\displaystyle \text{Question 1: Find } fog \text{ and }  gof, \text{ if }:  

\displaystyle \text{(i) } f(x) = e^x, \ \ g(x) = \log_e x

\displaystyle \text{(ii) }f(x) = x^2,  \ \ g(x) = \cos x

\displaystyle \text{(iii) }f(x) =|x| ,  \ \ g(x) = \sin x

\displaystyle \text{(iv) } f(x) = x+1 ,  \ \ g(x) = e^x

\displaystyle \text{(v) }f(x) =\sin^{-1} x ,  \ \ g(x) = x^2

\displaystyle \text{(vi) } f(x) = x+1,  \ \ g(x) = \sin x

\displaystyle \text{(vii) } f(x) = x+1, \ \ g(x) = 2x+3

\displaystyle \text{(viii) }f(x) =c, c\in R ,  \ \ g(x) = \sin x^2

\displaystyle \text{(ix) } f(x) =x^2 + 2 , \ \ g(x) = 1 - \frac{1}{1-x}

Answer:

\displaystyle \text{(i) } f(x) = e^x, \ \ g(x) = \log_e x

\text{Let } f: R \rightarrow (0, \infty); \text{ and } g: (0, \infty) \rightarrow R

\text{Let us calculate } fog,

\text{Clearly, the range of } g \text{ is a subset of the domain of } f.

fog: (0, \infty) \rightarrow R

(fog)(x) = f(g(x)) = f(\log_e x) = \log_e e^x = x

\text{Let us calculate } gof,

\text{Clearly, the range of } f \text{ is a subset of the domain of } g.

\Rightarrow  fog: R \rightarrow R \\ \\ (gof)(x) = g(f(x)) = g(e^x) = \log_e e^x = x

\\

\displaystyle \text{(ii) }f(x) = x^2,  \ \ g(x) = \cos x

f: R \rightarrow [0, \infty); g: R \rightarrow [-1, 1]

Let us calculate fog ,

Clearly, the range of g is not a subset of the domain of f.

\Rightarrow \text{Domain } (fog) = \{x: x \in \text{ domain of } g \text{ and } g (x) \in \text{ domain of } f\}

\Rightarrow \text{Domain } (fog) = \{x: x \in R \text{ and } \cos x \in R\}

\Rightarrow \text{Domain of } (fog) = R

(fog): R \rightarrow R

(fog)(x) = f(g(x)) = f(\cos x) = \cos^2 x

Let us calculate gof,

Clearly, the range of f is a subset of the domain of g 

\Rightarrow  fog: R \rightarrow R

(gof)(x) = g(f(x)) = g(x^2) = \cos x^2

\\

\displaystyle \text{(iii) }f(x) =|x| ,  \ \ g(x) = \sin x

f: R \rightarrow (0, \infty) ; g : R \rightarrow [-1, 1]

Let us calculate fog,

Clearly, the range of g is a subset of the domain of f.

\Rightarrow  fog: R \rightarrow R

(fog)(x) = f(g(x)) = f(\sin x) = |\sin x|

Let us calculate gof,

Clearly, the range of f is a subset of the domain of g.

\Rightarrow  fog : R \rightarrow R

(gof)(x) = g(f(x)) = g(|x|) = \sin |x|

\\

\displaystyle \text{(iv) } f(x) = x+1 ,  \ \ g(x) = e^x

f: R \rightarrow R ; g: R \rightarrow [ 1, \infty)

Let us calculate fog:

Clearly, range of g is a subset of domain of f.

\Rightarrow  fog: R \rightarrow R

(fog)(x) = f(g(x)) = f(e^x) = e^x + 1

Let us compute gof,

Clearly, range of f is a subset of domain of g.

