Question

answe it question number 70.

Answer 1

70. ) f-¹ g-¹ ( x ) = ??

g(x ) = ( sinx + 4)/( sinx - 2)

g(x).sinx -2g(x ) = sinx + 4

2g(x ) + 4 = sinx { g(x )-1 }

sinx = { 2g(x ) + 4}/{g(x ) -1}

x = sin-¹{ 2g(x ) +4 }/{ g(x) - 1}

g-¹( x ) = sin-¹ { (2x +4)/( x -1 )

domain of g-¹(x ) = ?

|2x +4 |/|x -1 | ≤ 1
case 1 :-
(2x +4)/( x -1) -1 ≤ 0
(2x +4 -x +1)/( x -1) ≤ 0
(x +5)/( x -1) ≤ 0

-5 ≤ x < 1

case 2 :-
(2x +4)/( x -1) +1 ≥ 0
(2x +4 + x -1)/(x -1) ≥ 0
(3x +3)/(x -1) ≥0
x> 1 or x ≤ -1

take common in both cases
-5 ≤ x ≤ -1

x € [ -5 , -1 ]


now ,

f(x ) = 2^(x⁴ -4x² )

y = 2^( x⁴ -4x²)
take log both sides
log2 y = x⁴ - 4x²

x⁴ - 4x² - log2 y = 0

let x² = { 4 + √( 16+ 4log2 y) }/2

x = ± √[{ 4 + √(16+ 4log2y)}/2 ]

but x ≥ 2
so,

f-¹(x) = √[ { 4+√(16 +4log2 x )}/2]

f-¹(x ) = √[ 2 + √(4 + log2 x ) ]

domain of f-¹( x ) € [ 1 , ∞)

now,

f-¹( g-¹(x ) ) =

domain of f-¹(g-¹(x ) ) =
1 ≤ sin-¹( 2x +4)/( x -1) < ∞

sin1 ≤ (2x +4)/(x -1) < sin∞ also -5 ≤ x ≤ -1


case 1 :-

sin1 ≤ (2x +4)/( x -1) and -5≤ x ≤ -1

(2x +4 -x.sin1 +sin1)/(x -1) ≥0

{x( 2 -sin1) +(sin1 +4) }/( x -1) ≥ 0

x≤ - ( sin1 +4)/( 2 -sin1) ,or x > 1
take common with -5 ≤ x ≤ -1

then,

-5 ≤ x ≤ -(sin1 +4)/( 2 - sin1)

x € [ -5, -(sin1 +4)/(2 -sin1) ]

case 2 :-
(2x +4)/( x -1) < sin∞

we can't say anything about what is the fixed value of sin∞ its may be 1 or -1 or between -1 to 1.
so, we are not able to solve it .


hence,
domain of f-¹(g-¹(x )) € [ -5 , -(sin1 +4)/( 2-sin1) ]

option (c) is correct .

Answer 2

according to your question  dude :- 

1)=> f => [2,∞]⇔[1,∞) , 

f(x) = 2^(x^4-4x^2)   

now we will find the inverse of f(x) 

so for f '(x) we will convert function in x so that we can find the inverse in terms of y 

so ,

y= 2^(x^4-4x^2) 

taking logarithm both sides so that equation will reduce 

logy = log 2^(x^4-4x^2) 

logy= (x^4-4x^2) log2 

x^4-4x^2=log(y)/log2 

x^2(x^2-2^2) = logy/log2 

x^2(x-2)(x+2) = logy/log2 

now we will let x ^2 = t  so we can easily solve this equation 

so it will reduce in the format of ======== : 

t^2 -4t -log y base 2 = 0             :=>  logy/log2 = log y base 2 using property 

of logarithm  

now we can solve this equation using shree dhan acharay method :-) 

