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Class 12 NCERT Solutions- Mathematics Part I – Chapter 1 Relations And Functions – Exercise 1.3
  • Last Updated : 19 Jan, 2021

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, 5), (5, 1)}. Write down gof.

Solution: 

f= {(1, 2), (3, 5), (4, 1)}

g= {(1, 3), (2, 3), (5, 1)}

f(1)= 2, g(2) = 3 => gof(1) = 3

f(3) = 5, g(5) = 1 => gof(3) = 1



f(4) =1, g(1) = 3 => gof(4) = 3 

=> gof = {(1,3), (3,1), (4,3)}

Question 2. Let f, g and h be functions from R to R. Show that (f+g) oh = foh + goh, (f * g) oh = (foh) * (goh).

Solution:

f: R-> R, g: R-> R, h: R-> R

(f+g) oh(x) = (f+g) oh(x)

= (f+g) [h(x)]

= f[h(x)] + g[h(x)]

= foh(x) + goh(x)

(f+g) oh = foh + goh  

(f * g) oh(x) = (f * g) oh(x)

= (f * g) [h(x)]

= f[h(x)] * g[h(x)]

= foh(x) * goh (x)

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

Question 3. Find gof and fog, if

(i) f(x) = |x| and g(x) = |5x – 2|

(ii) f(x) = 8x3  and g(x) = x1/3

Solution: 

(i) We have,        

f(x) = |x| and g (x) = | 5x – 2 |

gof(x) = g(f(x)) = g(|x|)



=> gof(x) = | 5 |x|-2 |

fog(x) = f(g(x)) = f(|5x-2|)

=> fog(x) = || 5x-2|| = | 5x -2 | 

(ii) We have,

f(x) = 8x3 and g(x) = x1/3

gof(x) = g(f(x)) = g(8x3)

=> gof(x) = (8x3)1/3 = 2x

fog(x) = f(g(x)) = f(x1/3)

=> fog(x) = 8(x1/3)3 = 8x

Question 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 ?

Solution:   

Given that,

f (x) = \frac{4x+3}{6x-4}x \neq \frac{2}{3}

Now, 

fof(x) = f(f(x)) = f(\frac{4x+3}{6x-4})

f(\frac{4(\frac{4x+3}{6x-4})+3}{6(\frac{4x+3}{6x-4})-4})

On simplifying by taking LCM = (6x-4)

fof(x) =  \frac{16x + 12 + 18x - 12}{24x + 18 -24x + 16}

=> fof (x) =  \frac{34x}{34}  = x 

=> fof(x) = IA (x) for all x \neq \frac{2}{3}

=> fof(x) = IA such that A = R –  {\frac{2}{3}}   which is the domain of f

=> f-1 = f          

Hence, proved.

Question 5. State with reason whether the following functions have inverse. Find the inverse, if it exists.

(i)  f : {1, 2, 3, 4} -> {10}

with f = {(1, 10), (2, 10), (3,10), (4,10)}

(ii) g: {5, 6, 7, 8} -> {1, 2, 3, 4}

with g = {(5, 4), (6, 3), (7, 4), (8, 2)}

(iii) h : {2, 3, 4, 5} -> {7, 9, 11, 13}

with h : {(2, 7), (3, 9), (4, 11), (5, 13)}

Solution:

(i) We have f(1) = f(2) = f(3) = f(4) = 10 which means that f is many-one 

and not one-one, therefore inverse of f does not exist.

(ii) Here g(5) = g(7) =4 i.e. g is many-one, so inverse of g does not exist.

(iii) Since range of h = {7, 9, 11, 13} = co-domain, therefore h is onto,

Also, each element of domain has a unique image in h, therefore h is one-one.

Now, since h is both one-one and onto,thus inverse of h exists.

h-1 = {(7, 2), (9, 3), (11, 4), (13, 5)}

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

Solution: 

Let x, y  \in    [-1, 1] 

f(x) =  \frac{x}{x+2}

f(y) = \frac{y}{y+2}

Now, 

Let f(x) = f(y) 

\frac{x}{x+2} = \frac{y}{y+2}     

=> x(x + 2) = y(x + 2)

=> x y + 2x = x y + 2y

=> 2x = 2y

=> x = y

=> f is one-one

Also,

X = [-1, 1] and,

Y = {  y \in R : y = \frac{x}{x+2} , x \in X  } = range of f.

=> f is onto

Since f is one-one and onto, therefore inverse of f exists.

Let y = f(x) => x =f-1(y)

=> y = \frac{x}{x+2}              

=> x y + 2y = x

=> 2y = x(1 – y)

=> x = \frac{2y}{1 - y}, y \ne 1

Therefore, f : Y-> X is defined by f(y) = \frac{2y}{1-y}, y \ne 1    .

Question 7. Consider f : R -> R is given by f(x) = 4x + 3 . Show that f is invertible. Find the inverse of f.

