XII Mathematics IIT JEE Advanced Study Package 2014 15

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    STUDY PACKAGE

    Target: IIT-JEE (Advanced)

    SUBJECT: MATHEMATICS-XII

    Chapter Pages Exercises

    1 Inverse Trigonometric

    Functions

    37 5

    2 Determinants and Matrices 53 10

    3 Continuity & Differentiability 30 8

    4 Applications of Derivative 51 15

    5 Integration 88 8

    6 Area Under Curves 20 5

    7 Differential Equations 34 8

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    STUDY PACKAGE

    Target: IIT-JEE (Advanced)

    SUBJECT: MATHEMATICS

    TOPIC: 18 XII M 1. Inverse

    Trigonometric Functions 

    Index:

    1. Key Concepts

    2. Exercise I to II

    3. Answer Key

    4. Assertion and Reasons

    5. 34 Yrs. Que. from IIT-JEE

    6. 10 Yrs. Que. from AIEEE

    1

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    (4) y = cos –1 

     

     

     

     

    + 2

    2

    x1

    x

    (5) y = tan –1  )1x( 2 −Answers (3) (– ∞, – 3] ∪ [ – 2, – 1] ∪ [0, ∞)

    (4) R(5) (– ∞, –1] ∪ [1, ∞)

    2. Properties of Inverse Trigonometric Functions:Property - 2(A)(i) sin (sin−1 x) = x, −1 ≤ x ≤ 1 (ii) cos (cos−1 x) = x, −1 ≤ x ≤ 1

    (iii) tan (tan−1 x) = x, x ∈ R (iv) cot (cot−1 x) = x, x ∈ R

    (v) sec (sec−1 x) = x, x ≤ −1, x ≥ 1 (vi) cosec (cosec−1 x) = x, x ≤ −1, x ≥ 1

    These functions are equal to identity function in their whole domain which may or may not be R. (See thegraphs on page 18)

    Solved Example # 3

    Find the value of cosec

       

          

          ππππ−−−−4

    3cotcot 1 .

    Solution.

    Let y = cosec

     

      

        π−4

    3cotcot 1 .......(i)

    ∵ cot (cot –1 x) = x, ∀∀∀∀ x ∈∈∈∈ R

    ∴ cot    

      

        π−4

    3cot 1  =

    4

    ∴ from equation (i), we get

    y = cosec  

      

       π

    4

    3

    y = 2 Ans.Self practice problems:

    Find the value of each of the following :

    (6) cos

     

      

        π−6

    sinsin 1 (7) sin

     

      

        π−4

    3coscos 1

    Answers (6)2

    3(7) not defined

    Property - 2(B)

    (i) sin−1 (sin x) = x, − ≤ ≤π π2 2

    x (ii) cos−1 (cos x) = x; 0 ≤ x ≤ π

    (iii) tan−1 (tan x) = x; − < <π π

    2 2x (iv) cot−1 (cot x) = x; 0 < x < π

    (v) sec−1 (sec x) = x; 0 ≤ x ≤ π, x ≠2

    π(vi) cosec−1 (cosec x) = x; x ≠ 0, − ≤ ≤

    π π

    2 2x

    These are equal to identity function for a short interval of x only.(See the graphs on page 19-20)

    Solved Example # 4

    Find the value of tan –1     

          

          ππππ4

    3tan

    Solution.

    Let y = tan –1   

      

        π4

    3tan

    Note   ∵ tan –1 (tan x) = x if x ∈    

      

        ππ−

    2,

    2

    3

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    4

    3π∉  

     

      

        ππ−

    2,

    2

    ∴ tan –1    

      

        π4

    3tan  ≠ 

    4

    4

    3π ∈ 

     

      

        ππ2

    3,

    2

    ∵ graph of y = tan –1 (tan x) is as :

    ∵ from the graph we can see that if2

    π < x <

    2

    3π,

    then y = tan –1 (tan x) can be written asy = x – π

    ∴ y = tan –1    

      

        π

    4

    3tan  =

    4

    3π – π ∴ y = –

    4

    ππππ123

    solved Example # 5Find the value of sin –1(sin7)

    Solution.

    Let y = sin –1 (sin 7)

    Note : sin –1 (sin 7) ≠ 7 as 7 ∉

      ππ−

    2,

    2

    ∵ 2π < 7 <2

    ∵ graph of y = sin –1 (sin x) is as :

    From the graph we can see that if 2π ≤ x ≤2

    5π then

    y = sin –1(sin x ) can be written as :y = x – 2π∴ sin –1 (sin 7) = 7 – 2ππππ

    Similarly if we have to find sin –1 (sin(–5)) then

    ∵  – 2π < – 5 < – 2

    3π4

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    ∴ from the graph of sin –1 (sin x), we can say thatsin –1 (sin(–5)) = 2π + (–5)

    = 2π – 5Self practice problems:

    (8) Find the value of cos –1 (cos 13)

    (9) Find sin –1 (sin θ), cos –1(cosθ), tan –1 (tanθ ), cot –1(cotθ) for θ ∈    

      

     π

    π3,

    2

    5

    Ans. (8) 13 – 4π(9) sin-1 (sinθ) = 3π – θ ; cos –1 (cos θ ) = θ – 2π ;

    tan –1 (tan θ) = θ – 3π ; cot –1 (cot θ) = θ – 2π

    Property - 2(C)(i) s i n s i n s i n s i n  −1 (−x) = − sin−1 x, −1 ≤ x ≤ 1 (ii) tan−1 (−x) = − tan−1 x, x ∈ R(iii) cos−1 (−x) = π − cos−1 x, −1 ≤ x ≤ 1 (iv) cot−1 (−x) = π − cot−1 x, x ∈ R

    The functions sin−1 x, tan−1 x and cosec−1 x are odd functions and rest are neither even nor odd.

    Solved Example # 6

    Find the value of cos –1 {sin( – 5)}

    Solution.Let y = cos –1 {sin(–5)}

      = cos –1 (– sin 5)   ∵∵∵∵ cos –1 (– x) = ππππ – cos –1x, |x| ≤≤≤≤ 1= π – cos –1 (sin 5)

    = π – cos –1

     

      

     −

    π

    52cos ..........(i)

    ∵  – 2π <    

      

     −

    π5

    2< – π

    ∵ graph of cos –1 (cos x ) is as :

    ∵ from the graph we can see that if – 2π ≤ x ≤ – πthen y = cos –1 (cosx) can be written as y = x + 2π

    ∴ from the graph cos –1

     

      

     −

    π5

    2cos  =

     

      

     −

    π5

    2+ 2π =  

     

      

     −

    π5

    2

    5

    ∴ from equation (i), we get

    ∴ y = π –     

      

     −

    π5

    2

    5

    ⇒ y = 5 –2

    3ππππ Ans.

    Self practice problems:

    Find the value of the following

    (10) cos –1 (– cos 4) (11) tan –1 

     

      

        π−

    8

    7tan

    (12) tan –1 

     

      

     −

    4

    1cot

    Answers. (10) 4 – π (11) 8π

    (12)     

         π− 24

    15

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    Property - 2(D)

    (i) c o s e c  c o s e c  c o s e c  c o s e c  −1 x = sin−1 1

    x;x ≤ −1, x ≥ 1 (ii) sec−1 x = cos−1

     1

    x;x ≤ −1, x ≥ 1

    (iii)

    =

    0x;x

    1tan

    0x;x

    1tan

    xcot1

    1

    1

    Solved Example # 7

    Find the value of tan

       

          

        −−−−−−−−3

    2cot

    1

    Solution

    Let y = tan

     

      

     −−3

    2cot

    1 ........(i)

    ∵∵∵∵ cot –1 (–x) = ππππ – cot –1x, x ∈∈∈∈ R∴ equation (i) can be written as

     y = tan

     

      

     −π   −

    3

    2cot 1

     y = – tan  

      

        −3

    2cot 1

    ∵∵∵∵ cot –1 x = tan –1  

     

     

     

     

     

    if x > 0

    ∴  y = – tan  

      

        −2

    3tan 1 ⇒ y = – 

    2

    3

    Self practice problems:

    Find the value of the followings

    (13) sec  

      

      

      

     −3

    2cos 1 (14) cosec

     

     

     

      

      

     −−

    3

    1sin

    1

    Answers. (13)2

    3(14) –  3

    Property - 2(E)

    (i) sin−1 x + cos−1 x = π2

    , −1 ≤ x ≤ 1 (ii) tan−1 x + cot−1 x = π2

    , x ∈ R

    (iii) cosec−1 x + sec−1 x = π

    2, x ≥ 1

    Solved Example # 8

    Find the value of sin (2cos –1x + sin –1x) when x =5

    1

    Solution.Let y = sin [2cos –1x + sin –1x]

    ∵ sin –1x + cos –1x =2π , |x| ≤≤≤≤ 1

    ∴ y = sin

    π+   −− xcos

    2xcos2 11

     = sin

    +

    π   − xcos2

    1

     = cos (cos –1x)   ∵ x =5

    1

    ∴ y = cos    

      

        −5

    1cos 1 ........(i)

    ∵ cos (cos –1x) = x if x ∈∈∈∈ [–1, 1]6

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    ∵5

    1 ∈ [–1, 1]

    ∴ cos    

      

        −5

    1cos 1  =

    5

    1∴ from equation (i), we get

    ∴ y =5

    1.

    Self practice problems:Solve the following equations

    (15) 5 tan –1x + 3 cot –1x = 2π(16) 4 sin –1x = π – cos –1x

    Answers. (15) x = 1 (16) x =2

    1

    Property - 2(F)

    (i) sin (cos−1 x) = cos (sin−1 x) =  2

    x1 − , −1 ≤ x ≤ 1

    (ii) tan (cot−1 x) = cot (tan−1 x) =x

    1, x ∈ R, x ≠ 0

    (iii) cosec (sec−1 x) = sec (cosec−1 x) =

    1x

    x

    2 −, x > 1

    Solved Example # 9

    Find the value of sin    

          

          −−−−4

    3tan 1 .

    Solution.

    Let y = sin  

      

        −4

    3tan 1 ..........(i)

    Note : To find y we use sin(sin –1 x) = x, – 1 ≤≤≤≤ x ≤≤≤≤ 1

    For this we convert tan –1x in sin –1 x

    Let   θ = tan –14

    3⇒ tanθ =

    4

    3and θ ∈  

     

      

        π

    2,0

    ⇒ sin θ =5

    3

    ∴ sin –1 (sin θ) = sin –1    

      

     

    5

    3..........(ii)

    ∵   θ ∈   

      

        π2

    ,0   ⇒ sin –1 (sin θ) = θ

    ∴ equation (ii) can be written as :

    ∴ θ = sin –1  

      

     

    5

    3∵   θ = tan –1  

     

      

     

    4

    3⇒ tan –1 

     

      

     

    4

    3 = sin –1 

     

      

     

    5

    3

    ∴ from equation (i), we get   ∴ y = sin         −

    53sin 1

    y =5

    3

    Solved Example # 10

    Find the value of tan 

       

       

       

        −−−−3

    5cos

    2

    1 1

    Solution.

