Function

Question 1

1. (a) 1. (b) 1. (c) 1. (d) 1. (e) 1. (f) 1. (g) 1. (h)

Question 2. (a) 2. (b)

Question 

3 (a) If \( f(x) = 3x \) and \( g(x) = 7x \), find \( f_og(x) \) and \( g_of(x) \). Solution: \begin{align} \text{Given,}&\\ f(x) & = 3x \\ g(x) & = 7 x \\ \text{Now,  } fog(x) & = f(g(x)) \\ & = f(7x) \\ & = 3(7x) \\ & = 21 x\\ \text{Also,  } gof(x) & = g(f(x)) \\ & = g(3x) \\ & = 7(3x) \\ & = 21x \end{align}

3. (b)

If \( f(x) = 2x + 1 \) and \( g(x) = x - 1 \), find \( f_og(x) \) and \(g_of(x) \). Solution: \begin{align} \text{Given,}&\\ f(x) & = 2x + 1  \\ g(x) & = x - 1 \\ \text{Now,  } fog(x) & = f(g(x)) \\ & = f( x - 1 ) \\ & = 2(x-1)+1 \\ & = 2x-2+1\\ & = 2x - 1 \\ \text{Also,  } gof(x) & = g(f(x)) \\ & = g(2x + 1) \\ & = 2x + 1 - 1\\ & = 2x \end{align}

3 (c)

If \( f(x) = 2x - 1 \) and \( h(x) = 3x+2 \), find \( f_oh(x) \) and \( h_of(x) \) Solution: \begin{align} \text{Given,}&\\ f(x) & = 2x - 1 \\ h(x) & = 3x+2 \\ \text{Now,  } f_oh(x) & = f(h(x)) \\ & = f( 3x+2) \\ & = 2(3x+2) - 1 \\ & = 6x + 4 - 1\\ & = 6x + 3\\ \text{Also,  } &\\ h_of(x) & = h(f(x)) \\ & = h(2x - 1) \\ & = 3(2x-1)+2\\ & = 6x-3+2\\ & = 6x - 1 \end{align}

3. (d) 

If \( g(x) = \frac{x+1}{2} \) and \( h(x) = \frac{2x-3}{2} \), find \(g_oh(x) \) and \( h_og(x) \). Solution: \begin{align} \text{Given,}&\\ g(x) & = \frac{x+1}{2} \\ h(x) & = \frac{2x-3}{2} \\ \text{Now,  } g_oh(x) & = g(h(x))\\ & = g\left( \frac{2x-3}{2}\right) \\ & = \frac{\dfrac{2x-3}{2}+1}{2} \\ & = \frac{\dfrac{2x-3+2}{2}}{2}\\ & = \frac{\dfrac{2x-1}{2}}{2}\\ & = \frac{2x-1}{4} \\ \text{Also,  } h_og(x) & = h(g(x)) \\ & = h\left(\frac{x+1}{2}\right) \\ & = \frac{2\left(\dfrac{x+1}{2} \right) - 3}{2}\\ & = \frac{x+1-3}{2}\\ & = \frac{x-2}{2} \end{align}

3 (e) 

If \( f(x) = \frac{2x-3}{x-1} \) and \( g(x) = \frac{4x+7}{3+2x} \) , find \( f_og(x) \) and \( g_of(x) \). Solution: \begin{align} \text{Given,}&\\ f(x) & = \frac{2x-3}{x-1} \\ g(x) & = \frac{4x+7}{3+2x} \\ \text{Now,  } f_og(x) & = f(g(x))\\ & = f\left( \frac{4x+7}{3+2x} \right) \\ & = \frac{2\left( \dfrac{4x+7}{3+2x} \right) - 3 }{\dfrac{4x+7}{3+2x} - 1 } \\ & = \frac{\dfrac{8x+14-3(3+2x)}{3+2x}}{\dfrac{4x+7-(3+2x)}{3+2x}} \\ & = \frac{8x+14-9-6x}{4x+7-3-2x}\\ & = \frac{2x+5}{2x+4} \\ \text{Also,  } g_of(x) & = g(f(x)) \\ & = g\left(\frac{2x-3}{x-1}\right) \\ & = \frac{4\left(\dfrac{2x-3}{x-1}\right) + 7 }{3+ 2\left(\dfrac{2x-3}{x-1}\right)}\\ & = \frac{\dfrac{8x-12+7(x-1)}{x-1}}{\dfrac{3(x-1)+2(2x-3)}{x-1}} \\ & = \frac{8x-12+7x-7}{3x-2+4x-6}\\ & = \frac{15x-19}{7x-9} \end{align}

