February 2021

Limits vsaqs questions and solutions

Limits (Q’s & Ans) || V.S.A.Q’S||

Limits (Q’s & Ans) || V.S.A.Q’S|| designed by the ‘Basics in Maths‘ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1B two marks questions and solutions are very useful in IPE examinations.


Limits  V.S.A.Q’S

Limits

Question 1

Find Limits 2

Sol:

Limits 1

= 9

Question 2

Compute Limits 3

Sol:

Limits 4

= a + a = 2a

Question 3

compute Limits 5 

Sol:

Limits 6

Question 4

Show thatLimits 7 = 1and Limits 8 = –1

Sol:

we know that if x > 0

                            Limits 9= –x if x < 0

As x → 0+ ⟹ x > 0

Limits 9 = x

Limits 10 =  1

As x → 0+ ⟹ x < 0

Limits 9 = –x

Limits 11  = –1

Question 5

If f (x) = Limits 12, then find Limits 13andLimits 14 . Does Limits 15 exist?

Sol:

Limits 16

 

1/5

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Hyperbolic Functions vsaqs questions and solutions

Hyperbolic Functions V.S.A.Q.’S || 2 Marks

Hyperbolic Functions V.S.A.Q.’Sdesigned by the ‘Basics in Maths‘ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1A  Hyperbolic Functionstwo marks questions and solutions are very useful in IPE examinations.


Hyperbolic Functions

Question 1

Prove that for any x∈ R, sinh (3x) = 3 sinh x + 4 sinh3 x

Sol:

sinh (3x) = sinh (2x + x)

                  = sinh 2x cosh x + cosh 2x sinh x

                 = (2 sinh x cosh x) cosh x + (1 + 2 sinh2 x) sinh x

                 = 2sinh x cosh2 x + sinh x + 2 sinh3 x

                 = 2 sinh x (1 + sinh2 x) + sinh x + 2 sinh3 x

                 = 2 sinh x + 2 sinh3 x+ sinh x + 2 sinh3 x

                 = 3 sinh x + 4 sinh3 x

Question 2

If cosh x =Hyperbolic Functions 2 , find the values of (i) cosh 2x and (ii) sinh 2x

Sol:

Given cosh x =Hyperbolic Functions 2

 Cosh 2x = 2 cosh2 x – 1

                 = 2. Hyperbolic Functions 2 – 1

                 =Hyperbolic Functions 3

Sinh2 2x = cosh2 2x – 1

                 = Hyperbolic Functions 4– 1

                 = Hyperbolic Functions 5 – 1

       =Hyperbolic Functions 6

  Sinh2 2x   = Hyperbolic Functions 7

Question 3

If cosh x = sec θ then prove that tanh2Hyperbolic Functions 8= tan2Hyperbolic Functions 9

Sol:

tanh2  =Hyperbolic Functions 10

              = Hyperbolic Functions 11

             =Hyperbolic Functions 12

             =Hyperbolic Functions 13

             = tan2Hyperbolic Functions 9

Question 4

If sinh x = 5, then show that x =Hyperbolic Functions 14

Sol:

Given, sinh x = 5

      ⟹ x = sinh-15    

We know that sinh-1x = Hyperbolic Functions 15 

        ⟹ x =  Hyperbolic Functions 16     

               x  =  Hyperbolic Functions 14

 

Question 5

Show that tanh-1 = Hyperbolic Functions 17  log3

Sol:

Given tanh-1

We know that tanh-1 x = Hyperbolic Functions 17 Hyperbolic Functions 18   

 tanh-1  = Hyperbolic Functions 17 Hyperbolic Functions 19  

                 = Hyperbolic Functions 17Hyperbolic Functions 20   

                 = Hyperbolic Functions 17  log3

Question 6

For x, y ∈ R prove that sinh (x + y) = sinh (x) cosh (y) + cosh (x) sinh (y)

Sol:

R.H.S = sinh (x) cosh (y) + cosh (x) sinh (y)

Hyperbolic Functions 21

= Hyperbolic Functions 22 

= sinh (x + y)

Question 7

For any x∈ R, prove that cosh4 x – sinh4 x = cosh 2x

Sol:

cosh4 x – sinh4 x = (cosh2 x)2 – (sinh2 x)2

                                 = (cosh2 x + sinh2 x) (cosh2 x – sinh2 x)

                                 = 1. cosh 2x

                                 = cosh 2x

Question 8

Prove thatHyperbolic Functions 24 = cosh x + sinh x

Sol:

Hyperbolic Functions 23 

= cosh x + sinh x

Question 9

If sin hx = ¾ find cosh 2x and sinh 2x.

Sol:

Given sin hx = ¾

We know that cosh2 x = 1 + sinh2 x

                                           = 1 + (3/4)2

                                           = 1 + 9/16

                                           = 25/16

cos hx = 5/4

cosh 2x = 2cosh2 x – 1

                = 2(25/16) – 1

                = 25/8 – 1

                = 17/8

Sinh 2x = 2 sinh x cosh x

                = 2 (3/4) (5/4)

                = 15/8

Question 10
Prove that (cosh x – sinh x) n = cosh nx – sinh nx

Sol:

  Hyperbolic Functions 25

Hyperbolic Functions 26

∴ (cosh x – sinh x) n = cosh nx – sinh nx

 


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Trigonometric Ratios vsaqs questions and solutions

Trigonometric Ratios(Qns.& Ans) V.S.A.Q.’S

Trigonometric Ratios(Qns.& Ans) V.S.A.Q.’S designed by the ‘Basics in Maths‘ team.These notes to do help intermediate First-year Maths students.

Inter Maths – 1A two marks questions and solutions are very useful in I.P.E examinations.


Trigonometric Ratios Up to Transformations

 Question 1

Find the value of sin2(π/10) + sin2(4π/10) + sin2(6π/10) + sin2(9π/10)

Sol:

 sin2(π/10) + sin2(4π/10) + sin2(6π/10) + sin2(9π/10)

= sin2(π/10) + sin2(π/2 – π/10) + sin2(π/2+ π/10) + sin2(π – π/10)

= sin2(π/10) + cos2(π/10) + cos2(π/10) + sin2(π/10)

= 1 + 1 = 2

 Question 2

If sin θ = 4/5 and θ not in the first quadrant, find the value of cos θ

Sol:

Given sin θ = 4/5 and θ not in the first quadrant

⇒ θ in the second quadrant

⇒ cos θ < 0

    cos2θ = 1 – sin2 θ

              =1 – (4/5)2

             = 1 – 16/25

∴cos θ   = – 3/5 (∵cos θ < 0)

 Question 3

If 3sin θ + 4 cos θ = 5, then find the value of 4 sin θ – 3cos θ

Sol:

Given, 3sin θ + 4 cos θ = 5

let 4 sin θ – 3cos θ = x

  (3sin θ + 4 cos θ )2 + (4 sin θ – 3cos θ)2 = 52 + x2

 9 sin2 θ + 16 cos2 θ + 12 sin θ cos θ + 16 sin2 θ + 9 cos2 θ – 12sin θ cis θ = 25 + x2

25 sin2 θ + 25 cos2 θ = 25 + x2

25 = 25 + x2

⇒ x2 = 0

 x = 0

∴ 4 sin θ – 3cos θ = 0

 Question 4

If sec θ + tan θ =Trigonometry up to Transformations 1, find the value of sin θ and determine the quadrant in which θ lies

Sol:

Given, sec θ + tan θ =  ———— (1)

 We know that sec2 θ – tan2 θ = 1

⇒ (sec θ + tan θ) (sec θ – tan θ) = 1

 sec θ – tan θ =Trigonometry up to Transformations 2

⇒ sec θ – tan θ = Trigonometry up to Transformations 3———— (2)

 (1) + (2)

⇒ (sec θ + tan θ) + (sec θ – tan θ) = Trigonometry up to Transformations 4

2sec θ =Trigonometry up to Transformations 5

 sec θ =Trigonometry up to Transformations 6

(1) – (2)

⇒ (sec θ + tan θ) – (sec θ – tan θ) = Trigonometry up to Transformations 7

2 tan θ = Trigonometry up to Transformations 8 ⇒ tan θ =Trigonometry up to Transformations 9

Now sin θ = tan θ ÷ sec θ =Trigonometry up to Transformatio

     Sin θ =Trigonometry up to Transformations 11

Since sec θ positive and tan θ is negative θ lies in the 4th quadrant.

 Question 5

Prove that cot (π/16). cot (2π/16). cot (3π/16).… cot (7π/16) = 1

Sol:

cot (π/16). cot (2π/16). cot (3π/16).… cot (7π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (4π/16). cot (5π/16) cot (6π/16) cot (7π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (π/4). cot (π/2 – 3π/16) cot (π/2 – 2π/16) cot (π/2 – π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (π/4). tan (3π/16) tan (2π/16) tan (π/16)

= [cot (π/16). tan (π/16)] [cot (2π/16). tan (2π/16)] [cot (3π/16). tan (3π/16]. cot (π/4)

= 1.1.1.1

 =1

 

 Question 6

If cos θ + sin θ = Trigonometry up to Transformations 12cos θ, then prove that cos θ – sin θ =  sin θ

Sol:

Given, cos θ + sin θ = Trigonometry up to Transformations 12cos θ

Sin θ = Trigonometry up to Transformations 12 cos θ – cos θ

           = ( Trigonometry up to Transformations 12 – 1) cos θ

( Trigonometry up to Transformations 12 + 1) sin θ = ( Trigonometry up to Transformations 12 + 1) ( Trigonometry up to Transformations 12 – 1) cos θ

Trigonometry up to Transformations 12 sin θ + sin θ = cos θ

∴ cos θ – sin θ = Trigonometry up to Transformations 12 sin θ

 Question 7

Find the value of 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ)

Sol:

2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ)

= 2[(sin2 θ)3 + (cos2 θ)3] – 3[(sin2 θ)2 + (cos2)2

= 2[(sin2 θ + cos2 θ)3 – 3 sin2 θ cos2 θ (sin2 θ + cos2 θ)] – 3[(sin2 θ + cos2 θ)2 – 2 sin2 θ cos2 θ]

= 2[1 – 3 sin2 θ cos2 θ] – 3 [1 – 2 sin2 θ cos2 θ]

= 2 – 6 sin2 θ cos2 θ – 3 + 6 sin2 θ cos2 θ

= – 1

 Question 8

If tan 200 = λ, then show thatTrigonometry up to Transformations 13    

Sol:

Given tan 200 = λ

Trigonometry up to Transformations 19 =Trigonometry up to Transformations 14

                               =Trigonometry up to Transformations 15

                              =Trigonometry up to Transformations 16

                              =Trigonometry up to Transformations 18

 Question 9

If sin α + cosec α = 2, find the value of sinn α + cosecn α, n∈ Z

Sol:

Given sin α + cosec α = 2

 ⇒ sin α + 1/ sin α = 2

 ⇒Trigonometry up to Transformations 17  = 2

      sin2 α + 1= 2 sin α

       sin2 α – 2 sin α + 1= 0

  (sin α – 1 )2 = 0

⇒ sin α – 1 = 0

sin α = 1 ⇒ cosec α = 1

 sinn α + cosecn α = (1)n + (1)n =1 + 1 =2

∴ sinn α + cosecn α = 2

 Question 10

Evaluate sin2 Trigonometry up to Transformations 20+ cos2 Trigonometry up to Transformations 21  – tan2   Trigonometry up to Transformations 22

Sol:

 Trigonometry up to Transformations 23

Trigonometry up to Transformations 24

 

 Question 11

Find the value of sin 3300. cos 1200 + cos 2100. Sin 3000

Sol:

 sin 3300. cos 1200 + cos 2100. Sin 3000

=sin (3600 – 300). cos (1800 – 600) + cos (1800 + 300). sin (3600 – 600)

= (– sin 300). (– cos 600) + (– cos300). (– sin600)

= sin 300.  cos 600 + cos300.  Sin600

= sin (600 + 300) = sin 900

=1

 Question 12

Prove that cos4 α + 2 cos2 α Trigonometry up to Transformations 25= (1 – sin4 α)

Sol:

cos4 α + 2 cos2 α Trigonometry up to Transformations 25

= cos4 α + 2 cos2 α (1 – cos2 α)

= (cos2 α)2 + 2 (1 – sin2 α) (sin2 α)

= (1 – sin2 α)2 + 2 sin2 α – 2sin4 α

= 1 + sin4 α – 2 sin2 α + 2 sin2 α – 2sin4 α

= 1 – sin4 α

 Question 13

Eliminate θ from x = a cos3 θ and y = b sin3 θ

Sol:

Given x = a cos3 θ and y = b sin3 θ

 cos3 θ = x/a and sin3 θ = y/b

 cos θ = (x/a)1/3 and sin θ = (y/b)1/3

we know that sin2 θ + cos2 θ = 1

 ⇒ [(y/b)1/3]2 + [(x/a)1/3]2 = 1

   (x/a)2/3 + (y/b)2/3 = 1

 Question 14

Find the period of the following functions

Sol:

(i) f(x) = tan 5x

we know that period of tan kx =Trigonometry up to Transformations 26

⇒ period of tan 5x =Trigonometry up to Transformations 27

(ii) f(x) =Trigonometry up to Transformations 28

we know that period of Trigonometry up to Transformations 29= Trigonometry up to Transformations 31

period of Trigonometry up to Transformations 28 = Trigonometry up to Transformations 32