\Rightarrow  fog: R \rightarrow R

(gof)(x) = g(f(x)) = g(x + 1) = e^x+1

\\

\displaystyle \text{(v) }f(x) =\sin^{-1} x ,  \ \ g(x) = x^2

\displaystyle f: [-1,1] \rightarrow \Big[  \frac{-\pi}{2}, \frac{\pi}{2} \Big] ; \    g : R \rightarrow [0, \infty) 

Let us compute the fog:

Clearly, the range of g is not a subset of the domain of f.

\text{Domain } (fog) = \{x: x \in \text{ domain of } g \text{ and } g (x) \in \text{ domain of } f\}

\text{Domain } (fog) = \{x: x \in R \text{ and } x^2 \in [-1, 1]\}

\text{Domain } (fog) = \{x: x \in R \text{ and } x \in [-1, 1]\}

\text{Domain of } (fog) = [-1, 1] fog: [-1,1] \rightarrow R

(fog)(x) = f(g(x)) = f(x^2) = \sin-1(x^2)

Let us compute the gof:

\text{Clearly, the range of } f \text{ is a subset of the domain of } g. \\ \\ fog: [-1, 1] \rightarrow R

(gof)(x) = g(f(x)) = g(\sin^{-1} x) = (\sin^{-1} x)^2

\\

\displaystyle \text{(vi) } f(x) = x+1,  \ \ g(x) = \sin x

f: R \rightarrow R; g: R \rightarrow [-1, 1]

Let us compute the fog

Set of the domain of f.

\Rightarrow fog: R \rightarrow R

(fog)(x) = f(g(x)) = f(\sin x) = \sin x + 1

Let us compute the gof,

Clearly, the range of f is a subset of the domain of g.

\Rightarrow fog: R \rightarrow R

(gof)(x) = g(f(x)) = g(x + 1) = \sin(x + 1)

\\

\displaystyle \text{(vii) } f(x) = x+1, \ \ g(x) = 2x+3

f: R \rightarrow R; g: R \rightarrow R

Let us compute the fog

Clearly, the range of g is a subset of the domain of f.

\Rightarrow fog: R\rightarrow R

(fog)(x) = f(g(x)) = f(2x + 3) = 2x + 3 + 1 = 2x + 4

Let us compute gof

Clearly, the range of f is a subset of the domain of g.

\Rightarrow fog: R \rightarrow R

(gof)(x) = g(f(x)) = g(x + 1) = 2(x + 1) + 3 = 2x + 5

\\

\displaystyle \text{(viii) }f(x) =c, c\in R ,  \ \ g(x) = \sin x^2

f: R \rightarrow {c}; g: R \rightarrow [0, 1]

Let us compute the fog

Clearly, the range of g is a subset of the domain of f.

fog: R \rightarrow R

(fog)(x) = f(g(x)) = f(\sin x^2) = c

Let us compute the gof,

Clearly, the range of f is a subset of the domain of g.

\Rightarrow fog: R \rightarrow R

(gof)(x) = g(f(x)) = g(c) = \sin c^2

\\

\displaystyle \text{(ix) } f(x) =x^2 + 2 , \ \ g(x) = 1 - \frac{1}{1-x}

f: R \rightarrow [2, \infty )

\text{For domain of } g: 1 - x \neq 0  \ \Rightarrow x \neq 1 \\ \\ \Rightarrow \text{ Domain of } g = R - {1}

\displaystyle g(x) = 1 - \frac{1}{1 - x} = \frac{1 - x - 1}{1 - x} = \frac{-x}{1 - x}

For range of g

\displaystyle y = \frac{-x}{1 - x} \\ \\ \Rightarrow y - xy = - x \Rightarrow y = xy - x \Rightarrow y = x(y - 1) \Rightarrow x = \frac{y}{y - 1}

\text{Range of } g = R - {1}

\text{Therefore, } g: R - {1} \rightarrow R - {1}

Let us compute the fog

Clearly, the range of g is a subset of the domain of f.

