i=1+i2+2i1-i2m×1-1-2i=1-1+2i1+1m×-2i=-2iim=-2im+1For this to be real, the smallest positive value of m will be 1. Thus, i1+1=i2=-1, which is real.
Page 13.35 Ex – 13.3
Q1.
Answer :
z=±z+Rez2+iz-Rez2 , if Im(z) >0z=±z+Rez2-iz-Rez2 , if Im(z) <0i z=-5+12i , Rez=-5 and z=25+144=13 Here, Im(z) >0∴ z=±z+Rez2+iz-Rez2 =±13-52+i13+52 =±4+i9 =±2+3iii z=-7-24i, Rez=-7, z=49+576=25 Here, Im(z) <0 z=±z+Rez2-iz-Rez2 =±25-72-i25+72 =±3-4iiii z=1-i , Rez=1, z=1+1=2Here, Im(z) <0∴ z=±z+Rez2-iz-Rez2 =±2+12-i2-12iv z=-8-6i, Rez=-8, z=64+36=10Here, Im(z) <0∴ z=±z+Rez2-iz-Rez2 =±10-82-i10+82 =±1-3iv z=8-15i, Rez=8, z=64+225=17Here, Im(z) <0∴ z=±z+Rez2-iz-Rez2 =±17+82-i17+82 =±125-3ivi -11-60-1=-11-60i, Rez=-11, z=121+3600=61Here, Im(z) <0 ∴ z=±z+Rez2-iz-Rez2 =±61-112-i61+112 =±5-6ivii z=1+43-1=1+43i, Rez=1, z=1+16×3=7 Here, Im(z) >0∴ z=±z+Rez2+iz-Rez2 =±7+12+i7-12 =±2+3iviii z=0+4i, Rez=0,……
Page 13.46 Ex – 13.4
Q1.
Answer :
i z=1+i r=z =1+1 =2Let tan α=ImzRez⇒tan α =11 ⇒ α =π4Since point (1,1) lies in the first quadrant, the argument of z is given by θ= α =π4Polar form =rcos θ+isin θ =2cosπ4+isinπ4ii z=3+ir=z =3+1 =4 =2Let tan α=ImzRez⇒tan α =13 ⇒ α =π6Since point (3,1) lies in the first quadrant, the argument of z is given by θ= α =π6
Polar form =r cosθ +isinθ =2 cos π6+isin π6
(iii) z=1-i r=z =1+1 =2Let tan α=ImzRez∴ tanα =-11 =π4⇒α =π4Since point (1,-1) lies in the fourth quadrant, the argument of z is given by θ=-α =-π4Polar form =rcos θ+isin θ =2cos-π4+isin-π4 =2cosπ4-isinπ4
(iv) 1-i1+iRationalising the denominator:1-i1+i×1-i1-i⇒ 1+i2-2i1-i2 ⇒-2 i2 ∵ i2=-1⇒-ir=z =0+1 =1Since point (0,-1) lies on the negative direction of the imaginary axis, the argument of z is given by 3π2.Polar form =rcos θ+isin θ =cos3π2+isin3π2 = cos2π-π2+isin2π-π2 =cosπ2-isinπ2
(v) 11+iRationalising the denominator: 11+i×1-i1-i⇒ 1-i1-i2 ⇒1-i2 ∵ i2=-1⇒12-i2r=z =14+14 =12Let tan α =Im(z)Re(z) ∴ tan α=12-12 =1 ⇒ α =π4Since point 12,-12 lies in the fourth quadrant, the argument is given by θ=-α=-π4Polar form =rcos θ+isin θ =12cos-π4+isin-π4 =12 cosπ4-isinπ4
(vi) 1+2i1-3iRationalising the denominator:1+2i1-3i×1+3i1+3i⇒ 1+3i+2i+6i21-9i2 ⇒-5+5i10 ∵ i2=-1⇒-12+i2r=z =14+14 =12Let tan α =Im(z)Re(z) Then, tan α=12-12 =1 ⇒ α =π4Since point -12,12 lies in the second quadrant, the argument is given byθ=π-α =π-π4 =3π4Polar form =rcos θ+isin θ =12cos3π4+isin3π4
(vii) sin 120°-i cos 120° 32+i2r=z =34+14 =1Let tan α=Im(z)Re(z)Then, tan α=1232 =13⇒α=π6Since point 32,12 lies in the first quadrant, the argument is given by θ=α=π6Polar form =rcosθ+i sinθ = cos π6+i sin π6
(viii) -161+i3Rationalising the denominator: -161+i3×1-i31-i3⇒ -16+163i1-3i2 ⇒-16
Q2.
Answer :
(i) Let z= 1+itan α ∵ tan α is periodic with period π. So, let us take α ∈[0,π2)∪( π2, π]Case I:When α∈[0,π2)z= 1+itan α ⇒z=1+tan2α =sec α ∵ 0<α<π2 =sec αLet β be an acute angle given by tan β=Im (z)Re(z)tan β=tan α =tan α⇒β=α As z lies in the first quadrant . Therefore, arg(z)=β=αThus, z in the polar form is given by z= sec α cosα+isin α Case II:z=1+i tan α ⇒z=1+tan2α =sec α ∵π2<α<π =-sec αLet β be an acute angle given by tan β=Im (z)Re(z)tan β=tan α =-tan α⇒tan β =tan π-α⇒β=π-αAs, z lies in the fourth quadrant . ∴ arg(z)=-β= α-πThus, z in the polar form is given by z= -sec α cosα-π+isin α-π
(ii) Let z= tan α-i ∵ tan α is periodic with period π. So, let us take α ∈[0,π2)∪( π2, π]Case I:z= tan α-i ⇒z=tan2+1 =sec α ∵ 0<α<π2 =sec αLet β be an acute angle given by tan β=Im (z)Re(z)tan β=1tan α = cot α =cot α =tan π2-α⇒β=π2-α We can see that Re(z) >0 and Im (z) <0.So, z lies in the fourth quadrant .∴ arg(z)=-β=α- π2Thus, z in the polar form is given by z= sec α cosα-π2+isin α-π2 Case II:z= tan α-i ⇒z=tan2+1 =sec α ∵π2<α<π =-sec αLet β be an acute angle given by tan β=Im (z)Re(z)tan β=1tan α = cot α =-cot α =tan α-π2⇒β=α- π2We can see that Re(z) <0 and Im (z) <0.So, z lies in the third quadrant .∴ arg(z)=π+β= π2+αThus, z in the polar form is given by z= -sec α cosπ2+α+isin π2+α
(iii) Let z= 1-sinα+icosα.∵sine and cosine functions are periodic functions with period 2π. So, let us take α ∈[0, 2π]Now, z=1-sinα+icosα⇒z=1-sinα2+cos2α=2-sinα=21-sinα⇒z=2cosα2-sinα22=2cosα2-sinα2Let β be an acute angle given by tanβ=ImzRez.Then,tanβ=cosα1-sinα=cos2α2-sin2α2cosα2-sinα22=cosα2+sinα2cosα2-sinα2⇒tanβ=1+tanα21-tanα2=tanπ4+α2Case I: When 0≤α<π2In this case, we have,cosα2>sinα2 and π4+α2∈[π4, π2)⇒z=2cosα2-sinα2and tanβ=tanπ4+α2=tanπ4+α2⇒β=π4+α2Clearly, z lies in the first quadrant.Therefore, argz=π4+α2Hence, the polar form of z is 2cosα2-sinα2cosπ4+α2+isinπ4+α2Case II: When π2<α<3π2In this case, we ……
Page 13.46 (Very Short Answers)
Q1.
Answer :
Let the square root of i be x+iy.⇒i=x+iy⇒i=x2+y2i2+2ixy⇒i=x2-y2+2ixy Squaring both the sidesComparing both the sides:x2-y2=0 …(i) and 2xy=1 …iiBy equation (ii), we find that x and y are of the same sign.From equation i, x=±y∴ xy=12, x2=12x=±12, y=±12∴i=±121+i
Page 13.47 (Very Short Answers)
Q2.
Answer :
Let -i=x+iySquaring both the sides-i=x2+y2i2+2ixy⇒2xy=-1 …iand x2-y2=0 …iiEquation ii shows that x and y are of opposite sign. From ii, x=±yFrom i, 2x-x=-12⇒x2=12⇒x=±12 Since x and y have opposite signs, y=-12 when x=12and vice versa∴-i=±121-i
Q3.
