If the function f : [1, infinity) -----[1, infinity) is defined by f(x) = 2^x ^(x-1),then

f^-1(x) is :
1). (1/2)^x(x-1)
2). (1/2) [ 1 + (1 + 4 log₂x)^1/2]

3).(1/2) [ 1 + (1 - 4 lo₂x)^1/2]

4). Not defined

Question

If the function f : [1, infinity) -----[1, infinity) is defined by
f(x) = 2x (x-1),then

f-1(x) is :
1) (1/2)x(x-1)
2) (1/2) [ 1 + (1 + 4 log2x)1/2]

3) (1/2) [ 1 + (1 - 4 lo2x)1/2]

4) Not defined

Anuradha Sharma · Accepted Answer

y = fx= 2xx-1Taking log we get, logy = xx-1log2logy =x2log2-xlog2x2log2-xlog2-logy=0It is a quadratic equation in x so using quadratic formula, x = log2 ±log22+4logylog22log2= log2 1±log22+4logylog2log22log2=121±log22+4logylog2log22=121±1+4logylog2=121±1+4log2yx=121±1+4log2y12But it is given that at y = 1 x = 1So we will neglect - value and take positive value only
x=121+1+4log2y12

If the function f : [1, infinity) -----[1, infinity) is defined by f(x) = 2x (x-1),then f-1(x) is :1). (1/2)x(x-1) 2). (1/2) [ 1 + (1 + 4 log2x)1/2] 3).(1/2) [ 1 + (1 - 4 lo2x)1/2] 4). Not defined

If the function f : [1, infinity) -----[1, infinity) is defined by f(x) = 2^x ^(x-1),then

f^-1(x) is :
1). (1/2)^x(x-1)
2). (1/2) [ 1 + (1 + 4 log₂x)^1/2]

3).(1/2) [ 1 + (1 - 4 lo₂x)^1/2]

4). Not defined