\Rightarrow fog: R - {1} \rightarrow R

\displaystyle (fog)(x) = f(g(x)) \\ \\ = f \Big(\frac{-x}{x - 1} \Big) = \Big(\frac{-x}{x - 1}\Big)^2 + 2 = \frac{x^2 + 2x^2 + 2 - 4x}{(1 - x)^2} = \frac{3x^2 - 4x + 2}{(1 - x)^2}

Now we have to compute gof

Clearly, the range of f is a subset of the domain of g.

\Rightarrow gof: R \rightarrow R

\displaystyle (gof)(x) = g(f(x)) = g(x^2 + 2) = 1 - \frac{1}{1 - (x^2 + 2)} = \frac{x^2 + 2}{x^2 + 1}

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\displaystyle \text{Question 2: Let } f(x) = x^2 + x + 1 \text{ and } g(x) = \sin x . \text{ Show that } fog \neq gof.

Answer:

\text{Given as } f(x) = x^2 + x + 1 \text{ and } g(x) = \sin x

(fog)(x) = f(g(x))  = f(\sin x)  = \sin^2 x + \sin x + 1

(gof)(x) = g(f(x))  = g(x^2 + x + 1)  = \sin(x^2+ x + 1)

\text{Therefore, } fog \neq gof.

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\displaystyle \text{Question 3: If  } f(x) = |x|, \text{ prove that } fof = f.

Answer:

\text{Given that } f(x) = |x|

(fof)(x) = f(f(x))  = f(|x|)  = ||x||  = |x|  = f(x)

\text{Therefore, } (fof)(x) = f(x), \ \forall \ x \in R \\ \\ \text{ Hence, } fof = f

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Question 4: If f(x) = 2x+5 \text{ and } g(x) = x^2 + 1 be two real functions, then describe each of the following functions:

(i) \ \ fog \hspace{1.0cm} (ii)  \ \ gof \hspace{1.0cm} (iii)  \ \ fof \hspace{1.0cm} (iv) \ \ f^2

Answer:

The function f(x) and g(x) are polynomials.

\Rightarrow f: R \rightarrow R \text{ and } g: R \rightarrow R.

\text{Therefore, } fog: R \rightarrow R \text{ and } gof: R \rightarrow R.

\text{(i) } (fog)(x) = f(g(x)) = f(x^2 + 1) = 2(x^2 + 1) + 5 = 2x^2 + 2 + 5 = 2x^2 +7

\text{(ii) } (gof)(x) = g(f(x)) = g(2x +5)  = (2x + 5)^2 + 1 = 4x^2 + 20x + 26

\text{(iii) } (fof)(x) = f(f(x)) = f(2x +5) = 2(2x + 5) + 5 = 4x + 10 + 5 = 4x + 15

\text{(iv) } f^2(x) = f(x) \times f(x) = (2x + 5) (2x + 5)  = (2x + 5)^2 = 4x^2 + 20x +25

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Question 5: If f(x) = \sin x \text{ and } g(x) = 2x be two real functions, then describe gof \text{ and } fog. Are these equal functions?

Answer:

\text{Given as } f(x) = \sin x \text{ and } g(x) = 2x

\text{We know }  f: R \rightarrow [-1, 1] \text{ and } g: R \rightarrow R

Clearly, the range of f is a subset of the domain of g.

gof: R \rightarrow R

(gof)(x) = g(f(x)) = g(\sin x) = 2 \sin x

Clearly, the range of g is a subset of the domain of f.

fog: R \rightarrow R

\text{Therefore, } (fog)(x) = f(g(x)) = f(2x) = \sin (2x)

\text{Clearly, } fog \neq gof

Hence they are not equal functions.

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Question 6: Let f, g, h be real functions given by f(x) = \sin x , g(x) = 2x \text{ and } h(x) = \cos x. \text{ Prove that } fog = go(fh).