    Let y = tan

     

     

     

     −

    3

    5cos

    2

    1 1..............(i)

    7

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    Let cos –1 3

    5= θ   ⇒   θ ∈  

     

      

        π

    2,0  and cos θ =

    3

    5

    ∴ equation (i) becomes

    y = tan  

      

      θ

    2............(ii)

    ∵ tan22

    θ=

    θ+

    θ−

    cos1

    cos1 =

    3

    51

    3

    51

    +

     =53

    53

    +

    − =

    4

    )53( 2−

    tan2

    θ= ±

     

     

     

        −

    2

    53.........(iii)

    ∵   θ ∈   

      

        π

    2,0  ⇒

    2

    θ ∈ 

     

      

        π

    4,0

    ∴ tan2

    θ > 0

    ∴ from equation (iii), we get

    tan2

    θ =

     

     

     

        −

    2

    53

    ∴ from equation (ii), we get

    ∴ y =

       

       

       

          −−−−

    2

    53Ans.

    Solved Example # 11

    Find the value of cos (2cos –1x + sin –1x) when x =5

    1

    Solution.Let y = cos [2cos –1x + sin –1x]

    ∵ sin –1x + cos –1x =2

    ππππ, |x| ≤≤≤≤ 1

    ∴ y = cos

    −π+   −− xcos

    2xcos2 11

     = cos

    +

    π   − xcos2

    1

     = – sin (cos –1x)   ∵ x =5

    1

    ∴ y = – sin    

      

        −5

    1cos

    1 ........(i)

    ∵∵∵∵ sin (cos –1x) = 2x1−−−− , | x | ≤≤≤≤ 1

    ∴ sin  

     

     

        −5

    1

    cos1

     = 25

    1

    1 −  = 5

    24

    ∴ from equation (i), we get

    y = –5

    24

    Aliter :  Let5

    1cos 1−  = θ   ⇒ cos θ =

    5

    1and θ ∈  

     

      

        π

    2,0

    ∴ sinθ =5

    24

    ∴ sin –1 (sin θ) = sin –1 

     

     

     

     

    5

    24..........(ii)

    ∵   θ ∈    

         π

    2,0 ⇒ sin –1 (sin θ) = θ

    8

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    ∴ equation (ii) can be written as

    θ = sin –1 

     

     

     

     

    5

    24∵   θ = cos –1

     

      

     

    5

    1

    ⇒ cos –1   

      

     

    5

    1 = sin –1 

     

     

     

     

    5

    24

    Now equation (i) can be written as

    y = – sin

     

     

     

     −5

    24sin 1 ........(iii)

    5

    24 ∈ [–1, 1]   ∴ sin

     

     

     

     −5

    24sin 1  =

    5

    24

    ∴ from equation (iii), we get

    y = –5

    24

    Self practice problems:

    Find the value of the followings :

    (17) tan

     

     

     

     −

    4

    41eccos 1 (18) sec

     

      

        −63

    16cot 1

    (19) sin

     

      

     −−4

    3cot

    2

    1 1 (20) tan

      π

    − 

      

     −45

    1tan2 1

    Answers : (17)5

    4(18)

    16

    65(19)

    5

    52(20)

    17

    7−

    3. Identities of Addition and Substraction:A.

    (i) sin−1 x + sin−1 y = sin−1 x y y x1 12 2

    − + −

      , x ≥ 0, y ≥ 0 & (x

    2 + y2) ≤ 1

    = π − sin−1 x y y x1 12 2

    − + −

      , x ≥ 0, y ≥ 0 & x

    2 + y2 > 1

    Note that: x2 + y2 ≤ 1 ⇒  0 ≤ sin−1 x + sin−1 y ≤   π2

    x2 + y2 > 1 ⇒  π

    2  0 & xy < 1

    = π + tan−1  x y

    x y

    +

    −1, x > 0, y > 0 & xy > 1

    = π2

    , x > 0, y > 0 & xy = 1

    Note that : xy < 1   ⇒ 0 < tan−1 x + tan−1 y <2

    π;xy > 1 ⇒

    2

    π< tan−1 x + tan−1 y < π

    B.

    (i) sin−1 x − sin−1 y = sin−1

    −−− 22 x1yy1x , x ≥ 0, y ≥ 0

    (ii) cos−1 x − cos−1 y = cos−1

    −−+ 22 y1x1yx , x ≥ 0, y ≥ 0, x ≤ y

    9

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    (iii) tan−1 x − tan−1y = tan−1yx1

    yx

    +

    −, x ≥ 0, y ≥ 0

    Note:  For x < 0 and y < 0 these identities can be used with the help of properties 2(C)i.e. change x and y to − x and − y which are positive .

    Solved Example # 12

    Show that sin –1 5

    3 + sin –1 

    17

    15= ππππ – sin –1 

    85

    84

    Solution.

    5

    3 > 0,

    17

    15 > 0 and

    2

    5

      

      +

    2

    17

    15 

      

      =

    7225

    8226 > 1

    ∴ sin –15

    3 + sin –1 

    17

    15= π – sin –1

     

     

     

     −+−

    25

    91

    17

    15

    289

    2251

    5

    3

    = π – sin –1    

      

     +

    5

    4.

    17

    15

    17

    8.

    5

    3

    = π – sin –1    

      

     

    85

    84

    Solved Example # 13Evaluate:

    cos –1 13

    12 + sin –1 

    5

    4 – tan –1

    16

    63

    Solution.

    Let z = cos –1 13

    12 + sin –1 

    5

    4 – tan –1

    16

    63

    ∵ sin –1 5

    4 =

    2

    ππππ – cos –1

    5

    4

    ∴ z = cos –113

    12 +

     

      

     −

    π   −

    5

    4cos

    2

    1 – tan –1

    16

    63.

    z =2

    π –

     

      

     −

      −−

    13

    12cos

    5

    4cos 11  – tan –1

    16

    63.........(i)

    5

    4 > 0,

    13

    12 > 0 and

    5

    4 <

    13

    12

    ∴ cos –15

    4 – cos –1

    13

    12 = cos –1 

    −−+×

    169

    1441

    25

    161

    13

    12

    5

    4 = cos –1 

     

      

     

    65

    63

    ∴ equation (i) can be written as

    z =2π

     – cos –1    

      

     

    6563  – tan –1  

     

      

     

    1663

    z = sin –1   

      

     

    65

    63 – tan –1  

     

      

     

    16

    63.........(ii)

    ∵ sin –1  

      

     

    65

    63 = tan –1 

     

      

     

    16

    63

    ∴ from equation (ii), we get

    ∴ z = tan –1    

      

     

    16

    63 – tan –1  

     

      

     

    16

    63⇒ z = 0 Ans.

    Solved Example # 1410

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    Evaluate tan –1 9 + tan –1 4

    5

    Solution.

    ∵ 9 > 0,4

    5 > 0 and 9

     

      

     

    4

    5 > 1

    ∴ tan –1 9 + tan –14

    5= π + tan –1

     

     

     

     

    +

    4

    5.91

    4

    59

    = π + tan –1

     (– 1)

    = π –4

    π.

     tan –1 9 + tan –1 4

    5 =

    4

    3π.

    Self practice problems:

    (21) Evaluate sin –1 5

    4 + sin –1 

    13

    5 + sin –1 

    65

    16

    (22) If tan –14 + tan –1 5 = cot –1λ  then find ‘λ’

    (23) Prove that 2 cos –1 13

    3 + cot –1 

    63

    16 +

    2

    1 cos –1 

    25

    7= π

    Solve the following equations

    (24) tan –1 (2x) + tan –1 (3x) =4

    π(25) sin –1x + sin –1 2x =

    3

    Answers. (21)2

    π(22)   λ = – 

    9

    19(24) x =

    6

    1(25) x =

    2

    1

    C.

    (i) sin−1  

      

      −   2x1x2 =

    ( )

    −−π

    2

    12121

    xifxsin2

    xifxsin2

    |x|ifxsin2

    1

    1

    1

    (ii) cos−1 (2 x2 − 1) =

    −π−

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    Let cos –1 x = θ   ⇒   θ ∈ [0, π] and x = cos θ∴   y = cos –1 (4 cos3θ – 3 cos θ )

    y = cos –1 (cos 3θ) ...........(i)

    Fig.: Graph of cos –1 (cos x)∵   θ ∈ [0, π]∴ 3θ ∈ [0, 3π]∴ to define y = cos –1 (cos 3θ), we consider the graph of cos –1 (cos x)

    in the interval [0, 3π]. Now, from the above graph we can see that(i) if 0 ≤ 3 θ ≤ π   ⇒ cos –1 (cos 3θ) = 3θ∴ from equation (i), we get

    y = 3θ if   θ ≤ 3θ ≤ π

    ⇒ y = 3θ if 0 ≤ θ ≤ 3

    π

    ⇒ y = 3 cos –1x if2

    1≤≤≤≤ x ≤≤≤≤ 1

    (ii) if π 

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    2

    2

    dx

    yd = – 2 / 32 )x1(

    x3

    ⇒ 2

    2

    dx

    yd< 0 if x ∈  

     

     

    1,

    2

    1⇒ concavity downwards if x ∈  

     

     

    1,

    2

    1

    (ii) if –2

    1≤ x <

    2

    1, y = 2π – 3cos –1 x.

    ∴dx

    dy =

    2x1

    3

    −  ⇒

    dx

    dy> 0 if x ∈ 

     

     

    2

    1,

    2

    1

    ⇒ increasing if x ∈    

     

    2

    1,

    2

    1and 2

    2

    dx

    yd = 2 / 32 )x1(

    x3

    (a) if x ∈    

     

    − 0,

    2

    1then 2

    2

    dx

    yd < 0

    ⇒ concavity downwards if x ∈    

     

    − 0,

    2

    1

    (b) if x ∈    

      

     

    2

    1,0  then 2

    2

    dx

    yd > 0

    ⇒ concavity upwards if x ∈   

     

     

     

    2

    1,0

    (iii) Similarly if – 1 ≤ x < – 2

    1 then

    dx

    dy < 0 and 2

    2

    dx

    yd > 0.

    ∴ the graph of y = cos –1 (4x3 – 3x) is as

    Self practice problems:

    (26) Define y = sin –1 (3x – 4x3) in terms of sin –1x and also draw its graph.

    (27) Define y = tan –1 

     

     

     

     

    −2

    3

    x31

    xx3 in terms of tan –1 x and also draw its graph.

    Answers

    (26) y = sin –1 (3x – 4x3) =

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    (27) y = tan –1 

     

     

     

     

    −2

    3

    x31

    xx3 =

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    Inverse Trigonometric FunctionsSome Useful Graphs

    1.

