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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 Ybe 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 functionf : [-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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