Question

4. (a) If \( f(x) = x+1 \) and \( g(x) = 5x -1 \), find \( f_og(2) \) and \( g_of(-1) \) . Solution: \begin{align} \text{Given,  } f(x) & = x+1 \\ g(x) & = 5x -1 \\ \text{Now,  } f_og(2) & = f(g(2)) \\ & = f( 5\times 2 - 1) \\ & = f(10 - 1) \\ & = f(9) \\ & = 9+1 \\ & = 10\\ \text{Also,  } gof(-1) & = g(f(-1)) \\ & = g(-1+1)\\ & = g(0) \\ & = 5(0) - 1 \\ & = 0 - 1 \\ & = -1 \end{align}

4. (b) 

If \( f(x) = 2x +5 \) and \( g(x) = 3 - 5x \), find \( f_og(0) \) and \( g_of(1/2) \). Solution: \begin{align} \text{Given,  } f(x) & = 2x +5 \\ g(x) & = 3 - 5x \\ \text{Now,  } f_og(0) & = f(g(0)) \\ & = f( 3 - 5 \times 0 ) \\ & = f(3 -0 ) \\ & = f(3) \\ & = 2\times 3 + 5 \\ & = 6 + 5 \\ & = 11 \\ \text{Also, } gof\left(\frac{1}{2} \right) & = g\left(f\left(\frac{1}{2} \right) \right) \\ & = g\left( 2\times \frac{1}{2} + 5 \right) \\ & = g ( 1 + 5 ) \\ & = g(6) \\ & = 3 - 5 \times 6 \\ & = 3 - 30 \\ & = -27 \end{align}

4. (c) 

If \( g(x) = \frac{x-5}{4} \) and \( h(x) = \frac{3x + 5}{2} \), find \( g_o h \left( \frac{1}{3} \right) \) and \( gh(5) \). Solution: \begin{align} \text{Given,  } g(x) & = \frac{x-5}{4} \\ h(x) & = \frac{3x + 5}{2} \\ \text{Now,  } g_o h \left( \frac{1}{3} \right) & = g \left( h \left( \frac{1}{3} \right) \right) \\ & = g\left( \frac{ 3\times \dfrac{1}{3} + 5 }{2} \right)\\ & = g\left( \frac{ 1+ 5 }{2} \right)\\ & = g\left( \frac{ 6}{2} \right)\\ & = g(3) \\ & = \frac{3-5}{4} \\ & = \frac{-2}{4} \\ & = \frac{-1}{2} \\ \text{ Also,  } gh(5) & = g(h(5)) \\ & = g\left( \frac{3\times 5 +5 }{2} \right) \\ & =  g\left( \frac{15 + 5  }{2} \right) \\ &  = g\left( \frac{20}{2} \right) \\ & = g(10) \\ & = \frac{10 - 5}{4} \\ & = \frac{5}{4} \\ \end{align}

4. (d) 

If \( f(x) = \frac{2x-4}{x} \) and \( g(x) = \frac{x-4}{3x} \), find \\ (i) \( f_og(-1) \) \hfill (ii)  \( fg(1) \) \hfill  (iii)\( gf(-2) \)  \hfill (iv) \( gf\left( \frac{1}{2} \right) \). Solution: \begin{align} \text{(i) } f_og(-1) & = f(g(-1)) \\ & = f \left(\frac{-1-4}{3(-1)} \right) \\ & = f \left(\frac{-5}{-3} \right) \\ & = f \left(\frac{5}{3} \right) \\ & = \frac{2\times \dfrac{5}{3} - 4}{\dfrac{5}{3} }\\ & = \frac{\dfrac{10-12}{3}}{\dfrac{5}{3}} \\ & = \frac{-2}{5} \\ \end{align} \begin{align} \text{(ii) } fg(1) & = f(g(1))  \\ & = f(g(1)) \\ & = f\left(\frac{1-4}{3\times 1} \right) \\ & = f\left( \frac{-3}{3} \right) \\ & = f(-1) \\ & = \frac{2(-1) + 4 }{-1} \\ & = \frac{-2-4}{-1} \\ & = \frac{-6}{-1} \\ & = 6 \end{align} \begin{align} \text{(iii) } gf(-2) & = g(f(-2)) \\ & = g\left( \frac{2(-2)-4}{-2} \right) \\ & = g\left(\frac{-4-4}{-2} \right) \\ & = g(4) \\ & = \frac{4-4}{3\times 4} \\ & = \frac{0}{12} \\ & = 0 \end{align}