Answer:

\displaystyle \text{Given as}  f(x) = \sin x, \ g(x) = 2x \text{ and } h(x) = \cos x

\displaystyle \text{We know, } f: R \rightarrow [-1, 1] \text{ and } g: R \rightarrow R

\displaystyle \text{Clearly, the range of } g \text{ is a subset of the domain of } f.

\displaystyle fog: R \rightarrow R \\ \\  (fh)(x) = f(x) h(x) = (\sin x) (\cos x) = \frac{1}{2} \sin (2x)

\displaystyle \text{Domain of } fh \text{ is } R.

\displaystyle \text{Since range of } \sin x \text{ is } \Big[-1, 1 \Big], -1 \leq \sin 2x \leq 1

\displaystyle \Rightarrow -\frac{1}{2} \leq \sin \frac{x}{2} \leq \frac{1}{2}

\displaystyle \text{The Range of } fh = \Big[-\frac{1}{2}, \frac{1}{2} \Big]

\displaystyle \text{Therefore, } (fh): R \rightarrow \Big[\frac{-1}{2}, \frac{1}{2} \Big]

\displaystyle \text{Clearly, range of } f h \text{ is a subset of } g.

\displaystyle \Rightarrow go(fh): R \rightarrow R

\displaystyle \Rightarrow \text{Domains of } fog \text{ and } go (fh) \text{ are the same. }

\displaystyle (fog)(x) = f(g(x))  = f(2x)  = \sin (2x)

\displaystyle (go(fh))(x) = g((f h)(x))  = g(\sin x \cos x)  = 2\sin x \cos x  = \sin (2x)

\displaystyle \Rightarrow (fog)(x) = (go(f h))(x), \ \forall \ x \in R

\displaystyle \text{Hence, } fog = go(fh)

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Question 7: Let f be any real function and let g be a function given by g(x) = 2x. Prove that gof = f + f.

Answer:

\text{Given, } f:R \rightarrow R

\text{Since } g(x) = 2x \text{ is a polynomial, } g : R \rightarrow R

\text{So, domains of } gof \text{ and } f+f \text{ are the same }

(gof)(x) = g(f(x)) = 2f(x)

(f+f) (x) = f(x) + f(x) = 2 f(x)

\Rightarrow (gof) (x) = ( f+ f) (x ) \ \forall \ x \in R

\text{Hence, } gof = f + f

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\text{Question 8: If } f(x) = \sqrt{1-x} \text{ and } g(x) = \log_e x \\ \\ \text{ are two real functions, then describe functions } fog \text{ and } gof.

Answer:

f(x) = \sqrt{1-x}

\text{For domain } 1 - x \geq 0

\Rightarrow x \leq 1

\Rightarrow \text{ domain of } f = ( -\infty, 1 ]

\Rightarrow f : ( - \infty , 1 ] \rightarrow ( 0, \infty)

g(x) = \log_e x

\text{Clearly, } g : ( 0, \infty) \rightarrow R

Computation of fog:

Clearly, the range of g is not a subset of the domain of f.

\Rightarrow \text{Domain } (fog) = \{  x: x \in \text{ Domain } (g) \text{ and } g(x) \in \text{ Domain of } f \}

\Rightarrow \text{Domain } (fog) = \{ x: x \in (0, \infty) \text{ and } \log_e x \in ( -\infty, 1 ] \}

\Rightarrow \text{Domain } (fog) = \{ x: x \in (0, \infty) \text{ and } x \in (0, e] \}

\Rightarrow \text{Domain } (fog) = \{ x: x \in (0, e] \}

\Rightarrow \text{Domain } (fog) = (0, e]

\Rightarrow fog : (0, e) \rightarrow R

(fog)(x) = f(g(x)) = f ( \log_e x) = \sqrt{1 - \log_e x}

Computation of gof:

Clearly, the range of f is a subset of the domain of g.