    (i) (ii)

    (iii) (iv)

    (v) (vi)

    y = sin − 1 x, x ≤ 1,

    ∈   ππ−2

    ,

    2

    y

    y

    x

    O

    π

    2

    − π

    2

    − 1 1

    y = cos −1 x, x ≤ 1, y ∈ [0, π]

    y

    x

    π

    2

    O− 1 1

    π

    y = sec −1 x, x ≥ 1,

     

      ππ 

     

      π∈ ,2

    U2

    ,0y

    − 1 O 1 x

    − π

    2

    π

    2

    y

    − ∞

    − 1 O 1 x

    y

    π

    −∞

    y = tan −1 x, x ∈ R,    

      

        ππ−∈2

    ,2

    y ,

    π

    2

    − π

    2

    O

    y

    x

    −∞

    y = cot −1 x, x ∈ R, y ∈ (0, π)

    O

    π

    − ∞  ∞

    y = cosec −1 x, x ≥ 1,

     

        π 

     

      π−∈2

    ,0U0,2

    y

    x

    y

    π

    2

    π

    2

    15

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    y = sin (sin −1 x) = cos (cos −1 x) = x, x ∈ [− 1, 1], y ∈ [− 1, 1]; y is aperiodic

    y

    −1 + 1 x

    −1

    1

      y  =   x

    O

    45º  )

    y = tan (tan -1  

    x) = cot (cot -1  

    x) = x, x ∈ R, y ∈ R; y is aperiodic

    x

    y

    O  )45º

      y   =   x

     ←                                 →

    y = cosec (cosec −1 x) = sec (sec −1 x) = x, x  ≥ 1, y  ≥ 1 ; y is aperiodic

    x

    y

    1

    − 1

    1

    − 1

      y   =

       x

      y   =

       x

    O

     ←          

               →

    Part - 2(A)(i)

    (ii)

    (iii)

    16

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    y = sec −1 (sec x ), y is periodic with period 2π; x  ∈ R −

    Ι∈π

    − n,2

    )1n2( ,  

     

      

     

    ∈   πππ

    ,22

    ,0 Uy

    y = cos −1 (cos x), x ∈ R, y ∈ [0, π], is periodic with period 2 π

    Part -2(B)(i)

    (ii)

    (iii)

    (iv)

    y = sin −1 (sin x), x ∈ R, ,y2

    ,2  

    ∈   ππ− is periodic with period 2 π

    x

    − π

    2

    π

    2

    y

    π

    2

    − π

    2

    − 3

    2

    π− 2 π   − π  π  2 π

    3

    2

    π   y   =    x

     y   =   π    −   x  

     y   =  −   (   π    + 

     x   )    y   =

       x   −   2  π

    O

      y   =

       2  π   + 

      x

    45º  )

    x

    y

    π

    π

    2

    − 2 π   − π  π  2 πO2

    π−

     π

    2

      y   =   x  +   2   π

     y    =   2    π    −   x  

     y    =   −   x     y 

      =   x

    y = tan −1 (tan x ), x  ∈ R −

    Ι∈π

    − n,2

    )1n2(   ,    

      

        ππ−∈2

    ,2

    y  is periodic with period π

    x

    y

    π

    π

    2

    − 2 π − π  π  2 πO

      y   =   x  +   2   π

     y    =   2    π    −   x  

     y    =   −   x     y 

      =   x

    π

    2−

     π

    2−

     3

    2

    π   3

    2

    π

    O x

    y

    π

    2

    − 3

    2

    π− 2 π

     2 π

    − π  ππ2

    − π

    2

    − π

    2

      y   =   x  +   π

      y   =

       x   −    π

       y   =

        x

    3

    2

    π

      y   =   x  +    2  π

      y   =   x   −    2  π

    17

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    (v) y = cot –1 (cot x), y is periodic with period π; x ∈ R – {nπ, n ∈ Ι}, y ∈  

     

        π

    2,0  ∪ 

     

     

    π

    π,

    2

    (vi) y = cosec –1 (cosec x), y is periodic with period 2π; x ∈ R – {nπ, n ∈ Ι}, y ∈  

      ππ−

    2,

    2 – {0}

    18

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    Part - 3(C)

    (i) g r a p h o f y = s i n  g r a p h o f y = s i n  g r a p h o f y = s i n  g r a p h o f y = s i n  −1  

      

      −   2x1x2

    (ii) graph of y = cos−1 (2 x2 − 1)

    Note : In this graph it is advisable not to check its derivability just by the inspection of the graphbecause it is difficult to judge from the graph that at x = 0 there is a shapr corner or not.

    (iii) graph of y = tan−1 2x1

    x2

    (iv) graph of y = sin−1  2

    1   2x

    x+

    (v) graph of y = cos−1 2

    2

    x1

    x1

    +

    19

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    KEY CONCEPTS

    (INVERSE TRIGONOMETRY FUNCTION)GENERAL DEFINITION(S):

    1. sin−1 x , cos−1 x , tan−1 x etc. denote angles or real numbers whose sine is x , whose cosine is x

    and whose tangent is x, provided that the answers given are numerically smallest available . These

    are also written as arc sinx , arc cosx etc .

    If there are two angles one positive & the other negative having same numerical value, then

    positive angle should be taken .

    2. PRINCIPAL VALUES AND DOMAINS OF INVERSE CIRCULAR FUNCTIONS :

    (i) y = sin−1 x where −1 ≤ x ≤ 1 ; − ≤ ≤π π2 2

    y   and sin y = x .

    (ii) y = cos−1 x where −1 ≤ x ≤ 1 ; 0 ≤ y ≤ π  and cos y = x .

    (iii) y = tan−1 x where x ∈ R ; − < 0 , y > 0 & xy < 1

    20

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    = π + tan−1x y

    xy

    +−1

     where x > 0 , y > 0 & xy > 1

    tan−1 x − tan−1y = tan−1x y

    xy

    −+1

     where x > 0 , y > 0

    P−−−−6 (i) sin−1 x + sin−1 y = sin−1 x y y x1 12 2− + − 

    where x ≥ 0 , y ≥  0 & (x2 + y2) ≤ 1

    Note that : x2 + y2 ≤ 1 ⇒ 0 ≤ sin−1 x + sin−1 y ≤ π2

    (ii) sin−1 x + sin−1 y = π − sin−1 x y y x1 12 2

    − + −   where x ≥ 0 , y ≥ 0 & x2 + y2 > 1

    Note that : x2 + y2 >1 ⇒  π

    2 < sin−1 x + sin−1 y < π

    (iii) sin–1x – sin–1y = 221 x1yy1xsin   −−−− where x > 0 , y > 0

    (iv) cos−1 x + cos−1 y = cos−1 22 y1x1yx   −−∓ where x ≥ 0 , y ≥ 0

    P−−−−7 If tan−1 x + tan−1 y + tan−1 z = tan−1x y z xy z

    xy y z zx

    + + −− − −

    1 if, x > 0, y > 0, z > 0 & xy + yz + zx < 1

    Note : (i) If tan−1 x + tan−1 y + tan−1 z = π  then x + y + z = xyz

    (ii) If tan−1 x + tan−1 y + tan−1 z =  π

    2  then xy + yz + zx = 1

    P−−−−8 2 tan−1 x = sin−1 21 2

    x

    x+= cos−1

    1

    1

    2

    2

    −+

    x

    x= tan−1

    2

    1 2x

    x−

    Note very carefully that :

    sin−12

    1 2x

    x+ =

    (   )

    2 1

    2 1

    2 1

    1

    1

    1

    tan

    tan

    tan

    ≤− >

    − + < −

    x if x

    x if x

    x if x

    ππ

    cos−11

    1

    2

    2

    −+

    x

    x =

    2 0

    2 0

    1

    1

    tan

    tan

    ≥− <

    x if x

    x if x

    tan−12

    1 2x

    x− =

    ( )

    >−π−−

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    8. (a) y = cos −1(cos  x),  x ∈ R, y ∈[0, π], periodic with period 2 π 8. (b) y = cos (cos −1  x) ,  =  x   =  x

     x ∈ [− 1 , 1] ,  y ∈ [− 1 , 1], y is aperiodic

    9. (a) y = tan (tan −1  x) ,  x ∈ R ,  y ∈ R , y is aperiodic 9. (b) y = tan −1 (tan x) , = x   =  x

     x ∈ R−  ( )2 12

    n n I− ∈

    π, y ∈ −

     

     

     

     

    π π2 2

    , ,

     periodic with period π

    10. (a) y = cot −1 (cot  x) , 10. (b) y = cot (cot −1  x) ,

      =  x   =  x

     x ∈ R− {n π} , y ∈ (0 , π) , periodic with π  x ∈ R ,  y ∈ R , y is aperiodic

    11. (a) y = cosec −1 (cosec  x), 11. (b) y = cosec (cosec −1  x) ,  =  x   =  x

    x ε R − { nπ , n ε I }, y ∈

         ∪

        

     

     π π2

    0 02

    , ,  x ≥ 1 ,  y ≥ 1, y is aperiodic

     y is periodic with period 2 π

    12. (a) y = sec −1 (sec  x) , 12. (b) y = sec (sec −1  x) ,  =  x   =  x

    y is periodic with period 2π ;    x ≥ 1 ;  y ≥ 1], y is aperiodic

     x ∈ R – ( )2 12

    n n I− ∈

    π y ∈

         ∪    

     

     

    0

    2 2, ,

    π ππ

    23

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    EXERCISE–1Q.1 Find the following

    (i) tan cos tan− −+

      − 

     

     

     

    1 11

    2

    1

    3 (ii) sin

    π3

    1

    2

    1−  − 

     

     

     

    −sin (iii) cos−1 cos7

    6

    π  

     

    (iv) tan−1 tan2

    3

    π  

     

       (v) cos tan

    − 

     

     

    1 3

    4 (vi) tan  

       +   −−

    2

    3cot

    5

    3sin 11

    Q.2 Find the following :

    (i) sin  π2

    3

    2

    1−  − 

     

     

     

    −sin (ii) cos cos

    −   − 

     

     

       +

    1 3

    2 6

    π (iii) tan−1

     

      

        π4

    3tan   (iv) cos−1

     

      

        π3

    4cos

    (v) sin cos−

    1 3

    5(vi) tan−1

     

      

     

    α+α2cos35

    2sin3+ tan−1

     

      

        α4

    tan  where −  π

    2< α < π

    2

    Q.3 Prove that:

    (a) 2 cos−13

    13+ cot−1

    16

    63 +

    1

    2cos−1

    7

    25= π   (b) cos cos sin− − −

     

      + −

       +1 1 1

    5

    13

    7

    25

    36

    325= π

    (c) arc cos 2

    3− arc cos 6 1

    2 3

    + =

    π6

    (d) Solve the inequality: (arc sec x)2 – 6(arc sec x) + 8 > 0

    Q.4 Find the domain of definition the following functions.

    ( Read the symbols [*] and {*} as greatest integers and fractional part functions respectively.)

