PPT_CLASS_12_MATHS_CH_2_Maths_XI_Inverse_trignometric_functions.ppt

INVERSE TRIGONOMETRIC FUNCTION
INVERSE TRIGONOMETRIC
FUNCTION

OBJECTIVE
PREVIOUS COMPETITIVE
QUESTIONS

1. If , z > 0, xy + yz + zx < 1 and if then x + y + z =
[E – 2014]
xyz 3xyz 3) 0
Solution:
𝒕𝒂𝒏−𝟏
( 𝒙+ 𝒚 +𝒛 − 𝒙𝒚𝒛
𝟏− 𝒙𝒚 − 𝒚𝒛 − 𝒛𝒙 )=𝝅
x+y+z – xyz = 0
x + y + z = xyz

2. [E– 2013]
𝟏¿
3
65
𝟐¿
−36
65
3) 𝟒¿−𝟏

Solution:
𝒄𝒐𝒔
− 𝟏5
13
= 𝜶 =
𝒙=
𝟓
𝟏𝟑
×
𝟑
𝟓
−
𝟏𝟐
𝟏𝟑
×
𝟒
𝟓
𝒙=
−𝟑𝟑
𝟔𝟓
3
x =
cos(+)

3. If then =
[E – 2012]
𝟏 ¿
π
2
𝟐 ¿
π
3
3) 0
Solution:
𝟏
𝟐
≤ 𝒙 ≤ 𝟏
Put x =1
𝒄𝒐𝒔
− 𝟏
𝒙 +𝒄𝒐𝒔
− 𝟏
(𝒙
𝟐
+
𝟏
𝟐
√𝟑− 𝟑 𝒙
𝟐
)=
𝝅
𝟑

4. = [E – 2012]
𝟏¿−𝟏 𝟐¿𝟏 3) 𝟒¿𝛑
√𝟓
𝟖
Solution: By verification put x = -1
then
=
(x = -1 Satisfied)
L. H. S =

5. + =
[E – 2010]
𝟏¿𝟏 𝟐¿𝟎 3) 𝟒¿𝟐
Solution:
A + B + C = then
Tan A tan B + tan B tan C +Tan C tan A =
Let

6. 4 =
[E – 2009]
𝟏¿
19 π
12
𝟐¿
35 π
12
3) 𝟒¿
43π
12
Solution:
=

7. If then x = [E – 2008]
𝟏¿𝟑 𝟐¿𝟓 3) 𝟒¿𝟏𝟏
Solution:
= 9 =
⇒ x =
= =
x =

8. If then x= [E – 2007]
𝟏¿√𝟓 /3 3) 𝟒¿𝟐/𝟑
Solution:
=
⇒ 5 − 5 x𝟐
=𝟒 x𝟐
⇒ 9x𝟐
=𝟓
/3

9. [E– 2005]
𝟏 ¿
π
3
𝟐 ¿
π
4
3) 𝟒¿𝟎
Solution:
=

10. [E– 2003]
𝟏¿−𝟏/𝟑 𝟐¿𝟎 3) 1/3 𝟒¿𝟒/𝟗
Solution:
Cos( = 0

11. [E– 2004]
𝟏¿{𝟏,𝟎} 𝟐¿{−𝟏,𝟏} 3) {0, } 𝟒¿{𝟐,𝟎}
Solution:
𝒙=𝟎( 𝒐𝒓 )𝟐 𝒙 − 𝒙𝟐
=𝟏− 𝒙𝟐

12. A value of x for which sin is
[JEE MAINS– 2013]
𝟏¿ −
1
2 𝟐¿𝟎 3) 1 𝟒¿
𝟏
𝟐
Solution:
𝟏
√𝟏+(𝟏+𝒙 )
𝟐
=
𝟏
√𝟏+ 𝒙𝟐
𝟏+𝒙𝟐
=𝟐+ 𝒙𝟐
+𝟐 𝒙
x =

