xyz

1. (i) , for all x ∈ [−1, 1]
(ii) , for all x ∈ R
(iii) for all x ∈ (−∞, −1] ∪ [1, ∞)
1
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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2. PROOF:
(i) Let sin−1 x = θ
where θ ∈ [−π/2, π/2] =θ
⇒−
⇒0
⇒ ∈ [0, π]
Now, =θ
⇒ x = sin θ
⇒x=
⇒ cos−1 x =
2
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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3. Now (π / 2 − θ) ∈ [0, π]
⇒ = –θ …(ii)
⇒ =
3
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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4. Example:
1. If sin−1 x = π/5, for some x ∈ (−1, 1), then find the value
of cos−1 x
Solution : Since sin−1 x + cos−1 x =
⇒ cos−1 x =
π π 3π
= − =
2 5 10
4
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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5. Example:
2. If sin , then find the value of x
Solution : sin(sin−1 1/5 + cos−1 x) = 1
⇒ sin−1
⇒
⇒
⇒x=
5
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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6. Example:
4. If sin−1 x + sin−1 y = , then find the value of
cos−1 x + cos−1 y
Solution :
sin−1 x + sin−1 y =
⇒
⇒
6
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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7. Example:
5. Solve sin−1 x ≤ cos−1 x
Solution :
cos−1 x ≥ sin−1 x
⇒
⇒
⇒ sin−1 x ≤
⇒ −1 ≤ x ≤ sin
⇒x ∈
7
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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8. Example:
6. Find the range of f(x) = sin−1 x + tan−1 x + cos−1 x
Solution :
Domain of tan−1 x is R
But domain of sin−1 x and cos−1 x is [−1, 1]
Hence domain of the function is [−1, 1].
For x ∈ [−1, 1], tan−1 x ∈
Also for x ∈ [−1, 1].
Thus f(x) = , where x ∈ [−1, 1].
Hence range is Or .
8
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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9. Example:
Find the least and greatest value of f(x) = (sin–1 x)2 + (cos–
1
x)2
Solution :
Here we first make the expression quadratic in either sin–1x or
cos– 1x
f(x) = (sin–1 x)2 + (cos–1 x)2
2
π 
= (sin−1 x)2 +  − sin−1 x ÷
2 
2
π
= 2(sin−1 x)2 − π sin−1 x +
4
=2
=2 +2
9
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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10. Now least value of f(x) occurs when = 0,
,
hence least value of f(x) is
The greatest value of f(x) occurs when ,
as for that becomes more negative which has
maximum square value
10
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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11. Example:
8. Solve : tan−1 x + 2 cot−1 x =
Solution :
tan−1 x + 2 cot−1 x =
⇒ tan−1 x + cot−1 x + cot−1 x =
⇒
⇒
⇒
⇒ x = cot
⇒x=
11
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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12. Example:
9. If , then find the
value of q
Solution :
3π
cos −1 p + cos −1 1 − p + cos −1 1 − q =
4
⇒
⇒
⇒
⇒
⇒
12
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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13. ⇒
⇒
13
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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14. Example:
10. sin−1 + cos−1 (1 + b + b2 + …) =
then prove that b =
Solution :
We know that , for all x ∈ [−1, 1]
Hence a − = 1 + b + b2 + ………
L.H.S. and R.H.S. are sum of infinite G.P.