Solution: \begin{align} \text{(iv) } gf\left(\frac{1}{2}\right) & = g\left(f\left(\frac{1}{2}\right)\right) \\ & = g\left( \frac{2\times \dfrac{1}{2} - 4 }{\frac{1}{2}} \right) \\ & = g\left( \frac{1 - 4}{\dfrac{1}{2} } \right) \\ & = g\left( \frac{-3}{\dfrac{1}{2} } \right) \\ & = g(-6) \\ & = \frac{-6-4}{3(-6) } \\ & = \frac{-10}{-18} \\ & = \frac{5}{9} \end{align}

4. (e) 

If \( f(x) = \frac{5-4x}{3+x} \) and \( g(x) = \frac{7 + 2x }{5-6x }, \) find 

(i) \( fg\left( \frac{1}{2} \right) \)  (ii) \( fg(-1) \)   (iii) \(gf(0)\)    (iv) \(gf(1)\)

Solution:

 \begin{align} \text{( i )  } fg\left( \frac{1}{2} \right) & = f\left(g\left( \frac{1}{2} \right) \right) \\ & = f \left( \frac{ 7 + 2\times \dfrac{1}{2}}{5-6\times \frac{1}{2} } \right)\\ & = f \left( \frac{7+1}{5-3} \right)\\ & = f \left( \frac{8}{2} \right) \\ & = f \left( 4 \right) \\ & = \frac{5-4\times 4 }{3+4} \\ & = \frac{5-16 }{3+4} \\ & = \frac{-11}{7} \end{align}

Solution: 

\begin{align} \text{(ii)  } fg(-1) & = f(g(-1)) \\ & = f \left( \frac{7 + 2(-1)}{5 - 6(-1) } \right)\\ & = f\left( \frac{7-2}{5+6} \right) \\ & = f\left( \frac{5}{11} \right) \\ & = \frac{5-4\times \dfrac{5}{11} }{3 + \dfrac{5}{11} } \\ & = \frac{ \dfrac{55-20}{11}}{\dfrac{33+5}{11}} \\ & = \frac{35}{38} \end{align}

Solution: 

\begin{align} \text{(iii)  } gf(0) & = g(f(0)) \\ & = g\left( \frac{5-4\times 0 }{3+0} \right) \\ & = g\left( \frac{5}{3} \right) \\ & = \frac{7+ 2\left( \dfrac{5}{3} \right) }{5 -6\left( \dfrac{5}{3} \right)  }\\ & = \frac{\dfrac{21+10}{3}}{\dfrac{15-30}{3} } \\ & = \frac{31}{-15} \\ & = \frac{-31}{15} \end{align}

Solution: 

\begin{align} \text{(iv)  }  gf(1) & = g(f(1))\\ & = g\left( \frac{5-4\times 1}{3+1} \right)\\ & = g\left( \frac{5-4}{4} \right) \\ & = g\left( \frac{1}{4} \right) \\ & = \frac{7+ 2\left(\dfrac{1}{4} \right)}{5-6\left(\dfrac{1}{4} \right) }\\ & = \frac{\dfrac{28}{4}}{\dfrac{30-6}{4}} \\ & = \frac{30}{24} \\ & = \frac{5}{4} \end{align}

Question  5
If \( f(x) = 4x - 5 \) and \( g(x) = \frac{3x}{x+1} ,  x \neq -1 \), then find
(i) \( f^2 (0) \) \hfill (ii) \( g^2 (4) \) \hfill (iii) \( f^2g(0) \) \hfill ( iv ) \( g^2 f(-3) \) 

Solution:
\begin{align} \text{(i)  } f^2(0) & = f((0)) \\ & = f(4\times 0 -5 ) \\ & = f(-5) \\ & =4(-5) - 5 \\ & = -20 - 5 \\ & = -25 \end{align}

Solution:

 \begin{align} \text{(ii)  } g^2(4) & = g(g(4)) \\ & = g\left( \frac{3\times 4}{4+1} \right)\\ & = g\left( \frac{12}{5} \right) \\ & = \frac{ 3\times \dfrac{12}{5} }{\dfrac{12}{5} + 1} \\ & = \frac{ \dfrac{36}{5}}{\dfrac{12+5}{5} }\\ & = \frac{36}{17} \end{align}

