Home
Class 12
MATHS
For x in RR - {0, 1}, let f1(x) =1/x, f2...

For `x in RR - {0, 1},` let `f_1(x) =1/x, f_2(x) = 1-x and f_3(x) = 1/(1-x)` be three given functions. If a function, `J(x)` satisfies `(f_2oJ_of_1)(x) = f_3(x)` then `J(x)` is equal to :

A

`f_3(x)`

B

`f_1(x)`

C

`f_2(x)`

D

`1/x f_3(x)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the function \( J(x) \) such that: \[ (f_2 \circ J \circ f_1)(x) = f_3(x) \] where the functions are defined as follows: - \( f_1(x) = \frac{1}{x} \) - \( f_2(x) = 1 - x \) - \( f_3(x) = \frac{1}{1 - x} \) ### Step-by-Step Solution: 1. **Substitute \( f_1(x) \) into the equation**: We start with the left-hand side of the equation: \[ (f_2 \circ J \circ f_1)(x) = f_2(J(f_1(x))) \] Substituting \( f_1(x) \): \[ f_1(x) = \frac{1}{x} \implies f_2(J(f_1(x))) = f_2(J(\frac{1}{x})) \] 2. **Express \( f_3(x) \)**: The right-hand side of the equation is: \[ f_3(x) = \frac{1}{1 - x} \] So we need: \[ f_2(J(\frac{1}{x})) = \frac{1}{1 - x} \] 3. **Find the inverse of \( f_2 \)**: We need to find \( J(\frac{1}{x}) \). First, let's find the inverse of \( f_2 \): \[ f_2(x) = 1 - x \implies y = 1 - x \implies x = 1 - y \implies f_2^{-1}(x) = 1 - x \] 4. **Apply the inverse function**: Now we apply the inverse of \( f_2 \) to both sides: \[ J\left(\frac{1}{x}\right) = f_2^{-1}\left(f_3(x)\right) \] Substituting \( f_3(x) \): \[ J\left(\frac{1}{x}\right) = f_2^{-1}\left(\frac{1}{1 - x}\right) = 1 - \frac{1}{1 - x} \] 5. **Simplify the right-hand side**: Now simplify \( 1 - \frac{1}{1 - x} \): \[ 1 - \frac{1}{1 - x} = \frac{(1 - x) - 1}{1 - x} = \frac{-x}{1 - x} \] Thus, we have: \[ J\left(\frac{1}{x}\right) = \frac{-x}{1 - x} \] 6. **Find \( J(x) \)**: To find \( J(x) \), we replace \( \frac{1}{x} \) with \( x \): Let \( y = \frac{1}{x} \) then \( x = \frac{1}{y} \): \[ J(y) = \frac{-\frac{1}{y}}{1 - \frac{1}{y}} = \frac{-1}{y(1 - \frac{1}{y})} = \frac{-1}{\frac{y - 1}{y}} = \frac{-y}{y - 1} \] Therefore, we can write: \[ J(x) = \frac{-x}{x - 1} \] ### Final Answer: Thus, the function \( J(x) \) is: \[ J(x) = \frac{-x}{x - 1} \]
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • JEE 2019

    CENGAGE ENGLISH|Exercise Chapter 2|7 Videos

  • JEE 2019

    CENGAGE ENGLISH|Exercise Chapter 3|5 Videos

  • JEE 2019

    CENGAGE ENGLISH|Exercise MCQ|92 Videos

  • INVERSE TRIGONOMETRIC FUNCTIONS

    CENGAGE ENGLISH|Exercise Archives (Numerical Value type)|2 Videos

  • LIMITS

    CENGAGE ENGLISH|Exercise Comprehension Type|4 Videos

Similar Questions

Explore conceptually related problems

Let f be a one-one function satisfying f'(x)=f(x) then (f^-1)''(x) is equal to

For x in R , the functions f(x) satisfies 2f(x)+f(1-x)=x^(2) . The value of f(4) is equal to

If a function satisfies f(x+1)+f(x-1)=sqrt(2)f(x) , then period of f(x) can be

Suppose f, g, and h be three real valued function defined on R. Let f(x) = 2x + |x|, g(x) = (1)/(3)(2x-|x|) and h(x) = f(g(x)) The domain of definition of the function l (x) = sin^(-1) ( f(x) - g (x) ) is equal to

The third derivative of a function f'''(x) vanishes for all f(0)=1,f'(1)=2 and f''=-1, then f(x) is equal to (-3//2)x^2+3x+9

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

For x inR - {1} , the function f(x ) satisfies f(x) + 2f(1/(1-x))=x . Find f(2) .

The function f(x) satisfies the equation f(x + 1) + f(x-1) = sqrt3 f(x) . then the period of f(x) is

Let a function f(x) satisfies f(x)+f(2x)+f(2-x)+f(1+x)=x ,AAx in Rdot Then find the value of f(0)dot

If f_1(x)=(1)/(x), f_(2) (x)=1-x, f_(3) (x)=1/(1-x) then find J(x) such that f_(2) o J o f_(1) (x)=f_(3) (x) (a) f_(1) (x) (b) (1)/(x) f_(3) (x) (c) f_(3) (x) (d) f_(2) (x)

Home
Profile