Matrix part 3.2 (1)

S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095 www.gtimchd.in ; [email protected]

1

(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]


2

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


3

Now ( / 2 ) [0, ]

1cos x = 2

– …(ii)

1 1sin x cos x = 2


4

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


5

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


6

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


7

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


8

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

.


9

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


10

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


11

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


12

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


13

1

1 q2

1

q2


14

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


15

3a 1a 3 1 b

3a – 3ab = a + 3

2a – 3ab = 3

b = 2a 3

3a


16

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


17

x2 + x + 1 = 1

x2 + x = 0

x = 0, 1


18

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


19

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


20

Example:

Solution :

Solve : tan– 1x > cot– 1x

tan– 1x > cot– 1x

tan– 1x > / 2 – tan– 1x

tan– 1x > /4

x > 1


21

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.


22

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


23

tan 1 x = 4

x = 1


24

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


25

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


26

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 .


27

2214 2 15

1| x | | x |

| x| = 16

x = ±16 which satisfy | x| 2 15 .


28

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


29

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

Matrix part 3.2 (1)

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Transcript of Matrix part 3.2 (1)