                                      =Trigonometry up to Transformations 33

(iii) f(x) = 2 sinTrigonometry up to Transformations 34 + 2 cosTrigonometry up to Transformations 35

period of sin Trigonometry up to Transformations 34 = Trigonometry up to Transformations 36 = 8

period of cos Trigonometry up to Transformations 35 = Trigonometry up to Transformations 37 = 6

period of given function is = LCM (8, 6) = 24

 

(iv) f(x) = tan (x + 4x + 9x +…. + n2x)

f(x) = tan (x + 4x + 9x +…. + n2x)

       = tan (1 + 4 + 9 + … + n2) x

= tanTrigonometry up to Transformations 38x

we know that period of tan kx =Trigonometry up to Transformations 26

Period of tan  =Trigonometry up to Transformations 39

                        = Trigonometry up to Transformations 40

 Question 15

Prove that sin2(52 ½)0 – sin2 (22 ½)0 =Trigonometry up to Transformations 41

Sol:

We know that sin2 A – sin2B = sin (A +B) sin (A – B)

 ⇒ sin2(52 ½)0 – sin2 (22 ½)0

= sin (52 ½+ 22 ½) sin (52 ½ – 22 ½)

 = sin 750 sin 300

 = Trigonometry up to Transformations 42      

∴ sin2(52 ½)0 – sin2 (22 ½)0 =Trigonometry up to Transformations 41

 

 Question 16

Prove that tan 700 – tan200 = 2 tan 500

Sol:

500 = 700 – 200

Tan 500 = tan (700 – 200)

We know that tan (A –B) =Trigonometry up to Transformations 43

  ⇒ Tan 500 =Trigonometry up to Transformations 44

 ⇒ tan 700 – tan 200 = tan 500 (1 + tan700 tan 200)

     tan 700 – tan 200 = tan 500 [1 + tan700 cot (900 – 200)]

     tan 700 – tan 200 = tan 500 [1 + tan700 cot 700]

     tan 700 – tan 200 = tan 500 [1 + 1]

∴ tan 700 – tan200 = 2 tan 500

 Question 17 

If sin α = Trigonometry up to Transformations 45, sin β =Trigonometry up to Transformations 46  and α, β are acute, show that α + β =Trigonometry up to Transformations 47

Sol:

Given sin α =    Trigonometry up to Transformations 45                                           sin β =Trigonometry up to Transformations 46

Trigonometry up to Transformations 48

 tan α = 1/3                                                          tan β = ½

tan (α + β) =Trigonometry up to Transformations 49

      Trigonometry up to Transformations 50 

  tan (α + β) = 1

∴ α + β =Trigonometry up to Transformations 47

 Question 18

Find tanTrigonometry up to Transformations 51 in terms of tan A

Sol:

 tan Trigonometry up to Transformations 51 =Trigonometry up to Transformations 52

                        =Trigonometry up to Transformations 53

 Question 19

Prove thatTrigonometry up to Transformations 54 = cot 360

Sol:

Trigonometry up to Transformations 54  = Trigonometry up to Transformations 55

(on dividing numerator and denominator by cos 90)

      = Trigonometry up to Transformations 56

   = tan (450 + 90)

    = tan 540

 = tan (900 – 360)

 = cot 360

 ∴Trigonometry up to Transformations 54  = cot 360      

 Question 20

Show that cos 420 + cos 780 + cos 1620 = 0

Sol:

cos 420 + cos 780 + cos 1620

= cos (600 – 180) + cos (600 + 180) + cos (1800 – 180)

=cos 600 cos180 + sin 600 sin 180 + cos 600 cos 180 – sin 600 sin 180 – cos 180

= 2 cos 600 cos 180 – cos 180

= 2 (1/2) cos 180 – cos 180

 = cos 180 – cos 180

= 0

 

 Question 21

Express Trigonometry up to Transformations 57sin θ + cos θ as a single of an angle

Sol:

Trigonometry up to Transformations 57sin θ + cos θ = 2(Trigonometry up to Transformations 58  sin θ + Trigonometry up to Transformations 59cos θ)

                                = 2(cos 300 sin θ + sin 300 cos θ)

                                = 2 sin (θ + 300)

 Question 22

Find the maximum and minimum value of the following functions

(i) 3 sin x –4 cos x

a= 3, b = –4 and c = 0

Trigonometry up to Transformations 60  

                                     = 5

∴ minimum value = –5 and maximum value = 5

(ii) cos (x + ) + 2  sin (x + ) – 3

a= 1, b = 2  and c = – 3

Trigonometry up to Transformations 61

 ∴ minimum value = –6 and maximum value = 0

 Question 23

Find the range of the function f(x) = 7 cos x – 24sin x + 5

Sol:

Given f(x) = 7 cos x – 24sin x + 5

a= 7, b = –24 and c = 5

Trigonometry up to Transformations 62

∴ Range = [–20, 30]    

 Question 24

Prove that sin2α + cos2 (α + β) + 2 sin α sin β cos (α + β) is independent of α

Sol:

sin2α + cos2 (α + β) + 2 sin α sin β cos (α + β)

= sin2α + cos (α + β) [ cos (α + β) +2 sin α sin β]

= sin2α + cos (α + β) [ cos α cos β – sin α sin β +2 sin α sin β]

=sin2α + cos (α + β) [ cos α cos β + sin α sin β]

=sin2α + cos (α + β) cos (α –β)

= sin2 α + cos2 α – sin2 β

=1 – sin2 β

= cos2 β

 Question 25

Simplify Trigonometry up to Transformations 63

Sol:

Trigonometry up to Transformations 63   =Trigonometry up to Transformations 64

                 =Trigonometry up to Transformations 65

                 = tan θ

 

Question 26

For what values of x in the first quadrantTrigonometry up to Transformations 66 is positive?

Sol:

Trigonometry up to Transformations 66 > 0 ⟹ tan 2x > 0

⟹ 0 < 2x < π/2 (∵ x is in first quadrant)

⟹ 0 < x < π/4

Question 27

If cos θ = Trigonometry up to Transformations 67 and π < θ < 3π/2, find the value of tan θ/2.

Sol:

cos θ = Trigonometry up to Transformations 67

π < θ < 3π/2 ⟹ π/2 < θ/2 < 3π/4

tan θ/2 < 0

tan θ/2 =Trigonometry up to Transformations 68

               =– Trigonometry up to Transformations 69 (tan θ/2 < 0)

              =–Trigonometry up to Transformations 70

           = – 2

Question 28

If A is not an integral multiple of π/2, prove that cot A – tan A = 2 cot 2A.

Sol:

cot A – tan A = Trigonometry up to Transformations 71

                         =Trigonometry up to Transformations 72

                          =Trigonometry up to Transformations 73

                          =Trigonometry up to Transformations 74

                           =Trigonometry up to Transformations 75

                           = 2 cot 2A

Question 29

Evaluate 6 sin 200 – 8sin3 200

Sol:

6 sin 200 – 8sin3 200 = 2 (3 sin 200 – 4sin3 200)

                                       = 2 sin 3(200)

                              = 2 sin 600

                              = 2Trigonometry up to Transformations 76 

                              =Trigonometry up to Transformations 77

Question 30

Express cos6 A + sin6 A in terms of sin 2A.

Sol:

cos6 A + sin6 A

= (sin2 A)3 + (cos2 A)3

= (sin2 A + cos2 A)3 – 3 sin2 A cos2 A (sin2 A + cos2 A)

= 1 – 3 sin2 A cos2 A

=1 – ¾ (4 sin2 A cos2 A)

 = 1 – ¾ sin22 A

 

Question 31

If 0 < θ < π/8, show that Trigonometry up to Transformations 79  = 2 cos (θ/2)

Sol:

Trigonometry up to Transformations 79

Trigonometry up to Transformations 80

Trigonometry up to Transformations 81

 =2 cos (θ/2)

Question 32

Find the extreme values of cos 2x + cos2x

Sol:

cos 2x + cos2x = 2cos2 x– 1 + cos2 x

                              =3cos2 x – 1

We know that – 1 ≤ cos x ≤ 1

 ⟹ 0 ≤ cos2 x ≤ 1

      3×0 ≤ 3×cos2 x ≤ 3×1

      0– 1 ≤3 cos2 x – 1≤ 3– 1

   – 1≤3 cos2 x – 1≤ 2

     Minimum value = – 1

     Maximum value = 2

Question 33

Prove that Trigonometry up to Transformations 82 = 4

Sol:

Trigonometry up to Transformations 82

Trigonometry up to Transformations 83

= 4

Question 34

Prove that sin 780 + cos 1320 =Trigonometry up to Transformations 84

Sol:

 sin 780 + cos 1320 = sin 780 + cos (900 + 420)

                                      = sin 780 – sin 420

                                      = 2 cosTrigonometry up to Transformations 85  sinTrigonometry up to Transformations 86

                                     = 2 cos 600 sin 180

                                      = 2Trigonometry up to Transformations 87

                                      =Trigonometry up to Transformations 84

Question 35

Find the value of sin 340 + cos 640 – cos40

Sol:

sin 340 + cos 640 – cos40

= sin 340 –2 sinTrigonometry up to Transformations 88  sinTrigonometry up to Transformations 89

= sin 340 – 2sin 340 sin 300

= sin 340 – 2 sin 340 (1/2)

=sin 340 – sin 340

=0

 

Question 36

Prove that 4(cos 660 + sin 840) =Trigonometry up to Transformations 90

Sol:

4(cos 660 + sin 840)  

=4(cos 660 + sin (900 – 60)  

=4(cos 660 + cos (60)  

= 4[ 2 cosTrigonometry up to Transformations 91  cos Trigonometry up to Transformations 92]

=4[ 2 cosTrigonometry up to Transformations 93  cosTrigonometry up to Transformations 94 ]

=8 cos 360 cos 300

= 8Trigonometry up to Transformations 95

=Trigonometry up to Transformations 90

Question 35

Prove that (tan θ + cot θ)2 = sec2 θ + cosec2 θ = sec2 θ. cosec2 θ

Sol:

tan θ + cot θ =    Trigonometry up to Transformations 96

                         =Trigonometry up to Transformations 97

                         =Trigonometry up to Transformations 98

                         = sec θ. cosec θ

(tan θ + cot θ)2   = sec2 θ. cosec2 θ

sec2 θ + cosec2 θ =Trigonometry up to Transformations 99

                               Trigonometry up to Transformations 100

                              = sec2 θ. cosec2 θ


 

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Product of Vectors (Qns.& Ans) V.S.A.Q.’S

Product of Vectors (Qns.& Ans) V.S.A.Q.’S

Product of Vectors (Qns.& Ans) V.S.A.Q.’S; These solutions were designed by the ‘Basics in Maths‘ team. These notes to do help intermediate First-year Maths students.

Inter Maths – 1A two marks questions and solutions are very useful in IPE examinations.


Product Of Vectors

Question 1

If a = 6i +2 j +3 k, b = 2i – 9 j+ 6k, then find the angle between the vectors a and b

Sol:

Given vectors are a = 6i +2 j +3 k, b = 2i – 9 j+ 6k

 If θ is the angle between the vectors Product of Vectors 1a and b, then cos θ = 

 a .b = (6i +2 j +3 k). (2i – 9 j+ 6k) = 6(2) + 2 (– 9) + 3(6)

          = 12 – 18 + 18 = 12

Product of Vectors 2      

              = 7

Product of Vectors 3      

              = 11

  ⟹ cos θ =Product of Vectors 4

 θ =Product of Vectors 5

Question 2

If a = i +2 j –3 k, b = 3ij+ 2k, then show that a + b and ab are perpendicular to each other.

Sol:

Given vectors are a = i +2 j –3 k, b = 3ij+ 2k

a + b = (i +2 j –3 k) + (3ij+ 2k) = 4i + jk

ab = (i +2 j –3 k) – (3ij+ 2k) = –2i +3 j – 5k

(a + b). (ab) = (4i + jk). (–2i +3 j – 5k)

                            = – 8 + 3 + 5

                           = 0

a + b and ab is perpendicular to each other.

Question 3

If a and b be non-zero, non-collinear vectors. If , then find the angle between a and b

Sol:

Given Product of Vectors 6

 Squaring on both sides

 Product of Vectors 7    

 (a + b) (a + b) = (a – b) (a – b)

  a2 + 2 a. b + b2 = a2 – 2 a.b + b2

 ⟹ 4 a.b = 0

  a.b = 0

∴ the angle between a and b  is 900

Question 4

IfProduct of Vectors 8 = 11, Product of Vectors 9= 23 and Product of Vectors 11 = 30, then find the angle between the vectors a and b and also find

Sol:

Given Product of Vectors 8= 11,  Product of Vectors 9 = 23 and Product of Vectors 11= 30

       Product of Vectors 11 = 30

Product of Vectors 12 = 900

Product of Vectors 13 = 900

(11)2 2 ×11×23 cos θ + (23)2 = 900

121 506 cos θ + 529   = 900

 650 506 cos θ = 900

cos θ = Product of Vectors 14 ⟹ θ =Product of Vectors 15

Product of Vectors 16   

                = (11)2 + 2 ×11×23 cos θ + (23)2

                = 121 + 2 ×11×23 ×  + 529

               = 400

 = 20

Question 5

If a = ijk and b = 2i – 3j + k, then find the projection vector of b on a and its magnitude.

