Let f-0- 1- -0-1- be defined by fx-1-x-1-x- 0- x- 1 and let g-0- 1- -0-1- be defined by gx-4x1-x-0- x- 1- Determine the composition functions fog1

f(x)=1 x1+x g(x)=4x(1 x) Therefore fog =f(g(x)) =1 4x(1 x)1+4x(1 x) Now fog(1) =1 4(1 1)1+4(1 1) =1 Hence fog(1)=1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Modulus Function

Let f:[0, 1...

Question

Let $f : [0, 1] \to [0, 1]$ be defined by $f (x) = \frac{1 - x}{1 + x}, 0 \leq x \leq 1$ and let $g : [0, 1] \to [0, 1]$ be defined by $g (x) = 4 x (1 - x), 0 \leq x \leq 1$ . Determine the composition functions fog(1)

Open in App

Solution

$f (x) = \frac{1 - x}{1 + x}$
$g (x) = 4 x (1 - x)$
Therefore
$f o g$
$= f (g (x))$
$= \frac{1 - 4 x (1 - x)}{1 + 4 x (1 - x)}$
Now
$f o g (1)$
$= \frac{1 - 4 (1 - 1)}{1 + 4 (1 - 1)}$
$= 1$
Hence
$f o g (1) = 1$

Suggest Corrections

Similar questions

Q. Let Q be the set of all rational numbers in [0, 1] and

f : [0, 1] \to [0, 1]

be defined by

f (x) = {\begin{matrix} x & f o r & x \in Q 1 - x & f o r & x \notin Q \end{matrix}

Then the set

S = {x \in [0, 1] : (f o f) (x) = x}

is equal to

Q. Let f defined on [0, 1] be twice differentiable such that | f"(x) | ≤ 1 for all x ∈ [0, 1]. If f(0) = f(1), then show that | f'(x) | < 1 for all x ∈ [ 0, 1].

Q. Let

f : (0, 1) \to R

be defined by

f (x) = \frac{b - x}{1 - b x}

, where

b

is a constant such that

0 < b < 1

, then:

Q. Let

f : R \to (0, 1)

be a function defined by

f (x) = \frac{e^{| x |} - e^{- x}}{e^{x} + e^{- x}}

Then

Let f(x) be a function defined on [0, 1] such that $f (x) = {\begin{matrix} x, & i f x ϵ Q 1 - x, & i f x / ϵ Q \end{matrix}$
Then, for all $x ϵ [0, 1], f (f (x)) =$

Join BYJU'S Learning Program

Let f:[0,1]→[0,1] be defined by f(x)=1−x1+x,0≤x≤1 and let g:[0,1]→[0,1] be defined by g(x)=4x(1−x),0≤x≤1. Determine the composition functions fog(1)

f(x)=1−x1+xg(x)=4x(1−x)Therefore fog=f(g(x))=1−4x(1−x)1+4x(1−x)Now fog(1)=1−4(1−1)1+4(1−1)=1Hence fog(1)=1

Let $f : [0, 1] \to [0, 1]$ be defined by $f (x) = \frac{1 - x}{1 + x}, 0 \leq x \leq 1$ and let $g : [0, 1] \to [0, 1]$ be defined by $g (x) = 4 x (1 - x), 0 \leq x \leq 1$ . Determine the composition functions fog(1)

$f (x) = \frac{1 - x}{1 + x}$
$g (x) = 4 x (1 - x)$
Therefore
$f o g$
$= f (g (x))$
$= \frac{1 - 4 x (1 - x)}{1 + 4 x (1 - x)}$
Now
$f o g (1)$
$= \frac{1 - 4 (1 - 1)}{1 + 4 (1 - 1)}$
$= 1$
Hence
$f o g (1) = 1$