Answer :
x+iy=a+ibc+idTaking modulus on both the sides, x+iy=a+ibc+id⇒x+iy=a+ibc+id⇒x2+y2=a2+b2c2+d2 ∵x+iy=x2+y2Squaring both the sides,x2+y2=a2+b2c2+d2Squaring again, we get,x2+y22=a2+b2c2+d2
Q4.
Answer :
π<θ<2π π2<θ2<π Dividing by 2z=1+cosθ+isinθ⇒z=1+cosθ2+sin2θ⇒z=1+cos2θ+2cosθ+sin2θ⇒z=1+1+2cosθ⇒z=21+cosθ⇒z=2×2cos2θ2⇒z=2cos2θ2⇒z=-2cosθ2 Since π2<θ2<π , cosθ2 is negative
Q5.
Answer :
i4n+1-i4n-12=i-1i2 ∵i4n=1, i-1=1i=i2-1i2=i2-12i=-1-12i=-2-2i =-1i=-ii2 ∵i2=-1=-i-1=i
Q6.
Answer :
i592+i590+i588+i586+i584i582+i580+i578+i576+i574=i4×148+i4×147+2+i4×147+i146×4+2+i4×146i4×145+2+i4×145+i4×144+2+i4×144+i4×143+2=1+i2+1+i2+1i2+1+i2+1+i2 ∵i4=1=1-1+1-1+1-1+1-1+1-1 ∵i2=1=1-1=-1
Q7.
Answer :
z=1-i r=z =1+1 =2Let tan α=ImzRez∴ tanα =-11 =π4⇒α =π4Since point (1,-1) lies in the fourth quadrant, the argument of z is given by θ=-α =-π4Polar form =rcos θ+isin θ =2cos-π4+isin-π4 =2cosπ4-isinπ4
Q8.
Answer :
Let z=-1+3i. Then , r=z=-12+32 =2Let tan α=Im(z) Re (z) = 3⇒α =π3Since the point representing z lies in the second quadrant. Therefore, the argument of z is given by θ=π-α =π-π3 =2π3So, the polar form is rcosθ+isinθ∴ z=2cos2π3+isin2π3
Q9.
Answer :
Let z=-iThen , Rez=0, Imz=-1Since, the point (0,-1) representing z=0-i lies on negative direction of imaginary axis.Therefore, arg (z) =-π2 or 3π2
Q10.
Answer :
1+i1-in=1+i1-i×1+i1+in=1+i2+2i1-i2n=1-1+2i1+1n⇒2i2n=inFor in to be real, the smallest positive value of n will be 2. As, i2=-1, which is real.
Page 13.47 (Multiple Choice Questions)
Q1.
Answer :
(b) 0
(1+ i) (1 + i2) (1 + i3) (1 + i4)
= (1+ i) (1 – 1) (1 – i) (1 + 1) (∵i2 = -1, i3 = -i and i4 = 1)
= (1 + i) (0) (1 – i) (2)
= 0
Q2.
Answer :
(a) π
Given:
3 + 2isinθ1 – 2i sinθ is a real number
On rationalising, we get,
3 + 2i sin θ1 – 2i sin θ×1 + 2i sin θ1 + 2i sin θ = (3 + 2i sin θ) (1 + 2i sin θ)(1)2 – (2i sin θ)2=3 + 2i sin θ + 6i sin θ + 4i2 sin2 θ1 + 4 sin2 θ=3 – 4 sin2 θ + 8i sin θ1 + 4sin2 θ ∵i2=-1=3 – 4 sin2 θ 1 + 4sin2 θ+ i8 sin θ1 + 4sin2 θ
For the above term to be real, the imaginary part has to be zero.
∴8sinθ1+4sin2θ=0⇒8sinθ=0
For this to be zero,
sin θ= 0
⇒ θ = 0, π, 2π, 3π…
But 0<θ<2π
Hence, θ=π
Q3.
Answer :
(c) a2 + b2
(1 + i)(1 + 2i)(1 + 3i) ……(1 + ni) = a + ib
Taking modulus on both the sides, we get:
1+i1+2i1+3i…….1+ni=a+ib1+i1+2i1+3i…….1+ni can be written as 1+i 1+2i 1+3i…….1+ni
12 + 12 × 12 + 22 × 12 + 32 × … × 1 + n2 = a2 + b2
⇒2 × 5 × 10 × … × 1 + n2 = a2 + b2
Squaring on both the sides, we get:
2×5×10×…×(1 + n2) = a2 + b2
Q4.
Answer :
(d) x – iy
a +ib = x + iySquaring on both the sides, we get,a + ib = x2 + (iy)2 + 2ixy⇒a +ib = (x2 – y2) + 2ixy∴ a = (x2 – y2) and b = 2xy∴ a -ib = (x2 – y2) – 2ixy⇒a -ib = x2 + i2y2 – 2ixy ∵i2=-1 Taking square root on both the sides, we get:a -ib = x – iy
Q5.
Answer :
(d) z =32, arg (z)=tan-112
z=cosπ4+isinπ6⇒z=12+12i⇒z=122+14⇒z=12+14⇒z=34⇒z=32tan α=Im(z)Re(z) =12⇒α=tan-112Since, the point z lies in the first quadrant .Therefore, arg(z) =α=tan-112
Q6.
Answer :
(d) cosπ2 -i sinπ2
(i25)3 = (i)75
= (i)4×18+ 3
= (i)3
= -i (∵ i4 = 1)
Let z=0-i Since, the point (0,-1) lies on the negative direction of imaginary axis.Therefore, arg (z) = -π2
Modulus, r = z = 1 = 1
∴ Polar form = r (cos θ + i sin θ)
= cos-π2+i sin-π2
= cosπ2 – i sin π2
Page 13.48 (Multiple Choice Questions)
Q7
Answer :
(d) 0
i+ i2+i3 +i4… i1000 i+ i2+i3 +i4 [∵ i2 =-1, i3=-i and i4=1]= i -1-i+1 = 0 Similarly, the sum of the next four terms of the series will be equal to 0. This is because the powers of i follow a cyclicity of 4.Hence, the sum of all terms, till 1000, will be zero.i+ i2+i3 +i4… i1000 = 0
Q8.
Answer :
(c) 2π3
z = -21 + i3
Rationalising z, we get,
z=-21 + i3×1 – i31 – i3⇒z=-2 + i231 + 3⇒z= -1 + i32 ⇒z= -12 + i32
tan α=Im(z)Re(z) =3⇒α=π3Since, z lies in the second quadrant .Therefore, arg (z) =π-π3 =2π3
Q9.
Answer :
(c) i cotθ2
a = cosθ +isinθ given ⇒1+a1-a = 1+cosθ + isinθ 1-cosθ – isinθ⇒1+a1-a = 1+cosθ +isinθ 1-cosθ – isinθ×1-cosθ + isinθ1-cosθ + isinθ
⇒1+a1-a = 1+isinθ2-cos2θ1-cosθ2-isinθ2
⇒1+a1-a = 1-sin2θ +2isinθ- cos2θ1+cos2θ -2cosθ + sin2θ
⇒1+a1-a = 1-sin2θ+cos2θ + 2isinθ1+sin2θ+cos2θ-2cosθ
⇒1+a1-a = 2isinθ2(1-cosθ)
⇒ 1+a1-a=2isinθ2cosθ22sin2θ2
⇒1+a1-a = icosθ2sinθ2
⇒1+a1-a=i cotθ2
Q10.
Answer :
(c) a2 + b2
(1 + i)(1 + 2i)(1 + 3i) ……(1 + ni) = a + ib
Taking modulus on both the sides, we get,
1+i1+2i1+3i……1+ni=a+ib1+i1+2i1+3i……1+ni can be wriiten as 1+i 1+2i 1+3i….1+ni∴12 + 12×12 + 22 ×12+ 32….×12 + n2 = a2+b2
⇒2×5 ×10….×1 + n2 = a2+b2
Squaring on both the sides, we get:
2×5 ×10….×(1 + n2) = a2+b2
2×5×10×…..(1 + n2) = a2 + b2
Q11.
AAnswer :
(a)a2+144a2+1
x+iy = a2+122a-i
Taking modulus on both the sides, we get:
x2+y2 = a2+124a2+1
Squaring both sides, we get,x2+y2=a2+144a2+1
Q12.
Answer :
(a)π4
Let z = (1+i)
tan α=Im(z) Re(z) =1⇒α=π4Since, z lies in the first quadrant .
Therefore, arg (z) = π4
Q13.