Sol:

Given vectors are a = ijk and b = 2i – 3j + k

 a.b = (ij k). (2i – 3j + k) = 2 + 3 – 1 = 4

Product of Vectors 18   =Product of Vectors 19

 The projection vector of b on a =Product of Vectors 17

                  = Product of Vectors 20  (i j k)

  The magnitude of the projection vector = Product of Vectors 21 =Product of Vectors 22

Question 6

If the vectors λ i – 3j + 5k and 2λ i – λ jk are perpendicular to each other, then find λ

Sol:

let a = λ i – 3j + 5k, b = 2λ i – λ jk

Given, that a and b are perpendicular to each other

a.b = 0

i – 3j + 5k). (2λ i – λ jk) = 0

2 λ2 + 3 λ – 5 = 0

2 λ2 + 5 λ – 2 λ – 5 = 0

λ (2 λ + 5) – 1 (2 λ + 5) = 0

(2 λ + 5) ((λ – 1) = 0

λ = 1 or λ = -5/2

Question 7

Find the Cartesian equation of the plane passing through the point (– 2, 1, 3) and perpendicular to the vector 3i + j + 5k

Sol:

let P (x, y, z) be any point on the plane

 ⟹ OP = xi + yj + zk

 OA = – 2i +j +3k

AP = OPOA = (xi + yj + zk) – (– 2i +j +3k)

 AP = (x + 2) i + (y – 1) j + (z – 3) k

 AP is perpendicular to the vector 3i + j + 5k

 ⟹ 3 (x + 2) + (y – 1) + 5(z – 3) = 0

  ⟹ 3x + 6 + y – 1 + 5z – 15 = 0

 ∴   3x + y + 5z – 10 = 0 is the required Cartesian equation of the plane

Question 8

Find the angle between the planes 2x – 3y – 6z = 5 and 6x + 2y – 9z = 4

Sol:

Given plane equations are: 2x – 3y – 6z = 5,6x + 2y – 9z = 4

 Vector equations of the above planes are: r. (2i – 3j – 6k) = 5 and r. (6i + 2j – 9k) = 4

 ⟹ n1 = 2i – 3j – 6k and n2 = 6i + 2j – 9k

 If θ is the angle between the planes r. n1 = d1 and r. n2 = d2, then

 Cos θ = Product of Vectors 23

  ⟹ Cos θ =Product of Vectors 24

 ⟹ θ = Product of Vectors 25

Question 9

a = 2ij + k, b = i – 3j – 5k. Find the vector c such that a, b, and c form the sides of a triangle.

Sol:

Given a = 2ij + k, b = i – 3j – 5k

 If a, b, and c form the sides of a triangle, then a + b + c = 0

 ⟹ a + b = – c

c = – (a + b)

       = – [(2ij + k) +( i – 3j – 5k)]

      = – (3i –4 j –4k)

     ∴ c = – 3i +4 j + 4k     

Question 10

Find the equation of the plane through the point (3, –2, 1) and perpendicular to the vector (4, 7, –4).

Sol:

 Let a = 3i – 2j + k and n = 4i + 7j – 4k

 The equation of the plane passing through point A(a) and perpendicular to the vector n is (ra). n = 0

 ⟹ [r – (3i – 2j + k)]. (4i + 7j – 4k) = 0

 ⟹ r. (4i + 7j – 4k)– [(3i – 2j + k). (4i + 7j – 4k)] = 0

r. (4i + 7j – 4k)– (12 – 14 – 4) = 0

r. (4i + 7j – 4k)– 6 = 0

r. (4i + 7j – 4k) = 6

Question 11

Find the unit vector parallel to the XOY-plane and perpendicular to the vector 4i – 3j + k

Sol:

The vector which is parallel to the XOY-plane is of the form xi + yj

The vector which is parallel to the XOY-plane and perpendicular to 4i – 3j + k

 is 3i + 4j

 its magnitude = Product of Vectors 26     = 5

∴ The unit vector parallel to the XOY-plane and perpendicular to the vector 4i – 3j + k =Product of Vectors 27

Question 12

If a + b + c = 0, Product of Vectors 18 = 3, Product of Vectors 9 = 5 and   Product of Vectors 30 = 7, then find the angle between a and b

Sol:

Given, a + b + c = 0,  = 3,  = 5 and    = 7

 a + b = – c

 (a + b)2 =Product of Vectors 29

Product of Vectors 28 

 32 + 52 + 2  cos θ = 72

  9 + 25 + 2.3.5 cos θ = 49

 34 + 30cos θ = 49

 30cos θ = 49 – 34

 30cos θ = 15

cos θ = 15/30 = 1/2

∴ θ = π/3

Question 13

If a = 2i – 3j + 5k, b = – i + 4j + 2k, then find a × b and unit vector perpendicular to both a and b

Sol:

 Given, a = 2i – 3j + 5k, b = – i + 4j + 2k             

 a × b =Product of Vectors 40

         = i (–6 – 20) – j (4 + 5) + k (8 – 3)

         = –26i – 9j + 5k

The unit vector perpendicular to both a and b    =   Product of Vectors 31

  =Product of Vectors 32

=Product of Vectors 33

Question 14

If a = i + j + 2k and b = 3i + 5jk are two sides of a triangle, then find its area.

Sol:

Given, a = i + 2j + 3k and b = 3i + 5jk

If a, b are two sides of a triangle, then area of the triangle =

a × b =Product of Vectors 41

         = i (–2 – 15) – j (–1 – 9) + k (5 – 6)

         = –17i + 10 j k

Product of Vectors 37 =Product of Vectors 35

Area of triangle =Product of Vectors 34 =

                              =Product of Vectors 36

Question 15

Find the area of the parallelogram for which the vectors a = 2i – 3j and b = 3ik are adjacent sides.

Sol:

Given, a = 2i – 3j and b = 3ik are adjacent sides of a parallelogram

The area of the parallelogram whose vectors a , b are adjacent sides =  

  a × b=Product of Vectors 38

              = i (3 – 0) – j (–2 – 0) + k (0 + 9)

             =3 i +2 j +9 k

Product of Vectors 37= Product of Vectors 42

                = Product of Vectors 39

∴ The area of the parallelogram =Product of Vectors 39

Question 16

Let a, b be two non-collinear unit vectors. If α = a – (a . b) b and β = a × b, then show that Product of Vectors 43

Sol:

Product of Vectors 44 = Product of Vectors 45

          = Product of Vectors 46

              = 1 – cos2 θ

             = sin2 θ

Product of Vectors 47 =

        = 1 + cos2 θ – 2cos2 θ

       = 1– cos2 θ

       = sin2 θ

Product of Vectors 43

Question 17

For any vector a, show thatProduct of Vectors 48     

Sol:

Let a = xi + yj + zk

       a × i =Product of Vectors 49

                = i ( 0 – 0) – j (0 – z) + k (0 – y)

           = – yk + zj

Product of Vectors 50   = y2 + z2        

Similarly, Product of Vectors 51  = x2 + z2      and      = y2 + x2        

Product of Vectors 50 + Product of Vectors 51+ Product of Vectors 51 = y2 + z2 + x2 + z2 + y2 + x2        

                                                        = 2(x2 + y2 +z2)

                                                        =2 Product of Vectors 53

Question 18

IfProduct of Vectors 54 = 2, Product of Vectors 55 = 3 and (p, q) =Product of Vectors 56 , then find

Sol:

Given Product of Vectors 54= 2, Product of Vectors 55 = 3 and (p, q) =Product of Vectors 56

Product of Vectors 58  = Product of Vectors 54Product of Vectors 55 sin (p, q)

                = 2 × 3 sin

                = 2 × 3×1/2

               = 3

Product of Vectors 57 = 32 = 9

Question 19

If 4i + j + pk is parallel to the vector i + 2j + 3k, then find p.

Sol:

Given 4i +  j + pk is parallel to the vector i + 2j + 3k

Product of Vectors 59   = Product of Vectors 60 =Product of Vectors 61

Product of Vectors 61 = 4

⇒ p = 12

Question 20

Compute a× (b + c) + b× (c + a) + c× (a + b)

Sol:

a× (b + c) + b× (c + a) + c× (a + b)

= a× b + a× c + b× c + b× a + c × a + c × b

= a× b + a× c + b× ca × b a × c b ×c

= 0

Question 21

Compute 2j× (3i – 4k) + (i + 2j) × k

Sol:

2j× (3i – 4k) + (i + 2j) × k

= 6(j × i) – 8(j × k) + (i × k) + 2(j × k)

 = 6k – 8i – j + 2i

 = 6i –j 6 j


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Addition of Vectors vsaqs questions and solutions

Addition of Vectors (Qns.& Ans) V.S.A.Q.’S

Addition of Vectors

These solutions were designed by the ‘Basics in Maths’ team. These notes to do help intermediate First-year Maths students.

Inter Maths – 1A two marks questions and solutions are very useful in IPE examinations.


Addition of Vectors

QUESTION 1

Find the unit vector in the direction of Addition of Vectors 17

Sol: Given vector isAddition of Vectors 17

 The unit vector in the direction of a vector Addition of Vectors 9 is given byAddition of Vectors 15 

Addition of Vectors 16

QUESTION 2

Find a vector in the direction of a where Addition of Vectors 19that has a magnitude of 7 units.

Sol: Given vector is Addition of Vectors 19

 The unit vector in the direction of a vector Addition of Vectors 9 is Addition of Vectors 20

 The vector having the magnitude 7 and in the direction of  is

  Addition of Vectors 21

QUESTION 3

Find the unit vector in the direction of the sum of the vectors, a = 2i + 2j – 5k and b = 2i + j + 3k

Sol Given vectors are a = 2i + 2j – 5k and b = 2i + j + 3k

a + b = (2i + 2j – 5k) + (2i + j + 3k) = 4i + 3j – 2k

Addition of Vectors 22

Addition of Vectors 23      

 QUESTION 4

Write the direction cosines of the vector Addition of Vectors 24

Sol: Given vector is Addition of Vectors 24

Addition of Vectors 25

    ∴ Direction cosines are Addition of Vectors 27

QUESTION 5

Show that the points whose position vectors are – 2a + 3b + 5c, a + 2b + 3c, 7 ac are collinear when a, b, c are non-collinear vectors

Sol: Let OA = – 2a + 3b + 5c, OB = a + 2b + 3c, OC = 7 acA

B = OB – OA = a + 2b + 3c (– 2a + 3b + 5c)

         AB = 3a b – 2c

 AC = OC – OA = 7 ac(– 2a + 3b + 5c)

 AC = 9a – 3b – 6c = 3(3a b – 2c)

          AC = 3 AB

   A, B and C are collinear

 QUESTION 6

ABCD is a parallelogram if L and M are middle points of BC and CD. Then find (i) AL and AM in terms of AB and AD (ii) 𝛌, if AM = 𝛌 AD – LM

Sol: Given, ABCD is a parallelogram and L and M are middle points of BC and CD

Addition of Vectors 1

(i) Take A as the origin

 M is the midpoint of CD

  AM =Addition of Vectors 2

          = AD + ½ AB (∵ AB = DC)

 L is the midpoint of BC

  AL =Addition of Vectors 3

      = AB + ½ AD ((∵ BC = AD)

(ii) AM = 𝛌 AD – LM

AM + LM= 𝛌 AD 

AD + ½ AB + AD + ½ AB – (AB + ½ AD) = 𝛌 AD 

AD + ½ AB + AD + ½ AB –  AB – ½ AD = 𝛌 AD 

3/2 AD = 𝛌 AD 

𝛌 = 3/2

QUESTION 7

If G is the centroid of the triangle ABC, then show that OG = Addition of Vectors 4  whenAddition of Vectors 9,Addition of Vectors 10 Addition of Vectors 42 are the position vectors of the vertices of triangle ABC.

Sol: OA = a, OB = b, OC = c and OD = d

Addition of Vectors 6

 D is the midpoint of BC

OD = Addition of Vectors 7

G divides median AD in the ratio 2: 1

OG =Addition of Vectors 8

OG =Addition of Vectors 4

QUESTION 8

If Addition of Vectors 9= Addition of Vectors 11, Addition of Vectors 10 =  Addition of Vectors 12  are collinear vectors, then find m and n.

Sol: Given Addition of Vectors 9 , Addition of Vectors 10 are collinear vectors

Addition of Vectors 9 = λAddition of Vectors 10

Addition of Vectors 13

 Equating like vectors

 2 = 4 λ; 5 = m λ; 1 = n λ

 λ =Addition of Vectors 14

 5 = Addition of Vectors 14m ⟹ m =10

1 = Addition of Vectors 14n ⟹ n = 2

∴ m = 10, n = 2

QUESTION 9

Let If  Addition of Vectors 28, Addition of Vectors 29. Find the unit vector in the direction of a + b.

Sol: Given vectors are Addition of Vectors 28 and  Addition of Vectors 29

     a + b = Addition of Vectors 33

    The unit vector in the direction of a + b = Addition of Vectors 30  

       = Addition of Vectors 31

        =  Addition of Vectors 32

QUESTION 10

If the vectors – 3i + 4j + λk and μi + 8j + 6k. are collinear vectors, then find λ and μ.

