S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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(iii) 1 1sec x cosec x2
for all x ( , 1] [1, )
(ii) 1 1tan x cot x2
, for all x R
(i) 1 1sin x cos x2
, for all x [ 1, 1]
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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PROOF:(i) Let sin1 x =
where [/2, /2]
2 2
0 2
2
[0, ]
=
x = sin
Now, 1sin x =
x = cos2
cos 1 x = 1 πcos cos θ
2
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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Now ( / 2 ) [0, ]
1cos x = 2
– …(ii)
1 1sin x cos x = 2
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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Example:
1. I f sin 1 x = / 5, for some x ( 1, 1), then find the value
of cos 1 x
Solution : Since sin 1 x + cos 1 x = 2
cos 1 x = 1sin x2
32 5 10
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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Example:
Solution :
2. I f sin 1 11sin cos x 1
5
, then find the value of x
sin(sin 1 1/ 5 + cos 1 x) = 1
sin 1 11cos x
5 2
1 11sin cos x
5 2
1 11sin sin x
5
x = 15
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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4. I f sin 1 x + sin 1 y = 23
, then find the value of
cos 1 x + cos 1 y
Example:
Solution :
sin 1 x + sin 1 y = 23
1 1 2cos x cos y
2 2 3
1 1 2cos x cos y
3 3
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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Example:
Solution :
5. Solve sin 1 x cos 1 x
cos 1 x sin 1 x
1 1sin x sin x2
12sin x2
sin 1 x 4
1 x sin4
x 1
1,2
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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6. Find the range of f (x) = sin 1 x + tan 1 x + cos 1 x
Example:
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 ,4 4
Also 1 1sin x cos x2
for x [ 1, 1].
Thus f(x) = 1tan x2
, where x [ 1, 1].
Hence range is ,4 2 4 2
Or 3
,4 4
.
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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Example:
Solution :
Find the least and greatest value of f(x) = (sin– 1 x)2 + (cos–
1x)2
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
1 2 1(sin x) sin x2
21 2 12(sin x) sin x
4
= 2 2
21 1sin x sin x2 8
= 2 2
1sin x4
+ 2 2
16
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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Now least value of f(x) occurs when 2
1sin x4
= 0,
hence least value of f(x) is 2
8
,
The greatest value of f(x) occurs when 1sin x2
,
as for that 1sin x4
becomes more negative which has
maximum square value
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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Example:
Solution :
8. Solve : tan 1 x + 2 cot 1 x = 23
tan 1 x + 2 cot 1 x = 23
tan 1 x + cot 1 x + cot 1 x =
23
1 2cot x
2 3
1 2cot x
3 2
1cot x6
x = cot 6
x = 3
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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Example:
Solution :
9. I f 1 1 1 3cos p cos 1 p cos 1 q
4
, then find the
value of q
1 1 1 3cos p cos 1 p cos 1 q
4
1 1 2 1 3cos p cos 1 ( p) cos 1 q
4
1 1 1 3cos p sin p cos 1 q
4
1 3cos 1 q
2 4
1cos 1 q4
1 q cos4
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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Example:
Solution :
10. sin 1 2 3a a
a ...3 9
+ cos 1 (1 + b + b2 + …) =
2
then prove that b = 2a 3
3a
We know that 1 1sin x cos x2
, for all x [ 1, 1]
Hence a 2 3a a
............3 9
= 1 + b + b2 + ………
L.H.S. and R.H.S. are sum of infinite G.P.a 1
a 1 b1
3
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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3a 1a 3 1 b
3a – 3ab = a + 3
2a – 3ab = 3
b = 2a 3
3a
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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11. The value of x for 1 1 2tan x(x 1) sin x x 1 / 2 is