Solution: 

\begin{align} \text{(iii)  } f^2g(0) & = f^2(g(0)) \\ & = f^2 \left( \frac{3\times 0 }{ 0 + 1 } \right)\\ & = f^2 \left( \frac{0}{1} \right) \\ & = f^2 ( 0 ) \\ & = f(f(0)) \\ & = f( 4\times 0 - 5 )\\ & = f(0-5)\\ & = f(-5) \\ & = -20 - 5 \\ & = -25 \end{align}

Solution: 

\begin{align} \text{(iv)  } g^2f(-3) & =g^2 f(( -3 ))\\ & = g^2 (4(-3)-5) \\ & = g^2 ( -12-5) \\ & =g^2 (-17) \\ & = g(g(-17)) \\ & = g\left( \frac{3(-17)}{-17+1} \right) \\ & = g\left( \frac{-51}{-16} \right)\\ & = g\left( \frac{51}{16} \right)\\ & = \frac{3\left(\dfrac{51}{16}\right) }{\dfrac{51}{16} + 1 }\\ & = \frac{\dfrac{153}{16}}{\dfrac{51+16}{16} } \\ & = \frac{153}{67} \end{align}

Question  6 

If \( f(x) = 7x -1,  g(x) = x -3 \), and \( h(x) = 5 - 7x \), find  \\ (i) \(fgh(x)\)  (ii) \(gfh(2) \)  (iii) \( fhg(-3) \)  ( iv ) \( hgf(0) \) 

Solution: \begin{align} \text{(i)  }  fgh(x) & = fg(h(x)) \\ & = fg(5-7x) \\ & = f(g(5-7x)) \\ & = f(5-7x-3)\\ & = f(2-7x)\\ & = 7(2-7x)-1 \\ & = 14 - 49 x - 1 \\ & = 13 - 49x \end{align}

Solution: 

\begin{align} \text{(ii)  }  ghf(2) & =gh(f(2)) \\ & = gf(5-7\times 2) \\ & = gf(5-14) \\ & = g(f(-9)) \\ & = g(7(-9) - 1 ) \\ & = g(-63 - 1 ) \\ & = g(-64) \\ & = -64 - 3 \\ & = -67 \end{align}

Solution:

 \begin{align} \text{(ii)  }  fhg(-3) & = fh(g(-3)) \\ & = fh(-3-3) \\ & = fh(-6) \\ & = f(h(-6)) \\ & = f(5-7(-6))\\ & = f(5+42)\\ & = f(47) \\ & = 7(47) - 1 \\ & = 329 - 1 \\ & = 328 \end{align}

Solution:

 \begin{align} \text{(iv)  }  hgf(0) & = hg(f(0)) \\ & = hg(7\times 0 -1 ) \\ & = hg(0-1) \\ & = hg(-1) \\ & = h(g(-1)) \\ & = h(-1-3) \\ & = h(-4) \\ & = 5-7(-4) \\ & = 5 + 28 \\ & = 33 \end{align}

Question  7

7. (a) If \( f(x) = 2x - 5, g (x) = x+1  \) and \( fg(a) = 3 \) , find the value of \( a \). 

Solution: \begin{align} \text{Given, } & fg(a) = 3 \\ \text{or, } & f(g(a)) = 3 \\ \text{or, } & f(a+1) = 3 \\ \text{or, } & 2(a+1) - 5 = 3 \\ \text{or, } & 2a + 2 -5 = 3 \\ \text{or, } & 2a - 3 =3 \\ \text{or, } & 2a = 3+3 \\ \text{or, } & 2a = 6 \\ \text{or, } & a =\frac{6}{2} \\ \therefore \ \  & a = 3 \end{align}

7. (b) If \( f(x) = 3x - 5 , g(x) = 5 - 2x \) and \( fg(b) = 4 \), find the value of \( b\). 

Solution: \begin{align} \text{Given, } & fg(b) = 4 \\ \text{or, } & f(g(b)) = 4 \\ \text{or, } & f(5-2b) = 4 \\ \text{or, } & 3(5-2b) + 5 = 4 \\ \text{or, } & 15-6b+5 = 4 \\ \text{or, } & 20-4 = 6b \\ \text{or, } & 6b = 16  \\ \text{or, } & b = \frac{16}{6} \\ \therefore \ \  & b = \frac{8}{3} \end{align}

7. (c) If \( f(x) = \frac{x} {2}, g(x) = \frac{ax+5}{2} \) and \(gf(3) = 0 \), find the value of \(a\).