Sol: let a = – 3i + 4j + λk, b = μi + 8j + 6k

     ⟹   a = tb

 – 3i + 4j + λk = t (μi + 8j + 6k)

 – 3i + 4j + λk = μt i + 8t j + 6t k

Equating like vectors

– 3 = μt; 4 = 8t, λ = 6t

4 = 8t

 t =Addition of Vectors 14

– 3 =Addition of Vectors 14 μ ⟹μ=– 6

λ = Addition of Vectors 14 6 ⟹ λ = 3

∴ μ=– 6, λ = 3

QUESTION 11

ABCD is a pentagon. If the sum of the vectors AB, AE, BC, DC, ED and AC is 𝛌 AC then find the value of 𝛌

Sol: Given, ABCD is a pentagon

        AB + AE + BC + + DC + ED + AC = 𝛌 AC

         (AB + BC) + (AE + DC + ED) + AC = 𝛌 AC

         AC + AC + AC = 𝛌 AC

         3 AC = 𝛌 AC

         𝛌 = 3

QUESTION 12

If the position vectors of the points A, B and C are – 2i + jk and –4i + 2j + 2k and 6i – 3j – 13k respectively and AB = 𝛌 AC, then find the value of 𝛌

Sol: Given, OA = – 2i + jk , OB = –4i + 2j + 2k and OC  = 6i – 3j – 13k

  AB = OB – OA = –4i + 2j + 2k – (– 2i + jk)

          = –4i + 2j + 2k +2ij + k

        = –2i + j + 3k

 AC = OC – OA = 6i – 3j – 13k – (– 2i + jk)

       = 6i – 3j – 13k +2ij + k

     = 8i –4 j –12k

     = – 4 (2i + j + 3k)

AC = – 4 AB

 Given AB = 𝛌 AC

    𝛌 = – 1/4

QUESTION 13

If OA = i + j +k, AB = 3i – 2j + k, BC = i + 2j – 2k, CD = 2i + j +3k then find the vector OD

Sol: Given OA = i + j +k, AB = 3i – 2j + k, BC = i + 2j – 2k, CD = 2i + j +3k

         OD = OA + AB + BC + CD

                = i + j +k + 3i – 2j + k + i + 2j – 2k + 2i + j +3k

                = 7i + 2j +4k

QUESTION 14

Let a = 2i +4 j –5 k, b = i + j+ k, c = j +2 k, then find the unit vector in the opposite direction of a + b + c

Sol:  Given, a = 2i +4 j –5 k, b = i + j+ k, c = j +2 k

         a + b + c = 2i +4 j –5 k + i + j+ k + j +2 k

                           = 3i +6j –2k

 The unit vector in the opposite direction of a + b + c is Addition of Vectors 34

 ⟹ Addition of Vectors 35

      =Addition of Vectors 36

QUESTION 15

Is the triangle formed by the vectors 3i +5j +2k, 2i –3j –5k, 5i – 2j +3k

Sol: Let a =3i +5j +2k, b = 2i –3j –5k, c = 5i – 2j +3k

  Addition of Vectors 37

 ∴ Given vectors form an equilateral triangle.

QUESTION 16

Using the vector equation of the straight line passing through two points, prove that the points whose position vectors are a, b,  (3a – 2b) are collinear.

Sol: the vector equation of the straight line passing through two points a, b is

         r = (1 – t) a+ t b

         3a – 2b = (1 – t) a+ t b

          Equating like vectors

          1 – t = 3 and t = – 2

∴ Given points are collinear.

QUESTION 17

OABC is a parallelogram If OA = a and OC = c, then find the vector equation of the side BC

Sol: Given, OABC is a parallelogram and OA = a, OC = c

Addition of Vectors 38

 The vector equation of BC is a line which is passing through C(c) and parallel to OA

 ⟹ the vector equation of BC is r = c + t a

QUESTION 18

If a, b, c are the position vectors of the vertices A, B, and C respectively of triangle ABC, then find the vector equation of the median through the vertex A

Sol: Given OA = a, OB = b, OC = c

Addition of Vectors 39

 D is mid of BC

  OD =Addition of Vectors 40 = Addition of Vectors 41

 The equation of AD is

  r = (1 – t) a + t ( Addition of Vectors 41 )

QUESTION 19

Find the vector equation of the line passing through the point 2i +3j +k and parallel to the vector 4i – 2j + 3k

Sol: Let a =2i +3j +k, b = 4i – 2j + 3k  

         The vector equation of the line passing through a and parallel to the vector b is r = a + tb

       r = 2i +3j +k + t (4i – 2j + 3k)

          = (2 + 4t) i + (3 – 2t) j + (1 + 3t) k

QUESTION 20

Find the vector equation of the plane passing through the points i – 2j + 5k, 2j –k and – 3i + 5j

Sol: The vector equation of the line passing through a, b and cis r = (1 – t – s) a + tb + sc

      r = (1 – t – s) (i – 2j + 5k) + t (2j –k) + s (– 3i + 5j)

              = (1 – t – 4s) i + (– 2 – 3t + 7s) j + (5 – 6t – 5s) k


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Matrices vsaqs questions and solutions

Matrices ( Qns & Solutions) || V.S.A.Q’S||

Matrices ( Qns & Solutions) || V.S.A.Q’S||

Matrices V.S.A.Q’s: This note is designed by the ‘Basics in Maths’ team. These notes to do help intermediate First-year Maths students.

Inter Maths – 1A two marks questions and solutions are very useful in IPE examinations.

These notes cover all the topics covered in the intermediate First-year Maths syllabus and include plenty of solutions to help you solve all the major types of Math problems asked in the

IPE examinations.  


Matrices

QUESTION 1

If A = Matrices 1, then show that A2 = –I

Sol: Given A = Matrices 1

Matrices 2

  ∴  A2 = –I

QUESTION 2

If A = Matrices 3, and A2 = 0, then find the value of k.

Sol: Given A = Matrices 3 and

 A2 = 0

⟹ A. A =0 ⟹  Matrices 4   = 0

Matrices 5 = 0

     8 + 4k = 0, – 2 – k = 0 and –4 + k2 = 0

    4k = –8; k = –2; k2 = 4

       k = –2; k = –2; k = ± 2

   ∴ k =– 2

QUESTION 3

Find the Trace of A, If A = Matrices 6

Sol: Given A =Matrices 6

       Trace of A = 1 – 1 + 1 = 1

QUESTION 4

If A =Matrices 12 , B = Matrices 13and 2X + A = B, then find X.

Sol: Given A =Matrices 12 , B = Matrices 13 and 2X + A = B

        2X = B – A

        2X =Matrices 13  – Matrices 12

              = Matrices 8

              =Matrices 9

           X =   Matrices 10 Matrices 9

         ∴ X =   Matrices 11

QUESTION 5

Find the additive inverse of A, If A =Matrices 14

Sol: Given A =Matrices 14

       Additive inverse of A = – A

    = –Matrices 14

   =Matrices 15

QUESTION 6

If Matrices 16, then find the values of x, y, z and a.

Sol: Given Matrices 16

 ⟹ x- 1 = 1 – x ; y – 5 =  – y ; z = 2 ; 1 + a = 1

 ⟹ x + x = 1 + 1; y + y = 5; z = 2; a =1– 1

  ⟹ 2x = 1; 2y = 5; z = 2; a = 0

∴ x = ½ ; y = 5/2; z = 2; a = 0

QUESTION 7

Construct 3 × 2 matrix whose elements are defined by aij =Matrices 17

Sol:

Let A= Matrices 18

a11 = Matrices 19

a11 = 1

a12 = Matrices 20

a12 =Matrices 22

a21 = Matrices 23

a21 =Matrices 24

a22 = Matrices 25

a22 = 2

a31 = Matrices 26

a31 = 0

a32 = Matrices 27

a32 =Matrices 28

 ∴ A =Matrices 29

QUESTION 8

If A = Matrices 30 and B = Matrices 31, do AB and BA exist? If they exist, find them. BA and AB commutative with respect to multiplication.

Sol: Given Matrices are A = Matrices 30 B =Matrices 31

       Order of A = 2 × 3 and Order of B = 3 × 2

AB and BA exist

 

 AB =   Matrices 30Matrices 31

    Matrices 32

BA =     Matrices 31Matrices 30

Matrices 33

 AB and  BA are not Commutative under Multiplication 

QUESTION 9

Define Symmetric and Skew Symmetric Matrices

Sol:

Symmetric Matrix: Let A be any square matrix, if AT = A, then A is called Symmetric Matrix

Skew Symmetric Matrix: Let A be any square matrix if AT = –A, then A is called Skew Symmetric Matrix

QUESTION 10

If A =Matrices 34 is a symmetric matrix, then find x.

Sol: Given, A = Matrices 34 is a symmetric matrix

       ⟹ AT = A

          Matrices 35     

          ⟹ x = 6

QUESTION 11

If A =Matrices 36 is a skew-symmetric matrix, then find x

Sol: Given A = Matrices 36is a skew-symmetric matrix

       ⟹ AT = – A

     

        ⟹ x = –x

        x+ x = 0 ⟹ 2x = 0

     ⟹ x = 0

QUESTION 12

If A =Matrices 38 and B = Matrices 39, then find (A BT) T

Sol: Given A = Matrices 38   B =Matrices 39

   BT = Matrices 40   

   (A BT) =  Matrices 38   Matrices 40

                = Matrices 41

(A BT) T = Matrices 42

QUESTION 13

If A =Matrices 43 and B =Matrices 44 , then find A + BT

Sol: Given A =Matrices 43  and B =Matrices 44

 BT =Matrices 45

A + BT = Matrices 43 + Matrices 45

 Matrices 47            

QUESTION 14

If A = Matrices 48, then show that AAT = ATA = I

Sol: Given A =Matrices 48

  AT =Matrices 49

AAT =

= Matrices 51

 = Matrices 52 

ATA =Matrices 49Matrices 48

        =

       =Matrices 52

∴ AAT = ATA = I

QUESTION 15

Find the minor of – 1 and 3 in the matrixMatrices 54

Sol: Given Matrix is

       minor of – 1 = Matrices 55 = 0 + 15 = 15

     minor of 3 = Matrices 56 = – 4 + 0 = – 4

QUESTION 16

Find the cofactors 0f 2, – 5 in the matrixMatrices 57

Sol: Given matrix is

 Cofactor of 2 = (–1)2 + 2 Matrices 58= –3 + 20 = 17

  Cofactor of – 5 = (–1)3 + 2  Matrices 59= –1(2 – 5) = –1(–3) = 3

QUESTION 17

If ω is a complex cube root of unity, then show that Matrices 62= 0(where 1 + ω+ω2 = 0)

Given matrix is    Matrices 62

   R1 → R1 + R2 + R3

   Matrices 73 Matrices 74

Matrices 74 = 0 (∵ 1 + ω+ω2 = 0)

QUESTION 18

If A = Matrices 63and det A = 45, then find x.

Sol: Given A = Matrices 63

Det A = 45

Matrices 64= 45

   ⟹ 1(3x + 24) – 0 (2x – 20) + 0 (– 12 – 15) = 45

 ⟹ 3x + 24 = 45

        3x = 45 – 24

        3x = 21

         x = 7

QUESTION 19

Find the adjoint and inverse of the following matrices

(i)

A =Matrices 65

Adj A =Matrices 66

A-1 = Matrices 67

       =Matrices 68

   ∴ A-1 =Matrices 69

(ii)

A =Matrices 70

Adj A =Matrices 71

A-1 =Matrices 72

 ∴ A-1 = Matrices 71   

QUESTION 20

Find the inverse of Matrices 75 (abc ≠ 0)

Sol: Let A =Matrices 75

        Det A = a (bc – 0) – 0(0 – 0) + 0(0 – 0)

        Det A = abc ≠ 0

Cofactor matrix of A =Matrices 76

Adj A = (Cofactor matrix of A) T

           =Matrices 76

A-1 =Matrices 77

  A-1 = Matrices 85 Matrices 76

 ∴ A-1Matrices 78

QUESTION 20

Find the rank of the following matrices.

(i) Matrices 79

Let A =Matrices 79

Det A = 1 (0 – 2) – 2(1 – 0) + 1(– 1 – 0)

           = – 2– 2– 1

           = – 5 ≠ 0

∴ Rank of A = 3

(ii)Matrices 80

Let A =Matrices 80

Det A = – 1 (24 – 25) + 2(18 – 20) + – 3(15 – 16)

           = – 1– 4 + 3

           = – 0

Sub matrix of A = Matrices 81

      Let B =Matrices 81

       Det B = – 4 + 6 = 2 ≠ 0

          ∴ Rank of A = 2

(iii)Matrices 82

Let A =Matrices 82

Sub matrix of A = Matrices 86

Det of Sub matrix of A = – 1 – 0 = – 1 ≠ 0

      ∴ Rank of A = 2

(iv)Matrices 83

Let A =Matrices 83

Sub matrix of A =Matrices 84

Det of Sub matrix of A =1 (1 – 0) – 0(0 – 0) + 0(0 – 0)

                                           = 1≠ 0

      ∴ Rank of A = 3


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The Plane vsaqs questions and solutions

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The Plane: These solutions were designed by the ‘Basics in Maths’ team. These notes to do help intermediate First-year Maths students.
Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations.
These notes cover all the topics covered in the intermediate First-year Maths syllabus and include plenty of solutions to help you solve all the major types of Math problems asked in the IPE examinations.  


The Plane

Question 1

Find the equation of the plane if the foot of the perpendicular from the origin to the plane is (2, 3, – 5).

Sol:

The plane passes through A and is perpendicular to OA, then the line segment OA is normal to the plane.