    (i) f(x) = arc cos2

    1

    x

    x+(ii) cos (sin ) sinx

    x

    x+

      +−12

    1

    2

    (iii) f  (x) = sin log ( )−   − 

     

     

      − −1 10

    3

    24

    xx

    (iv) f(x) =1

    1 41

    52

    1−

    −  + −−

    sin

    log ( )cos ( { })

    x

    xx  , where {x} is the fractional part of x .

    (v) f (x) =  ( )   ( )33 2

    52 3

    16

    12− +

      −    + − +− −x

    xx xcos log sin log

    (vi) f  (x) = log10

    (1 − log7

     (x2 − 5 x + 13)) + cos−13

    2 92+

     

     

     

    sin   πx

    (vii) f(x) = ( ) ( )ex x

    n x xsin

    tan [ ]−

    −+ −

     + −

    12 1

    21  

    (viii) f(x) = sin(cos )x  + ln (− 2 cos2 x + 3 cos x + 1) + ex

    x

    cos sin

    sin

    −   + 

     

     

    1 2 1

    2 2

    Q.5 Find the domain and range of the following functions .

    (Read the symbols [*] and {*} as greatest integers and fractional part functions respectively.)

    (i) f (x) = cot−1(2x − x²) (ii) f (x) = sec−1 (log3

     tan x + logtan x

    3)24

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    (iii) f(x) = cos−12 1

    1

    2

    2

    x

    x

    ++

     

     

     

     

    (iv) f (x) = (   )tan log− − + 

     

     

     

    14

    5

    25 8 4x x

    Q.6 Find the solution set of the equation, 3 cos−1 x = sin−1 1 4 12 2− −   x x( ) .

    Q.7 Prove that:

    (a) sin–1 cos (sin−1 x) + cos–1 sin (cos–1 x) =2

    , | x | ≤ 1

    (b) 2 tan−1 (cosec tan−1x − tan cot−1x) = tan−1x (x ≠ 0)

    (c) tan−122 2

    mn

    m n− 

     

      + tan−1

    22 2

    pq

    p q− 

     

     

      = tan−1

    22 2

    MN

    M N− 

     

      where M = mp − nq,  N = np + mq,

    1M

    Nand1

    p

    q;1

    m

    n b > c > 0 then find the value of : cot–1  

      

     

    −+ba

    1ab + cot–1

     

      

     

    −+cb

    1bc + cot–1

     

      

     

    −+ac

    1ca.

    Q.12 Solve the following equations / system of equations:

    (a) sin−1x + sin−1 2x =  π3

    (b) tan−11

    1 2+ x+ tan−1

    1

    1 4+ x= tan−1

    22x

    (c) tan−1(x−1) + tan−1(x) + tan−1(x+1) = tan−1(3x) (d) sin−11

    5+ cos−1x =

      π4

    (e) cos−1x

    x

    2

    2

    1

    1

    −+

    + tan−12

    12x

    x   −=

    2

    3

    π  (f) sin−1x + sin−1y =

    2

    3

    π & cos−1x − cos−1y = π

    3

    (g) 2 tan−1x = cos−11

    1

    2

    2

    −+

    a

    a  − cos−1

    1

    1

    2

    2

    −+

    b

    b (a > 0, b > 0).

    Q.13 Let l1 be the line 4x + 3y = 3 and l2 be the line y = 8x. L1 is the line formed by reflecting l1 across theline y = x and L

    2  is the line formed by reflecting l

    2  across the x-axis. If θ is the acute angle between

    L1 and L

    2 such that tan θ =

    b

    a, where a and b are coprime then find (a + b).

    Q.14 Let y = sin–1(sin 8) – tan–1(tan 10) + cos–1(cos 12) – sec–1(sec 9) + cot–1(cot 6) – cosec–1(cosec 7).

    If y simplifies to aπ +b  then find (a – b).

    Q.15 Show that : sin sin cos cos tan tan cot cot− − − − 

     

     

      +

       

     

     

      + −

     

     

     

      + −

     

     

     

     

     

     

     

    1 1 1 133

    7

    46

    7

    13

    8

    19

    8

    π π π π =

    13

    7

    π

    25

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    (b) sec−1x

    a − sec−1

    x

    b= sec−1b − sec−1a a ≥ 1; b ≥ 1, a ≠ b.

    (c) tan−1x

    x

    −+

    1

    1 + tan−1

    2 1

    2 1

    x

    x

    −+  = tan

    −1 23

    36

    Q.8 Express2

    3βcosec2

    αβ−1tan

    2

    1 +

    2

    3αsec2

    βα−1tan

    2

    1as an integral polynomial in α & β.

    Q.9 Find the integral values of K for which the system of equations ;

    arc x arc yK

    arc y arc x

    cos ( sin )

    ( sin ) . ( cos )

    + =

    =

    22

    244

    16

    π

    π possesses solutions & find those solutions.

    Q.10 If the value of ∑=

    ∞→  

     

     

     

     

    +++−+n

    2k 

    1

    n )1k (k 

    )2k )(1k (k )1k (1cosLim  is equal to

    120π, find the value of k.

    Q.11 If X = cosec . tan−1 . cos . cot−1 . sec . sin−1 a & Y = sec cot−1 sin tan−1 cosec cos−1 a ;where 0 ≤ a ≤ 1 . Find the relation between X & Y . Express them in terms of ‘a’.

    Q.12 Find all values ofk  for which there is a triangle whose angles have measure tan–1  

     

     

     

    2

    1

    , tan–1  

     

     

     

    + k 21

    ,

    and tan–1  

      

      + k 22

    1.

    Q.13 Prove that the equation ,(sin−1x)3 + (cos−1x)3 = α π3 has no roots for α < 132

    and α >8

    7

    Q.14 Solve the following inequalities :(a) arc cot2 x − 5 arc cot x + 6 > 0 (b) arc sin x > arc cos x (c) tan2 (arc sin x) > 1

    Q.15 Solve the following system of inequations

    4 arc tan2

    x – 8arc tanx + 3 < 0 & 4 arc cotx – arc cot2

     x – 3 > 0

    Q.16 Consider the two equations in x ; (i) sin cos

    − 

     

     

     

    1x

    y = 1 (ii) cos 

    sin− 

     

     

    1x

    y = 0

    T h e s e t s X  T h e s e t s X  T h e s e t s X  T h e s e t s X   1, X2 ⊆  [−1, 1] ; Y1, Y2  ⊆

      I − {0} are such thatX1  : the solution set of equation (i) X2  : the solution set of equation (ii)Y1  : the set of all integral values of y for which equation (i) possess a solutionY2  : the set of all integral values of y for which equation (ii) possess a solutionLet : C1 be the correspondence : X1 → Y1 such that x C1 y for x ∈ X1 , y ∈ Y1 & (x

     , y) satisfy (i).

    C2 be the correspondence : X2 → Y2  such that x C2 y for x ∈ X2 , y ∈ Y2 & (x , y) satisfy (ii).

    State with reasons if C1 & C2 are functions ? If yes, state whether they are bijjective or into?

    Q.17 Given the functions f(x) =( )( )

    excos sin− +1

    , g(x) = cosec−14 2

    3

    −  

     

     

    cosx & the function h(x) = f(x)

    defined only for those values of x, which are common to the domains of the functions f(x) & g(x).Calculate the range of the function h(x).

    Q.18 (a) If the functions f(x) = sin−12

    1 2x

    x+& g(x) = cos−1

    1

    1

    2

    2

    +

    x

    x are identical functions, then compute

    their domain & range .

    (b) If the functions f(x) = sin−1 (3x − 4x3) & g(x) = 3 sin−1 x are equal functions, then compute the

    maximum range of x.  27

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     INVERSE TRIGONOMETRY 

    EXERCISE–1

    Q 1. (i)1

    3, (ii) 1, (iii) 

    5

    6

    π, (iv) −

    3, (v) 

    4

    5, (vi)

    17

    6Q 2. (i) 

    1

    2, (ii) −1, (iii) −

     π4

    , (iv) 2

    3

    π, (v) 

    4

    5, (vi) α

    Q.3 (d) (–∞, sec 2) ∪ [1, ∞)Q 4. (i)   −1/3 ≤ x ≤ 1 (ii)  {1, −1} (iii)  1 < x < 4

    (iv) x ∈(−1/2 , 1/2), x ≠ 0 (v) (3/2 , 2]

    (vi) {7/3, 25/9} (vii) (−2, 2) − {−1, 0, 1} (viii) {xx = 2n π +  π6 , n ∈ I}

    Q5. (i) D : x ε R R : [π /4 , π)

    (ii) D: x ∈    

      

        π+ππ2

    n,n   −

      π+π=

    4nxx   n ∈ I ; R :

      π π π3

    2

    3 2,

     −

     

    (iii) D : x ∈ R R : 02

    , π

     

      (iv) D : x ∈ R R :

     

        ππ−4

    ,2

    Q 6. 3

    21,

    Q 8.π3

    Q.11   π

    Q.12 (a)  x = 12

    37

     (b)  x = 3 (c) x = 0 , 12

    , −  12  (d)  x =3

    10

    (e)  x = 2 3−   or 3   (f)  x =1

    2, y = 1 (g)  x =

    a b

    a b

    −+1

    Q.13 57 Q.14 53 Q 19. x = 1 ; y = 2 & x = 2 ; y = 7 Q.202

    171±

    EXERCISE–2Q 4.  − π Q5. 6 cos2x –

    9

    2

    π, so a = 6, b = –

    9

    2

    Q 6. (a) 

    π2  (b) 

    π4  (c) arc cot

    2 5n

    n

    +

    (d) arc tan (x + n) − arc tan x (e)  π4

    Q 7. (a)  x = n² − n + 1 or x = n (b)  x = ab (c)  x = 43

    Q 8. (α2 + β2) (α + β)

    Q 9.  K = 2 ; cos4

    2π,1 & cos

    4

    2π, −1 Q 10. 720 Q.11 X = Y= 3 2− a

    Q 12.  k =4

    11Q 14. (a)  (cot 2 , ∞) ∪ (− ∞ , cot 3) (b)

    2

    21,

    H GO

    QP  (c) 2

    21,

     

         ∪ − −

     

        1

    2

    2,

    Q15.  tan , cot1

    2

    1  

     

    Q16. C1 is a bijective function, C2 is many to many correspondence, hence it is not a function

    Q17.  [eπ /6 , eπ] Q 18.(a)  D : [0, 1] , R : [0, π /2] (b)  − ≤ ≤1

    2

    1

    2x   (c)  D : [− 1, 1] , R : [0, 2]

    Q.194

    3πQ.20 x ∈ (–1, 1)

    EXERCISE–3

    Q.1 C Q.2  π Q.3  x ∈{− 1, 0, 1} Q.4  x =1

    3Q.5  B Q.7  D Q.8  A

    29

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    EXERCISE–4 (Inv. Trigono.)Part : (A) Only one correct option1. If cos