Answer :
(b) 8Let z=2i1+i⇒z=2i1+i×1-i1-i⇒z=2i1-i1-i2⇒z=2i1-i1+1 ∵i2=-1⇒z=2i1-i2⇒z=i-i2⇒z=i+1Now, zn=1+inFor n =2,z2=1+i2 =1+i2+2i =1-1+2i =2i …(1) Since this is not a positive integer,For n=4,z4=1+i4 =1+i22 =2i2 Using (1) =4i2 =-4 …(2)This is a negative integer.For n=8,z8=1+i8 =1+i42 =-42 Using (2) =16This is a positive integer.Thus, z=2i1+in is positive for n=8.Therefore, 8 is the least positive integer such that 2i1+in is a positive integer.
Q14.
Answer :
(a) zz
z2zz = z2z2 (∵ zz = z2)Let z =a+ib⇒ z = a2+ b2Let z = a-ib⇒z = a2+ b2∴ z2zz= z2z2 = zz
Q15.
Answer :
(b) 2i
a = 1 + i
On squaring both the sides, we get,
a<sup>2</sup> = (1 + i)<sup>2</sup>
⇒a<sup>2</sup> = 1 + i<sup>2</sup> + 2i
⇒a<sup>2</sup> = 1-1 + 2i (∵ i<sup>2</sup> = -1)
⇒a<sup>2</sup> = 2i
Q16.
Answer :
(d) none of these
x + iy13 = a + ibCubing on both the sides, we get:x + iy = a + ib3⇒x + iy = a3 + ib3 + 3a2bi + 3aib2⇒x+ iy = a3 + i3b3 + 3a2ib + 3i2ab2⇒x + iy = a3 – ib3 + 3a2ib – 3ab2 (∵ i2 = -1, i3 = -i)⇒x + iy = a3 – 3ab2 + i-b3 + 3a2b∴ x = a3 – 3ab2 and y = 3a2b- b3or ,xa = a2- 3b2 and yb = 3a2 – b2⇒xa+ yb = a2- 3b2 + 3a2 – b2 ⇒xa+ yb = 4a2- 4b2
Q17.
Answer :
(b) -6
-2×-3 = 2×3×-1×-1= 6×i×i = 6×i2= -6 ∵ i2 = -1
Q18.
nswer :
(d) 240°
1-i31+i3Rationalising the denominator, 1-i31+i3×1-i31-i3=1+3i2-23 i1-3i2=-2-23 i4 ∵ i2=-1=-12-i32tan α=Im (z)Re (z)Then, tan α =-32-12 =3 ⇒α=60°Since the points -12,-32 lie in the third quadrant, the argument is given by:θ=180°+60° =240°
Q19.
Answer :
(a) 1
Let z = 1+i1-i
Rationalising the denominator:
z = 1+i1-i×1+i1+i
⇒z=1+i2+2i1-i2
⇒z= 1-1+2i1+1
⇒z=2i2⇒z=i
⇒z4=i4Since i2=-1, we have:⇒z4=i2×i2⇒z4=1
Page 13.49 (Multiple Choice Questions)
Q20.
Answer :
(a) 0
Let z = 1+2i1-1-i2
⇒z= 1+2i1-1+i2 -2i
⇒z= 1+2i1-1-1-2i
⇒z =1+2i1+2i
⇒ z=1Since point (1,0) lies on the positive direction of real axis, we have: arg (z)=0
Q21.
Answer :
(a) 113
Let z= 12+3i2⇒z= 14+9i2+12i ⇒z = 14-9+12i ⇒z= 1-5+12i
⇒z = 1-5+12i×-5-12i-5-12i
⇒z = -5-12i25+144⇒z= -5169-12i169
⇒ z = 251692+1441692
⇒z= 1169
⇒z=113
Q22.
Answer :
(b) 126
Let z = 11-i2+3i⇒z = 12+i -3i2 ⇒z=12+i+3
⇒z= 15+i×5-i5-i
⇒z= 5-i25-i2
⇒z= 5-i25+1
⇒z= 5-i26
⇒ z=526-i26
⇒z = 25676+1676
⇒z=126
Q23.
Answer :
(c) 2 sinθ2
∵z= 1 -cosθ + i sinθ⇒z = 1-cosθ2 + sin2θ⇒z = 1+cos2θ – 2cosθ+ sin2θ⇒z = 1 + 1 – 2cosθ⇒z = 21-cosθ⇒z = 4sin2θ2⇒z = 2sinθ2
Q24.
Answer :
(c) 100
∵x + iy = (1+i)(1+2i)(1+3i)Taking modulus on both the sides:x + iy=(1+i)(1+2i)(1+3i)⇒x + iy= 1+i×1+2i×1+3i⇒x2+y2=12+1212+2212+32⇒x2+y2 = 2510 ⇒x2+y2= 100Squaring both the sides, ⇒x2+y2 =100
Q25.
Answer :
(b) 12
z = 11-cosθ-isinθz = 11-cosθ-isinθ×1-cosθ+isinθ1-cosθ+isinθ⇒z = 1-cosθ+isinθ1-cosθ2 -isinθ2⇒z= 1-cosθ+isinθ1+cos2θ -2cosθ +sin2θ⇒z=1-cosθ+isinθ1+1-2cosθ⇒z= 1-cosθ+isinθ2(1-cosθ)⇒Re(z) =1-cosθ21-cosθ=12
Q26.
Answer :
(c) 5385
x+iy=3+5i7-6i⇒x+iy=3+5i7-6i×7+6i7+6i⇒x+iy = 21+53i+30i249-36i2⇒x+iy = 21-30+53i49+36⇒x+iy =-985+i5385On comparing both the sides: y = 5385
Q27.
Answer :
(a) 1
1-ix1+ix = a+ibTaking modulus on both the sides, we get:1-ix1+ix = a+ib⇒12+x212+x2= a2+b2⇒a2+b2 = 1Squaring both the sides, we get: a2+b2 = 1
Q28.
Answer :
(b) 2aba2-b2
z=a+iba-ib×a+iba+ib⇒z=a2+i2b2+2abia2-i2b2⇒z=a2-b2+2abia2+b2⇒z=a2-b2a2+b2+i2aba2+b2⇒Rez=a2-b2a2+b2, Imz=2aba2+b2tan α=ImzRez =2aba2-b2α= tan-12aba2-b2Since, z lies in the first quadrant . Therefore, arg (z) =α= tan-12aba2-b2tan θ=2aba2-b2
Q29.
Answer :
(d) amp (z) = 3π4
z=1+7i2-i2⇒z=1+7i4+i2-4i⇒z=1+7i4-1-4i ∵ i2=-1⇒z=1+7i3-4i⇒z=1+7i3-4i×3+4i3+4i⇒z=3+4i+21i+28i29-16i2⇒z=3-28+25i9+16⇒z=-25+25i25⇒z=-1+itan α=ImzRez =1⇒α=π4Since, z lies in the second quadrant . Therefore, amp (z) = π-α =π-π4 = 3π4
Q30.
Answer :
(c) -π2
Let z= 1i⇒z = 1i×ii⇒z = ii2⇒z =-i
Since , z 0,-1 lies on the negative imaginary axis . Therefore, arg (z) = -π2
Q31.
Answer :
(a) -π2
Let z = 1-i1+i⇒z = 1-i1+i×1-i1-i⇒z = 1+i2-2i1-i2⇒z = 1-1-2i1+1⇒z = -2i2⇒z =-iSince, z lies on negative direction of imaginary axis . Therefore, arg (z) = -π2
Page 13.50 (Multiple Choice Questions)
Q32.
Answer :
(c) π6
Let z = 1+i33+i⇒z =1+i33+i×3-i3-i⇒z = 3+2i-3i23-i2⇒z =3 +3 +2i4⇒z=23+2i4⇒z=32+12itan α=Im(z)Re(z) =13⇒α=π6Since, z lies in the first quadrant . Therefore, arg(z) = tan-1 13=π6
Q33.
Answer :
(a) 12(1+i)
i5+i6+i7+i8+i91+i=i-1-i+1+i1+i As, i5=i, i6=-1, i7=-i, i8=1, i9=i=ii+1=ii+1×i-1i-1=ii-1i2-1=i2-i-2=121+i
Q34.
Answer :
(c) -i
Let z= 1+2i+3i21-2i+3i2⇒z = 1+2i-31-2i-3⇒z = -2+2i-2-2i×-2+2i-2+2i⇒z = -2+2i2-22-2i2⇒z = 4+4i2-8i4+4⇒z = 4-4-8i8⇒z= -8i8⇒z = -i
Q35.