Solution: \begin{align} \text{Given, } & gf(3)= 0 \\ \text{or, } & g(f(3)) = 0 \\ \text{or, } & g\left( \frac{3}{2} \right) = 0 \\ \text{or, } & \frac{ 2\times \dfrac{3}{2} + 5}{2} = 0 \\ \text{or, } & \frac{3a}{2} + 5 = 0 \\ \text{or, } & \frac{3a}{2} = -5 \\ \text{or, } & 3a = -10  \\ \therefore \ \  & a = \frac{-10}{3} \end{align}

7. (d) If \( f(x) = \frac{kx+5}{2}, g(x) = 3x - 5 \) and \( fg(5) = - 1 \), find the value of \(k \). 

Solution: \begin{align} \text{Given, } & fg(5) = - 1\\ \text{or, } & f(g(5)) = 3 \\ \text{or, } & f(3\times 5 - 5) = 3 \\ \text{or, } & f( 15 - 5) = -1  \\ \text{or, } & f(10) = -1 \\ \text{or, } & 10k + 5 = -2 \\ \text{or, } & 10k = -2 - 5 \\ \text{or, } & 10k = -7 \\ \text{or, } & a =\frac{-7}{10} \\ \therefore \ \  & a = -\frac{7}{10} \\ \end{align}

7. (e) If \( f(x) = \frac{5-3ax}{4-x}, g(x) = \frac{7+4x}{x} \) and \( fg(-1) = 5 \), find the value of \( a \). Solution: \begin{align} \text{Given, } & fg(-1) = 5 \\ \text{or, } & f(g(-1)) = 5 \\ \text{or, } & f\left( \frac{7+4(-1)}{-1} \right) = 5 \\ \text{or, } & f\left( \frac{7-4}{-1} \right) = 5 \\ \text{or, } & f\left( \frac{3}{-1} \right) = 5 \\ \text{or, } & f\left( -3 \right) = 5 \\ \text{or, } & \frac{5-3a(-3)}{4-(-3)} = 5 \\ \text{or, } & \frac{5+9a}{7} = 5  \\ \text{or, } & 5 + 9a = 35 \\ \text{or, } &  9a = 35  - 5\\ \text{or, } &  9a = 30 \\ \text{or, } &  a = \frac{30}{9}  \\ \therefore \ \  & a = \frac{10}{3} \\ \end{align}

Question

8. (a) If \( f:R\to R \) and \( g:R \to R \) are functions such that \( f(x) = x^2 \) and \( g(x) = x^3 \), show that \( f_og(x) = g_of(x) \). 

Solution: \begin{align} \text{Given, }  f(x) & = x^2 \\ g(x) & = x^3 \\ \text{Now, } \\ f_og(x) & = f(g(x)) \\ & = f( x^3 ) \\ & = (x^3 )^2 \\ & = x^6\\ \text{Also, } \\ gof(x) & = g(f(x)) \\ & =g(x^2 )\\ & = (x^2 )^3 \\ & = x^6 \\ \therefore \ \  & f_og(x) = g_of(x) = x^6 \end{align}

8. (b) If \( f(x) = 10x \) and \( g(x) = 3x \), show that \( f_og(5) = g_of(5) \). 

Solution: \begin{align} \text{Given, } f(x) & = 10x\\ g(x) & = 3x\\ \text{Now, } \\ f_og(5) & = f(g(5)) \\ & = f(3\times 5 ) \\ & = f(15) \\ & = 10 \times 15 \\ & = 150 \text{Also, } \\ g_of(x) & = g(f(x)) \\ & =g(10 \times 5 )\\ & = g(50) \\ & = 3\times 50 \\ & = 150 \\ \therefore \ \   f_og(5) & = g_of(5) =150 \end{align}

