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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
3π
∴ 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
3π
∵
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
5π
∵ 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
3
+
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
2π
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
k
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
3π
, 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,
F
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
3π
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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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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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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D E T . &
M A
T R I C E S / P a g e
: 2
o f
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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D E T . &
M A
T R I C E S / P a g e
: 4
o f
5 4
= (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
++++++
++++++
++++++
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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 −−−+− −−−−
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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
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