13. Let x(0,1). The set of all x such that is the interval
[JEE MAINS– 2013]
𝟏¿
(𝟏
𝟐
,
𝟏
√𝟐) 𝟐¿
( 𝟏
√𝟐
,𝟏
) 3) (0,1) 𝟒¿(𝟎, √𝟑
𝟐 )
Solution: 𝑮𝒊𝒗𝒆𝒏𝒔𝒊𝒏−𝟏
𝒙>𝒄𝒐𝒔−𝟏
𝒙
𝟐𝒔𝒊𝒏
−𝟏
𝒙>
𝝅
𝟐
𝒔𝒊𝒏
−𝟏
𝒙 >
𝝅
𝟒
-

x >
 x 
KEY :
2

14. The number of solutions of the equations (in principal
values ) is [JEE MAINS– 2013]
3 1 3) 2
Solution:
𝒔𝒊 𝒏− 𝟏
𝒙=𝒔𝒊𝒏−𝟏
( 𝟐 𝒙
𝟏+𝒙
𝟐 )
𝒙 =
𝟐 𝒙
𝟏+ 𝒙
𝟐
𝒙 (𝟏+𝒙𝟐
)− 𝟐𝒙=𝟎
𝒙 ( 𝒙𝟐
−𝟏)=𝟎  x = {0, 1}

15. If s =
[JEE MAINS– 2013]
𝟏¿
𝟐𝟎
𝟒𝟎𝟏+𝟐𝟎𝒏
𝟐¿
𝒏
𝒏
𝟐
+𝟐𝟎𝒏+𝟏
𝟑¿
𝟐𝟎
𝒏
𝟐
+𝟐𝟎𝒏+𝟏
𝟒¿
𝒏
𝟒𝟎𝟏+𝟐𝟎𝒏

Solution:
¿𝒕𝒂𝒏−𝟏
(𝒏+𝟐𝟎)− 𝒕𝒂𝒏−𝟏
𝒏
𝒔=𝒕𝒂𝒏
−𝟏
( 𝟐𝟎
𝒏
𝟐
+𝟐𝟎𝒏+𝟏 )

𝒕𝒂𝒏𝒔=
𝟐𝟎
𝒏
𝟐
+𝟐𝟎𝒏+𝟏
KEY :
3

16. The value of cot [A – 2008]
𝟏¿𝟓/𝟏𝟕 𝟐¿𝟔/𝟏𝟕 3) 𝟒¿𝟒/𝟏𝟕
Solution:
Cot
Cot = 6/17

17. If = then the value of x is
[A – 2007]
𝟏¿𝟏 𝟐¿𝟑 3) 𝟒¿𝟓
Solution:
=
⇒ x = 3

18. If = then 4is equal to
[A– 2005]
𝟏¿𝟐𝒔𝒊𝒏𝟐𝜶 𝟐¿𝟒 3) 𝟒¿−𝟒𝒔𝒊𝒏𝟐
𝜶
Solution:
⇒ 4
⇒ 2x = y cos
⇒ 4
⇒ x = cos
⇒ x = cos .cos + s. s
= cos . + s.

19. The equation has a solution [A– 2003]
in all values of  𝟐¿|𝜶|≤
1
√2
3)
𝟒¿
−1
√2
<|𝜶|<
1
√2
Solution:
𝒔𝒊𝒏−𝟏
𝒙=𝒔𝒊𝒏−𝟏
𝟐𝜶√𝟏−𝜶𝟐
Put  = sin
 = sin2 
= 2 

If
− π
4
≤ 𝜽 ≤
π
4
− 1
√2
≤ 𝜶 ≤
1
√2
KEY:2

20. then sinx =
[A– 2002]
𝟏 ¿ 𝒕𝒂𝒏
𝟐 𝜶
𝟐
𝟐¿𝒄𝒐𝒕𝟐
(𝜶
𝟐 ) 3) tan  𝟒¿ 𝒄𝒐𝒕
𝛂
𝟐
Solution:
𝒄𝒐𝒕−𝟏
√𝒄𝒐𝒔 𝜶 −𝒕𝒂𝒏−𝟏
√𝒄𝒐𝒔 𝜶=𝐱
𝒕𝒂𝒏
− 𝟏
( 𝟏
√cos α )−𝒕𝒂𝒏
−𝟏
√cosα=𝒙
 sinx =
1
𝟐√cos α
x
1