Question

9. (a) If \( f(3x+2) = 4x - 5 \), then find \( f_of(3) \). Solution: \begin{align} f(3x+2) & = 4x - 5 \\ \text{Let } 3x + 2 & = a \\ \text{or, } 3x & = a -2 \\ \text{or, } x & = \frac{a-2}{3} \\ \text{Then, } \\ f(a) & = f4\left(\frac{a-2}{3}\right) - 5 \\ & = \frac{4a-8-15}{3} \\ & = \frac{4a-23}{3} \\ f(x) & = \frac{4x-23}{3} \\ \text{Now, } \\ f_of(3) & = f(f(3)) \\ & =f \left( \frac{4\times 3 -23 }{3} \right) \\ & = f\left( \frac{12-23}{3} \right) \\ & = f\left( \frac{-11}{3} \right) \\ & = \frac{4\left(\dfrac{-11}{3}\right) - 23}{3}\\ & = \frac{\dfrac{-44-69}{3}}{3} \\ & = \frac{-113}{9} \end{align}

9. ( b) If \( g(x-1) = 2x - 3 \), then find \( g_og(-2) \). Solution: \begin{align} g(x-1) & = 2x - 3 \\ \text{Let } x -1  & = a \\ \text{or, } x & = a +1 \\ \text{Then, } \\ g(a) & = 2(a+1) - 3 \\ & = 2a + 2 - 3 \\ & = 2a - 1 \\ f(x) & = 2x - 1\\ \text{Now, } \\ g_og(-2) & = g((-2)) \\ & = g( 2(-2) - 1 ) \\ & = g(-4-1)\\ & =g(-5) \\ & = 2(-5) -1 \\ & = -10 - 1\\ & = -11 \end{align}

Question

10. (a) If \( f(x) = 5x -1 \) and \( f_og(x) = 2x + 9 \), find \(g(x) \). 

Solution: \begin{align} & f(x) = 5x -1  \\ & f_og(x)  = 2x + 9 \\ & g(x) = ?\\ \text{Now, }\\ & f_og(x)  = 2x + 9\\ \text{or, }&  f(g(x))  = 2x + 9  \\ \text{or, }& 5g(x) - 1  = 2x + 9 \\ \text{or, }& 5g(x)  = 2x + 9 +1 \\ \text{or, }& 5g(x) = 2x+ 10\\ \therefore &  g(x)  = \frac{2x+10}{5} \end{align}

10. (b) If \( g(x) = 3x+2 \) and \( g_of(x) = 7x-5\), find \( f(x) \). 

Solution: \begin{align} & g(x) = 3x+2  \\ & g_of(x) = 7x-5 \\ & g(x) = ?\\ \text{Now, }\\ & g_of(x) = 7x-5\\ \text{or, }&  g(f(x)) = 7x - 5  \\ \text{or, }& 3f(x) + 2 = 7x - 5 \\ \text{or, }& 3f(x) = 7x -5 -2 \\ \text{or, }& 3f(x) = 7x - 7\\ \therefore &  f(x)  = \frac{7x - 7}{3} \end{align}

10. (c) If \( f(x) = 4x+5, f_og(x) = 8x + 13 \) and \( gf(x) = 28 \), find the value of x 

Solution: \begin{align} & gf(x) = 4x+5  \\ & f_og(x) = 8x + 13\\ & gf(x) = 28\\ & x = ? \\ \text{Now, }\\ & f_og(x) = 8x + 13\\ \text{or, }&  f(g(x)) = 8x + 13  \\ \text{or, }& 4g(x) + 5 = 8x + 13 \\ \text{or, }& 4g(x) = 8x + 13 - 5 \\ \text{or, }& 4g(x) = 8x + 8\\ \text{or, }& g(x) = \frac{8x+8}{4} \\ \text{or, }& g(x) = \frac{8(x+1)}{4} \\ \therefore & g(x) = 2(x+1)\\ \text{Now, }\\ gf(x)  = 28 \\ \text{or, }& g(f(x)) =28 \\ \text{or, }& g(4x+5) = 28 \\ \text{or, }& 2(4x+5+1) = 28 \\ \text{or, }& 4x+6 = \frac{28}{2} \\ \text{or, }& 4x+6 = 14 \\ \text{or, }& 4x = 14-6\\ \text{or, }& 4x = 8\\ \text{or, }& x =\frac{8}{4} \\ \therefore & x = 2 \end{align}

Question

11. (a) If \( f(x) = \frac{2x+3}{2} \) and \( g_of(x) = x-2 \), find \( g(x) \). 