  Dr’s of OA = (2, 3, – 5)

  The equation of the plane passing through the point (x1, y1, z1) and dr’s (a, b, c) is

   a(x – x1) + b (y – y1) + c (z – z1) = 0

  ⟹ 2(x – 2) + 3 (y – 3) – 5 (z + 5) = 0

        2x – 4 + 3y – 9 – 5z – 25 = 0

        2x + 3y – 5z – 38 = 0

Question 2

Find the equation of the plane passing through the points (0, – 1, – 1), (4, 5, 1) and (3, 9, 4)

Sol:

The equation of the plane passing through the points (x1, y1, z1) (x2, y2, z2) (x3, y3, z3) isPlane 1

The plane passing through the points (0, – 1, – 1), (4, 5, 1) and (3, 9, 4) is

Plane 2   =0

Plane 3  = 0

   x (30 – 20) – (y + 1) (20 – 6) + (z + 1) (40 – 18) = 0

   x (10) – (y + 1) (14) + (z + 1) (22) = 0

  10x – 14y – 14 + 22z + 22 = 0

  10x – 14y + 22z + 8 = 0

  2(5x – 7y + 11z + 4) = 0

 ∴ the equation of the plane is 5x – 7y + 11z + 4 = 0

Question 3

Find the equation to the plane parallel to the ZX-plane and passing through (0, 4, 4).

Sol:

Equation of ZX-plane is y = 0

 The equation of the plane parallel to the ZX-plane is y = k

 But it is passing through (0, 4, 4)

 ⟹ y = 4

Question 4

Find the equation to the plane passing through the point (α, β, γ) and parallel to the plane axe + by + cz + d = 0.

Sol:

The equation of the plane parallel to the plane ax + by + cz + d = 0 is ax + by + cz + k = 0

        But it is passing through the point (α, β, γ)

        a α + b β + c γ + k = 0

  ⟹ k = – a α – b β – c γ

   The equation of the plane is ax + by + cz – a α – b β – c γ = 0

    ⟹ a(x – α) + b (y – β)+ c (z – γ) = 0

Question 5

Find the angle between the planes 2x – y + z = 6 and x + y + 2z = 7.

Sol: If θ is the angle between the planes a1 x + b1 y + c1 z + d1 = 0 and a2x + b2 y + c2 z + d2 = 0, then cos θ = Plane 4

            Cos θ = Plane 5

                        =Plane 6

                          =Plane 7

            Cos θ = cos 600

                    θ = 600 =

Question 6

Reduce the equation x + 2y – 2z – 9 = 0 to the normal form and hence find the dc’s of the normal to the plane.

Sol: Given plane is x + 2y – 2z – 9 = 0

         x + 2y – 2z = 9

         dividing into both sides byPlane 9 

         the normal form is Plane 10

        dc’s of the normal to the plane arePlane 11

Question 7

Suppose a plane makes intercepts 2, 3, 4 on X, Y, Z axes respectively. Find the equation of the plane in the intercept form.

 Sol: Given a = 2, b = 3, c = 4

        The equation of the line in the intercept form isPlane 12

        ⟹plane 21

Question 8

Express x – 3y + 2z = 9 in the intercept form

Sol: Given plane is x – 3y + 2z = 9

       ⟹ Plane 13

      ⟹ Plane 14

       It is in the form of Plane 12

      a = 9, b = – 3, c = 9/2

Question 9

Find the direction cosine of the normal to the plane x + 2y + 2z – 4 = 0.

Sol: Given plane is x + 2y + 2z – 4 = 0

        We know that Dr’s of the normal to the plane ax + by + cz + d = 0 are (a, b, c)

        ⟹ dc’s of the normal to the plane =Plane 15

         ⟹ dr’s of the normal to the plane x + 2y + 2z – 4 = 0 are (1, 2, 2)

          ⟹ dc’s of the normal to the plane are Plane 16

                      =Plane 17

Question 10

Find the midpoint of the line joining the points (1, 2, 3) and (–2, 4, 2)

Sol: Given points are A (1, 2, 3), B (–2, 4, 2)

       The midpoint of AB =Plane 18

                                    =Plane 19

                                    =Plane 20


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3D Coordinates vsaqs questions and solutions

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These solutions were designed by the ‘Basics in Maths‘ team. These notes to do help intermediate First-year Maths students.

Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations.


3D Coordinates

 

1. Show that the points A (– 4, 9, 6), B (– 1, 6, 6), and C (0, 7, 10) form a right-angled isosceles triangle.

Sol:

Distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is PQ = 3d - 1

3d - 2

∴ The points A (– 4, 9, 6), B (– 1, 6, 6) and C (0, 7, 10) form a right-angled isosceles triangle.

2. Show that the locus of the point whose distance from Y – axis is thrice its distance from (1, 2, – 1) is 8x 2 + 9y2 + 8 z2 – 18x – 36y + 18z + 54 = 0

Sol:

Let P (x, y, z) be the locus of the point

        A (0, y, 0) be any point on Y – axis

         B = (1, 2, – 1)

     Condition is PA = 3PB

                             PA2 = (3PB)2

                             PA2= 9 PB

       ⟹ x2 + z2 = 9[(x – 1)2 + (y – 2)2 + (z + 1)2]

            x2 + z2 = 9[x2 – 2x + 1 + y2 – 4y + 4 + z2 + 2z + 1]

           x2 + z2 = 9x2 – 18x + 9 + 9y2 – 36y + 36 +9 z2 + 18z + 9

          ∴ 8x 2 + 9y2 + 8 z2 – 18x – 36y + 18z + 54 = 0

3. A, B, and C are three points on OX, OY, and OZ respectively, at distances a, b, c (a≠0, b≠0, c≠0) from the origin ‘O’. Find the coordinate of the point which is equidistant from A, B, C, and O

Sol:

Let P (x, y, z) be the required point

    O = (0, 0, 0)   A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c)

Given, AP = BP = CP = OP

             AP = OP

        ⟹   AP2 =OP2

 (x – a )2 + y2 + z2 = x2 + y2 + z2

  x2 + a2 – 2ax + y2 + z2 = x2 + y2 + z2

  a2 – 2ax = 0

  a (a – 2x) = 0

    a – 2x = 0 (∵ a≠0)

    a = 2x ⟹ a/2

    Similarly, y = b/2 and z = c/2

∴ P = (a/2, b/2, c/2)

4. Show that the points A (3, – 2, 4), B (1, 1, 1) and C (– 1, 4, – 2) are collinear

Sol:

Given points are A (3, – 2, 4), B (1, 1, 1), and C (– 1, 4, – 2)

3d - 3

5.Find x if the distance between (5, – 1, 7), (x, 5, 1) is 9 units.

Sol:

 Let A = (5, – 1, 7), B = (x, 5, 1)

   Given AB = 9

    ⟹ AB2 = 81

          (5 – x)2 + (– 1 – 5)2 + (7 – 1)2 = 81

         (5 – x)2 + 36 + 36 = 81

         (5 – x)2 + 72 = 81

         (5 – x)2 = 81 – 72 = 9

        (5 – x) = ± 3

     5 – x = 3 or 5 – x = – 3

    5 – 3 = x or 5 + 3 = x

     x = 2 or x = 8

6.If the point (1, 2, 3) is changed to point (2, 3, 1) through the translation of axes. Find a new origin.

Sol:

Given (x, y, z) = (1, 2, 3) and (X, Y, Z) = (2, 3, 1)

x = X + h, y = Y + k, z = Z + l

h = x – X, k = y – Y, l = z – Z

h = 1 – 2, k = 2 – 3, l = 3 – 1

h = – 1, k = – 1, l = 2

New origin is (– 1, – 1, 2)

7.By section formula, find the point which divides the line joining the points (2, – 3, 1) and (3, 4, – 5) in the ratio 1 : 3.

Sol:

If a point P divides the line segment joining the points (x1, y1, z1), (x2, y2, z2) in the ratio, then

3d - 7

Let P divides the line joining the points (2, – 3, 1) and (3, 4, – 5) in the ratio 1 : 3

3d - 8

8.Find the ratio in which the line joining two points (7, 0, – 1) and (– 2, 3, 5) is divided by the point (1,2,3).

Sol:

Let A = (7, 0, – 1), B = (– 2, 3, 5) and P = (1,2,3)

Suppose P divides AB in the ratio k : 1

3d - 5

9.Using section formula, verify whether the points A (2, –4, 3), B (–4, 5, 6), C (4, –7, 2) are collinear or not.

Sol:

Given Points are A (2, –4, 3), B (–4, 5, 6), C (4, –7, 2)

Let C divides AB in the ratio k : 1

3d - 9

2 – 4k = 4 (k + 1)

 2 – 4k = 4k + 4

– 4k– 4k = 4 – 2

– 8k = 2

K = -1/4

C divides AB in the Ratio 1 : 4 externally

∴ A, B, C are collinear

10.Find the centroid of the triangle whose vertices are (5, 4, 6), (1, –1, 3) and (4, 3, 2)

Sol:

The centroid of the triangle whose vertices are (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is

3d - 10

11. Find the centroid of the tetrahedron whose vertices are (2, 3, –4), (–3, 3, –2), (–1, 4, 2) and (3, 5, 1)

Sol:

The centroid of the tetrahedron whose vertices are (x1, y1, z1), (x2, y2, z2) (x3, y3, z3), and (x4, y4, z4) is 

3d - 11

 the centroid of the tetrahedron whose vertices are (2, 3, –4), (–3, 3, –2), (–1, 4, 2) and (3, 5, 1) is

3d - 12

12. Find the ratio in which the YZ-plane divides the line joining A(2, 4, 5) and B (3, 5, – 4)

Sol:

Let P be any point on the YZ-plane

   P = (0, y, z)

 Let P divides AB int eh ratio k:1

3d - 13

   YZ-plane divides AB in the ratio – 2:3

13. Find the fourth vertex of the parallelogram whose consecutive vertices are (2, 4, – 1), (3, 6, – 1) and (4, 5, 1).

Sol:  

let A = (2, 4, – 1), B = (3, 6, – 1), C = (4, 5, 1) and D = (x, y, z)

            ABCD is a parallelogram

            Midpoint of AC = Midpoint of BD

     3d - 14

        ⟹ 3 + x = 6, 6 + y = 9, – 1 + z = 0

        x = 3, y = 3, z = 1

∴ The fourth vertex D = (3, 3, 1)

14. A (5, 4, 6), B (1, –1, 3), and C (3, 3, 1) are three points. Find the coordinates of the point at which the bisector of ∠BAC meets the side BC.

Sol:

We know that the bisector of ∠BAC divides BC in the ratio AB: AC

3d - 15

If D is a point where the bisector of ∠BAC meets BC

 ⟹ D divides BC in the ratio 5:3

3d - 16

15. If M (α, β, γ) is the midpoint of the line joining the points (x1, y1, z1) and B, then find B

Sol:

Let B (x, y, z) be the required point

 M is the midpoint of AB

⟹ (α, β, γ) =3d - 17

  ⟹ 2 α = x + x1; 2 β = y + y1; 2 γ = z + z1

         x =2 α – x1; x =2 β – y1; x =2 γ – z1

    ∴ B = (2 α – x1, 2 β – y1, 2 γ –)

16. If H, G, S, and I respectively denote the orthocenter, centroid, circumcenter, and incentre of a triangle formed by the points (1, 2, 3), (2, 3, 1) and (3, 1, 2), then find H, G, S, I.

Sol:

Let A = (1, 2, 3), B= (2, 3, 1), C = (3, 1, 2)

3d - 18

⟹ ∆ ABC is an equilateral triangle

 We know that, in an equilateral triangle orthocenter, centroid, circumcenter, and incentre are the same

Centroid G = 3d - 19

                      = (2, 2, 2)

∴ H = (2, 2, 2), S = (2, 2, 2), I = (2, 2, 2)

17. Find the incentre of the triangle formed by the points (0, 0, 0), (3, 0, 0) and (0, 4, 0).

Sol:

Let A = (0, 0, 0), B = (3, 0, 0) and C = (0, 4, 0)

3d - 20

∴ I = (1, 1,0)

18. Find the ratio in which point P (5, 4, – 6) divides the line segment joining the points A (3, 2, – 4) and B (9, 8, –10). Also, find the harmonic conjugate of P

Sol:

Harmonic Conjugate: If P divides AB in the ratio m: n, then the Harmonic Conjugate of P (i.e., Q) divides AB in the ratio –m : n.

Given points are A (3, 2, – 4), B (9, 8, –10), and P (5, 4, – 6)

P divides AB in the ratio is x2 – x : x – x2

                                                     = 3 – 5 : 5 – 9

                                                      = 1 : 2 (internally)

Let Q be the harmonic conjugate of P

  ⟹ Q divides AB in the ratio –1 : 2

      Q = 3d - 21

         = (–3, –4, –2)

Q (–3, –4, –2) is the hormonic conjugate of P (5, 4, – 6)

19. If (3, 2, – 1), (4, 1,1), and (6, 2, 5) are three vertices and (4, 2, 2) is the centroid of the tetrahedron, find the fourth Vertex.

Sol:

Given vertices of Tetrahedron are A (3, 2, – 1), B (4, 1,1), C (6, 2, 5) and centroid G = (4, 2, 2)

Let the fourth vertex is D = (x, y, z)

 The centroid of the tetrahedron whose vertices are (x1, y1, z1), (x2, y2, z2) (x3, y3, z3), and (x4, y4, z4) is

3d - 22

13 + x = 16, 5 + y = 8, 5 + z = 2

x = 3, y = 3, z = 3

∴ the fourth vertex D = (3, 3, 3)  

20. Show that the points A(3, 2, –4), (5, 4, –6), and C(9, 8, –10) are collinear and find the ratio in which B divides AC.