21. [A– 2002]
𝟏¿
𝟏
𝟐
𝒄𝒐𝒔
− 𝟏
(𝟑
𝟓 ) 𝟐¿
𝟏
𝟐
𝒔𝒊𝒏
−𝟏
(𝟑
𝟓)
3) 𝟒 ¿ 𝒕𝒂𝒏−𝟏
(𝟏
𝟐 )
Solution:
Apply

22. If for 0<then x = [AP EAM– 2015]
𝟏 ¿
𝟏
𝟐
𝟐¿𝟏 3) 𝟒¿− 𝟏
Solution: 𝒙
𝟏+
𝒙
𝟐
=
𝒙 𝟐
𝟏 +
𝒙
𝟐
𝟐
𝒙 +
𝒙𝟑
𝟐
=𝒙
𝟐
+
𝒙𝟑
𝟐

x - = 0
x(1-x) = 0
x = 0,1
 x=1 ,
x≠0
KEY:2

23. If then a value of x is
[TS EAM– 2015]
𝟏¿
𝟏
√𝟔
𝟐¿
−𝟏
√𝟏𝟐
𝟑¿
𝟐
√𝟔
𝟒¿
−𝟐
√𝟔
Solution: 𝟏
√𝟓
=
𝒙
√𝟏− 𝒙𝟐
𝟏 − 𝒙𝟐
=𝟓 𝒙𝟐
𝟏=𝟔 𝒙𝟐 

24. If the value of y is
[JEE MAINS– 2015]
𝟏¿
𝟑𝒙 −𝒙𝟑
𝟏−𝟑 𝒙
𝟐
𝟐¿
𝟑 𝒙+𝒙𝟑
𝟏− 𝟑𝒙
𝟐 𝟑¿
𝟑𝒙 −𝒙𝟑
𝟏+𝟑 𝒙
𝟐
𝟒¿
𝟑𝒙+𝒙𝟑
𝟏+𝟑𝒙
𝟐
Solution:
𝒕𝒂𝒏−𝟏
𝒚 =𝒕𝒂𝒏−𝟏
𝒙 +𝟐𝒕𝒂𝒏− 𝟏
𝒙
𝒕𝒂𝒏−𝟏
𝒚 =𝟑𝒕𝒂𝒏−𝟏
𝒙
𝒕𝒂𝒏
−𝟏
𝒚 =𝒕𝒂𝒏
−𝟏
(𝟑𝒙 − 𝒙
𝟑
𝟏−𝟑 𝒙
𝟐 ) 

25. If f (x) = 2
[J.M.O.L– 2015]
𝟏 ¿
𝝅
𝟐 𝟐¿𝛑 𝟑¿𝒕𝒂𝒏
−𝟏
( 𝟔𝟓
𝟏𝟓𝟔) 𝟒¿𝟒 𝒕𝒂𝒏−𝟏
(𝟓)
Solution:
𝒇 (𝒙)=𝟐𝒕𝒂𝒏−𝟏
𝒙+ 𝝅−𝟐𝒕𝒂𝒏−𝟏
𝒙
𝒇 ( 𝒙)=𝝅
𝒇 (𝟓 )=𝝅

𝟐𝟔.𝐓𝐡𝐞 𝐯𝐚𝐥𝐮𝐞𝐨𝐟 𝐜𝐨𝐭
(∑
𝐧=𝟏
𝟏𝟗
𝐜𝐨𝐭
−𝟏
(𝟏+∑
𝐩=𝟏
𝐧
𝟐𝐩))𝐢𝐬𝐜𝐨𝐭
(∑
𝐧=𝟏
𝟏𝟗
𝐜𝐨𝐭
−𝟏
(𝟏+∑
𝐩=𝟏
𝐧
𝟐𝐩))
𝟏¿
𝟏𝟗
𝟐𝟎
𝟐¿
𝟐𝟎
𝟏𝟗
𝟑 ¿
𝟏𝟗
𝟐𝟏
𝟒 ¿
𝟐𝟏
𝟏𝟗
Solution:
𝒄𝒐𝒕(∑
𝒏=𝟏
𝟏𝟗
𝒄𝒐𝒕
−𝟏
(𝟏+𝟐
𝒏(𝒏+𝟏)
𝟐 ))
[B. Arch
2016]

