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Q. 14 Let $f: R - \left\{-\frac{4}{3}\right\} \rightarrow R$ be a function defined as $f(x) = \frac{4x}{3x + 4}$. The inverse of $f$ is the map $g : Range\;f \rightarrow R - \left \{-\frac{4}{3} \right \}$given by

(A) $g(y) = \frac{3y}{3 -4y}$

(B) $g(y) = \frac{4y}{4 -3y}$

(C) $g(y) = \frac{4y}{3 -4y}$

(D) $g(y) = \frac{3y}{3 -4y}$

Let f inverse   Let y be element of range f. Then there is     such that                                                                                                                            Now , define     as                                                                                                                                                                                 ...

Q.  13 If $f : R \rightarrow R$ be given by $f(x) = (3 - x^3)^{\frac{1}{3}}$, then $fof(x)$ is

(A) $x^{\frac{1}{3}}$

(B) $x^3$

(C) $x$

(D) $(3 - x^3)$

Thus, is x. Hence, option c is correct answer.

Q. 12 Let $f : X \rightarrow Y$ be an invertible function. Show that the inverse of $f^{-1}$ is $f$, i.e.,
$(f^{-1})^{-1} = f$

To prove:  Let    be a invertible function.   Then there is      such that     and      Also,                 and                              and          Hence,     is invertible function and f is inverse of . i.e.

11 Consider $f : \{1, 2, 3\} \rightarrow \{a, b, c\}$ given by $f (1) = a$,$f (2) = b$  and $f (3) = c$. Find $f^{-1}$ and show that  $(f^{-1})^{-1} = f$.

,  and Let there be a function g such that   i.e.      ,        and       Now , we have And,          , here                    Hence, f exists and  is g.       ,       and    Let inverse of  be h such that   And        ,  here   Thus,    It is noted that h=f. Hence,.

Q.10 Let $f : X \rightarrow Y$be an invertible function. Show that f has a unique inverse.
(Hint: suppose $g_1$ and $g_2$ are two inverses of $f$. Then for all $y \in Y$,
$fog_1 (y) = I_Y (y) = fog_2 (y)$. Use one-one ness of f).

Let be an invertible function Also, suppose f has two inverse  For , we have                                                  [f is invertible implies f is one - one]                                              [g is one-one] Thus,f has a unique inverse.

Q .9 Consider $f : R_+ \rightarrow [- 5, \infty)$ given by $f (x) = 9x^2 + 6x - 5$. Show that $f$ is invertible with $f^{-1} (y) = \left (\frac{(\sqrt{y + 6}) - 1}{3} \right )$

One- one: Let                                                                                   Since, x and y are positive.                f is one-one. Onto: Let for    ,                                                                                                                                                                                                                          ...

Q. 8 Consider $f : R_+\rightarrow [4,\infty)$ given by $f (x) = x^2 + 4$. Show that $f$ is invertible with the
inverse $f^{-1}$ of $f$ given by $f^{-1} = \sqrt{y-4}$ , where $R_+$ is the set of all non-negative
real numbers.

One- one: Let                                                                 f is one-one. Onto: Let for    ,                                                                                                                 for  there is   such that                                             f is onto. Since, f is one-one and onto so it is invertible. Let      by   Now,                     ...

Q. 7  Consider $f : R \rightarrow R$ given by $f (x) = 4x + 3$. Show that f is invertible. Find the
inverse of $f$.

is given by   One-one : Let                                    f is one-one function. Onto: So, for  there is     ,such that                                                                               f  is onto. Thus, f is one-one and onto so  exists. Let,  by  Now,                                                                                                                        ...

Q. 6 Show that $f : [-1, 1] \rightarrow R$, given by $f(x) = \frac{x}{(x + 2)}$ is one-one. Find the inverse of the function$f : [-1, 1] \rightarrow Range f$

One -one:                                                                f is one-one. It is clear that   is onto. Thus,f is one-one and onto so inverse of f exists. Let g be inverse function of f in   let y be an arbitrary element of range f Since,  is onto ,so           for                                            ,

Q. 3 State with reason whether following functions have inverse

(iii)   $h : \{2, 3, 4, 5\}\rightarrow \{7, 9, 11, 13\}$ with
$h = \{(2, 7), (3, 9), (4, 11), (5, 13)\}$

(iii)    with         From the definition, we can see the set  have distant values under h.  h is one-one . For every element y of set ,there exists an element x  in  such that    h is onto Thus, h is one-one and onto so h has an inverse function.

Q. 5 State with reason whether following functions have inverse

(ii)  $g : \{5, 6, 7, 8\} \rightarrow \{1, 2, 3, 4\}$ with
$g = \{(5, 4), (6, 3), (7, 4), (8, 2)\}$

(ii)   with       From the definition, we can conclude :  g is not one-one. Hence, function g does not have inverse function.

Q5.State with reason whether following functions have inverse

(i) $f : \{1, 2, 3, 4\} \rightarrow\{10\}$ with
$f = \{(1, 10), (2, 10), (3, 10), (4, 10)\}$

(i)  with     From the given definition,we have:  f is not one-one. Hence, f do not have inverse function.

Q.4  If $f(x) = \frac{4x + 3}{6x - 4}, x \neq \frac{2}{3}$ show that $fof (x) = x$,  for all $x \neq\frac{2}{3}$. What is the inverse of $f$?

,  for all   Hence,the given function is invertible and the inverse of  is  itself.

Q.3 Find gof and fog, if

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

The solution is as follows (ii)  and

Q.3 Find $gof$ and $fog$, if

(i) $f (x) = | x |$  and  $g(x) = \left | 5x-2 \right |$

and

Q.2 Let $f$, $g$ and $h$ be functions from $R$ to $R$. Show that
$\\(f + g) o h = foh + goh\\ (f \cdot g) o h = (foh) \cdot (goh)$

To prove :                                                                                                                   Hence,      To prove:                                                                                                                                      Hence,

Q.1 Let $f : \{1, 3, 4\}\rightarrow \{1, 2, 5\}$and $g : \{1, 2, 5\} \rightarrow \{1, 3\}$ be given by
$f = \{(1, 2), (3, 5), (4, 1)\}$ and $g = \{(1, 3), (2, 3), (5, 1)\}$. Write down $gof$.

Given :                    and                                   and                                                                  Hence,    =
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