a 1
⇒ =
 a 1− b
1− − ÷
 3
14
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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15. 3a 1
⇒ =
a+3 1−b
⇒ 3a – 3ab = a + 3
⇒ 2a – 3ab = 3
⇒b=
15
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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16. Example:
11.The value of x for is
(A) zero (B) one (C) two (D) infinite
[IITJEE 1999]
Solution :
tan−1 x(x + 1) + sin−1 x2 + x + 1 = π / 2
–1
sin x + cos– 1 x = π/2
Write in terms of
⇒ =
1
⇒ cos −1 + sin−1 x2 + x + 1 = π / 2
x2 + x + 1
1
⇒ = x2 + x + 1
x2 + x + 1
16
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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17. ⇒ x2 + x + 1 = 1
⇒ x2 + x = 0
⇒ x = 0, −1
17
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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18. Example:
Solve : cot−1 x + tan−1 3 =
Solution :
We have cot−1 x + tan−1 3 =
⇒ cot−1 x =
⇒ cot−1 x =
⇒x=3
18
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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19. Example:
2. Solve : sec– 1x > cosec– 1x
Solution :
sec– 1x > π/2 – sec– 1x
⇒ sec– 1 x > π/4
But sec– 1 x ∈ [0, π] – {π/2}
⇒ π/4 < sec– 1x < π/2 or π/2 < sec– 1x < π
⇒ < x < ∞ or – ∞ < x < – 1
19
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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20. Example:
Solve : tan– 1x > cot– 1x
Solution :
tan– 1x > cot– 1x
⇒ tan– 1x > π/2 – tan– 1x
⇒ tan– 1x > π/4
⇒ x>1
20
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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21. Example:
Solve : 2 cos–1 x + sin−1 x =
Solution :
Given equation is 2 cos−1 x + sin−1 x =
⇒ cos−1 x + (cos−1 x + sin−1 x) =
⇒ cos−1 x + =
⇒ cos−1 x = 4π/3
But cos−1 x ∈ [0, π]
Hence equation has no real roots.
21
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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22. Example:
Solve : (tan−1 x)2 + (cot−1 x)2 =
Solution :
Lets first make the quadratic in tan– 1x
(tan−1 x)2 + (cot−1 x)2 =
⇒ (tan−1 x)2 +
−1 2 π −1 π 2 5 π2
⇒ 2(tan x) − 2 × tan x + =
2 4 8
⇒ 2(tan−1 x)2 − π tan−1 x −
⇒ tan−1 x = ,
But tan−1 x ∈( – π/2, π/2
22
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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23. ⇒ tan−1 x = −
⇒ x = −1
23
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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24. Example:
Find the value of : cos (2 cos−1 x + sin−1 x) at x = ,
where 0 ≤ cos−1 x ≤ π and −π/2 ≤ sin−1 x ≤ π/2 [IITJEE 1981]
Solution :
cos (2 cos−1 x + sin−1 x)
= cos
= − sin (cos−1 x)
= − sin (sin– 1 )
=−
At x = , value is − =−
24
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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25. Example:
Find the minimum value of (sec−1 x)2 + (cosec−1 x)2
Solution :
Let I = (sec−1 x)2 + (cosec−1 x)2
= (sec−1 x + cosec−1 x)2 − 2 sec−1 x⋅ cosec−1 x
π2 π 
= − 2 sec x  − sec −1 x ÷
−1
4 2 
π2
+ 2 ( sec −1 x ) − π sec −1 x
2
=
4
= −
2
 −1 π π2 π2
= 2  sec x − ÷ + ⇒1≥
 4 8 8
25
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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26. Example:
Solve .
Solution :
14 2 15 π
sin−1 + sin−1 =
|x| |x| 2
⇒ =
=
2
−1
 2 15 
= sin 1− ÷
 |x| 
for 0 ≤ ≤ 1 or |x| ≥ .
26
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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27. 2 2
 14   2 15 
⇒ =1−  ÷
 | x |÷
  |x| 
⇒ |x| = 16
⇒ x = ±16 which satisfy |x| ≥ .
27
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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28. ⇒ x2 −
Example:
If + for 0 <
|x| < , then x equals
(A) 1/2 (B) 1 (C) −1/2 (D) −1
[IITJEE 2001]
Solution :

−1 x2 x 3  −1  2 x4 x6  π
sin  x − + + .... ÷ + cos  x − + + .... ÷ =
 2 3   2 4  2
−1  2 x 4 x6  π −1  x2 x 3 
⇒ cos  x − + ... ÷ = − sin  x − + .... ÷
 2 4  2  2 4 
−1  2 x4 x6  −1  x2 x 3 
⇒ cos  x − + ... ÷ = cos  x − + .... ÷
 2 4   2 4 
28
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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29. ⇒ x2 −
Both sides we have G.P. of infinite terms.
x x2
∴ =
 x  − x2 
1− − ÷ 1−  ÷
 2  2 
2x 2x2
⇒ =
2 + x 2 + x2
⇒ 2x + x3 = 2x2 + x3
⇒ x(x − 1) = 0
⇒ x = 0, 1 but 0 < |x| <
⇒ x=1
29
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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