Ex 1.3

EXERCISE 1.3

Question 1: Let f :{1,3,4} →{1,2,5}and g: {1,2,5} →{1,3} be given by f = {(1,2),(3,5),(4,1)} and g = {(1,3),(2,3),(5,1)}. Write down gof.

Solution: The functions f :{1,3,4} →{1,2,5}and g: {1,2,5} →{1,3} are f = {(1,2),(3,5),(4,1)} and g = {(1,3),(2,3),(5,1)}.

gof(1) = g [ f (1)] = g(2) = 3 [as f(1) = 2 and g(2)=3]

gof(3) = g [ f (3)] = g(5) = 1 [as f(3) = 5 and g(5)=1]

gof(4) = g [ f (4)] = g(1) = 3 [as f(4) = 1 and g(1)=3]

∴ gof = {(1,3),(3,1),(4,3)}

Question 2:

Let f, g, h be function form R to R. Show that

(f + g) oh = foh + goh

(f . g) oh = (foh) . (goh)

Solution:

$\displaystyle \begin{array}{l}\left[ {\left( {f+g} \right)oh} \right]x\\=\left( {f+g} \right)\left[ {h\left( x \right)} \right]=f\left[ {h\left( x \right)} \right]+g\left[ {h\left( x \right)} \right]\\=\left( {foh} \right)\left( x \right)+goh\left( x \right)\\=\left\{ {\left( {foh} \right)+\left( {goh} \right)} \right\}\left( x \right)=RHS\end{array}$

$\displaystyle \therefore \left\{ {\left( {f+g} \right)oh} \right\}\left( x \right)=\left\{ {\left( {foh} \right)+\left( {goh} \right)} \right\}\left( x \right)$ for all x ∈ R

Hence, $\displaystyle \left( {f+g} \right)oh=foh+goh$

LHS = $\displaystyle \begin{array}{l}\left[ {\left( {f.g} \right)oh} \right]\left( x \right)\\=\left( {f.g} \right)\left[ {h\left( x \right)} \right]=f\left[ {h\left( x \right).g\left( x \right)} \right]\\=\left( {foh} \right)\left( x \right).\left( {goh} \right)\left( x \right)\\=\left\{ {\left( {foh} \right).\left( {goh} \right)} \right\}\left( x \right)\end{array}$ = RHS

∴ $\displaystyle \left[ {\left( {f.g} \right)oh} \right]\left( x \right)=\left\{ {\left( {foh} \right).\left( {goh} \right)} \right\}\left( x \right)$ for all x ∈ R

Hence, $\displaystyle \left( {f.g} \right)oh=\left( {foh} \right).\left( {goh} \right)$

Question 3:

Find gof and fog, if

i. f (x) = ∣ x ∣ and g (x) = ∣5x – 2 ∣

ii. f (x) = 8x³ and g (x) = $\displaystyle {{x}^{{\frac{1}{3}}}}$

Solution:

(i) f (x) = ∣ x ∣ and g (x) = ∣5x – 2 ∣

∴ $\displaystyle gof\left( x \right)=g\left( {f\left( x \right)} \right)=g\left( {\left| x \right|} \right)=\left| {5\left| x \right|-2} \right|$

$\displaystyle fog\left( x \right)=f\left( {g\left( x \right)} \right)=f\left( {\left| {5x-2} \right|} \right)=\left| {5x-2} \right|$

(ii) $\displaystyle f\left( x \right)=8{{x}^{3}}$ and $\displaystyle g\left( x \right)={{x}^{{\frac{1}{3}}}}$

∴ $\displaystyle gof\left( x \right)=g\left( {f\left( x \right)} \right)=g\left( {8{{x}^{3}}} \right)={{\left( {8{{x}^{3}}} \right)}^{{\frac{1}{3}}}}=2x$

$\displaystyle fog\left( x \right)=f\left( {g\left( x \right)} \right)=f\left( {{{x}^{{\frac{1}{3}}}}} \right)=8\left( {{{x}^{{\frac{1}{3}}}}} \right)=8x$

Question 4:

If $\displaystyle f\left( x \right)=\frac{{\left( {4x+3} \right)}}{{\left( {6x-4} \right)}},x\ne \frac{2}{3}$ , show that $\displaystyle fof\left( x \right)=x$ for all $\displaystyle x\ne \frac{2}{3}$ . What is the reverse of f ?

Solution:

$\displaystyle \left( {fof} \right)f\left( x \right)=f\left( {f\left( x \right)} \right)=f\left( {\frac{{\left( {4x+3} \right)}}{{\left( {6x-4} \right)}}} \right)$