a=1 b= 4 c = -log y base 2 

so formula is to find its root is 

t= [-b +,-√b²-4ac ] / 2a 

so t = [ -(-4 +,-√4^2-4×1×log y base 2 ] /2×1 

so we will get two values of t 

t = [4+,-√16-4 log y base 2 ] /2×1 which means 

t= [4+,-√4(4-log y base 2 ] /2 

t= [ 4 +,-2√ 4 - log y base 2 ]/2 

now we will divide 2 by the above quantity 

t= [ 2+,- √4-log y base 2]  
 

t= [ 2+ √4-log y base 2]   and  t= [ 2- √4-log y base 2]  

now buddy we will replace t with x^2 

x^2 = [ 2+ √4-log y base 2]   and  x^2= [ 2- √4-log y base 2]  

which means 
 
x = √[ 2+ √4-log y base 2]   and  x= √[ 2- √4-log y base 2]  

but dude in x= √[ 2- √4-log y base 2]  is not applicable because 4 >2  and 

whole value is in root and it is a negative value so we cant add it in our 

now we will nagelete this value x= √[ 2- √4-log y base 2 

only valid value is x = √[ 2+ √4-log y base 2] now we will replace x with y so 

that we can say it is f '(x) which means 

f ' (x) =  x = √[ 2+ √4-log y base 2] 

now we need to find the value of g(x) = [sin x + 4 ] / [ sin x - 2 ] 

we can find its inverse : follow my steps what i am doing 

y = [sin x + 4 ] / [ sin x - 2 ] 

y [ sin x - 2 ] = sin x + 4 

y sin x - 2y  = sin x + 4 

y sin x - sin x = 2 y + 4 

sin x  ( y - 1 ) = 2 ( y + 2 )

sin x =  2 ( y + 2 ) /  ( y - 1 ) 

now dude i will use the property of inverse trignometric function  to find directly x 

so , 

x = > sin^-1 [ 2 ( y + 2 ) /  ( y - 1 )  ] 

and x is g ' (x)  so we will replace x with g '(x) 

g ' (x) = sin^-1 [ 2 ( y + 2 ) /  ( y - 1 )  ] 

here y is ≠ 1 if we put y = 1 then function gets not defined because it will give  
2 ( 1+ 2 ) / 0 ======= which is not possible 

now we get both  f ' (x) and g ' (x) 

now we will find f ' g ' (x) 

g ' (x) = sin^-1 [ 2 ( y + 2 ) /  ( y - 1 )  ]  and  = f ' (x) =  x = √[ 2+ √4-log y base 2] 

now i will put whole f(x) in g(x) but but but i will take intersection without finding it because it will make our answer so long i will first find the domaion of f ' (x) 

first I will find the domain of f ' (x) = √[ 2+ √4-log y base 2] 
in first cae 

so 2 +√4 - logy base 2 ≥  0 
now in second case 4- log y base 2 ≥ 0 
which means log y base 2 ≤ 4  

y  ≥ 2 ^4  using property of log i did it 
means y ≤ 16 

now in last log y base 2 > 0
 y > 1 
 which means y ∈ (1, 16]  

now we will consider g ' (x) = sin^-1 [ 2 ( y + 2 ) /  ( y - 1 )  ]
domain of  sin^-1 x is  [-1 ,1] 
 
-1≤ [ 2 ( y + 2 ) /  ( y - 1 )  ] ≤ 1 
in case first 
 
-1≤ [ 2 ( y + 2 ) /(y-1) ] 
2 ( y+2) ≤ 1-y
3y  3 

hence y ≤ 1 which means y ≤ sin 1  when you will solve sin inverser 1 u get y = sin (sin pi/2 which means but it is in inverse  = sin pi/2 

now in case 2 

[ 2 ( y + 2 ) /  ( y - 1 )  ] ≤ 1 which means 

2y + -y + 4 + 1 
y ≥-5 

now taking both part common 
but first i will put values of sinx in g (x) 

g'(X) =[sin(1) + 4 ]/[ 2 - sin (1) ]  and now we get g '( x ) ≥ -5 

now we can say that domain of f ' g ' (x) is [  -5  , [sin(1) + 4 ]/[ 2 - sin (1) ] ] 

hope this helps you my buddy thanks for the tough question really it builds my concept good 

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