Answer :
(b) -2
i592+i590+i588+i586+i584i582+i580+i578+i576+i574-1=i4×148+i4×147+2+i4×147+i4×146+2+i4×146i4×145+2+i4×145+i4×144+2+i4×144+i4×143+2-1 ∵ i4=1 and i2=-1=1+i2+1+i2+1i2+1+i2+1+i2-1=1-1-1 =-2
Q36.
Answer :
(c) -8
Using a4+b4 =a2+b22-2a2b2(1+i)4+(1-i)4 =1+i2+1-i22-21+i21-i2 =1+i2+2i+1+i2-2i2-21+i2+2i1+i2-2i =1-1+2i+1-1-2i2-21-1+2i1-1-2i=0-22i-2i ∵ i2 =-1=8i2=-8
QUADRATIC EQUATIONS
Page 14.6 Ex. 14.1
Q1.
Answer :
Given: x2 + 1 = 0
x2 + 1 = 0⇒ x2 – 1i2 = 0⇒ (x + i) (x – i) = 0 [ ( a2 – b2) = (a + b) (a – b)]
⇒ (x + i) = 0 or (x – i) = 0
⇒ x =-i or x = i
Hence, the roots of the equation are i and -i.
Q2.
Answer :
Given: 9×2 + 4 = 0
9×2 + 4 = 0⇒ (3x)2 + 22 = 0⇒ (3x)2 – (2i)2 = 0⇒ (3x + 2i) (3x – 2i) = 0 [(a2 – b2) = (a + b) (a – b)]
⇒ (3x + 2i) = 0 or, (3x – 2i) = 0
⇒ 3x =-2i or 3x = 2i
⇒ x = -2i3 or x = 2i3
Hence, the roots of the equation are 2i3 and -2i3.
Video Previous Next
Questions
Q3.
Answer :
Given: x2 + 2x + 5 = 0
x2 + 2x + 5 = 0⇒x2 + 2x + 1 + 4 = 0⇒ (x + 1)2 – (2i)2 = 0 [(a + b)2 = a2 + b2 + 2ab]⇒ (x + 1 + 2i) (x + 1 – 2i) = 0 [ a2 – b2 = (a + b) (a – b)]
⇒ (x + 1 + 2i) = 0 or, (x + 1 – 2i) = 0
⇒ x = -(1+ 2i) or, x = -1 + 2i
Hence, the roots of the equation are -1 + 2i and -1 – 2i.
Q4.
Answer :
We have:
4×2 – 12x + 25 = 0⇒ 4×2 – 12 x + 9 + 16 = 0⇒ (2x)2 + 32 – 2×2x×3 – (4i)2 = 0⇒(2x- 3)2 – (4i)2 = 0⇒ (2x – 3 + 4i) (2x – 3 -4i) = 0 [a2 – b2 = (a + b) (a – b)]
⇒ (2x – 3 + 4i) = 0 or, (2x – 3 – 4i) = 0
⇒ 2x = 3 – 4i or, 2x = 3 + 4i
⇒x = 32 – 2i or, x = 32 + 2i
Hence, the roots of the equation are 32 – 2i and 32 + 2i.
Q5.
Answer :
We have:
x2 + x + 1 = 0⇒ x2 + x + 14 + 34 = 0⇒ x2 + 122 +2×x×12- 3i22 = 0⇒x + 122 – 3i22 = 0⇒ x + 12 + 3i2 x + 12 – 3i2 = 0
⇒x + 12 + 3i2 = 0 or, x + 12 – 3i2 = 0
⇒x = -12 – 3i2 or, x =-12 + 3i2
Hence, the roots of the equation are -12 – i32 and -12 + i32.
Q6.
Answer :
We have:
4×2 + 1 = 0⇒ (2x)2 – i2 = 0⇒ (2x)2 – (i)2 = 0⇒ (2x + i) (2x – i) = 0
⇒ (2x + i) = 0 or (2x – i) = 0
⇒ 2x = -i or 2x = i
⇒ x =-i2 or x = i2
Hence, the roots of the equation are 12i and -12i.
Q7.
Answer :
We have:
x2 – 4x + 7 = 0⇒ x2 – 4x + 4 + 3 = 0⇒ x2 – 2×x×2 + 22 – (3i)2 = 0⇒ (x – 2)2 – (3i)2 = 0⇒ (x – 2 + 3i) (x – 2 – 3i) = 0
⇒ (x – 2 + 3i) = 0 or, (x – 2 – 3i) = 0
⇒ x = 2 – 3i or, x = 2 + 3i
Hence, the roots of the equation are 2 ± i3.
Q8.
Answer :
We have:
x2 + 2x + 2 = 0⇒ x2 + 2x + 1 + 1 = 0⇒ x2 + 2×x×1 + 12 – (i)2 = 0⇒ (x + 1)2 – (i)2 = 0⇒ (x + 1 + i) (x + 1 – i) = 0
⇒ (x + 1 + i) = 0 or (x + 1 – i) = 0
⇒ x = -1 – i or x = -1 + i
Hence, the roots of the equation are -1 + i and -1 – i.
Q9.
Answer :
Given: 5×2 – 6x + 2 = 0
Comparing the given equation with general form of the quadratic equation ax2+ bx + c = 0, we get a = 5, b =-6 and c = 2.
Substituting these values in α = -b + b2 – 4ac2a and β = -b – b2 – 4ac2a, we get:
α = 6 + 36 – 4×5× 22×5 and β= 6 – 36 – 4×2×52×5
⇒ α = 6 + -410 and β = 6 – -410
⇒ α = 6 + 4i210 and β= 6 – 4i210
⇒ α = 6 + 2i10 and β = 6 – 2i10
⇒α = 2 ( 3 + i)10 and β = 2 ( 3 – i)10
⇒ α = 35 + 15i and β = 35 – 15i
Hence, the roots of the equation are 35 ± 15i.
Q10.
Answer :
Given: 21×2 + 9x + 1 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 21, b = 9 and c = 1.
Substituting these values in α = -b + b2 – 4ac2a and β= -b – b2 – 4ac2a, we get:
α = -9 + 81 – 4×21×12×21 and β= -9 – 81 – 4×21×12×21
⇒ α = -9 + 3i42 and β = -9 – 3i42
⇒α = -942 + 3i42 and β=- 942 – 3i42
⇒ α = -314 + 3i42 and β=-314 – 3i42
Hence, the roots of the equation are -314 ± i342.
Q11.
Answer :
We have:
x2 – x + 1 = 0⇒ x2 – x + 12 + 34 = 0⇒x2 -2× x ×12+ 122 – 34i2 = 0⇒ x – 122 – i322 =0⇒ x – 12 + i32 x – 12 – i32 = 0
⇒x – 12 + i32 = 0 or x – 12 – i32 = 0
⇒x = 12 – i32 or x = 12 + i32
Hence, the roots of the equation are 12 ± i32.
Q12.
Answer :
We have:
x2 + x + 1 = 0⇒ x2 + x + 14 + 34 = 0⇒ x + 122 – 34i2 = 0⇒ x + 12 2 – i322 = 0⇒ x + 12 + i32 x + 12 – i32 = 0
⇒ x + 12 + i32 = 0 or x + 12 – i32 = 0
⇒ x =-12 – i32 or x =-12 + i32
Hence, the roots of the equation are -12 ± i32.
Q13.
Answer :
Given: 17×2 – 8x + 1 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 17, b =-8 and c = 1.
Substituting these values in α= -b + b2 – 4ac2a and β = -b – b2 – 4ac2a, we get:
α=8 + 64 – 4×17×12×17 and β = 8 – 64 – 4×17 ×12×17
⇒ α = 8 + 64 – 6834 and β = 8 – 64 – 6834
⇒ α = 8 + -434 and β = 8 – -434
⇒ α = 8 + 4i234 and β= 8 – 4i234
⇒α = 8 + 2i34 and β = 8 – 2i34
⇒ α = 4 + i17 and β= 4 – i17
⇒ α = 417 + 117i and β= 417 – 117i
Hence, the roots of the equation are 417 ± 117i.
Q14.
Answer :
Given: 27×2 – 10x + 1 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 27, b =-10 and c = 1.
Substituting these values in α = -b + b2 – 4ac2a and β = -b – b2 – 4ac2a, we get:
α= 10 + 100 – 4×27×12×27 and β = 10 – 100 – 4×27×12×27
⇒ α = 10 + 100 – 10854 and β = 10 – 100 – 10854
⇒ α = 10 + -854 and β = 10 – -854
⇒ α = 10 + 8i254 and β = 10 – 8i254
⇒ α = 10 + i2254 and β= 10 – i2254
⇒ α = 2(5 + i2) 54 and β = 2(5 – i2)54
⇒ α = 527 + 227i and β= 527 – 227i
Hence, the roots of the equation are 527 ± 227i.