Sol:

Given points are A(3, 2, –4), (5, 4, –6) and C(9, 8, –10)

3d - 23

∴ points A (3, 2, –4), (5, 4, –6), and C (9, 8, –10) are collinear

      B divides AB in the ratio is AB: BC =  3d - 24


 

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Straight Lines vsaqs questions and solutions

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This note is designed by the ‘Basics in Maths’ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations.

These notes cover all the topics covered in the intermediate First-year Maths syllabus and include plenty of solutions to help you solve all the major types of Math problems asked in the IPE examinations.  


Straight Lines

QUESTION1.

Prove that the points (1, 11), (2, 15), and (– 3, – 5) are collinear, and find the equation of the line containing them.

Sol:

 Let A (1, 11), B (2, 15), and C (– 3, – 5)

 The slope of the line segment joining the points (x1, y1) and (x2, y2) is Straight Lines 10

  Slope of AB  =Straight Lines 1 = 4

  Slope of BC = Straight Lines 2= Straight Lines 3 = 4

QUESTION2.

Find the condition for the points (a, 0), (h, k), and (0, b) to be collinear.

Sol:

Let A (a, 0), B (h, k) and C (0, b)

The slope of the line segment joining the points (x1, y1) and (x2, y2) is Straight Lines 10Given that A, B, and C are collinear points

  The slope of AB = The slope of BC

          ⟹    Straight Lines 4

          ⟹    Straight Lines 5

          ⟹ – hk = (h – a) ( b – k)

                  – hk = hb – hk – ab + ak

           ⟹ 0 = hb + ak – ab

           ⟹ hb + ak = ab or Straight Lines 6

QUESTION3.

Find the equations of the straight lines which makes intercepts whose sum is sum is 5 and product is 6.

Sol:

The equation of the line in the intercept form is Straight Lines 7

   Given that, a + b = 5 and ab = 6

         ⟹   b = 5 – a

           a(5 – a) = 6

         5a – a2 = 6

          a2 – 5a + 6 = 0

        a – 3a – 2a + 6 = 0

       a (a – 3) – 2(a – 3) = 0

       (a – 3) (a – 2) = 0

       a = 3 or a = 2  

case (i) if a = 3 ⟹ b = 2

         Straight Lines 8     ⟹ 2x + 3y – 6 = 0

case (ii) if a = 2 ⟹ b = 3

               Straight Lines 9= 1    ⟹ 3x + 2y – 6 = 0

QUESTION4.

Find the equation of the straight line which makes an angle 1350 with the positive X – axis measured countered clockwise and passing through the point (– 2, 3).

Sol:

Slope of the line m = tan 1350 = – 1

         The point is (– 2, 3)

 The equation of the straight line in slope point form is (y – y1) = m (x – x1)

The equation of the line passing through the point (– 2, 3) with slope – 1 is

 y – 3 = – 1 (x + 2)

 y – 3 =– x – 2

x + y – 1 = 0

QUESTION5.

Find the equation of the straight line passing through the points (1, – 2) and (– 2, 3).

Sol:

Given points are (1, – 2), (– 2, 3)

 The equation of the straight line in two points form is (y – y1) = Straight Lines 10(x – x1)

 The equation of required straight line is

        (y + 2) = Straight Lines 11 (x – 1)

         (y + 2) =Straight Lines 12  (x – 1)

         – 3 (y + 2) = 5 (x – 1)

          – 3y – 6 = 5x – 5

         5x + 3y + 1 = 0

QUESTION6.

Find the slopes of the line x + y = 0 and x – y = 0

Sol:

The slope of the line ax + by + c = 0 is Straight Lines 13

The slope of the line x + y = 0 is  = – 1

The slope of the line x – y = 0 is  = 1

QUESTION7.

Find the angle which the straight-line y = Straight Lines 19x – 4 makes with the Y-axis.

Sol:

Given equation is y = Straight Lines 19x – 4

        Compare with y = mx + c

          m = Straight Lines 19  ⟹ tan θ =Straight Lines 19
         θ = Straight Lines 17

Straight Lines 20

      Angle made by the line with X-axis isStraight Lines 17

      Angle made by the line with Y-axis is Straight Lines 18

QUESTION8.

Find the equation of the reflection of the line x = 1 in the Y-axis.

Sol:

Given equation is     x = 1

 Reflection about the Y-axis is x =– 1

 Required equation of the line is x + 1 = 0

QUESTION9.

Write the equations of the straight lines parallel to X-axis and (i) at a distance of 3 units above the X-axis and (ii) at a distance of 4 units below the X-axis.

Sol:

(i) The equation of the straight line parallel to X-axis which is at a distance of 3 units above the X-axis is y = 3

        ⟹ y – 3 = 0

 (ii) The equation of the straight line parallel to X-axis which is at a distance of 4   units below the X-axis is y = – 4

         ⟹ y + 4 = 0

QUESTION10.

Write the equations of the straight lines parallel to the Y-axis and (i) at a distance of 2 units from the Y-axis to the right of it (ii) at a distance of 5 units from the Y-axis to the left of it.

Sol:

(i) The equation of the straight line parallel to the Y-axis which is at a distance of 2 units from the Y-axis to the right of it is x = 2

                                                                          ⟹ x – 2 = 0

 (ii) The equation of the straight line parallel to the Y-axis which is at a distance of 5 units from the Y-axis to the left of it is x =– 5

                              ⟹ x + 5 = 0

QUESTION 11.

Find the value of x, if the slope of the line passing through (2, 5) and (x, 3) is 2.

Sol:

Given the slope of the line passing through (2, 5) and (x, 3) is 2.

 The slope of the line segment joining the points (x1, y1) and (x2, y2) is Straight Lines 10

       ⟹ Straight Lines 21 = 2

       ⟹ 2(x – 2) = – 2

       x – 2 = – 1  

       x = 1

QUESTION 12.

Find the equation of the straight line passing through (– 4, 5) and making non – zero intercepts on the coordinate axes whose sum is zero.

Sol:

The equation of a line in the intercept form is Straight Lines 7

         Given a = b

         ⟹Straight Lines 22 = 1

         ⟹ Straight Lines 23

              x + y = a

     but it is passing through (– 4, 5)

           – 4+ 5 = a

             a = 1

    ∴ The equation of the required straight line is x + y = 1

         x + y – 1 = 0

QUESTION13.

Find the equation of the straight line passing through (– 2, 5) and cutting off equal and non – zero intercepts on the coordinate axes.

Sol:

The equation of a line in the intercept form is Straight Lines 7

         Given a + b = 0

                     b = – a

         ⟹Straight Lines 24

         ⟹ Straight Lines 25 

                     x – y = a

    but it is passing through (– 2, 4)

           – 2– 4 = a

             a = –6

    ∴ The equation of the required straight line is x – y = –6

         x + y + 6 = 0

QUESTION14.

Find the equation of the straight line whose distance from the origin is 4, if the normal ray from the origin to the straight line makes an angle of 1350 with  the positive direction of X – axis

Sol:

The equation of the straight line in the normal form is x cos α + y sin α = p

        p = 4 and α = 1350

        x cos 1350 + y sin 1350 = 4

        x Straight Lines 26 + yStraight Lines 27  = 4

       Straight Lines 29 = 4

  – x + y = 4Straight Lines 28

   x – y + 4Straight Lines 28 = 0

QUESTION15.

Write the parametric equation of the straight line passing through the point (3, 2) and making an angle 1350 with the positive direction of the X – axis.

Sol:

Given θ = 1350 and (h, k) = (3, 2)

  The parametric equations are: x = h + r Cos θ, y = k +r Sin θ

x = 3 + r Cos 1350, y = 2 + r Sin 1350

         x = 3 – Straight Lines 30 , y = 2 +Straight Lines 30

QUESTION16.

Transform the following equation into normal form.

(i)  x + y +1 = 0

 Sol: 

Given equation is x + y +1 = 0

    x + y =– 1

  – x – y = 1

  Dividing on both sides by Straight Lines 31

Straight Lines 32

(ii) x + y = 2

 Sol: 

 Given equation is x + y = 2

  Dividing on both sides by Straight Lines 33

   Straight Lines 34

QUESTION 17.

If the area of the triangle formed by the straight lines x = 0, y = 0 and 3x + 4y =a (a > 0) is 6 square units. Find the value of ‘a’.

Sol:

Given equation is 3x + 4y = a

 The area of the triangle formed by the straight-line ax + by + c = 0 with the coordinate axes is Straight Lines 35

        ⟹Straight Lines 36 = 6

            Straight Lines 37  = 6

            a2 = 24 ⟹ a = 12(∵ a > 0)

QUESTION18.

Find the sum of the squares of the intercepts of the line 4x – 3y = 12

Sol:

Given equation is 4x – 3y = 12    

    Straight Lines 38         

       It is in the form of Straight Lines 7   

        a = 3 and b =– 4

the sum of the squares of intercepts = a2 + b 2

      = 32 + (– 4)2

      = 9 + 16 = 25

QUESTION 19.

Find the angle made by the straight – line y = – x + 3 with the positive direction of the X-axis measured in the counter-clockwise direction.

Sol:

Give equation of straight line is y =  Straight Lines 39 x + 3

 It is in the form of y = mx + c

                                   m =Straight Lines 39

                                 tan θ = Straight Lines 39

                         ⟹ θ =Straight Lines 40

QUESTION 20.

Find the equation of the straight line in the symmetric form, given the slope and point on the line (2, 3).

Sol:

Equation of the line in the symmetric form is Straight Lines 41

 Given Point (x1, y1) = (2, 3) and slope m =   Straight Lines 19 

tan θ = Straight Lines 19

⟹ θ = 600

 ∴ The equation of the line is Straight Lines 42

QUESTION 21.

If the product of the intercepts made by the straight-line x tan α + y sec α = 1 (0≤ α < π/2) on the coordinate axes is equal to sin α, find α.

Sol:

Given equation is x tan α + y sec α = 1

Straight Lines 43

It is in the form of Straight Lines 7

 a = cot α, b = cos α

product of intercepts = sin α

⟹ ab = sin α

       cot α.cos α = sin α

       (cos α)/ (sin α). cos α = sin α

       ⟹ cos2 α = sin2 α

        ⟹ tan2 α = 1

              tan α = 1(∵ 0≤ α < π/2)

     ∴ α = 450

QUESTION22.

If the sum of the reciprocals of the intercepts made by a variable straight–line on the axes of coordinates is a constant, then prove that the line always passes through a fixed point.

Sol:

The equation of the line in the intercept form is Straight Lines 7

Let the sum of the reciprocals of intercepts is k

  ⟹

            Straight Lines 44

   ∴ the line always passes through a fixed point (1/k, 1/k).

QUESTION23.

Find the ratio in which the straight-line 2x + 3y – 20 = 0 divides the join of the points (2, 3) and (2, 10).

Sol:

Given equation is L ≡ 2x + 3y – 20 = 0

      L11 = 2(2) + 3(3) = 4 + 9– 20 = – 7

      L22 = 2(2) + 3(10) = 4 + 30 – 20 = 14

 We know that the line L = 0 divides the line joining the points (x1, y1) and (x2, y2) in the ratio – L11 : L22

                      = – (– 7) : 14

                      = 1 : 2

QUESTION24.

State whether (3, 2) (– 4, – 3) are on the same or opposite sides of the straight-line 2x – 3y + 4 = 0.

Sol:

Given equation is L ≡ 2x – 3y + 4 = 0

 L11 = 2(3) – 3(2) + 4 = 6 – 6 + 4 = 4 > 0

 L22 = 2(– 4) – 3(–3) + 4 = – 8 + 9 + 4 = 5 > 0

     L11, L22 has the Same sign

    ∴ Given points lies the same side of the line 2x – 3y + 4 = 0.

QUESTION25.

Find the ratio in which (i) X-axis and (ii) Y-axis divide the line segment Straight Lines 45 joining A (2, –3) and B (3, – 6)

Sol:

(i) X – axis divide Straight Lines 45in the ratio – y1 : y2  = – ( –3) : – 6 = – 1 : 2

(ii) Y – axis Straight Lines 45 in the ratio – x1 : x2 =  – 2 : 3 = – 2 : 3

QUESTION26.

Find the equation of the straight line passing through the point of intersection of the lines x + y + 1 = 0 and 2x – y + 5 = 0 and containing the point (5, – 2).

Sol:

 Given equations are

         x + y + 1 = 0 ……… (1)

         2x – y + 5 = 0……. (2)

       The equation of the line passing through the point of intersection of the lines L1 = 0 and L2 = 0 is L1 + λ L2 = 0

     The equation of the line passing through the point of intersection of the lines (1) and (2) is

      x + y + 1 + λ (2x – y + 5) = 0

     but the line passes through (5, – 2)

      5 – 2 + 1 + λ (2(5) – (– 2) + 5) = 0

      4 + λ (10 + 2 + 5) = 0

       4 + λ (17) = 0

         λ = –4 /17

∴ the equation of required line is

        x + y + 1 + (–4 /17) (2x – y + 5) = 0

       17(x + y + 1) –4 (2x – y + 5) = 0

        17x + 17y + 17 – 8x + 4y – 20 = 0

        9x + 21y – 3 = 0

        3x + 7y – 1 = 0

QUESTION27.

If a, b, and c are in arithmetic progression, then show that the equation ax + by + c = 0 represents a family of concurrent lines and find the point of concurrency.

Sol:

Given, a, b, and c are in A.P. and the equation is ax + by + c = 0

  a, b, and c are in A.P.

