answe it question number 70.
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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 .
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 .
abhi178:
i hope i am right ...
Answered by
4
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
C H E E R R S !
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
C H E E R R S !
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