Show that if f - - - 75- 35 is defined by fx - 3x-45x-7 and g - - - 35- - 75 is defined by gx - 7x- 45x-3- then fog - IA and gof - IB- where A - - - 35- B - - - 75 IA x - x- - x - A- IB x - x- - x - B are called identity functions on sets A and B- respectively-

Finding gof. Given: f(x) = 3x+ 45x 7 & g(x) = 7x+45x 3 g(x) = 7x+45x 3 g(f(x)) = 7f(x) nbsp; +45f(x) 3 gof = nbsp;7 ( 3x+45x 7 )+45 ( (3x+4)(5x 7) ) ...

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Standard XII

Mathematics

Integration by Substitution

Show that if ...

Question

Show that if $f : R - {\frac{7}{5}} \to R - {\frac{3}{5}}$ is defined by $f (x) = \frac{3 x + 4}{5 x - 7}$ and $g : R - {\frac{3}{5}} \to R - {\frac{7}{5}}$ is defined by $g (x) = \frac{7 x + 4}{5 x - 3},$ then $f o g = I_{A}$ and $g o f = I_{B},$ where $A = R - {\frac{3}{5}}, B = R - {\frac{7}{5}} I_{A} (x) = x, \forall x \in A, I_{B} (x) = x, \forall x \in B$ are called identity functions on sets $A$ and $B$ , respectively.

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Solution

Finding $g o f$ .
Given:
$f (x) = \frac{3 x + 4}{5 x - 7} & g (x) = \frac{7 x + 4}{5 x - 3}$
$g (x) = \frac{7 x + 4}{5 x - 3}$
$g (f (x)) = \frac{7 f (x) + 4}{5 f (x) - 3}$
$g o f = \frac{7 (\frac{3 x + 4}{5 x - 7}) + 4}{5 (\frac{(3 x + 4)}{(5 x - 7)}) - 3}$
$g o f = \frac{\frac{7 (3 x + 4) + 4 (5 x - 7)}{5 x - 7}}{\frac{5 (3 x + 4) - (5 x - 7)}{5 x - 7}}$
$g o f = \frac{7 (3 x + 4) + 4 (5 x - 7)}{5 (3 x + 4) - 3 (5 x - 7)}$
$g o f = \frac{21 x + 28 + 20 x - 28}{15 x + 20 - 15 x + 21} = \frac{41 x}{41}$
$g o f = x$
Thus,
$g o f = x$
$g o f = I_{V}$ where $B = R - {\frac{7}{5}}$

Finding $f o g$ .
$f (x) = \frac{3 x + 4}{5 x - 7} & g (x) = \frac{7 x + 4}{5 x - 3}$
$f (g (x)) = \frac{3 g (x) + 4}{5 g (x) - 7}$
$f o g = \frac{3 (\frac{7 x + 4}{5 x - 3}) + 4}{5 (\frac{(7 x + 4)}{(5 x - 3)}) - 7}$
$f o g = \frac{\frac{3 (7 x + 4) + 4 (5 x - 3)}{5 x - 3}}{\frac{5 (7 x + 4) - 7 (5 x - 3)}{5 x - 3}}$
$f o g = \frac{3 (7 x + 4) + 4 (5 x - 3)}{5 (7 x + 4) - 7 (5 x - 3)}$
$f o g = \frac{21 x + 12 + 20 x - 12}{35 x + 20 - 34 x + 21}$
$f o g = \frac{41 x}{41}$
$f o g = x$
Thus,
$f o g = x = I_{A}$
where $A = R - {\frac{3}{5}}$

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Similar questions

Q. Show that, if

f : R - {\frac{7}{5}} \to R - {\frac{3}{5}}

is defined by f (x) =

\frac{3 x + 4}{5 x - 7}

and

g : R - {\frac{3}{5}} \to R - {\frac{7}{5}}

is defined by g(x) =

\frac{7 x + 4}{5 x - 3}

,
then log =

l_{A}

and gof =

l_{B}

, where

l_{A}

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are called identity functions on sets A and B.

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g (x) = \frac{x}{x^{2} + 1} .

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$1)$ Functions $f, g : R \to R$ are defined, respectively, by $f (x) = x^{2} + 3 x + 1, g (x) = 2 x - 3,$ find

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find

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Integration by Substitution

Standard XII Mathematics

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Show that if f:R−{75}→R−{35} is defined by f(x)=3x+45x−7 and g:R−{35}→R−{75} is defined by g(x)=7x+45x−3, then fog=IA and gof=IB, where A=R−{35},B=R−{75}IA(x)=x,∀x ∈ A,IB(x)=x,∀x ∈ B are called identity functions on sets A and B, respectively.