      –1 λ + cos –1µ + cos

      –1 v = 3π then λµ + µv + vλ is equal to

    (A) – 3 (B) 0 (C) 3 (D) – 12. Range of f(x) = sin –1 x + tan –1 x + sec –1 x is

    (A)  

      

        ππ4

    3,

    4(B)

      ππ4

    3,

    4(C)

      ππ

    4

    3,

    4(D) none of these

    3. The solution of the equation sin−1  

      

        π4

    tan  − sin−1

     

     

     

     

    x

    3 −

     π6

     = 0 is

    (A) x = 2 (B) x = − 4 (C) x = 4 (D) none of these

    4. The value of sin –1 [cos{cos –1 (cosx) + sin –1 (sin x)}], where x ∈    

      

      ππ

    ,2

     is

    (A)2

    π(B)

    4

    π(C) –

    4

    π(D) –

    2

    π

    5. The set of values of k f or which x2 − kx + sin –1 (sin 4) > 0 for all real x is(A) {

     0

     } (B) (−2, 2) (C) R (D) none of these

    6. sin –1 (cos(sin –1x)) + cos –1 (sin (cos –1x)) is equal to

    (A) 0 (B)4

    π(C)

    2

    π(D)

    4

    7. cos−11

    21 1

    4

    2 2 2x xx

    + − −

    .  = cos –1

    x

    2  − cos –1x holds for

    (A) | x | ≤ 1 (B) x ∈ R (C) 0 ≤ x ≤ 1 (D) −1 ≤ x ≤ 08. tan –1 a + tan –1 b, where a > 0, b > 0, ab > 1, is equal to

    (A) tan –1   

      

     

    −+ab1

    ba(B) tan –1 

     

      

     

    −+ab1

    ba – π

    (C) π + tan –1  

      

     

    −+ab1

    ba(D) π – tan –1

     

      

     

    −+ab1

    ba

    9. The set of values of ‘x’ for which the formula 2 sin –1x = sin –1 (2x 2x1− ) is true, is

    (A) (– 1, 0) (B) [0, 1] (C)

    −23,

    23 (D)

    2

    1,2

    1

    10. The set of values of ‘a’ for which x2 + ax + sin –1 (x2 – 4x + 5) + cos –1 (x2 – 4x + 5) = 0 has at least one solution is

    (A) (– ∞, – π2 ] ∪ [   π2 , ∞) (B) (– ∞, – π2 ) ∪ (   π2 , ∞)(C) R (D) none of these

    11. All possible values of p and q for which cos –1  p  + cos –1  p1−  + cos –1  q1−  =4

    3π holds, is

    (A) p = 1, q =2

    1(B) q > 1, p =

    2

    1(C) 0 ≤ p ≤ 1, q =

    2

    1(D) none of these

    12. If [cot –1x] + [cos –1x] = 0, where [.] denotes the greatest integer function, then complete set of values of ‘x’ is

    (A) (cos1, 1] (B) (cot 1, cos 1) (C) (cot1, 1] (D) none of these13. The complete solution set of the inequality [cot –1x]2 – 6 [cot –1 x] + 9 ≤ 0, where [.] denotes greatest integer

    function, is(A) (– ∞, cot 3] (B) [cot 3, cot 2] (C) [cot 3, ∞) (D) none of these

    14. tan    

      

      +π   − xcos

    2

    1

    4

    1 + tan

     

      

      −π   − xcos

    2

    1

    4

    1, x ≠ 0 is equal to

    (A) x (B) 2x (C)x

    2(D)

    2

    x

    15. If1

    2sin−1  

     

      

     

    θ+θ2cos45

    2sin3 =

    π4

    , then tan θ is equal to

    (A) 1/3 (B) 3 (C) 1 (D) − 130

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    16. If u = cot−1 αtan  − tan−1 αtan , then tan    

      

      −π

    2

    u

    4 is equal to

    (A) αtan (B) αcot (C) tan α (D) cot α

    17. The value of cot−1

    +−−

    ++−

    xsin1xsin1

    xsin1xsin1,

    2

    π< x < π, is:

    (A) π −2

    x(B)

    2

    π +

    2

    x(C)

    2

    x(D) 2

     π −

    2

    x

    18. The number of solution(s) of the equation, sin−1x + cos−1 (1 − x) = sin−1 (− x), is/are(A) 0 (B) 1 (C) 2 (D) more than 2

    19. The number of solutions of the equation tan −1  

      

     

    +1x2

    1+ tan −1

     

      

     

    + 1x41

     = tan −1  

      

     2x

    2 is

    (A) 0 (B) 1 (C) 2 (D) 3

    20. If tan−121

    1

    + + tan−1

    3.21

    1

    +  + tan−1

    4.31

    1

    +  + .......+ tan−1

    )1n(n1

    1

    ++  = tan−1 θ, then θ is equal to

    (A)2n

    n

    + (B) 1n

    n

    + (C) n

    1n +(D)

    n

    1

    21. If cot−1n

    π >

    π6

    , n ∈ N, then the maximum value of ‘ n

     ‘ is:

    (A) 1 (B) 5 (C) 9 (D) none of these22. The number of real solutions of (x, y) where, y = sin x, y = cos –1 (cos x), −2π ≤ x ≤ 2π, is:

    (A) 2 (B) 1 (C) 3 (D) 4

    23. The value of cos    

      

        −8

    1cos

    2

    1 1  is equal to

    (A) 3/4 (B) – 3/4 (C) 1/16 (D) 1/4Part : (B) May have more than one options correct

    24.   α, β and γ  are three angles given by

    α = 2tan –1 ( 2  − 1), β = 3sin –11

    2+ sin –1 − 

     

     

     

    1

    2and γ  = cos –1 

    1

    3. Then

    (A) α > β (B) β > γ  (C) α  γ 25. cos−1x = tan−1x then

    (A) x2 =

     

     

     

        −2

    15(B) x2 =

     

     

     

        +2

    15(C) sin (cos−1x) =

     

     

     

        −2

    15(D) tan (cos−1x) =

     

     

     

        −2

    15

    26. For the equation 2x = tan (2 tan  −1 a) + 2 tan (tan −1 a + tan −1 a3), which of the following is invalid?(A) a2 x + 2a = x (B) a2 + 2 ax + 1 = 0 (C) a ≠ 0 (D) a ≠ −  1, 1

    27. The sumn =

    ∞∑

    1

    tan  −12n2n

    n424 +−  is equal to:

    (A) tan −1 2 + tan −1 3 (B) 4 tan −1 1 (C) π /2 (D) sec −1 ( )− 228. If the numerical value of tan (cos –1 (4/5) + tan –1 (2/3)) is a/b then

    (A) a + b = 23 (B) a – b = 11 (C) 3b = a + 1 (D) 2a = 3b29. If α satisfies the inequation x2 – x – 2 > 0, then a value exists for

    (A) sin –1 α (B) cos –1 α (C) sec –1 α (D) cosec –1 α

    30. If f (x) = cos−1x + cos−1

    xx

    2

    1

    23 3

    2+ −

     then:

    (A) f  

      

     

    3

    2=

    3

    π(B) f  

     

      

     

    3

    2 = 2 cos−1

    3

    2 –

    3

    π

    (C) f    

      

     

    3

    1 =

    3

    π(D) f  

     

      

     

    3

    1= 2 cos−1

    3

    1 –

    3

    πm

    31

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    EXERCISE–81. Find the value of the following :

    (i) sin

     

      

     −−

    π   −2

    1sin

    3

    1(ii) tan

     

      

     −+   −−

    3

    1tan

    2

    1cos 11

    (iii) sin –1 

     

     

     

     −2

    3sincos 1

    2. Solve the equation : cot –1x + tan –1 3 =2

    π

    3. Solve the equation : tan –1    

      

     

    −−

    2x

    1x + tan –1 

     

      

     

    ++

    2x

    1x =

    4

    π

    4. Solve the following equations :

    (i) tan –1   

      

     

    +−

    x1

    x1 =

    2

    1 tan –1x , (x > 0)

    (ii) 3tan –1   

      

     

    + 32

    1 – tan –1  

     

      

     

    x

    1 = tan –1 

     

      

     

    3

    1

    5. Find the value of tan

     

     

     

     

    +−

      

     

    +−−

    2

    21

    2

    1

    y1

    y1cos

    2

    1

    x1

    x2sin

    2

    1, if x > y > 1.

    6. If x = sin (2 tan –12) and y = sin    

      

        −3

    4tan

    2

    1 1 then find the relat ion between x and y .

    7. If arc sinx + arc siny + arc sinz = π then prove that:(x, y, z > 0)

    (i) xyz2z1zy1yx1x 222 =−+−+−(ii) x4 + y4 + z4 + 4 x2y2z2 = 2 (x2 y2 + y2 z2 + z2x2)

    8. Solve the following equations :

    (i) sec−1a

    x− sec−1

    b

    x= sec−1b − sec−1a a ≥ 1; b ≥ 1, a ≠ b .

    (ii)x1

    1sin

    1x

    1xsin

    x1

    xsin 111

    +

    =

    +

    −−

    +

    −−−

    (iii) Solve for x, if (tan –1x)2 + (cot –1 x)2  =8

    5 2π

    9. If α = 2 tan –1  

      

     

    −+

    x1

    x1& β = sin –1

     

     

     

     

    +−

    2

    2

    x1

    x1for 0 < x < 1, then prove that α + β = π. What the value of α + β will be if

    x > 1 ?

    10. If X = cosec tan−1 cos cot−1 sec sin−1 a & Y = sec cot−1 sin tan−1 cosec cos−1 a; where 0 ≤ a ≤ 1. Find the relationbetween X & Y. Express them in terms of 'a'.

    11. Solve the following inequalities:(i) cos −1 x > cos −1 x2 (ii) sin –1 x > cos –1 x(iii) tan –1 x > cot –1 x. (iv) sin –1 (sin 5) > x2 – 4x.

    (v) tan2

    (arc sin 

    x) > 1 (vi) arccot2

    x − 5 arccot 

    x + 6 > 0(vii) tan −1 2 x ≥ 2 tan −1 x

    12. Find the sum of each of the following series :

    (i) cot –1 12

    31 + cos –1 

    12

    139 + cot –1 

    12

    319 + ... + cot –1 

     

      

      −12

    5n3 2 .

    (ii) tan−13

    1+ tan−1

    9

    2+ ..... + tan−1

    1n2

    1n

    21

    2−

    ++ ..................... ∞

    13. Prove that the equation, (sin−1x)3 + (cos−1x)3 = α π3 has no roots for α <32

    1.