𝒄𝒐𝒕
(∑
𝒏=𝟏
𝟏𝟗
𝒕𝒂𝒏−𝟏
( 𝟏
𝟏+(𝒏+𝟏)(𝒏)))
𝒄𝒐𝒕
(∑
𝒏=𝟏
𝟏𝟗
𝒕𝒂𝒏
−𝟏
(𝒏+𝟏)−𝒕𝒂𝒏
−𝟏
𝒏
)
𝒄𝒐𝒕
(𝒕𝒂𝒏
−𝟏
(𝟏𝟗
𝟐𝟏 ))=
𝟐𝟏
𝟏𝟗
KEY :
4

27. The value of x which satisfies
[J.M.O.L– 2015]
𝟏 ¿ −
𝟏
𝟐
𝟐¿
𝟏
𝟐 𝟑¿−𝟏 𝟒 ¿ 𝟏
Solution:
𝒔𝒊𝒏(𝒄𝒐𝒕−𝟏
) 𝒙=𝒄𝒐𝒔 (𝒕𝒂𝒏−𝟏
(𝟏+𝒙))
𝒄𝒐𝒕−𝟏
𝒙=𝜶,𝒕𝒂𝒏− 𝟏
(𝟏+ 𝒙)=𝜷
𝒙=𝒄𝒐𝒕 𝜶 𝒕𝒂𝒏 𝜷=𝟏+𝒙

𝒔𝒊𝒏 𝜶=
𝟏
√𝟏+𝒙
𝟐
𝒄𝒐𝒔 𝜷=
𝟏
√𝟐 +𝟐 𝒙+𝒙
𝟐
𝟐+𝟐 𝒙 +𝒙𝟐
=𝟏+𝒙𝟐
𝟐 𝒙 =− 𝟏
𝒙 =−
𝟏
𝟐
KEY :
1

28. If x = sin(2) and y = sin , then
[AP EAM – 2016]
𝟏¿ 𝒙>𝒚 𝟐¿ 𝒙=𝒚 𝟑¿ 𝒙=𝟎=𝒚 𝟒 ¿ 𝒙 < 𝒚
Solution:
𝒙=𝒔𝒊𝒏(𝟐𝒕𝒂𝒏−𝟏
(𝟐))
¿ 𝒕𝒂𝒏−𝟏
(𝟐 )=𝜶
𝒚 =𝒔𝒊𝒏(𝟏
𝟐
𝒕𝒂𝒏
− 𝟏 𝟒
𝟑)
¿ 𝒕𝒂𝒏
−𝟏 𝟒
𝟑
=𝜷
𝒕𝒂𝒏 𝜶=𝟐
¿ 𝒕𝒂𝒏 𝜷=
𝟒
𝟑
5
3
4


𝝅
𝟒
<𝜶 <
𝝅
𝟐
𝒙=𝒔𝒊𝒏𝟐𝜶
¿
𝟐 (𝟐 )
𝟏+𝟒
=
𝟒
𝟓
𝝅
𝟒
< 𝜷<
𝝅
𝟐
𝒚 =𝒔𝒊𝒏
𝜷
𝟐
𝒚 =
√𝟏−𝒄𝒐𝒔 𝜷
𝟐
𝒚 =
√𝟏 −
𝟑
𝟓
𝟐
⇒
𝟏
√𝟓
𝒙 >𝒚
KEY :
1

Thank you…

PPT_CLASS_12_MATHS_CH_2_Maths_XI_Inverse_trignometric_functions.ppt

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PPT_CLASS_12_MATHS_CH_2_Maths_XI_Inverse_trignometric_functions.ppt