Q15.
Answer :
Given: 17×2 + 28x + 12 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 17, b = 28 and c= 12.
Substituting these values in α = -b + b2 – 4ac2a and β = -b – b2- 4ac2a, we get:
α = -28 + 784 – 4×17×1234 and β = -28 – 784 – 4×17×1234
⇒ α = -28 + 784 – 81634 and β = -28 – 784 – 81634
⇒ α = -28 + -3234 and β = -28 – -3234
⇒α = -28 + 32i234 and β= -28 – 32i234
⇒ α = -28 + 42 i34 and β = -28 – 42 i34
⇒ α = -14 + 22 i17 and β = -14 – 22 i17
Hence, the roots of the equation are -1417 ± 2217i.
Q16.
Answer :
Given: 21×2 – 28x + 10 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 21, b =-28 and c = 10.
Substituting these values in α = -b + b2 – 4ac2a and β = -b – b2 – 4ac2a, we get:
α = 28 + 784 – 4×21×102×21 and β = 28 – 784 – 4×21×102×21
⇒ α = 28 + -5642 and β = 28 – -5642
⇒α = 28 +2i1442 and β = 28 – 2i1442
⇒ α = 14 + i1421 and β = 14 – i1421
⇒ α = 23 + 1421i and β = 23 – 1421i
Hence, the roots of the equation are 23 ± 1421.
Q17.
Answer :
Given: 8×2 – 9x + 3 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 8, b=-9 and c = 3.
Substituting these values in α= -b + b2- 4ac2a and β = -b – b2 – 4ac2a, we get:
α = 9 + 81 – 4×8×32×8 and β= 9 – 81 – 4×8×32×8
⇒ α = 9 + 81 – 9616 and β = 9 – 81 – 9616
⇒ α = 9 + -1516 and β= 9 – -1516
⇒α = 9 + 15i216 and β = 9 – 15i216
⇒ α = 9 + i1516 and β = 9 – i1516
⇒ α = 916 – 1516i and β = 916 + 1516i
Hence, the roots of the equation are 916 ± 1516i.
Q18.
Answer :
Given: 13×2 + 7x + 1 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 13, b = 7 and c = 1.
Substituting these values in α = -b + b2 – 4ac2a and β= -b – b2 – 4ac2a, we get:
α = -7 + 49 – 4×13×12×13 and β= -7 – 49 – 4×13×12×13
⇒ α = -7 + 49 – 5226 and β = -7 – 49 – 5226
⇒ α = -7 + -326 and β = -7 – -326
⇒ α = -7 + i326 and β = -7 – i326
⇒α = -726 + 326i and β = -726 – 326i
Hence, the roots of the equation are -726 ± 326i.
Q19.
Answer :
Given: 2×2 + x +1 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 2, b = 1 and c = 1.
Substituting these values in α = -b + b2 – 4ac2a and β = -b – b2 – 4ac2a, we get:
α = -1 + 1 – 4×2×12×2 and β = -1 – 1 – 4×2×12×2
⇒ α = -1 + -74 and β = -1 – -74
⇒ α = -1 + i74 and β = -1 – i74
⇒ α =-14 + 74i and β =-14 – 74i
Hence, the roots of the equation are -1 ± i74.
Q20.
Answer :
Given: 3×2 – 2x + 33 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 3, b = -2 and c = 33.
Substituting these values in α = -b + b2 – 4ac2a and β = -b – b2 – 4ac2a, we get:
α = 2 + 2 – 4×3×3323 and β = 2 – 2 – 4×3×3323
⇒α = 2 + -3423 and β = 2 – -3423
⇒α = 2 + i3423 and β = 2 – i3423
Hence, the roots of the equation are 2 ± i3423.
Q21.
Answer :
Given: 2×2 + x + 2 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 2, b = 1 and c =2.
Substituting these values in α = -b + b2 -4ac2a and β = -b – b2 – 4ac2a, we get:
α = -1 + 1 – 4×2×222 and β = -1 – 1 – 4×2×222
⇒ α =-1 + -722 and β = -1 – -722
⇒α = -1 + i722 and β = -1 – i722
Hence, the roots of the equation are -1 ± i722.
Q22.
Answer :
Given equation: x2 + x + 12 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 1, b = 1 and c = 12.
Substituting these values in α = -b + b2 – 4ac2a and β = -b – b2 – 4ac2a, we get:
α = -1 + 1 – 4×122 and β = -1 – 1 – 4×122
⇒α = -1 + 1 – 222 and β = -1 – 1 – 222
⇒ α = -1 + i22 – 12 and β =-1 – i22 – 12
Hence, the roots of the equation are -1 ± i22 – 12.
Q23.
Answer :
Given equation: x2 + x2 + 1 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 1, b= 12 and c = 1.
Substituting these values in α = -b + b2 – 4ac2a and β = -b – b2 – 4ac2a, we get:
α = -12 + 12- 4×1×12 and β = -12 – 12 – 4×1×12
α = -12 + -722 and β = -12 – -722
α = -12 + i722 and β = -12 – i722
α = -1 + i722 and β = -1 – i722
Hence, the roots of the equation are -1 ± i722.
Q24.
Answer :
Given: 5×2 + x + 5 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 5 , b = 1 and c = 5.
Substituting these values in α= -b + b2 – 4ac2a and β = -b – b2 – 4ac2a, we get:
α = -1 + 1 – 4×5 × 525 and β = -1 – 1 – 4×5 × 525
α = -1 + -1925 and β = -1 – -1925
α = -1 + i1925 and β = -1 – i1925
Hence, the roots of the equation are -1 ± i1925.
Q25.
Answer :
-x2 +x- 2 = 0⇒ x2 – x + 2 = 0⇒ x2 – x + 14 + 74 = 0⇒ x2 -2× x×12 + 122 – 74i2 = 0⇒ x – 122 – i722 = 0⇒ x – 12 + i72 x – 12 – i72 = 0
⇒ x – 12 + 72i = 0 or, x – 12 – 72i = 0
⇒x = 12 – 72i or, x = 12 + 72i
Hence, the roots of the equation are 12 ± 72i.
Q26.
Answer :
x2 – 2x + 32 = 0⇒ x2 – 2x + 1 + 12 = 0⇒ x -12 – 12i2 = 0⇒ x – 1 + 12i x – 1 – 12i = 0
⇒ x – 1 – 12i = 0 or, x – 1 +12i = 0
⇒ x = 1 + 12i or, x = 1 – 12i
Hence, the roots of the equation are 1 ± 12i.
Q27.
Answer :
Given: 3×2 – 4 x + 203 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx = c = 0, we get a = 3, b=-4 and c = 203.
Substituting these values in α = -b + b2 – 4ac2a and β = -b – b2 – 4ac2a, we get:
⇒ α = 4 + 16 – 4×3×2036 and β = 4 – 16 – 4×3×2036
⇒ α = 4+ -646 and β = 4 – -646
⇒ α = 4 + 8i6 and β = 4 – 8i6
⇒α = 2 + 4i3 and β= 2 – 4i3
Hence, the roots of the equation are 2 ± 4i3.
Page 14.10 Ex. 14.2
Q1.
Answer :
i x2+10ix-21=0⇒x2+7ix+3ix-21=0⇒xx+7i+3ix+7i=0⇒x+7ix+3i=0⇒x+7i=0 or x+3i=0⇒x=-7i, -3iSo, the roots of the given quadratic equation are -3i and -7i.
ii x2+1-2i x-2i=0⇒ x2+x-2ix-2i=0⇒xx+1-2ix+1=0⇒x+1x-2i=0⇒x+1=0 or x-2i=0⇒x=-1, 2iSo, the roots of the given quadratic equation are -1 and 2i.
iii x2-23+3i x+63i=0⇒ x2-23x-3i x+63i=0⇒xx-23-3ix-23=0⇒x-23x-3i=0⇒x-23=0 or x-3i=0⇒x=23, 3iSo, the roots of the given quadratic equation are 23 and 3i.
iv 6×2-17ix-12=0⇒6×2-9ix-8ix-12=0⇒3x2x-3i-4i2x-3i=0⇒2x-3i3x-4i=0⇒2x-3i=0 or 3x-4i=0⇒x=32i, 43iSo, the roots of the given quadratic equation are 32i and 43i.
Q2.