Solution: \begin{align} & f(x) = \frac{2x+3}{2}\\ & g_of(x) = x-2 \\ & g(x) = ? \\ \text{Now,} & \\ & g_of(x) = x-2 \\ & \text{or, }  g(f(x)) = x - 2 \\ & \text{or, } g\left(\frac{2x+3}{2} \right) = x - 2 \\ & \text{Let, } \frac{2x+3}{2} = a \\ & \text{or, } 2x + 3 = 2a \\ & \text{or, } 2x= 2a - 3 \\ & \text{or, } x = \frac{2a -3 }{2} \\ \text{ Then, }&  \\ g(a) & = \frac{2a-3}{2} - 2 \\ & = \frac{2a-3-4}{2} \\ & = \frac{2a-7}{2} \\ \therefore f(x) & = \frac{2x-7}{2} \end{align}

11. ( b) If \( g(x) = \frac{3x+1}{4} \) and \(f_og(x) = 5x + 7 \), find \( g(x) \) 

Solution: \begin{align} \text{Given, }& \\ g(x) & = \frac{3x+1}{4} \\ f_og(x) & = 5x + 7 \\ f(x) & = ?  \\ \text{Now, } & \\ & f_og(x) = 5x + 7 \\ & \text{or, } f(g(x)) = 5x + 7 \\ & \text{or, } f \left( \frac{3x+1}{4} \right) = 5x + 7 \\ & \text{Let, }  \frac{3x+1}{4} = a \\ & \text{or, } 3x+1 = 4a \\ & \text{or, } 3x = 4a -1 \\ & \text{or, } x = \frac{4a-1}{3} \\ \text{Now, }&  \\ f(a)  & = 5\left( \frac{4a-1}{3} \right) + 7 \\ & = \frac{20a - 5 + 21}{3} \\ & = \frac{20a + 16}{3} \\ \therefore f(x) & = \frac{20x + 16}{3} \\ \end{align}

11. (c) If \( f(x) = 4x +5, f_og(x) = 8x + 13 \) and \( gf(x) = 28 \), find the value of \( x \). 

Solution: \begin{align} \text{Given, }& \\ f(x) & = 4x +5 \\ f_og(x) & = 8x + 13 \\ gf(x) & = 28 \\ x & = ? \\ \text{Now, }& \\ & f_g(x) = 8x + 13 \\ & \text{or, } f(g(x)) = 8x + 13\\ & \text{or, } 4g(x) + 5 = 8x + 13 \\ & \text{or, } 4g(x) = 8x + 13 -5 \\ & \text{or, } 4 g(x) = 8x + 8 \\ & \text{or, } g(x) = \frac{8x+8}{4} \\ & \text{or, } g(x) = \frac{8(x+1)}{4} \\ & \text{or, } g(x) = 2(x+1) \\ \text{Also, }& \\ & gf(x) = 28 \\ & \text{or, } g(f(x)) = 28 \\ & \text{or, } g( 4x+ 5 ) = 28 \\ & \text{or, } 2(4x+5+1) = 28 \\ & \text{or, } 2( 4x + 6 ) = 28 \\ & \text{or, }4x + 6 = \frac{28}{2} \\ & \text{or, } 4x + 6 = 14 \\ & \text{or, } 4x = 14 - 6 \\ & \text{or, } 4x = 8 \\ & \text{or, } x = \frac{8}{4} \\ \therefore  x &  = 2 \end{align}

Question  12 

If \( f(x) = x -1 , g(x) = 2x+3, fg(x) = 5 \)  and \( gf(x) = 3 \), find the value of \( x \). 

Solution: \begin{align} \text{Given, } & \\ f(x) & = x -1 \\ g(x) & = 2x+3 \\ fg(x) & = 5  \\ \text{Now, } & \\ & fg(x) = 5 \\ & \text{or, } f(g(x)) = 5 \\ & \text{or, } f(2x+3) = 5 \\ & \text{or, } 2x+3 - 1 = 5 \\ & \text{or, } 2x + 2 = 5 \\ & \text{or, } 2x = 5 - 2 \\ & \text{or, } 2x = 3 \\ & \therefore  x = \frac{3}{2} \\ \text{Also, } & \\ & gf(x) =3 \\ & \text{or, } g(f(x)) = 3 \\ & \text{or, } g(x - 1 ) = 3 \\ & \text{or, } 2(x-1) + 3 = 3 \\ & \text{or, } 2x - 2  + 3 = 3 \\ & \text{or, } 2x + 1 = 3 \\ & \text{or, } 2x = 3 - 1 \\ & \text{or, } 2x = 2 \\ & \text{or, } x = \frac{2}{2} \\ & \therefore   x = 1 \end{align}