\Rightarrow gof : (-\infty, 1] \rightarrow R

\displaystyle \Rightarrow (gof)(x) = g(f(x)) \\ \\ = g(\sqrt{1-x}) = \log_e \sqrt{1-x} = \log_e (1-x)^{\frac{1}{2}} = \frac{1}{2} \log_e (1-x)

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\displaystyle \text{Question 9: If } f : \Big( \frac{-\pi}{2}, \frac{\pi}{2} \Big) \rightarrow R \text{ and } g : [-1, 1 ]\rightarrow R \text{ be defined as } \\ \\ f(x) = \tan x \text{ and } g(x) = \sqrt{1-x^2} \text{ respectively. Describe } fog \text{ and } gof.

Answer:

g(x) = \sqrt{1-x^2}

\Rightarrow x^2 \geq 0,  \ \forall \ x \in [-1, 1]

\Rightarrow -x^2 \leq 0,   \ \forall \ x \in [-1, 1]

\Rightarrow  1-x^2 \leq 1,   \ \forall \ x \in [-1, 1]

We know that 1 - x^2 \geq 0

\Rightarrow  0 \leq 1-x^2 \leq 1

\Rightarrow \text{Range of } g(x) = [0, 1]

\displaystyle \text{So, } f: \Big( \frac{\pi}{2}, \frac{\pi}{2} \Big) \rightarrow R \text{ and } g : [-1, 1] \rightarrow [0, 1]

Computation of fog:

Clearly, the range of g is a subset of the domain of f.

\text{So, } fog : [-1, 1 ] \rightarrow R

(fog) (x) = f(g(x)) = f( \sqrt{1-x^2}) = \tan \sqrt{1-x^2}

Computation of gof:

Clearly, the range of f is not a subset of the domain of g.

\displaystyle \Rightarrow \text{Domain } (gof) = \Big\{  x \in \text{ domain of } f \text{ and } f(x) \in \text{ domain of } g     \Big\}   

\displaystyle \Rightarrow \text{Domain } (gof) = \Big\{  x \in \Big( \frac{-\pi}{2}, \frac{\pi}{2} \Big) \text{ and } \tan x \in [-1, 1] \Big)     \Big\}   

\displaystyle \Rightarrow \text{Domain } (gof) = \Big\{ x \in \Big( \frac{-\pi}{2}, \frac{\pi}{2} \Big) \text{ and } x \in  \Big( \frac{-\pi}{4}, \frac{\pi}{4} \Big)     \Big\}   

\displaystyle \Rightarrow \text{Domain } (gof) = \Big\{  x \in  \Big( \frac{-\pi}{4}, \frac{\pi}{4} \Big)    \Big\}   

\text{So, } (gof)(x) = g(f(x)) = g( \tan x) = \sqrt{1-x^2}

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\displaystyle \text{Question 10: If } f(x) = \sqrt{x+3} \text{ and } g(x) = x^2 + 1 \text{ be two real functions, } \\ \\ \text{ then find } fog \text{ and  } gof.

Answer:

f(x) = \sqrt{x+3}

\text{For Domain } x + 3 \geq 0

\Rightarrow x \geq -3

\text{Domain of  } f = [-3, \infty )

\text{Since } f \text{ is a square root function, range of } f = [0, \infty )

f: [-3, \infty) \rightarrow [0, \infty)

g(x) = x^2 + 1 \text{ is a polynomial }

\Rightarrow g: R \rightarrow R

Computation of fog:

Range of g is not a subset of the domain of f

\text{Domain } (fog) = \{ x : x \in \text{ domain of  } g \text{ and } g(x) \in \text{ domain of } f(x)        \}

\Rightarrow \text{Domain } (fog) = \{ x : x \in R \text{ and } x^2 + 1 \in [-3, \infty )       \}

\Rightarrow \text{Domain } (fog) = \{ x : x \in R \text{ and }   x^2 + 1 \geq -3     \}

\Rightarrow \text{Domain } (fog) = \{ x : x \in R \text{ and }  x^2 + 4 \geq 0     \}