Answer :
i) x2-32+2i x+62i=0⇒ x2-32 x-2i x+62i=0⇒xx-32 -2ix-32=0⇒x-32x-2i=0⇒x-32=0 or x-2i=0⇒x=32, 2iSo, the roots of the given quadratic equation are 32 and 2i.
i)i x2-5-ix+18+i=0Comparing the given equation with the general form ax2+bx+c=0, we geta=1, b=-5-i and c=18+ix=-b±b2-4ac2a⇒x=5-i±5-i2-418+i2⇒x=5-i±5-i2-418+i2⇒x=5-i±-48-14i2⇒x=5-i±i48+14i2⇒x=5-i±i49-1+2×7×i2⇒x=5-i±i7+i22⇒x=5-i±i7+i2⇒x=5-i+i7+i2 or x=5-i-i7+i2⇒x=2+3i, 3-4iSo, the roots of the given quadratic equation are 2+3i and 3-4i.
iii) 2+i x2-5-i x+21-i=0Comparing the given equation with the general form ax2+bx+c=0, we geta=2+i, b=-5-i and c=21-ix=-b±b2-4ac2a⇒x=5-i±5-i2-42+i21-i22+i⇒x=5-i±-2i22+i⇒x=5-i±-2i22+i …iLet x+iy=-2i. Then,⇒x+iy2=-2i⇒x2-y2+2ixy=-2i ⇒x2-y2=0 and 2xy=-2 …iiNow, x2+y22=x2-y22+4x2y2⇒ x2+y22=4⇒x2+y2=2 …iiiFrom ii and iii⇒x=±1 and y=±1As, xy is negative from ii⇒x=1, y=-1 or, x=-1, y=1⇒x+iy=1-i or, -1+i⇒-2i=±1-iSubstituting this value in i, we get⇒x=5-i±1-i22+i⇒x=1-i, 45-25iSo, the roots of the given quadratic equation are 1-i and 45-25i.
iv) x2-2+i x-1-7i=0Comparing the given equation with the general form ax2+bx+c=0, we geta=1, b=-2+i and c=-1-7ix=-b±b2-4ac2a⇒x=2+i±2+i2+41-7i2⇒x=2+i±7-24i2 … iLet x+iy=7-24i. Then,⇒x+iy2=7-24i⇒x2-y2+2ixy=7-24i ⇒x2-y2=7 and 2xy=-24 … iiNow, x2+y22=x2-y22+4x2y2⇒ x2+y22=49+576=625⇒x2+y2=25 … iii From ii and iii⇒x=±4 and y=±3As, xy is negative From ii⇒x=-4, y=3 or, x=4, y=-3⇒x+iy=-4+3i or, 4-3i⇒7-24i=±4-3iSubstituting these values in i, we get⇒x=2+i±4-3i2⇒x=3-i, -1+2iSo, the roots of the given quadratic equation are 3-i and -1+2i.
v) ix2-4x-4i=0⇒ix2+4ix-4=0⇒x2+4ix-4=0⇒x+2i2=0⇒x+2i=0⇒x=-2iSo, the roots of the given quadratic equation are -2i and -2i.
vi) x2+4ix-4=0⇒x2+2×x×2i+2i2=0⇒x+2i2=0⇒x+2i=0⇒x=-2iSo, the roots of the given quadratic equation are -2i and -2i.
vii) 2×2+15 ix-i=0Comparing the given equation with the general form ax2+bx+c=0, we geta=2, b=15 i and c=-ix=-b±b2-4ac2a⇒x=-15 i±15 i2+8i4⇒x=-15 i±8i-154 … iLet x+iy=8i-15. Then,⇒x+iy2=8i-15⇒x2-y2+2ixy=8i-15 ⇒x2-y2=-15 and 2xy=8 … iiNow, x2+y22=x2-y22+4x2y2⇒ x2+y22=225+64=289⇒x2+y2=17 … iii From ii and iii⇒x=±1and y=±4As, xy is positive From ii⇒x=1, y=4 or, x=-1, y=-4⇒ x+iy=1+4i or, -1-4i⇒8i-15=±1-4iSubstituting these values in i, we get,⇒x=-15 i±1+4i4 ⇒x= 1+4-15i4 , -1-4+15i4So, the roots of the given quadratic equation are 1+4-15i4 and -1-4+15i4.
viii) x2-x+1+i=0Comparing the given equation with the general form ax2+bx+c=0, we geta=1, b=-1 and c=1+ix=-b±b2-4ac2a⇒x=1 ±1-41+i2⇒x=1±-3-4i2 … iLet x+iy=-3-4i. Then,⇒x+iy2=-3-4i⇒x2-y2+2ixy=-3-4i ⇒x2-y2=-3 and 2xy=-4 … iiNow, x2+y22=x2-y22+4x2y2⇒ x2+y22=9+16=25⇒x2+y2=5 … iii From ii and iii⇒x=±1 and y=±2As, xy is negative From ii⇒x=1, y=-2 or, x=-1, y=2⇒x+iy=1-2i or -1+2i⇒-3-4i=±1-2i Substituting these values in i, we get⇒x=1±1-2i 2⇒x=1-i, iSo, the roots of the given quadratic equation are 1-i and i.
ix) ix2-x+12i=0⇒ix2+ix+12=0⇒x2+ix+12=0⇒x2+4ix-3ix+12=0⇒xx+4i-3ix+4i=0⇒x+4ix-3i=0⇒x+4i=0 or x-3i=0⇒x=-4i , 3iSo, the roots of the given quadratic equation are -4i and 3i.
x) x2-32-2i x-2i=0Comparing the given equation with the general form ax2+bx+c=0, we geta=1, b=-32-2i and c=-2ix=-b±b2-4ac2a⇒x=32-2i±32-2i2+42i2⇒x=32-2i±14-82i2 … iLet x+iy=14-82i. Then,⇒x+iy2=14-82i⇒x2-y2+2ixy=14-82i ⇒x2-y2=14 and 2xy=-82 … iiNow, x2+y22=x2-y22+4x2y2⇒ x2+y22=196+128=324⇒x2+y2= 18 … iii From ii and iii⇒x=±4 and y=±2As, xy is negative From ii⇒x=-4, y=2 or, x=4, y=-2⇒x+iy=4-2 i or, -4+2 i⇒14-82i=±4-2 iSubstituting these values in i, we get⇒x=32-2i±4-2 i2⇒x=32+4-i2+2 2, 32-4-i2-2 2So, the roots of the given quadratic equation are 32+4-i2+2 2 and 32-4-i2-2 2 .
xi) x2-2+i x+2 i=0Comparing the given equation with the general form ax2+bx+c=0, we geta=1, b=-2+i and c=2ix=-b±b2-4ac2a⇒x=2+i±2+i2-42i2⇒x=2+i±1-22 i2 ⇒x=2+i±22-12-22 i2⇒x=2+i±2-i22⇒x=2+i±2-i2⇒x=2, i So, the roots of the given quadratic equation are 2 and i.
xii) 2×2-3+7i x+9i-3=0Comparing the given equation with the general form ax2+bx+c=0, we geta=2, b=-3+7i and c=9i-3x=-b±b2-4ac2a⇒x=3+7i±3+7i2-89i-34⇒x=3+7i±-16-30i4 … iLet x+iy=-16-30i. Then,⇒x+iy2=-16-30i⇒x2-y2+2ixy=-16-30i ⇒x2-y2=-16 and 2xy=-30 … iiNow, x2+y22=x2-y22+4x2y2⇒ x2+y22=256+900=1156⇒x2+y2= 34 … iii From ii and iii⇒x=±3 and y=±5As, xy is negative From ii⇒x=-3, y=5 or, x=3, y=-5⇒x+iy=3-5 i or, -3+5 i⇒14-82i=±3-5 iSubstituting these values in i, we get⇒x=3+7i±3-5 i4⇒x=3+i2, 3iSo, the roots of the given quadratic equation are 3+i2 and 3i .
Page 14.11 (Very Short Answers)
Q1.
Answer :
(x -1)2 + (x – 2)2 + (x – 3)2 = 0⇒x2 + 1 – 2x + x2 + 4 – 4x + x2 + 9 – 6x = 0⇒ 3×2 – 12 x + 14 = 0
Comparing the given equation with the general form of the quadratic equation ax2 + bx + c = 0, we get a = 3 , b=-12 and c = 14.
D=b2-4ac=-122-4×3×14=144-168=-24Since the value of D is less than 0, the given equation has no real roots.
Q2.
Answer :
Given: x2 – px + q = 0
Also, a and b are the roots of the given equation.