Solution:

It is given that,

f(x) = 4x + 3     where f : R -> R

Let, 

f(x) = f(y) 

=> 4x + 3 = 4y + 3

=> 4x = 4y

=> x = y

=> f is one-one function 

Also,

Let y  = 4x + 3     where y \in    R

=> x = \frac{y-3}{4}  \in R

Since for any  y \in R  . there exists x = \frac{y-3}{4} \in R  such that

f(x) = f(\frac{y-3}{4})  = 4 (\frac{y-3}{4})  +3 = y

=> f is onto

Since f is both one-one and onto, therefore f-1 exists

=> f-1(y) = \frac{y-3}{4}

Question 8. Consider f : R+ -> [4, \infty  ) given by f(x) = x2 + 4. Show that f is invertible with the inverse f-1 of f  given by f-1(y) = \sqrt{ y -4} , where R+ is the set of all non-negative real numbers.

Solution: 

Let f(x) = f(y)

=> x + 4 = y + 4

=> x2 = y2

=> x = y          [ x,y \in   R+ ]

=> f is one-one

Let y = x2 + 4             where y \in  [4, \infty )

=>  x2 = y – 4 \geq  4        [ y \geq 4 ]

=> x = \sqrt{y -4}

Therefore, for any y \in R  , there exists x =  \sqrt{y-4}  \in R

=> f is onto

Since, f is both one-one and onto, f-1 exists for every y \geq 4 ,

=> f-1(y) = \sqrt{y-4}

Question 9. Consider R+ -> [ -5, \infty ) given by f (x) = 9x2 + 6x -5. Show that f is invertible with f-1 (y) = (\frac{(\sqrt{(y+6)}-1}{3})

Solution: 

Let f(x) = f(y)

=> 9x2 + 6x -5 = 9y2 + 6y – 5

=> 9x2 + 6x = 9y2 + 6y

=> 9(x2 – y2) + 6 (x – y) = 0

=> (x – y) [9 (x + y) + 6] = 0

=> x – y =0

=> x = y

=> f is one-one

Now, let y = 9x2 + 6x – 5

=> 9x2 + 6x – 5 (x + y) = 0

=> x = \frac{-6 \pm \sqrt{6*6 + 4*9(5+y)} }{18} = \frac{\sqrt{(y+6)}-1}{3}    

=> f(x) = f(\frac{\sqrt{(y+6)}-1}{3}) = 9 (\frac{\sqrt{(y+6)}-1}{3})^{2}+(\frac{\sqrt{(y+6)}-1}{3}) -5

On simpliying, we have f (x) = y 

=> f is onto

Since f is both one-one and onto. f-1 exists

f-1(y) = \frac{\sqrt{(y+6)}-1}{3}

Question 10. Let f : X -> Y be an invertible function. Show that f has unique inverse. 

Solution:

We have,

f : X -> Y is an invertible function

Let g and h be two distinct inverses of f.

Then, for all y \in  Y,

fog (y) = I (y) = foh (y)

=> f g (y)) = f(h (y))

=> g(y) = h(y)        [f is one-one]

=> g = h                  [g is one-one]

which contradicts our supposition. 

Hence, f has a unique inverse.

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

Solution:

Given that,

f(1) = a, f(2) = b, f(3) = c

We have,

f = {(1, a), (2, b), (c, 3)}

which shows that f is both one-one and onto and thus f is invertible.

Therefore,

f-1 = {(a, 1), (b, 2), (c, 3)} 

Also,

(f-1)-1 = {(1, a), (2,b), (3, c)}

=> (f-1)-1 = f 

Hence proved.

Question 12. Let f: X -> Y be an invertible function. Show that the inverse of f-1 is f, i.e., (f-1 )-1 = f.

Solution:

Since, f is an invertible function,

=> f is both one-one and onto

Also,

Let g : Y -> X , where g is a one-one and onto function such that

gof (x) = Ix and fog (y) = Iy => g = f-1

=> f-1 o (f-1)-1 = I

=> f o [f-1 o (f-1)-1] = f o I

=> (f o f-1) o (f-1)-1 = f 

=> I o (f-1)-1 = f

Hence, (f-1)-1 = f

Question 13.  If f : R -> R given by f (x) = (3 – x3)1/3 ,then fof (x) is :

(A) x1/3                     (B) x3                     (C) x.                   (D) (3 – x3)

Solution:

Answer: (C)

We have,

f(x) = (3 – x3)1/3           where f : R -> R

Now,

fof(x) = f(f(x))

=> fof(x) = f((3 – x3)1/3

=> fof(x) = [3 – ((3 – x3)1/3 )3]1/3  

=> fof(x) = [3 – (3 – x3)]1/3              

=> fof(x) = (x3)1/3

=> fof(x) = x

Hence, option C is correct.

Question 14.  Let f : R -{ -\frac{4}{3}  } -> R be a function defined as f(x) = \frac{4x}{3x+4}  . The inverse of f is the map g : Range f -> R – { -\frac{4}{3}  } 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}{4-3y}

Solution:

Answer: (B)

Let y = f(x)

=> y = \frac{4x}{3x+4}

=> 3xy + 4y = 4x

=> x( 4 – 3y) = 4y

=> x = \frac{4y}{4-3y}   

f-1(y) = g (y) = \frac{4y}{4-3y}

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