    14. (i) Find all positive integral solutions of the equation, tan−1 x + cot−1 y = tan−1 3.(ii) If 'k' be a positive integer, then show that the equation:

    tan−1

     x + tan−1

     y = tan−1

     k has no non−zero integral solution.32

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    35

    15. If f is a polynomial function satisfying 2 + f(x).f(y) = f(x) + f(y) + f(xy) ∀ x, y∈RStatement-1: f(2) = 5 which implies f(5) = 26Statement-2: If f(x) is a polynomial of degree 'n' satisfying f(x) + f(1/x) = f(x). f(1/x), then

    f(x) = 1 xn + 1

    16. Statement-1: The range of the function sin-1

     + cos-1

    x + tan-1

    x is [π /4, 3π /4]

    Statement-2: sin-1

    x, cos-1

    x are defined for |x| ≤ 1 and tan-1x is defined for all 'x'.

    17. A function f(x) is defined as f(x) =0 where x is rational

    1 where x is irrational

    Statement-1 : f(x) is discontinuous at xll x∈RStatement-2 : In the neighbourhood of any rational number there are irrational numbers and in thevincity of any irrational number there are rational numbers.

    18. Let f(x) = sin ( ) ( )2 3 x cos 3 3 xπ + πStatement-1 : f(x) is a periodic functionStatement-2: LCM of two irrational numbers of two similar kind exists.

    19.  Statements-1: The domain of the function f(x) = cos-1

    x + tan-1

    x + sin-1

    x is [-1, 1]

    Statements-2: sin-1

    x, cos-1

    x are defined for |x| ≤ 1and tan-1

    x is defined for all x.20.  Statement-1 :  The period of f(x) = = sin2x cos [2x] – cos2x sin [2x] is 1/2

    Statements-2: The period of x – [x] is 1, where [⋅] denotes greatest integer function.

    21.  Statements-1: If the function f : R → R be such that f(x) = x – [x], where [⋅] denotes the greatest integerless than or equal to x, then f 

    -1(x) is equals to [x] + x

    Statements-2: Function ‘f ’ is invertible iff is one-one and onto.

    22.  Statements-1 : Period of f(x) = sin 4π {x} + tan π [x] were, [⋅] & {⋅} denote we G.I.F. & fractional partrespectively is 1.Statements-2: A function f(x) is said to be periodic if there exist a positive number T independent of x

    such that f(T + x) = f(x). The smallest such positive value of T is called the period or fundamentalperiod.

    23.  Statements-1: f(x) =

    x 1

    x 1

    +

    −  is one-one function

    Statements-2:x 1

    x 1

    +

    −is monotonically decreasing function and every decreasing function is one-one.

    24.  Statements-1: f(x) = sin2x (|sinx| - |cosx|) is periodic with fundamental period π /2Statements-2: When two or more than two functions are given in subtraction or multiplication form we

    take the L.C.M. of fundamental periods of all the functions to find the period.25.  Statements-1: e

    x = lnx has one solution.

    Statements-2: If f(x) = x ⇒ f(x) = f −1(x) have a solution on y = x.26.  Statements-1: F(x) = x + sinx. G(x) = -x

    H(x) = F(X) + G(x), is a periodic function.

    Statements-2: If F(x) is a non-periodic function & g(x) is a non-periodic function then h(x) = f(x) ±

    g(x) will be a periodic function.

    27.  Statements-1:x 1, x 0

    f(x)x 1, x 0

    + ≥= 

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    36

    29.  Statement–1 : Let f : [1, 2] ∪  [5, 6] →  [1, 2] ∪  [5, 6] defined asx 4, x [1, 2]

    f(x)x 7, x [5, 6]

    + ∈= 

    − + ∈  then the

    equation f(x) = f −1

    (x) has two solutions.

    Statements-2: f(x) = f −1

    (x) has solutions only on y = x line.

    30.  Statements-1: The function px q

    rx s

    ++

    (ps − qr ≠ 0) cannot attain the value p/r.

    Statements-2: The domain of the function g(y) =q sy

    ry p

    − is all real except a/c.

    31.  Statements-1: The period of f(x) = sin [2] xcos [2x] – cos2x sin [2x] is 1/2Statements-2: The period of x – [x] is 1.

    32.  Statements-1: If f is even function, g is odd function thenb

    g(g ≠ 0) is an odd function.

    Statements-2: If f(–x) = –f(x) for every x of its domain, then f(x) is called an odd function and if f(–x)

    = f(x) for every x of its domain, then f(x) is called an even function.

    33.  Statements-1: f : A → B and g : B → C are two function then (gof)–1 = f –1 og–1.

    Statements-2: f : A → B and g : B → C are bijections then f –1 & g–1 are also bijections.

    34.  Statements-1: The domain of the function2f (x) log sin x=  is (4n + 1)

    2

    π, n ∈ N.

    Statements-2: Expression under even root should be ≥ 0

    35.  Statements-1: The function f : R → R given 2af (x) log (x x 1)= + +  a > 0, a ≠ 1 is invertible.

    Statements-2: f is many one into.

    36.  Statements-1: φ(x) = sin (cos x) x 0,2

    π ∈

     is a one-one function.

    Statements-2: '(x) x 0,2

    π φ ≤ ∀ ∈

    37.  Statements-1: For the equation kx2 + (2 − k)x + 1 = 0 k ∈ R − {0} exactly one root lie in (0, 1).Statements-2: If f(k 1) f(k 2) < 0 (f(x) is a polynomial) then exactly one root of f(x) = 0 lie in (k 1, k 2).

    38.  Statements-1: Domain of2

    1 1 xf (x) sin is { 1, 1}2x

    −   += −

     

    Statements-2:1

    x 2x

    + ≥  when x > 0 and1

    x 2x

    + ≤ −  when x < 0.

    39.  Statements-1: Range of f(x) = |x|(|x| + 2) + 3 is [3, ∞)

    Statements-2: If a function f(x) is defined ∀ x ∈ R and for x ≥ 0 if a ≤ f(x) ≤ b and f(x) is even functionthan range of f(x) f(x) is [a, b].

    40.  Statements-1: Period of {x} = 1. Statements-2: Period of [x] = 1

    41.  Statements-1: Domain of f = φ. If f(x) =

    1

    [x] x−

    Statements-2: [x] ≤ x ∀ x∈ R42. Statements-1: The domain of the function sin

    –1x + cos

    –1x + tan

    –1x is [–1, 1]

    Statements-2: sin–1

    x, cos–1

    x are defined for |x| ≤ 1 and tan–1x is defined for all ‘x’

    ANSWER KEY1. A 2. D 3. A  4. C  5. A  6. A 7. A

    8. C 9. B 10. C 11. A 12. C 13. B 14. A

    15. A 16. A 17. A 18. A 19. A 20. A 21. D

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    22. A 23. A 24. A 25. D 26. C 27. D  28. C

    29. C 30. A 31. A 32. A 33. D 34. A  35. C

    36. A 37. C 38. A 39. A 40. A 41. A  42. A

     SSOOL L UUTTIIOONNSS 

    4.  f(f(x)) =1 1 x 1

    11 f (x) x1

    1 x

    −= =

    − −−

    ∴ f(f(f(x))) =1 1

    xx 11 f(f(x))

    1x

    = =−−

    Ans. C

    5. f(1 + x) = f(1 – x) ... (1)

    f(4 + x) = f(4 – x) ... (2)

    x → 1– x in (1) ⇒  f(1 – x) = f(x) ... (3)x → 4 – x in (2) ⇒ f(2 – x) f(8 – x) = f(x) ... (4)(1) and (4) ⇒ f(2 – x) = f(8 – x) .... (5)Use x → x – x in (5), we getf(x) = f(6 + x)

    ⇒  f(x) is periodic with period 6Obviously 6 is not necessary the fundamental period. Ans. A

    6.  L.C.M. of {π, 1} does not exist

    ∴ (A) is the correct option.7.  (a)

    Clearly both are true and statement – II is correct explantion of Statement – I .

    8.  (c)

    2xf (x)

    4 x−′   =−

    ∴ f(x) is increasing for – 2 ≤ x ≤ 0 and decreasing for 0 ≤ x ≤ 2.

    9. Suppose a > b. Statement – II is true as( )

    2

    b af (x)

    b x

    −′   =

    +, which is always negative and hence monotonic

    in its continuous part. Alsox blim f (x)

    +→−= ∞  and

    x blim f (x)

    −→−= −∞ . Moreover

    x xlimf (x) 1 and lim f(x) 1

    →∞ →−∞= + = − − . Hence range of f is R – {1}.

    F is obviously one–one as f(x1) = f(x2) ⇒ x1 = x2.However statement – II is not a correct reasoning for statement – I

    Hence (b) is the correct answer.10.  Statement – I is true, as period of sin x and cos πx are 2π and 2 respectively whose L.C.M does not exist.

    Obviously statement – II is falseHence (c) is the correct answer.

    11.  Graph of f(x) is symmetric about the line x = 0 if f(- x) = f(x) i.e. if f(0 – x) = f(0 + x)

    ∴  Graph of y = f(x) is symmetric about x = 1, if f(1 + x) = f(1 – x).Hence (a) is the correct answer.

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    12.  Period of sin( )

      ( )x

    2 n 1 !n 1 !

    π= −

    Period of cos ( )x

    2 n !n!

    π=

    ⇒  Period of f(x) = L.C.M of 2(n – 1)! And 2(n)! = 2(n!)

    Now, f(x) = | copsx | | sin x | 3 1 | sin 2x | 3+ + = + +

    ∴ f(x) is periodic function with period =2

    π.

    Hence (c) is the correct answer.

    13.  tan(|tan–1

    x|) = |x|, since |tan–1

    x| = tan–1

    |x|Obviously cos|x| and |x| meets at exactly two points

    ∴ (B) is the correct option.14. (A)Since cos n is also even function. Therefore solution of cosx = f(x) is always sym. also out y–axis.

    19.  (a) Both A and R are obviously correct.20.  (a) f(x) = x [x]

    f(x + 1) = x + 1 − ([x] + 1) = x – [x]So, period of x – [x] is 1.Let f(x) = sin (2x – [2x])

    1 1 1f x sin 2 x 2 x

    2 2 2

    + = + − +

    = sin (2x + 1 – [2x] – 1)

    = sin (2x – [2x])So, period is 1/2

    21.  f(1) = 1 – 1 = 0 f(0) = 0

    ∴ f is not one-one

    ∴ f -1(x) is not defined Ans. (D)

    22.  Clearly tan π[x] = 0 ∀ x∈R and period of sin 4 π {x} = 1. Ans. (A)

    23.  f(x) =x 1

    x 1

    +

    −f ′(x) =

    2 2

    (x 1) (x 1) 20

    (x 1) (x 1)

    − − + −= <

    − − 

    So f(x) is monotonically decreasing & every monotonic function is one-one.

    So ‘a’ is correct.

    24.  f(x) = sin2x (|sinx| -|cosx|) is periodic with period π /2 because f(π /2 + x) = sin 2 (π /2 + x) (|sin (π /2 + x)|

    -|cos (π /2 + x)|)

    = sin (π + 2x) (|cosx| - |sinx|) = -sin2x (|cosx| - |sinx|)= sin2x (|sinx| - |cosx|)

    Sometimes f(x + r) = f(x) where r is less than the L.C.M. of periods of all the function, but according todefinition of periodicity, period must be least and positive, so ‘r’ is the fundamental period.