(A) zero (B) one (C) two (D) infinite [I I TJ EE 1999]
Example:
Solution :1 1 2tan x(x 1) sin x x 1 / 2
sin – 1 x + cos– 1 x = / 2
Write 1 1 tan x(x ) in terms of 1cos
1tan x(x 1) = 1
2
1cos
x x 1
1 1 2
2
1cos sin x x 1 / 2
x x 1
2
2
1x x 1
x x 1
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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x2 + x + 1 = 1
x2 + x = 0
x = 0, 1
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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Example:
Solution :
Solve : cot 1 x + tan 1 3 = 2
We have cot 1 x + tan 1 3 = 2
cot 1 x = 1tan 32
cot 1 x = 1cot 3
x = 3
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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Example:
Solution :
2. Solve : sec– 1x > cosec– 1x
sec– 1x > / 2 – sec– 1x
sec– 1 x > / 4
But sec– 1 x [0, ] – {/ 2}
/ 4 < sec– 1x < / 2 or / 2 < sec– 1x <
2 < x < or – < x < – 1
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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Example:
Solution :
Solve : tan– 1x > cot– 1x
tan– 1x > cot– 1x
tan– 1x > / 2 – tan– 1x
tan– 1x > /4
x > 1
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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Example:
Solution :
Solve : 2 cos– 1 x + sin 1 x = 116
Given equation is 2 cos 1 x + sin 1 x = 116
cos 1 x + (cos 1 x + sin 1 x) = 116
cos 1 x + 2
= 116
cos1 x = 4/3
But cos1 x [0, ]
Hence equation has no real roots.
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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Example:
Solution :
Solve : (tan 1 x)2 + (cot 1 x)2 = 25
8
Lets first make the quadratic in tan– 1x
(tan 1 x)2 + (cot 1 x)2 = 25
8
(tan 1 x)2 + 2 2
1 5tan x
2 8
2 21 2 1 5
2(tan x) 2 tan x2 4 8
2(tan 1 x)2 tan 1 x 23
08
tan 1 x = 4
, 34
But tan 1 x ( – / 2, / 2
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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tan 1 x = 4
x = 1
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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Find the value of : cos (2 cos 1 x + sin 1 x) at x = 15
,
where 0 cos 1 x and / 2 sin 1 x / 2 [I I TJ EE 1981]
Example:
Solution :
cos (2 cos 1 x + sin 1 x)
= cos 1cos x2
= sin (cos 1 x)
= sin (sin– 1 21 x )
= 21 x
At x = 15
, value is 1 24
125 25
= 2 6
5
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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Find the minimum value of (sec 1 x)2 + (cosec 1 x)2 Example:
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 2
1 12sec x sec x4 2
2
21 12 sec x sec x4
= 22
21 12 sec x 2 sec x4 4 4
2
8
2 2 212 sec x 1
4 8 8
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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Solve 1 114 2 15sin sin
| x | | x | 2
.
Example:
Solution :
1 114 2 15sin sin
| x | | x | 2
1 14sin
| x | = 1 2 15
sin2 | x |
= 1 2 15cos
| x |
2
1 2 15sin 1
| x |
for 0 2 15| x |
1 or | x| 2 15 .
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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2214 2 15
1| x | | x |
| x| = 16
x = ±16 which satisfy | x| 2 15 .
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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I f 2 3
1 x xsin x ....
2 4
+ 4 6
1 2 x xcos x ...
2 4 2
for 0 <
| x| < 2 , then x equals (A) 1/ 2 (B) 1 (C) 1/ 2 (D) 1
[I I TJ EE 2001]
Example:
Solution :2 3 4 6
1 1 2x x x xsin x .... cos x ....
2 3 2 4 2
4 6 2 31 2 1x x x x
cos x ... sin x ....2 4 2 2 4
4 6 2 3
1 2 1x x x xcos x ... cos x ....
2 4 2 4
x2
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]
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x2 4 6 2 3x x x x
.... x ....2 4 2 4
Both sides we have G.P. of infinite terms. 2
2
x xx x1 12 2
2
2
2x 2x2 x 2 x
2x + x3 = 2x2 + x3
x(x 1) = 0
x = 0, 1 but 0 < | x| < 2
x = 1
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