   ⟹ b = (a + c)/2

  Substitute b value in the equation ax + by + c = 0

   ⟹  ax + [ (a + c)/2]y + c = 0

   ax + ay/2 + cy/2 + c = 0

  a(x + y/2) + c(y/2 + 1) = 0

 dividing on both sides by a

  ⟹ (x + y/2) + (c/a) (y/2 + 1) = 0

 It is in the form of L1 + λ L2 = 0

  ∴ given equation represent a family of concurrent lines

  x + y/2 = 0 and y/2 + 1 = 0 are the lines

  y/2 + 1 = 0 ⟹ y/2 = – 1

y =– 2 ⟹ x = 1

The point of concurrency is (– 2, 1)    

QUESTION28.

Find the point of intersection of the line 7x + y + 3 = 0 and x + y = 0

Sol:

Given equations are 7x + y + 3 = 0 and x + y = 0

Straight Lines 46

Straight Lines 47

 ∴ The point of intersection is Straight Lines 48

QUESTION29.

Transform the following equations into the form L1 + λ L2 = 0 and find the point of concurrency of the family of straight lines represented by the equation.

(i) (2 + 5k) x – 3 (1 + 2k) y + (2 – k) = 0

     Given equation is

  (2 + 5k) x – 3 (1 + 2k) y + (2 – k) = 0

 2x + 5kx – 3y – 6ky + 2 – k = 0

 2x – 3y + 2 + k (5x – 6y – 1) = 0

 It is in the form of L1 + λ L2 = 0 L1 ≡ 2x – 3y + 2 = 0 and L2≡ 5x – 6y – 1 = 0

By solving above equations, we get point of concurrency

Straight Lines 49

∴ The point of concurrency is (5, 4)

(ii) ( k + 1) x + (k + 2) y + 5 = 0

 Given equation is  (k + 1) x + (k + 2) y + 5 = 0

  kx + x + ky + 2y + 5 = 0

 x + 2y + 5 + k (x + y) = 0

 It is in the form of L1 + λ L2 = 0

  L1 ≡ x + 2y + 5 = 0 and L2≡ x + y = 0

By solving the above equations, we get point of concurrency

Straight Lines 50

 ∴ The point of concurrency is (5, – 5)

QUESTION30.

Find the value of ‘p’, if the straight lines x + p = 0, y + 2 = 0 and 3x + 2y + 5 = 0 are concurrent.

Sol:

Given that x + p = 0 —— (1), y + 2 = 0 —— (2) and 3x + 2y + 5 = 0 —— (3)   are concurrent.

       From (1) x + p = 0 ⟹ x = – p

        From (2) y + 2 = 0 ⟹ y = – 2

       Point of intersection of (1) and (2) is (– p, – 2)

       From (3)

         3y + 2x + 5 = 0

         3(– p) + 2 (– 2) + 5 = 0

           – 3p – 4 + 5 = 0

            – 3p + 1 = 0

                3p = 1

                  p = 1/3

QUESTION31.

Find the area of the triangle formed by the following straight lines and the coordinate axes.

(i) x – 4y + 2 = 0

The area of the triangle formed by the straight-line ax + by + c = 0 with the coordinate axes isStraight Lines 35

 The area of the triangle formed by the straight-line x – 4y + 2 = 0    with the coordinate axes is  = Straight Lines 51 

(ii) 3x – 4y + 12 = 0   

The area of the triangle formed by the straight-line 3x – 4y + 12 = 0 with the coordinate axes is  = Straight Lines 52

QUESTION32.

Find the angle between the lines 2x + y + 4 = 0 and y – 3x =7

Sol:

Given Equations are 2x + y + 4 = 0 and y – 3x =7

  If θ is the angle between the lines a1x + b1y + c = 0 and a2x + b2y + c = 0, then  cos θ = Straight Lines 53

⟹   cos θ = Straight Lines 54

                  =Straight Lines 55

                  =Straight Lines 56

          cos θ =

∴ θ = 450

QUESTION33.

Find the perpendicular distance from the point (– 3, 4) to the straight line 5x – 12y = 2

Sol:

Given equation is 5x – 12y = 2 ⟹ 5x – 12y – 2 = 0

The perpendicular distance from the point (x1, y1) to the straight-line ax + by + c = 0 is Straight Lines 58

The perpendicular distance from the point (– 3, 4) to the straight-line 5x – 12y – 2 = 0 is

Straight Lines 59

 = 5

QUESTION34.

Find the distance between the parallel lines 3x + 4y – 3 = 0 and 6x + 8y – 1 = 0

Sol:

Given equations are 3x + 4y – 3 = 0 and 6x + 8y – 1 = 0

3x + 4y – 3 = 0 ⟹ 2(3x + 4y – 3) = 2(0) (multiplying on both sides by 2)

         ⟹ 6x + 8y – 6 = 0

 The distance between two parallel lines ax + by + c1 =0 and ax + by + c2 =0 is Straight Lines 60

The distance between two parallel lines 6x + 8y – 6 = 0 and 6x + 8y – 1 = 0 is

Straight Lines 61

QUESTION35.

Find the foot of the perpendicular from (– 1, 3) on the straight line 5x – y – 18 = 0

Sol:

Let (h, k) be the foot of the perpendicular from (– 1, 3) on the straight line 5x – y – 18 = 0

If (h, k) is the foot of the perpendicular from (x1, y1) on the straight-line ax + by + c = 0, then Straight Lines 62

Straight Lines 63

Straight Lines 64

Straight Lines 84

      h+ 1 = 5 and k – 3 = – 1

      h = 4 and k = 2

   ∴ (h, k) = (4, 2)

QUESTION36.

Find the image of (1, – 2) with respect to the straight line 2x – 3y + 5 = 0

Sol:

Let  (h, k) be the image of (1, – 2) with respect to the straight line 2x – 3y + 5 = 0

If (h, k) is the image of the point (x1, y1) with respect to the straight line ax + by + c = 0, then Straight Lines 85

If (h, k) is the image of (1, – 2) with respect to the straight line 2x – 3y + 5 = 0,        

Straight Lines 86

Straight Lines 87

Straight Lines 88

      h – 1 = – 4 and k + 2 = 6

      h =– 3 and k = 4

   ∴ (h, k) = (– 3, 4)

QUESTION37.

If 2x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining (3, –4) and (α, β), find α + β.

Sol:

Given 2x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining (3, –4) and (α, β)

     ⟹ (α, β) is the image of (3, –4) concerning the straight line 2x – 3y – 5 = 0

    ⟹Straight Lines 70

      Straight Lines 71     

        α – 3 = – 4 and β + 4 = 6

        α = – 1 and β = 2

         α + β = – 1 + 2 = 1

QUESTION38.

Find the incenter of the triangle whose vertices are (1, ), (2, 0) and (0, 0)

Sol:

Let O = (0, 0), A = (1,Straight Lines 72 ) and B = (2, 0)

Straight Lines 73

 a = AB = Straight Lines 74   

b = OB = Straight Lines 75

c = OA = Straight Lines 76   

in centre of a triangle with sides a, b and c, whose vertices (x1, y1), (x2, y2) and (x3, y3) is

In centre of ∆ OAB =Straight Lines 77

                                    =Straight Lines 78

                                    =Straight Lines 79 = Straight Lines 80

QUESTION39.

Find the ortho centre of the triangle whose sides are given by x + y + 10 = 0, x – y – 2 = 0 and 2x + y – 7 = 0.

Sol:

Given equations are

x + y + 10 = 0 …… (1)

x – y – 2 = 0 …… (2)

2x + y – 7 = 0. …… (3)

   Slope of the line (1) is – 1

   Slope of the line (2) is 1

   The straight lines (1) and (2) are perpendicular lines

   The point of intersection of (1) and (2) is the orthocenter

 x + y = – 10 ⟹ y = – 10 – x

  from (2) x – y – 2 = 0

                  x – (– 10 – x) – 2 = 0

                 x + 10 + x – 2 = 0

               2x + 8 = 0 ⟹ x = – 4

            ⟹ y = – 10 + 4 = – 6

                ∴ Orthocenter is (– 4, – 6)

QUESTION40.

Find the circum centre of the triangle whose sides are x = 1, y = 1 and x + y = 1

Sol:

Given equations are x = 1, y = 1 and x + y = 1

Straight Lines 81

       A = (1, 0), B = (1, 1) and C = (0, 1)

     AB and BC are perpendicular lines

      Circum centre is the midpoint of hypotenuse

      i.e., midpoint of AC

  ∴ Circum centre =  

                                =Straight Lines 83


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Functions vsaqs questions and solutions - 1

Functions (2M Questions &Solutions)|| V.S.A.Q.’S||

Functions (2M Questions &Solutions)|| V.S.A.Q.’S||

Functions (2M Questions &Solutions): This note is designed by the ‘Basics in Maths’ team. These notes to do help intermediate First-year Maths students.

Inter Maths – 1A two marks questions and solutions are very useful in IPE examinations.

These notes cover all the topics covered in the intermediate First-year Maths syllabus and include plenty of solutions to help you solve all the major types of Math problems asked in the

IPE examinations.  


Functions

QUESTION 1

Find the Domain of the following real-valued functions.

(i) f(x) =Functions VSAQS 12 

Sol:   given f(x) = Functions VSAQS 12

         It is defined when 6x – x2 – 5 ≠ 0

              ⇒ x2 – 6x + 5 ≠ 0

                  x2 – 5x – x + 5 ≠ 0 

                  x (x – 5) –1(x – 5) ≠ 0

                     (x – 5) (x – 1) ≠ 0

                      x ≠ 5 or x ≠ 1

                      ∴ domain = R – {1, 5}

(ii) f(x) = Functions VSAQS 13

Sol: Given f(x) =Functions VSAQS 13

  It is defined when 3 + x ≥ 0, 3 – x ≥ 0 and x ≠ 0

 ⇒ x ≥ –3, x ≤ 3 and x ≠ 0

       ⇒   –3≤ x, x ≤ 3 and x ≠ 0

       ⇒   –3≤ x ≤ 3 and x ≠ 0

              x ∈ [–3, 3] – {0}

        ∴ domain = [–3, 3] – {0}

(iii) f(x) = Functions VSAQS 14 

Sol: Given f(x) =Functions VSAQS 14

 It is defined when x + 2 ≥ 0, 1 – x > 0 and 1 – x ≠ 0

 ⇒ x ≥ –2, x < 1 and x ≠ 0 

  x ∈ [–2, ∞) ∩ (– ∞, 1) – {0}

 ⇒ x ∈ [–2, 1) – {0}

    ∴ domain = [–2, 1) – {0}

(iv) f(x) =Functions VSAQS 15

         Given f(x) =Functions VSAQS 15

         It is defined when 4x – x2 ≥ 0

               ⇒ x2 – 4x ≤ 0

                   x (x – 4) ≤ 0

                  (x – 0) (x – 4) ≤ 0

                  x ∈ [0, 4]

               ∴ domain = [0, 4]

(v) f(x) = log (x2 – 4x + 3)

Given f(x) = log (x2 – 4x + 3)

 It is defined when x2 – 4x + 3 > 0

      ⇒   x2 – 3x – x + 3 > 0

             x (x – 3) –1(x – 3) > 0

            (x – 3) (x – 1) > 0

          x ∈ (–∞, 1) ∪ (3, ∞)

          x ∈ R – [1, 3]

         ∴ domain = R – [1, 3]

(vi) f(x) = Functions VSAQS 16

    Given f(x) =Functions VSAQS 16 It is defined when x2 – 1 ≥ 0 and x2 – 3x + 2 > 0  

(x + 1)(x – 1) ≥ 0 and x2 – 2x – x + 2 > 0  

(x + 1) (x – 1) ≥ 0 and x (x – 2) (x – 1) > 0 

x∈ (–∞, –1) ∪ (1, ∞) and x ∈ (–∞, –1) ∪ (2, ∞)

            ∴ domain = R – (–1, 2]

(vii)  f(x) = Functions VSAQS 17

Given f(x) = Functions VSAQS 17

It is defined when Functions VSAQS 18 – x > 0

                                 ⇒Functions VSAQS 18 > x

                                 ⇒ x < 0

 ∴ domain = (–∞, 0)  

(viii) f(x) =Functions VSAQS 19

           Given f(x) =Functions VSAQS 19

        It is defined when Functions VSAQS 18 +x ≠0

                                 ⇒ Functions VSAQS 18 ≠ – x

                                   ⇒ x > 0

         ∴ domain = (0, ∞)          

QUESTION 2

If f : R→ R , g : R → R defined by f (x ) = 4x – 1, g(x) = x2 + 2 then find (i) (gof) (x) (ii) (gof) (Functions VSAQS 1)  (iii) (fof) (x)  (iv) go(fof) (0).