\Rightarrow \text{Domain } (fog) = \{ x : x \in R \text{ and }  x \in R      \}

\Rightarrow \text{Domain } (fog) = R

(fog)(x) = f(g(x)) = f( x^2 +1) = \sqrt{x^2+1+3} = \sqrt{x^2+4}

Computation of gof:

Range of f is a subset of the domain of g.

gof: [-3, \infty) \rightarrow R

(gof)(x) = g(f(x)) = (\sqrt{x+3})^2 + 1 = x+ 3 + 1 = x+ 4

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Question 11: Let f be a real function given by f(x) = \sqrt{x-2}. Find each of the following:

(i) \ \ fof  \hspace{1.0cm}  (ii) \ \ fofof \hspace{1.0cm}   (iii) \ \ ( fofof)(38)  \hspace{1.0cm}  (iv) \ \ f^2

Also, show that fof \neq f^2

Answer:

f(x) = \sqrt{x-2}

\text{For Domain } x - 2 \geq 0 \Rightarrow x \geq 2

\text{Domain of } f = [2, \infty )

\text{Since } f \text{ is a square root function, range of } f = ( 0, \infty)

\text{So, } f: [2, \infty) \rightarrow (0, \infty)

\text{(i)  } fof

Range of f is not a subset of the domain of f.

\Rightarrow \text{Domain } (fof) = \{ x : x \in  \text{ domain of } f \text{ and } f(x) \in \text{ domain of } f         \}

\Rightarrow \text{Domain } (fof) = \{ x : x \in [2, \infty) \text{ and } \sqrt{x-2} \in [2, \infty)          \}

\Rightarrow \text{Domain } (fof) = \{ x : x \in  [2, \infty) \text{ and } \sqrt{x-2} \geq 2        \}

\Rightarrow \text{Domain } (fof) = \{ x : x \in  [2, \infty) \text{ and } x-2 \geq 4        \}

\Rightarrow \text{Domain } (fof) = \{ x : x \in [2, \infty) \text{ and }  x \geq 6       \}

\Rightarrow \text{Domain } (fof) = \{ x : x \geq 6          \}

\Rightarrow \text{Domain } (fof) = [6, \infty)

fof: [6, \infty) \rightarrow R

(fof)(x) = f(f(x)) = f(\sqrt{x-2}) = \sqrt{ \sqrt{x-2} -2  }

\\

\text{(ii)  } fofof = (fof)of

\text{We have, } f: [2, \infty) \rightarrow (0, \infty) \text{ and } fof : [6, \infty) \rightarrow R

Range of f is not a subset of the domain of fof.

\Rightarrow \text{Domain } ((fof)of) = \{ x : x \in  \text{ domain of } f \text{ and } f(x) \in \text{ domain of } f of        \}

\Rightarrow \text{Domain } ((fof)of) = \{ x : x \in [2, \infty) \text{ and } \sqrt{x-2} \in [6, \infty)          \}

\Rightarrow \text{Domain } ((fof)of) = \{ x : x \in  [2, \infty) \text{ and } \sqrt{x-2} \geq 6        \}

\Rightarrow \text{Domain } ((fof)of) = \{ x : x \in  [2, \infty) \text{ and } x-2 \geq 36        \}

\Rightarrow \text{Domain } ((fof)of) = \{ x : x \in [2, \infty) \text{ and }  x \geq 38       \}

\Rightarrow \text{Domain } ((fof)of) = \{ x : x \geq 38          \}

\Rightarrow \text{Domain } ((fof)of) = [38, \infty)

fof: [38, \infty) \rightarrow R

((fof)of)(x) = (fof)(f(x)) = (fof)(\sqrt{x-2}) = \sqrt{\sqrt{\sqrt{x-2}-2}-2}

\text{(iii) We have } (fofof)(x) = \sqrt{\sqrt{\sqrt{x-2}-2}-2}

\therefore (fofof)(38) = \sqrt{\sqrt{\sqrt{38-2}-2}-2} = \sqrt{ \sqrt{6-2} -2  } = \sqrt{2-2} = 0