    So ‘f’ is correct.

    27.  (D) If f(x) is an odd function, then f(x) + f(−x) = 0 ∀ x ∈ Df 28.  (C) For one to one function if x1 ≠ x2

    ⇒ f(x1) ≠ f(x2) for all x1, x2 ∈ Df   3 1>

    but f ( 3) f (1)< and 3 > 1

    f(5) > f(1) f(x) is one-to-one but non-monotonic

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    29.  (C)3 11 11 3

    , and ,2 2 2 2

      both lie on y = f(x) then they will also lie on y = f −1

    (x) ⇒  there are two

    solutions and they do not lie on y = x.

    30.  If we take y =px q

    rx s

    +

    + then x =

    q sx

    rx p

    −⇒ x does not exist if y = p/r

    Thus statement-1 is correct and follows from statement-2 (A)

    31. f(x) = sin(2x – [2x] f(x + 1/2) =1

    sin 2x 1 2 x2

    + − +

    = sin (2x + 1 – [2x] – 1] = sin (2x – [2x].) i.e., period is 1/2.f(x) = x – [x]

    f(x + 1) = x + 1 – ([x] + 1) = x – [x]i.e., period is 1. (A)

    32.  (A) Let h(x) =f(x)

    g(x)

    h(–x) =f ( x) f (x) f (x)

    h(x)g( x) g( x) g(x)

    −= = = −− − −  

    ∴ h(x) =f 

    g is an odd function.

    33.  (D) Assertion : f : A → B, g : B → C are two functions then (gof)–1 ≠ f –1 og–1 (since functions neednot posses inverses. Reason : Bijective functions are invertibles.

    34.  (A) for f(x) to be real log2(sin x) ≥ 0

    ⇒  sin x ≥ 2º   ⇒  sin x = 1 ⇒ x = (4n + 1)2

    π, n ∈ N.

    35.  (C) f is injective since x ≠ y (x, y ∈ R)

    ⇒  { } { }2 2a alog x x 1 log y y 1+ + ≠ + +⇒

     

    f(x) ≠ f(y)

    f is onto because (   )2alog x x 1 y+ + =   ⇒y ya a

    x2

    −−= .

    40.  Since {x} = x – [x]

    ∴ {x + 1} = x + 1 – [x + 1]= x + 1 – [x] – 1

    = x – [x] = [x]

    Period of [x] = 1 Ans (A)

    41.  f(x) =1

    [x] x−[x] – x ≠ 0

    [x] ≠ x → [x] > x It is imposible or [x] ≤ x

    So the domain of f is φ

    because reason [x] ≤ x Ans. (A)

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    STUDY PACKAGE

    Target: IIT-JEE (Advanced)

    SUBJECT: MATHEMATICS

    TOPIC: 19 XII M 2. Determinants and

    Matrices 

    Index:

    1. Key Concepts

    2. Exercise I to X

    3. Answer Key

    4. Assertion and Reasons

    5. 34 Yrs. Que. from IIT-JEE

    6. 10 Yrs. Que. from AIEEE

    1

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       5   4

    1 .1 .1 .1 . D e f i n i t i on :D e f i n i t i o n :D e f i n i t i o n :D e f i n i t i o n :Let us consider the equations a

    1x + b

    1y = 0, a

    2x + b

    2y = 0

    ⇒  –1

    1

    b

    a =

    x

    y = –

    2

    2

    b

    a⇒

    1

    1

    b

    a =

    2

    2

    b

    a⇒ a

    1b

    2 – a

    2b

    1 = 0

    we express this eliminant as 22

    11

    ba

    ba

     = 0

    The symbol22

    11

    ba

    ba is called the determinant of order two.

    I ts value is given by: D = a1b

    2 − a

    2b

    12 .2 .2 .2 . Ex pa ns ion of D eter mina nt :Expans ion of Determinant :Expans ion of Determinant :Expans ion of Determinant :

    The symbol

    333

    222

    111

    cba

    cba

    cba

     is called the determinant of order three.

    Its value can be found as:

    D = a1

    33

    22

    cb

    cb − a

    233

    11

    cb

    cb + a

    322

    11

    cb

    cbOR

    D = a1

    33

    22

    cb

    cb− b

    133

    22

    ca

    ca+ c

    133

    22

    ba

    ba... & so on.

    In this manner we can expand a determinant in 6 ways using elements of ; R1, R

    2, R

    3 or C

    1, C

    2, C

    3.

    3 .3 .3 .3 . M i n o r s :M i n o r s :M i n o r s :M i n o r s :The minor of a given element of a determinant is the determinant of the elements which remain afterdeleting the row & the column in which the given element stands. For example, the minor of a

    1 in

    333

    222

    111

    cba

    cba

    cba

     is33

    22

    cb

    cb & the minor of b

    2 is

    33

    11

    ca

    ca.

    Hence a determinant of order two will have “4 minors” & a determinant of order three will have “9minors”.

    4 .4 .4 .4 . Cofac t or :Cofac tor :Cofac tor :Cofac tor :Cofactor of the element a

    ijis C

    ij = (−1)i+j. M

    ij; Where i & j denotes the row & column in which the

    particular element lies.Note that the value of a determinant of order three in terms of ‘Minor’ & ‘Cofactor’ can be written as:D = a

    11M

    11 − a

    12M

    12 + a

    13M

    13OR D = a

    11C

    11 + a

    12C

    12 + a

    13C

    13 & so on.

    5.5.5.5. Transpose of a Determinant :Transpose of a Determinant:Transpose of a Determinant:Transpose of a Determinant:The transpose of a determinant is a determinant obtained af ter interchanging the rows & columns.

    D =

    321

    321

    321T

    333

    222

    111

    ccc

    bbb

    aaa

    D

    cba

    cba

    cba

    =⇒

    6.6.6.6. S ym me tr ic , S ke w- Sy mm et ri c, As ym me tr ic D et er mi na nt s:Symmetr ic , Skew-Symmetr ic , Asymmetr ic Determinants :Symmetr ic , Skew-Symmetr ic , Asymmetr ic Determinants :Symmetr ic , Skew-Symmetr ic , Asymmetr ic Determinants :(i) A determinant is symmetric if it is identical to its transpose. Its ith row is identical to its i th

    column i.e. aij = a

     ji for all values of '

     i ' and '

      j '

    (ii) A determinant is skew-symmetric if it is identical to its transpose having sign of each elementinverted i.e. a

    ij = – a

     ji for all values of '

     i ' and '

      j

     '. A skew-symmetric determinant has all elements

    zero in its principal diagonal.

    (iii) A determinant is asymmetric if it is neither symmetric nor skew-symmetric.7 .7 .7 .7 . Prop ert ies of Det erminants:Propert ies of Determinants :Propert ies of Determinants :Propert ies of Determinants :

    (i) The value of a determinant remains unaltered, if the rows & columns are inter changed,

    i.e. D =

    321

    321

    321

    333

    222

    111

    ccc

    bbb

    aaa

    cba

    cba

    cba

    =  = D′

    (ii) If any two rows (or columns) of a determinant be interchanged, the value of determinantis changed in sign only. e.g.

    Let D =

    333

    222

    111

    cba

    cba

    cba

     & D′ =

    333

    111

    222

    cba

    cba

    cba

    Then D′ = − D.

    NOTE : A skew-symmetric deteminant of odd order has value zero.

    Determinant

    2

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    = (a – b) (b – c)

    100

    ccbcbbaba

    ccbba32222

    2

    ++++

    ++

    = (a – b) (b – c) [ab2 + abc + ac2 + b3 + b2C + bc2 – a2b – a2c – ab2 – abc – b3 – b2c]= (a – b) (b – c) [c(ab + bc + ca) – a(ab + bc + ca)]= (a – b) (b – c) (c – a) (ab + bc + ca) Use of factor theorem.

    USE OF FACTOR THEOREM TO FIND THE VALUE OF DETERMINANTIf by putting x = a the value of a determinant vanishes then (x − a) is a factor of the determinant.

    Example : Prove that

    abcabc

    cba

    cba

    222  = (a – b) (b – c) (c – a) (ab + bc + ca) by using factor theorem.

    Solution. Let a = b

    ⇒ D =

    abacbc

    cba

    cba222

     = 0 Hence (a – b) is a factor of determinant

    Similarly, let b = c, D = 0c = a, D = 0

    Hence, (a – b) (b – c) (c – a) is factor of determinant. But the given determinant is of fifthorder so

    abcabc

    cba

    cba222

     = (a – b) (b – c) (c – a) (λ (a2 + b2 + c2) + µ (ab + bc + ca))

    Since this is an identity so in order to find the values of λ and µ. Leta = 0, b = 1, c = – 1

     – 2 = (2) (2λ – µ)(2λ – µ) = – 1. ........(i)Let a = 1, b = 2, c = 0

    200

    041

    021

     = (–1) 2 (– 1) (5λ + 2µ)

    ⇒ 5λ + 2µ = 2 .......(ii)from (i) and (ii) λ = 0 and µ = 1

    Hence

    abcabc

    cba

    cba222  = (a – b) (b – c) (c – a) (ab + bc + ca).

    Self Practice Problems

    1. Find the value of ∆ =

    0cbca

    bc0ba

    acab0

    −−

    −−

    −−

    . Ans. 0

    2. Simplify2

    22

    2

    aabacacbc

    abbbaaab

    acbccbabb

    −−−

    −−−

    −−−

    . Ans. 0

    3. Prove thatbacc2c2

    b2acbb2

    a2a2cba

    −−

    −−

    −−

     = (a + b + c)3.

    4. Show that

    abc1

    cab1

    bca1

     = (a – b) (b – c) (c – a) by using factor theorem .

    8 .8 .8 .8 . M ult i pli ca tio n O f T wo De te rmi nan ts :Mul t ip l ica t ion Of Two Determinants :Mul t ip l ica t ion Of Two Determinants :Mul t ip l ica t ion Of Two Determinants :

    22122212

    21112111

    22

    11

    22

    11

    mbmaba

    mbmaba

    m

    m

    ba

    ba

    ++

    ++=×

    333

    222

    111

    cba

    cba

    cba

     ×

    333

    222

    111

    nm

    nm

    nm

     =

    332313332313332313

    322212322212322212

    312111312111312111

    ncnbnamcmbmacba

    ncnbnamcmbmacba

    ncnbnamcmbmacba

    ++++++

    ++++++

    ++++++

    4

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       5   4Self Practice Problems

    1. Find the value of ∆

    222

    222

    222

    cab2ab

    abca2c

    bcabc2

    Ans. (3abc – a3 – b3 – c3)2

    2. If A, B, C are real numbers then find the value of ∆ =

    1)CBcos()CAcos(

    )BCcos(1)BAcos(

    )ACcos()ABcos(1

    −−

    −−

    −−

    . Ans.0

    9 .9 .9 .9 . Su mm ation of D et ermin antsSummat ion of DeterminantsSummat ion of DeterminantsSummat ion of Determinants

    Let ∆(r) =

    321

    321

    bbb

    aaa

    )r(h)r(gf(r)

    where a1, a

    2, a

    3, b

    1, b

    2, b

    3 are constants indepedent of r, then

    ∑=

    ∆n

    1r

    )r(  =

    321

    321

    n

    1r

    n

    1r

    n

    1r

    bbb

    aaa

    )r(h)r(g)r(f ∑∑∑===

    Here function of r can be the elements of only one row or column. None of the elements other then that

    row or column should be dependent on r. If more than one column or row have elements dependent onr then first expand the determinant and then f ind the summation.