Sol: Given f(x) = 4x – 1, g(x) = x2 + 2

(i)  (gof) (x) = g (f (x))

= g (4x – 1)

= (4x – 1)2 + 2

= 16x2 – 8x + 1 + 2

= 16x2 – 8x + 3

(ii) (gof)  (Functions VSAQS 1)= (g (f (Functions VSAQS 1))

=  g(4(Functions VSAQS 1) – 1)

= g (a + 1 – 1)

= g(a)

= a2 + 2

(iii) (fof) (x) = f (f (x))

= f (4x – 1)

= 4 (4x – 1) – 1

= 16x – 4 – 1

= 16x – 5

(iv) go(fof) (0) = g(fof) (0)

= g (f (f (0)))

= g (f (– 1))

= g (– 4 – 1)

= g (– 5)

= (– 5)2 + 2

= 25 + 2 = 27

QUESTION 3

If f and g are real valued functions defined by f(x) = 2x – 1 and g(x) = x2, then find (i) (3f – 2g) (x) (ii) (fg)(x) (iii) Functions VSAQS 3 (x) (iv) (f + g+ 2) (x)

Sol: Given f(x) = 2x – 1 and g(x) = x2

(i) (3f – 2g) (x) = 3f(x) – 2 g(x)

= 3(2x – 1) – 2(x2)

= 6x – 3 – 2x2

(ii) (fg)(x) = f(x) g(x)

= (2x – 1) (x2)

= 2x3 – 3x2

(iii)  Functions VSAQS 2

(iv) (f + g+ 2) (x) = f(x) + g(x) + 2

= 2x – 1 + x2 + 2

= x2 + 2x  + 1

QUESTION 4

If f = {(4, 5), (5, 6), (6, – 4)} g = {(4, – 4), (6, 5), (8,5)}, then find (i) f + g (ii) f – g (iii) 2f +4g (iv) f + 4 (v) fg (vi) f/g (vii)Functions VSAQS 4 (viii)Functions VSAQS 5   (ix) f2 (x) f3

Sol: Given f = {(4, 5), (5, 6), (6, – 4)}

  g = {(4, – 4), (6, 5), (8,5)} 

The domain of f ∩ The Domain of g = {4, 6}

(i) (f + g) (4) = f (4) + g (4)

                     = 5 – 4 = 1

     (f + g) (6) = f (6) + g (6)

                     =– 4 + 5 = 1

           ∴ f + g = {(4, 1), (6, 1)}                    

(ii) (f – g) (4) = f (4) – g (4)

                   = 5 – (– 4) = 5 + 4 = 9

(f – g) (6) = f (6) – g (6)

                   = – 4– 5   = – 9

         ∴ f – g   = {(4, 9), (6, – 9}

(iii) (2f +4g) (4) = 2 f (4) + 2 g (4)

                         = 2(5) + 4 (– 4)

                         = 10 – 16

                          =– 6

    (2f +4g) (6) = 2 f (6) + 2 g (6)

                         = 2(– 4) + 4 (5)

                         = – 8 + 20

                          =12

     ∴ (2f +4g) = {(4, – 6), (6, 12)}

(iv) (f + 4) (4) = f (4) + 4 = 5 + 4 = 9

       (f + 4) (5) = f (5) + 4 = 6 + 4 = 10

      (f + 4) (6) = f (6) + 4 = – 4 + 4 = 0

      ∴ (f + 4) = {(4, 9), (5, 10), (6, 0)}

(v) fg (4) = f (4) g (4) = (5) (– 4) =– 20

      fg (6) = f (6) g (6) = (– 4) (5) =– 20

      ∴ fg = {(4, – 20), (6, – 20)}

(vi) f/g (4) = f(4)/g(4) = 5/ – 4 = – 5/4

       f/g (6) = f(6)/g(6) = – 4/ 5

     ∴ f/g = {(4, – 5/4), (6, – 4/ 5)}

Functions VSAQS 6

Functions VSAQS 7

(ix) f2(4) = (f (4))2 = (5)2 = 25

       f2(5) = (f (5))2 = (6)2 = 36  

       f2(6) = (f (6))2 = (– 4)2 = 16

     ∴ f2 = {(4, 25), (5, 36), (6, 16)}

(x) f3(4) = (f (4))3 = (5)3 = 125

     f3(5) = (f (5))3 = (6)3 = 216  

     f3(6) = (f (6))3 = (– 4)3 = –64

   ∴ f3 = {(4, 125), (5, 216), (6, –64)}

QUESTION 5

If A = {0, π/6, π/4, π/3, π/2} and f: A→ B is a surjection defined by f(x) = cos x, then find B.

 Sol: Given A = {0, π/6, π/4, π/3, π/2}

            f(x) = cos x

          f (0) = cos (0) = 1

          f(π/6) = cos (π/6) =Functions VSAQS 8

          f(π/4) = cos (π/4) =Functions VSAQS 9

          f(π/3) = cos (π/3) = 1/2

          f(π/2) = cos (π/2) = 0

∴ B = {1,Functions VSAQS 8 ,Functions VSAQS 9 , ½, 0}

QUESTION 6

If A = {–2, –1, 0, 1, 2} and f: A→ B is a surjection defined by f(x) =x2 + x + 1, then find B.

Sol: Given A = {–2, –1, 0, 1, 2} and f(x) = x2 + x + 1

   f (–2) = (–2)2 + (–2) + 1= 4 – 2 + 1 = 3

    f (–1) = (–1)2 + (–1) + 1= 1 – 1 + 1 = 1

    f (0) = (0)2 + (0) + 1= 0 + 0 + 1 = 1

    f (1) = (1)2 + (1) + 1= 1 + 1 + 1 = 3

    f (2) = (2)2 + (2) + 1= 4 + 2 + 1 = 7

        ∴ B = {1, 3, 7}

QUESTION 7

If A = {1, 2, 3, 4} and f: A→ B is a surjection defined by f(x) =Functions VSAQS 10 then find B.

 Sol: Given A = {1, 2, 3, 4} and f(x) =Functions VSAQS 10

    Functions VSAQS 11

QUESTION 8

If f(x) = 2, g(x) = x2, h(x) = 2x ∀ x ∈ R, then find (fo(goh)) (x)

Sol: Given f(x) = 2, g(x) = x2, h(x) = 2x

(fo(goh)) (x) = fo (g (h (x))

                          = f (g (h (x))

                           = f(g(2x)

                           = f((2x)2)

                           = f(4x2) = 2

QUESTION 9

If f: R→ R, g: R → R defined by f (x) = 3x – 2, g(x) = x2 +1 then find (i) (gof-1) (2) (ii) (gof) (x – 1) 

Sol: Given f: R→ R, g: R → R defined by f (x) = 3x – 2, g(x) = x2 +1

(i) Let y = f(x) ⟹ x = f-1(y)

      y = 3x – 2

     y + 2 = 3x

    x = (y + 2)/3

f-1(y) = (y + 2)/3

∴ f-1(x) = (x + 2)/3

  Now 

(gof-1) (2) = g(f-1(2))

                = g ((2 + 2)/3)

               = g (4/3)

              = (4/3)2 + 1 = 16/9 + 1 = 25/9

(ii) (gof) (x – 1) = g (f (x – 1))

                        = g [ 3(x – 1) – 2)]

                        = g (3x – 3 – 2)

                        =g (3x – 5)

                        = (3x – 5)2 + 1

                        = 9x2 – 30x + 25 + 1

                       = 9x2 – 30x + 26

QUESTION 10

If f: N→ N defined by f (x) = 2x + 5, Is onto? Explain with reason.

Sol: Given f (x) = 2x + 5

 Let y = f(x) ⟹ x = f-1(y)

y = 2x + 5

2x = y – 5

 x = (y – 5)/2 ∉ N

∴ f(x) is not onto 

QUESTION 11

Find the inverse of the following functions

(i) If a, b ∈ R, f: R→ R defined by f(x) = ax + b

    Given function is f(x) = ax + b

   Let y = f(x) ⟹ x = f-1(y)

      y = ax + b

     y – b = ax

    x = (y – b)/a

f-1(y) = (y – b)/a

∴ f-1(x) = (x – b)/a

(ii) f: R→ (0, ∞) defined by f(x) = 5x

    Given function is f(x) = 5x

   Let y = f(x) ⟹ x = f-1(y)

        y = 5x

        x =Functions VSAQS 20

   f-1(y) =Functions VSAQS 20

    ∴ f-1(x) =Functions VSAQS 21

(iii) f: (0, ∞) → R defined by f(x) = Functions VSAQS 22

    Given function is f(x) =Functions VSAQS 22

     Let y = f(x) ⟹ x = f-1(y)

    y =Functions VSAQS 22

   x = 2y

  f-1(y) = 2y

  ∴ f-1(x) = 2x

(iv) f: R→ R defined by f(x) = e4x + 7

     Given function is f(x) = e4x + 7

   Let y = f(x) ⟹ x = f-1(y)

  y = e4x + 7

  4x + 7 =Functions VSAQS 23

  4x = Functions VSAQS 23  – 7

  x = ( Functions VSAQS 23 – 7)/4

  f-1(y) = (Functions VSAQS 23  – 7)/4

 ∴ f-1(x) = ( Functions VSAQS 24 – 7)/4

(v) f: R→ R defined by f(x) = (2x + 1)/3

   Given function is f(x) = (2x + 1)/3

  Let y = f(x) ⟹ x = f-1(y)

 y= (2x + 1)/3

 3y = 2x + 1

 2x = 3y – 1

  x = (3y – 1)/2

 f-1(y) = (3y – 1)/2

 ∴ f-1(x) = (3x – 1)/2

QUESTION 12

If f: R→ R defined by f(x) =Functions VSAQS 25 , then show that f (tan θ) = cos 2θ

Sol: Given function is f(x) =Functions VSAQS 25

        f (tan θ) =Functions VSAQS 26

   ∴  f (tan θ) = cos 2θ                           

QUESTION 13  

If f: R – {±1} → R defined by f(x) =Functions VSAQS 27 , then show that Functions VSAQS 28= 2f (x)

Sol: Given f(x) =Functions VSAQS 27

         Functions VSAQS 28 =Functions VSAQS 29

                          = Functions VSAQS 30

                          = Functions VSAQS 31

= Functions VSAQS 32

                          =Functions VSAQS 33

                           = 2Functions VSAQS 27

    ∴ Functions VSAQS 28= 2f (x)

QUESTION 14

If the function f: R→ R defined by f(x) =Functions VSAQS 35 , then show that f (x + y) + f (x – y) = 2 f(x) f(y).

Sol: Given function is f(x) =Functions VSAQS 35

f (x + y) + f (x – y) =  Functions VSAQS 36

 = Functions VSAQS 37 

=Functions VSAQS 38

=Functions VSAQS 39 

=Functions VSAQS 40

=2 Functions VSAQS 41

∴ f (x + y) + f (x – y) = 2 f(x) f(y)

QUESTION 15

If f(x) = cos (log x), then show thatFunctions VSAQS 42 = 0

Sol: Given function is f(x) = cos (log x)

Functions VSAQS 43

 = cos (log x) cos (log y) – cos (log x) cos (log y)

= 0

Hence proved

QUESTION 16

Find the range of the following real-valued functions

Sol:

(i) f(x) =Functions VSAQS 44

     Given function is f(x) =Functions VSAQS 44

    Let y = Functions VSAQS 44

    ⟹ ey = Functions VSAQS 45

       ey > 0

∴ The range of f(x) is R

(ii) f(x) =Functions VSAQS 46

     It is defined when x – 2 ≠ 0

     ⟹ x ≠ 2

    Domain = R – {2}                 

    let y = Functions VSAQS 46

          = Functions VSAQS 52

         y = x + 2

  if x = 2 ⟹ y = 4

 ∴ Range of f(x) is R – {4}

QUESTION 17

Find the domain and range of the following real-valued functions

Sol:

(i) f(x) =Functions VSAQS 47

It is defined when 1 + x2 ≠ 0

  ⟹ x2 ≠ – 1

    x ∈ R

   ∴ domain of f(x) is R

  Let y = Functions VSAQS 47

      y (1 + x2) = x

     y + x2y = x

     x2 y – x + y = 0

   x = Functions VSAQS 48

It is defined when 1 – 4y2 ≥ 0 and 2y ≠ 0

    ⟹ 4y2 – 1 ≤ 0 and y ≠ 0     

        (2y – 1) (2y + 1) ≤ 0 and y ≠ 0 

        (y – 1/2) (y + 1/2) ≤ 0 and y ≠ 0 

  –  ½≤ y ≤ ½ and y ≠ 0 

    ∴ Range of f(x) is [– ½, ½] – {0}    

(ii) f(x) =  Functions VSAQS 49

      It is defined when 9 – x2 ≥ 0

     ⟹ x2 – 9 ≤ 0

       (x + 3) (x – 3) ≤ 0

   –  3 ≤ y ≤ 3  

∴ the domain of f (x) is [– 3, 3]

  Let y =Functions VSAQS 49

       y2 = 9 – x2

       x2 = 9 – y2

       x =Functions VSAQS 50

      It is defined when 9 – y2 ≥ 0

     ⟹ y2 – 9 ≤ 0

       (y + 3) (y – 3) ≤ 0

     –  3 ≤ y ≤ 3  

         y ∈ [– 3, 3]

  ∴ Range of f (x) is [0, 3] (∵ y ≥ 0)

(iii) f(x) =Functions VSAQS 51

     clearly, x ∈ R

    ∴ domain of f(x) is R

        Let y = Functions VSAQS 51

       If x = 0, then y = 1

       If x = – 1, then y = 1

       If x = 1, then y = 3

      If x = – 2, then y = 3

      If x = 2, then y = 5

      ∴ Range of f (x) is [1, ∞)

(iv) f(x) = [x]

clearly Domain = R  and Range = Z     

QUESTION 18.

Define (i) One – One function (ii) Onto function (iii) Bijection (iv) Even and Odd functions

     (i) One – One Function: one – one, if every element of A has a unique image in B.

    (ii) Onto function: A function f: A→ B is said to be onto if ∀ y ∈ B there exists x ∈ A such that f(x) = y.

    (iii) Bijection: A function f: A→ B is said to be a Bijection if it is both one-one and onto.

    (iv) Even and Odd functions:

     If f(–x) = f(x), then f(x) is even function

          If f(– x)

= – f(x), then f(x) is odd function

 


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