\text{(iv) We have } fof = \sqrt{ \sqrt{x-2} -2  }

f^2(x) = f(x) \times f(x) = \sqrt{x-2} \times \sqrt{x-2} = x-2

\text{Therefore } fof \neq f^2

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\displaystyle \text{Question 12: } f(x) =  \Bigg\{ \begin{array}{ll}  1+x, & 0 \leq x \leq 2 \\ 3-x, & 2 < x \leq 3 \end{array}  . \ \ \ \text{    Find } fof.  

Answer:

f(x) =  \Bigg\{ \begin{array}{ll}  1+x, & 0 \leq x \leq 2 \\ 3-x, & 2 < x \leq 3 \end{array}

This can be written as

f(x) = \Bigg\{ \begin{array}{ll}  1+x, & 0 \leq x \leq 1 \\ 1+x, & 1 < x \leq 2 \\ 3-x &  2 < x \leq 3 \end{array}

\text{When } 0 \leq x \leq 1

\text{Then, } f(x) = 1 + x

\text{Now when } , 0 \leq x \leq 1 \text{ then, } 1 \leq x+1 \leq 2

\text{Then, } f(f(x)) = 1 + ( 1 + x ) = 2 + x \ \ \ [ \because 1 \leq f(x) < 2 ]

\text{When, } 1 < x \leq 2

\text{Then, } f(x) = 1 + x

\text{Now when, } 1 < x \leq 2 \text{ then } 2 < x +1 \leq 3

\text{Then } f(f(x)) = 3 - ( 1 + x) = 2 - x \ \ \ [ \because 2 \leq f(x) < 3 ]

\text{When, } 2 < x \leq 3

\text{Then, } f(x) = 3- x

\text{Now when, } 2 < x \leq 3 \text{ then } 0 \leq 3-x < 1

\text{Then, } f(f(x)) = 1 + (3-x) = 4 - x \ \ \ [ \because 0 \leq f(x) < 1 ]

f(f(x)) = \Bigg\{ \begin{array}{ll}  2+x, & 0 \leq x \leq 1 \\ 2-x, & 1 < x \leq 2 \\ 4-x &  2 < x \leq 3 \end{array}

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\displaystyle \text{Question 13: If } f, g : R \rightarrow R \text{ be two functions defined as } f(x) = |x| + x \\ \\ \text{ and } g(x) = |x| - x \text{ for all } x \in R. \text{ Then, find } fog \text{ and } gof. \text{ Hence, find } \\ \\ fog(-3) , fog(5) \text{ and } gof(-2).  

Answer:

\text{Given } f(x) = |x| + x

\text{and } g(x) = |x| - x \ \forall \ x \in R

fog = f(g(x)) = |g(x) | + g(x) = ||x|-x|+(|x|-x)

Therefore,

f(g(x)) =  \Bigg\{ \begin{array}{ll}  0, & x \geq 0 \\ \\ 4x & x < 0 \end{array} 

f(g(x)) =  \Bigg\{ \begin{array}{ll}  4x, & x > 0 \\ \\ 0 & x \geq 0 \end{array} 

gof = g(f(x)) = |f(x)|-f(x) = ||x|+x| - ( |x|+x)

g(f(x)) =  \Bigg\{ \begin{array}{ll}  0, & x \geq 0 \\ \\ 0 & x < 0 \end{array} 

Therefore, g(f(x)) = gof = 0

\text{Now, } fog(-3) = 4 ( -3) = - 12  \hspace{1.0cm} ( \text{ since } fog = 4x \text{ for } x < 0 )

fog(5) = 0   \hspace{1.0cm} ( \text{ since } fog = 0 \text{ for } x \geq 0 )

gof(-2) = 0  \hspace{1.0cm} ( \text{ since } gof = 0 \text{ for } x < 0 )