    Example : Evaluate ∑=

    n

    1r 2212n

    ycosx

    2n1r2

    1nn2

    2

    rCr

    −−

    θ

    +

    Solution :   ∑=

    n

    1r

    rD  =

    2212n

    ycosx

    2n)1r2(

    1nn2

    2

    n

    1r

    rn

    1r

    C

    n

    1rr

    −−

    θ

    +

    ===

    ∑∑∑

    =

    2212n

    ycosx

    2212n

    1nn2

    2

    1nn2

    −−

    θ

    −−

    +

    +

    = 0

    Example : Dr =

    012

    113

    CCC r2n

    1r2n

    2r2n   −

    −−

    −−

    evaluate ∑=

    n

    2r

    rD

    Solution : ∑=

    n

    2r

    rD  = ∑=

    n

    2r 012

    113

    CCC r2n

    1r2n

    2r2n   −

    −−

    −−

    =

    012

    113

    C....CCC....CCC....CC 2n2n

    32n

    22n

    2n2n

    22n

    12n

    2n2n

    12n

    02n

    −−−−

    −−−−

    −−−− +++++++++

    =

    012

    113

    n12122 2n2n2n −−−   −−−

    C1 → C

    1 – 2 × C

    2=

    010

    111

    n1212222 2n2n1n2n −−−+−   −−−−

    6

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    = (–1)11

    n12222 2n1n2n −−+−   −−−

    = 2n – 1 – n – 3

    Example : If ∆r =

    211r

    r3r2

    011r

    −−+

    +

    , find ∑=

    ∆n

    1r

    r

    Solution. On expansion of determinent, we get

    Dr = (r –1) (3 – r) + 7 + r2 + 4r = 8r + 4   ⇒   ∑=∆

    n

    1r

    r  = 4n (n + 2)

    Self Practice Problem

    1. Evaluate ∑=

    n

    1r

    rD

    n3n3z)1r(

    2n4y)1r(

    6x1r

    23

    2

    −−

    −−

    Ans. 0

    1 0 .1 0 .1 0 .1 0 . I nteg ra tion of a det er mi nan tIntegrat ion of a determinantIntegrat ion of a determinantIntegrat ion of a determinant

    Let   ∆(x) =

    222

    111

    cba

    cba

    )x(h)x(g)x(f

    where a1, b

    1, c

    1, a

    2, b

    2, c

    2 are constants independent of x. Hence

    ∫  ∆b

    a

    dx)x(  =

    222

    111

    b

    a

    b

    a

    b

    a

    cba

    cba

    dx)x(hdx)x(gdx)x(f ∫∫∫

    Note : If more than one row or one column are function of x then first expand the determinant and thenintegrate it.

    Example : If f(x) =

    xcos210

    1xcos21

    01xcos

    , then find ∫π 2 / 

    0

    dx)x(f

    Solution. Here f(x) = cos x (4 cos2x – 1) –2 cos x

      = 4 cos

    3

    x – 3 cos x = cos 3x

    so ∫π 2 / 

    0

    dxx3cos =

    2 / 

    03

    x3sin  π

     = –

    3

    1

    Example : If ∆ =32

    222

    xxx

    346

    321   −γ −β−α

    , then find ∫  ∆1

    0

    dx)x(

    Solution. ∫  ∆1

    0

    dx)x(  =

    ∫∫∫

    −γ −β−α

    1

    0

    3

    1

    0

    2

    1

    0

    222

    dxxdxxdxx

    346

    321

    =

    4

    1

    3

    1

    2

    1346

    321 222 −γ −β−α

     =12

    1

    346

    346

    321 222 −γ −β−α

     = 0

    1 1 .1 1 .1 1 .1 1 . D iff er en ti at io n of D et er mi na nt :D i f fe rent ia t ion o f Determinant :D i f fe rent ia t ion o f Determinant :D i f fe rent ia t ion o f Determinant :

    Let ∆(x) =

    )x(h)x(h)x(h

    )x(g)x(g)x(g

    )x(f)x(f)x(f

    321

    321

    321

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    then ∆′(x) =

    )x(h)x(h)x(h

    )x(g)x(g)x(g

    )x(f)x(f)x(f

    321

    321

    321   ′′′

     +

    )x(h)x(h)x(h

    )x(g)x(g)x(g

    )x(f)x(f)x(f

    321

    321

    321

    ′′′  +

    )x(h)x(h)x(h

    )x(g)x(g)x(g

    )x(f)x(f)x(f

    321

    321

    321

    ′′′

    Example : If f(x) =2

    432

    aa1

    xx2x6

    123

    , then find the value of f ′′(a).

    Solution. f′(x) =2

    32

    aa1x4x6x12

    123

    f′′(x) =2

    2

    aa1

    x12x1212

    123

    ⇒ f′′(a) = 122

    2

    aa1

    aa1

    123

     = 0.

    Example : Let α be a repeated root of quadratic equation f(x) = 0 and A(x), B(x) and C(x) be polynomial ofdegree 3, 4 and 5 respectively, then show that

    )(C)(B)(A

    )(C)(B)(A

    )x(C)x(B)x(A

    α′α′α′

    αααdivisible by f(x).

    Solution. Let g(x) =)(C)(B)(A)(C)(B)(A

    )x(C)x(B)x(A

    α′α′α′ααα

    ⇒ g′(x) =)(C)(B)(A

    )(C)(B)(A

    )x(C)x(B)x(A

    α′α′α′

    ααα

    ′′′

    Since g(α) = g′(α) = 0⇒ g(x) = (x – α)2 h(x) i.e. α  is the repeated root of g(x) and h(x) is any polynomial

    expression of degree 3. Also f(x) = 0 have repeated root α. So g(x) is divisible by f(x).Example : Prove that F depends only on x

    1, x

    2 and x

    3

    F =

    23123221

    22211

    21

    131211

    bxbxbxbxbxbx

    axaxax

    111

    ++++++

    +++

    and simplif y F.

    Solution :1da

    dF =

    23123221

    22211

    21

    131211

    bxbxbxbxbxbx

    axaxax

    000

    ++++++

    +++

     +

    23123221

    22211

    21 bxbxbxbxbxbx

    111

    111

    ++++++

     +

    000

    axaxax

    111

    131211   +++  = 0

    Hence F is independent of a1.

    Similarly1db

    dF =

    2db

    dF = 0.

    Hence F is independent of b1 and b

    2 also.

    So F is dependent only on x1, x2, x3

    Put a1 = 0, b

    1 = 0, b

    2 = 0   ⇒ F =

    23

    22

    21

    321

    xxx

    xxx

    111

    = (x1 – x

    2) (x

    2 – x

    3) (x

    3 – x

    1).

    Example : If)x1(nxcos

    xsinex

    + = A + Bx + Cx2 + ....., then find the value of A and B.

    Solution : Put x = 0 in

    )x1(nxcos

    xsinex

    + = A + Bx + Cx2 + .......

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    ⇒01

    01 = A A = 0.

    Differentiating the given determinant w.r.t x, we get

    )x1(nxcos

    xcosex

    + +

    x1

    1xsin

    xsinex

    +−  = B + 2 C x + ......

    Put x = 0, we get

    01

    11 +

    10

    01 = 0

    ⇒ B = –1 + 1 = 0∴ A = 0, B = 0

    Self Practice Problem

    1. Ifx11x

    11xx2

    x1xx

    +

    +−

     = ax3 + bx2 + cx + d. Find

    (i) d Ans. [– 1](ii) a + b + c + d Ans. [– 5](iii) b Ans. [– 4]

    1 2 .1 2 .1 2 .1 2 . C ra me r' s R ul e: S ys te m o f L in ea r E q ua ti on sCramer's Rule : System of L inear Equat ionsCramer's Rule : System of L inear Equat ionsCramer's Rule : System of L inear Equat ions(i) Two Variables

    (a) Consistent Equations: Definite & unique solution. [ intersecting lines ](b) Inconsistent Equation: No solution. [ Parallel l ine ](c) Dependent equation: Inf inite solutions. [ Identical l ines ]

    Let a1x + b1y + c1 = 0 & a2x + b2y + c2 = 0 then:

    2

    1

    2

    1

    2

    1

    c

    c

    b

    b

    a

    a≠= ⇒ Given equations are inconsistent &

    2

    1

    2

    1

    2

    1

    c

    c

    b

    b

    a

    a==   ⇒ Given equations are dependent

    (ii) Three Variables

    Let, a1x + b

    1y + c

    1z = d

    1............ (I)

    a2x + b

    2y + c

    2z = d

    2............ (II)

    a3x + b

    3y + c

    3z = d

    3............ (III)

    Then, x =D

    D

    1, Y =

    D

    D

    2, Z =

    D

    D

    3.

    Where D =

    333

    222

    111

    cba

    cba

    cba

    ; D1 =

    333

    222

    111

    cbd

    cbd

    cbd

    ; D2 =

    333

    222

    111

    cda

    cda

    cda

     & D3 =

    333

    222

    111

    dba

    dba

    dba

    (iii) Consistency of a system of Equations(a) If D ≠ 0 and alteast one of D

    1, D

    2, D

    3 ≠ 0, then the given system of equations are consistent and

    have unique non trivial solution.(b) If D ≠ 0 & D

    1= D

    2= D

    3 = 0, then the given system of equations are consistent and have triv ial

    solution only.(c) If D = D

    1= D

    2= D

    3 = 0, then the given system of equations have either infinite solutions or no

    solution.

    (Refer Example & Self Practice Problem with*)(d) If D = 0 but atleast one of D

    1, D

    2, D

    3 is not zero then the equations are inconsistent and have no solution.

    (e) If a given system of linear equations have Only Zero Solution for all its variables then the given equationsare said to have TRIVIAL SOLUTION.

    (iv) Three equation in two variables :If x and y are not zero, then condition for a

    1x + b

    1y + c

    1 = 0 ; a

    2x + b

    2y + c

    2 = 0 &

    a3x + b

    3y + c

    3 = 0 to be consistent in x and y is

    333

    222

    111

    cba

    cba

    cba

     = 0.

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       5   4Example: Find the nature of solution for the given system of equations.

    x + 2y + 3z = 12x + 3