Sum of the roots = a + b = p …(1)
Product of the roots = ab = q …(2)
Now, 1a + 1b = b+aab = pq [Using equation (1) and (2)]
Hence, the value of 1a + 1b is pq.
Q3.
Answer :
Given equation: x2 – px + 16 = 0
Also, α and β are the roots of the equation satisfying α2 + β2 = 9.
From the equation, we have:
Sum of the roots = α +β = –p1 = p
Product of the roots = αβ =161 = 16
Now, α + β2 = α2 + β2 + 2αβ⇒ p2 = 9 + 32⇒ p2 = 41⇒ p =41
Hence, the value of p is 41.
Q4.
Answer :
Irrational roots always occur in conjugate pairs.
If 2+3 is a root and 2-3 is its conjugate root.⇒2+3+2-3=-p⇒4=-9⇒p=-4Also, 2+32-3=q⇒4-3=q⇒q=1
Q5.
Answer :
Given: x2 + ax + 8 = 0.
Let α and β are the roots of the equation.
Sum of the roots = α + β =-a1 =-a.
Product of the roots = αβ = 81 = 8
Given: α – β = 2
Then, α+ β2 – α – β2 = 4αβ⇒ α + β2 – 22 = 4×8⇒ α + β2 – 4 = 32⇒ α + β2 = 32 + 4 = 36⇒α + β = ± 6
α + β =-a = ±6
∴ a = ±6
Q6.
Answer :
Given:(a-b)x2+(b-c)x+(c-a)=0⇒x2+b-ca-bx+c-aa-b=0⇒x2-c-aa-bx-x+c-aa-b=0 ∵b-ca-b=-c+a-a+ba-b=-c-aa-b-1⇒xx-c-aa-b-1x+c-aa-b=0⇒x-c-aa-bx-1=0⇒x-c-aa-b=0 or x-1=0⇒x=c-aa-b or x=1Thus, roots of the equation are c-aa-b and 1.
Now,
α+β=-b-ca-b⇒1+β=-b-ca-b⇒β=-b-ca-b-1=c-aa-b
Q7.
Answer :
Given: x2 – x + 1 = 0
Also, a and b are the roots of the equation.
Then, sum of the roots = a + b = – -11 = 1
Product of the roots = ab = 11 = 1
∴ a + b2 = a2 + b2 + 2ab⇒ 12 = a2 + b2 + 2×1⇒ a2+ b2 = 1 – 2 =-1⇒ a2 + b2 =-1
Q8.
Answer :
Let α and β be the real roots of the quadratic equation ax2 + bx + c= 0.
On squaring these roots, we get:
α= α2 and β= β2
⇒α (1 – α) =0 and β1-β=0
⇒ α = 0, α = 1 and β=0, 1
Three cases arise:
(i) α = 0, β = 0(ii) α = 1, β = 0(iii) α = 1, β = 1
(i) α = 0, β =0∴α+β = 0 and αβ = 0
So, the corresponding quadratic equation is,
x2 – (α+β)x +αβ = 0⇒x2 = 0
(ii) α= 0, β = 1α+β = 1αβ = 0
So, the corresponding quadratic equation is,
x2 – (α+β)x +αβ = 0⇒x2 -x+0= 0⇒x2 -x= 0
(iii) α = 1, β = 1α+β = 2αβ =1
So, the corresponding quadratic equation is,
x2 – (α+β)x +αβ = 0⇒x2 -2x+1= 0
Hence, we can construct 3 quadratic equations.
Q9.
Answer :
Given equation: x2 + lx + m = 0
Also, α and β are the roots of the equation.
Sum of the roots = α + β = -l1 = -l
Product of the roots = αβ = m1 = m
Now, sum of the roots = -1α – 1β =-α + βαβ =- -lm = lm
Product of the roots = 1αβ = 1m
∴ x2-Sum of the rootsx+Product of the roots=0⇒x2 – lmx + 1m = 0⇒ mx2 – lx + 1= 0
Hence, this is the equation whose roots are -1α and -1β.
Q10.
Answer :
Given: x2 – a(x + 1) – c = 0 or x2 – ax -a – c = 0
Also, α and β are the roots of the equation.
Sum of the roots = α + β = –a1 = a
Product of the roots = αβ = -(a+c)1 = -(a + c)
∴ (1 + α) (1 + β) = 1 + β + α + αβ = 1 + (α + β) + αβ = 1 + a – a – c = 1- c
Page 14.11 (Multiple Choice Questions)
Q1.
Answer :
(b) 2±12
Since the equation has real roots.⇒D=0⇒b2-4ac=0⇒k2-41k+2=0⇒k2-4k-8=0⇒k=4±16-41-821⇒k=4±2122⇒k=2±12
Q2.
Answer :
(b) 0
Let p=x⇒p2+p-6=0⇒p2+3p-2p-6=0⇒p+3p-2=0⇒p=-3, 2Also, x=p⇒x=2, or x=-3Modulus can not be negative,∴x=2⇒x=±2⇒x=2 or -2Sum of the roots of x is 0
Q3.
Answer :
(c) −1
Given equation: x2 + x + 1 = 0
Also, a and b are the roots of the given equation.
Sum of the roots = a + b =-Coefficient of xCoefficient of x2=- 11 =-1
Product of the roots = ab = Constant term Coefficient of x2 =11= 1
∴ (a + b)2 = a2 + b2 + 2ab⇒ (-1)2 = a2 + b2 + 2×1⇒ 1 – 2 = a2 + b2⇒ a2 + b2 =-1
Page 14.12 (Multiple Choice Questions)
Q4.
Answer :
(d) −3/7
Given equation: 4×2 + 3x + 7 = 0
Also, α and β are the roots of the equation.
Sum of the roots = α + β =-Coefficient of xCoefficient of x2=-34
Product of the roots = αβ =Constant termCoefficient of x2= 74
∴ 1α + 1β = α + βαβ = -3474 =-37
Q5.
Answer :
(d) −1, −3
The given equation is log3(x2+4x+12)=2.
⇒x2+4x+12=32=9⇒x2+4x+3=0⇒x+1x+3=0⇒x=-1, -3
Q6.
Answer :
(a) 2
x2+2×2-x+12-55=0⇒x2+2x+1-12-x+12-55=0⇒x+12-12-x+12-55=0⇒x+122+1-3x+12-55=0⇒x+122-3x+12-54=0Let p=x+12⇒p2-3p-54=0⇒p2-9p+6p-54=0⇒p+6p-9=0⇒p=9 or p=-6
Rejecting p=-6⇒x+12=9⇒x2+2x-8=0⇒x2+4x-2x-8=0⇒x+4x-2=0⇒x=2, x=-4
Q7.
Answer :
(c) b / ac
Given equation: ax2 + bx + c = 0
Also, α and β are the roots of the given equation.
Then, sum of the roots = α + β =-ba
Product of the roots = αβ = ca
∴1aα + b + 1aβ + b = aβ + b + aα + b(aα + b) (aβ + b) = a(α + β) + 2ba2αβ + abα + abβ + b2 = a(α + β) + 2ba2αβ + abα + β + b2 =a-ba + 2ba2ca + ab-ba + b2 = bac
Q8.
Answer :
(a) q2-p2
Given: α and β are the roots of the equation x2 + px + 1 = 0.
Also, γ and δ are the roots of the equation x2 + qx + 1 = 0.
Then, the sum and the product of the roots of the given equation are as follows:
α + β =-p1 =-pαβ = 11 = 1γ + δ =-q1 =-qγδ = 11 = 1
Moreover, (γ + δ)2 = γ2 + δ2 +2γδ⇒γ2 + δ2 = q2 – 2
∴ (α – γ) (α + δ) (β – γ) (β + δ) = (α – γ) (β – γ) (α + δ) (β + δ) =αβ -αγ – βγ + γ2 αβ + αδ + βδ + δ2 =αβ -γα + β + γ2 αβ +δ α + β + δ2 = (1 -γ(-p) + γ2 ) (1 + δ(-p) + δ2 ) = (1 +γp + γ2 ) (1 – δp + δ2 ) =1 -pδ + δ2+ pγ – p2γδ + pγδ2 + γ2- pδγ2 + γ2δ2 =1 -pδ + pγ+ δ2 – p2γδ + pγδ2 + γ2- pδγ2 + γ2δ2 =1 – p(δ – γ) – p2γδ+ pγδ (δ – γ) + (γ2 + δ2) + 1 =1 – p2γδ+ pγδ (δ – γ) – p(δ – γ) + (γ2 + δ2) + 1 =1 – p2+ (δ – γ) p (γδ – 1)+ q2 – 2 + 1 =-p2 + (δ- γ) p (1 – 1) + q2 =q2 – p2
Q9.
Answer :
(b) 2
Given equation: |2x – x2 – 3| = 1
(i) 2x – x2 – 3 = 1⇒ 2x – x2 – 4 = 0⇒x2 – 2x + 4 = 0⇒ (x – 2)2 = 0
⇒ x = 2, 2
(ii)
-2x + x2 + 3 = 1⇒ x2 – 2x + 2 = 0
⇒ x2 – 2x + 1 + 1 = 0⇒ (x – 1)2 – i2 = 0⇒ (x – 1 + i) (x – 1 – i) = 0
⇒ x = 1 – i , 1 + i
Hence, the real solutions are 2, 2.
Q10.
Answer :
(c) 2
x2 + |x – 1| = x2+x-1 , x≥1 =x2-x+1 , x<1
(i)
x2 + x – 1 = 1⇒x2+x-2=0⇒x2+2x-x-2=0⇒xx+2-1x+2=0⇒x+2x-1=0⇒x+2=0 or, x-1=0⇒x=-2 or x=1
Since -2 does not satisfy the condition x≥1
(ii)
x2 – x + 1 = 1⇒ x2 – x = 0⇒x2 – x= 0⇒ x ( x – 1) = 0⇒ x= 0 or, (x – 1) = 0⇒ x = 0, x = 1
x = 1 does not satisfy the condition x < 1
So, there are two solutions.
Q11.
Answer :
(a) k ∈ [1/3,3]
k=x2-x+1×2+x+1⇒kx2+kx+k=x2-x+1⇒k-1×2+k+1x+k-1=0
For real values of x, the discriminant of k-1×2+k+1x+k-1=0 should be greater than or equal to zero.
∴if k≠1k+12-4k-1k-1≥0 ⇒k+12-2k-12≥0⇒k+1+2k-2k+1-2k+2≥0⇒3k-1-k+3≥0⇒3k-1k-3≤0⇒13≤k≤3 i.e. k∈13, 3-1 …(i)
And if k=1, then,
x=0, which is real …(ii)
So, from (i) and (ii), we get,
k∈13,3
Q12.
Answer :
(b) 1
Given equation: x2 – bx + c = 0
Let α and α+1 be the two consecutive roots of the equation.
Sum of the roots = α + α + 1 = 2α + 1
Product of the roots = α (α + 1)=α2 + α
So, sum of the roots= 2α+1=-Coeffecient of xCoeffecient of x2=b1=bProduct of the roots= α2+α=Constant termCoeffecient of x2=c1=cNow, b2-4c=2α+12-4α2+α=4α2+4α+1-4α2-4α=1
Q13.
Answer :
(a) and (c)
Let α be the common roots of the equations x2-11x+a=0 and x2-14x+2a=0.
Therefore,
α2 – 11α + a = 0 … (1)
α2 – 14α + 2a = 0 … (2)
Solving (1) and (2) by cross multiplication, we get,
α2-22a+14a=αa-2a=1-14+11⇒α2=-22a+14a-14+11, α=a-2a-14+11⇒α2=-8a-3=8a3, α=-a-3=a3⇒a32=8a3⇒a2= 24a⇒a2-24a=0⇒aa-24=0⇒a=0 or a=24
Q14.
Answer :
(c) 5, −7
The given equation is kx2+1=kx+3x-11×2 which can be written as.
kx2 + 11×2-kx – 3x+1 = ⇒k+11×2-k+3x+1=0
For equal and real roots, the discriminant of k+11×2-k+3x+1=0.
∴k+32-4k+11=0⇒k2+2k-35=0⇒k-5k+7=0⇒k=5, -7
Hence, the equation has real and equal roots when k = 5 , -7.
Q15.
Answer :
(b) −1
Let α be the common roots of the equations, x2 + 2x + 3λ = 0 and 2×2 + 3x + 5λ =0
Therefore,
α2 + 2α + 3λ = 0 … (1)
2α2 + 3α + 5λ = 0 … (2)
Solving (1) and (2) by cross multiplication, we get
α210λ-9λ=α6λ-5λ=13-4⇒α2=-λ, α=-λ⇒-λ=λ2⇒λ=-1
Q16.
Answer :
(a) 49/4
It is given that, 4 is the root of the equation x2+px+12=0.
∴16+4p+12=0⇒p=-7
It is also given that, the equation x2+px+q=0 has equal roots. So, the discriminant of
x2+px+q=0 will be zero.
∴p2-4q=0⇒4q=-72=49⇒q=494
Q17.
Answer :
(a) p = 1, q = −2
It is given that, p and q (p ≠ 0, q ≠ 0) are the roots of the equation x2+px+q=0.
∴Sum of roots=p+q=-p⇒2p+q=0 … (1)
Product of roots=pq=q⇒qp-1=0⇒p=1, q=0 but q≠0
Now, substituting p = 1 in (1), we get,
2+q=0⇒q=-2
Q18.
Answer :
(c) m∈(-4,-3]
The roots of the quadratic equation x2-(m+1)x+m+4=0 will be real, if its discriminant is greater than or equal to zero.
∴m+12-4m+4≥0⇒m-5m+3≥0⇒m≤-3 or m≥5 … (1)
It is also given that, the roots of x2-(m+1)x+m+4=0 are negative.
So, the sum of the roots will be negative.
∴ Sum of the roots < 0
⇒m+1<0⇒m<-1 … (2)
and product of zeros >0
⇒m+4>0⇒m>-4 …(3)
From (1), (2) and (3), we get,
m∈(-4,-3]
Page 14.13 (Multiple Choice Questions)
Q19.
Answer :
(b) 1
(x +2) (x – 5)(x -3) (x + 6) = (x – 2)(x + 4)⇒ (x2 – 3x – 10) (x + 4) = (x2 + 3x – 18) (x – 2)⇒ x3 + 4×2 – 3×2 – 12x – 10x – 40 = x3 – 2×2 + 3×2 – 6x -18x + 36⇒x2 – 22x – 40 = x2 – 24x + 36⇒ 2x = 76⇒ x = 38
Hence, the equation has only 1 root.
Q20.
Answer :
(b) −3/7
Given equation: 4×2 + 3x + 7 = 0
Also, α and β are the roots of the equation.
Then, sum of the roots = α + β =-Coefficient of xCoefficient of x2=-34
Product of the roots = αβ =Constant termCoefficient of x2= 74
∴ 1α + 1β = α + βαβ = -3474 =-37
Q21.
Answer :
(d) qx2-px+1=0
Given equation: x2 + px + q = 0
Also, α and β are the roots of the given equation.
Then, sum of the roots = α + β =-p
Product of the roots = αβ = q
Now, for roots -1α , -1β, we have:
Sum of the roots = -1α – 1β = – α + βαβ =–pq = pq
Product of the roots = 1αβ = 1q
Hence, the equation involving the roots -1α,-1β is as follows:
x2-α+βx+αβ=0
⇒ x2 – pqx + 1q = 0⇒ qx2 -px + 1 = 0
Q22.
Answer :
(b) p2-4q=1
Given equation: x2 – px + q = 0
Also α and β are the roots of the equation such that α – β = 1.
Sum of the roots = α + β =-Coefficient of xCoefficient of x2= –p1 = p
Product of the roots = αβ =Constant termCoefficient of x2= q
∴ (α + β)2 – (α – β)2 = 4αβ⇒p2 – 1 = 4q⇒p2 – 4q = 1
Q23.
Answer :
(c) 1 − c
Given equation: x2 – p(x + 1) – c = 0 or x2 – px-p- c=0
Also α and β are the roots of the equation.
Sum of the roots = α + β = p
Product of the roots = αβ = -(c + p)
Then, (α + 1) (β + 1) = αβ + α + β + 1 = -(c + p) + p + 1 = -c – p + p + 1 = 1- c
Q24.
Answer :
(d) 7
The roots of the quadratic equation x2+5x+k=0 will be imaginary if its discriminant is less than zero.
∴25-4k<0⇒k>254
Thus, the minimum integral value of k for which the roots are imaginary is 7.
Q25.
Answer :
(b) x2-2x+2=0
We know that, imaginary roots of a quadratic equation occur in conjugate pair.
It is given that, 1 + i is one of the roots.
So, the other root will be 1-i.
Thus, the quadratic equation having roots 1 + i and 1 – i is,
x2-1+i+1-ix+1+i1-i=0⇒x2-2x+2=0
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