Mathematics MISCELLANEOUS EXAMPLES AND MISCELLANEOUS EXERCISE
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Class 11 Chapter 2 - Relations & Functions

MISCELLANEOUS EXAMPLES AND MISCELLANEOUS EXERCISE

Miscellaneous Examples

Q 3088301207

Let `R` be the set of real numbers. Define the real function `f: R → R` by `f(x) = x + 10` and sketch the graph of this function.



Solution:

Here `f(0) = 10, f(1) = 11, f(2) = 12, ...,`
`f(10) = 20`, etc., and `f(–1) = 9, f(–2) = 8, ..., f(–10) = 0` and so on.
Therefore, shape of the graph of the given
function assumes the form as shown in Fig
Q 3038401302

Let `R` be a relation from `Q` to `Q` defined by `R = {(a,b): a,b ∈ Q` and `a – b ∈ Z}`. Show that

(i) `(a,a) ∈ R` for all `a ∈ Q`
(ii) `(a,b) ∈ R` implies that `(b, a) ∈ R`
(iii) `(a,b) ∈ R` and `(b,c) ∈ R` implies that `(a,c) ∈ R`

Solution:

(i) Since, `a – a = 0 ∈ Z`, if follows that `(a, a) ∈ R.`
(ii) `(a,b) ∈ R` implies that `a – b ∈ Z`. So,` b – a ∈ Z.` Therefore, `(b, a) ∈ R`
(iii) `(a, b)` and `(b, c) ∈ R` implies that `a – b ∈ Z. b – c ∈ Z.` So, `a – c = (a – b) + (b – c) ∈ Z.` Therefore, `(a,c) ∈ R`
Q 3008401308

Let `f = {(1,1), (2,3), (0, –1), (–1, –3)}` be a linear function from Z into Z. Find f(x).

Solution:

Since f is a linear function, `f (x) = mx + c`. Also, since `(1, 1), (0, – 1) ∈ R,`
`f (1) = m + c = 1` and `f (0) = c = –1`. This gives `m = 2` and `f(x) = 2x – 1.`
Q 3038501402

Find the domain of the function `f(x) = ( x^2+3x+5)/(x^2-5x+4)`



Solution:

Since `x^2 –5x + 4 = (x – 4) (x –1),` the function `f (x)` is defined for all real numbers except at `x = 4` and `x = 1.` Hence the domain of `f` is `R – {1, 4}.`
Q 3018501409

The function f is defined by `f(x) = { tt ( ( 1-x , x < 0 ) , (1 , x = 0) , (x+1 , x>0) )` Draw the graph of `f (x).`





Solution:

Here, `f(x) = 1 – x, x < 0,` this gives
`f(– 4) = 1 – (– 4) = 5;`
`f(– 3) =1 – (– 3) = 4,`
`f(– 2) = 1 – (– 2) = 3`
`f(–1) = 1 – (–1) = 2;` etc, and `f(1) = 2, f (2) = 3, f (3) = 4`
` f(4) = 5` and so on for `f(x) = x + 1, x > 0.`
Thus, the graph of f is as shown in Fig

Miscellaneous Exercise

Q 2665767665

The relation `f` is defined by

`f(x) = { tt ( (x^2 , 0 le x le 3 ), (3x , 3 le x le 10 ) )`

The relation `g` is defined by

`g(x) = { tt ( (x^2 , 0 le x le 2 ), (3x , 2 le x le 10 ) )`

Show that `f` is a function and `g` is not a function.
Class 11 Exercise 2.mis Q.No. 1
Solution:

(i) `f (x) = x^2`, is defined in the interval

`0 le x le 3`

Also, `f(x) = 3x` is defined in the interval

`3 le x le 10`

At `x = 3`, from `f(x) = x^2,f (3) = 3^2 = 9` and

from `g(x) = 3x, g(3) = 3 xx 3 = 9`

` :. f` is defined at `x = 3`, Hence, `f` is function.

(ii) `g(x) = x^2` is defined in the interval

`0 le x le 2`

`g(x) = 3x` is defined in the interval

`2 le x le 10`

But at `x = 2, g(x) = x^2 => g(2) = 2^2=4` and

`g(x)=3x => g(2)=3 xx 2=6`

At `x = 2`, relation on `g` has two values

`:.` Relation `g` is not a function.
Q 2685767667

If `f(x) = x^2` , find `(f(1.1)- f (1) )/(1.1-1)`
Class 11 Exercise 2.mis Q.No. 2
Solution:

`f(x) = x^2 => f( 1.1 )= ( 1.1 )^2 = 1.21 ; f (1 )= 1^2 = 1`

`:. (f( 1.1) -f(1) )/(1.1 -1) = (1.21-1)/(1.1-1) = (0.21)/(0.1) = 2.1`
Q 2605767668

Find the domain of the function

`f(x) = (x^2 +2x +1 )/(x^2 -8x +12 )`
Class 11 Exercise 2.mis Q.No. 3
Solution:

`f(x ) = (x^2 +2x +1 )/(x^2 -8x +12) = ((x+1)^2)/((x-2)(x-6))`


The function is not defined at `x = 2, 6`

Domain of `f= { x : x in R` and `x ≠ 2, x ≠ 6}`

`=R-{2,6}`
Q 2615767669

Find the domain and the range of the real
function `f` defined by `f(x) = sqrt ( x-1 )`
Class 11 Exercise 2.mis Q.No. 4
Solution:

(i) `f(x) = sqrt (x-1)`, `f` is not defined for `x-1 < 0`
or `x < 1`; Domain of `f(x) = { x: x in R, x ge 1}`

(ii) Let `f(x) = y = sqrt (x-1)`

`:.y` is well defined for all values of `x ge 1 `.

Range `= [0, oo)`
Q 2635867762

Find the domain and the range of the real
function `f` defined by `f(x) = | x-1 |`
Class 11 Exercise 2.mis Q.No. 5
Solution:

(i) `f(x) = | x- 1| ` is defined for all real values of

`x` ; `:.` Domain of `f= {x : x in R} = R .`

(ii) `f(x) = | x- 1 |` can acquire only non-negative

values.; `:.` Range `= {y : y in R, y ge 0}`
Q 2645867763

Let `f = { (x, x^2/(1+x^2) ) : x in R }` be a function

from `R` into `R`. Determine the range of `f`.
Class 11 Exercise 2.mis Q.No. 6
Solution:

Let `y = f(x) = x^2/(1+x^2) ; f(x )` is positive for all

values of `x`

when `x= 0 ,y= 0`. Also `text (denominator) > text ( numerator )`


`:.` Range of `f= { y : y in R` and `y in [0 ,1 l)}`
Q 2655867764

Let `f, g : R -> R` be defined, respectively by

`f(x) = x+ 1 ,g(x) =2x-3`. Find `f+ g, f-g` and `f/g`
Class 11 Exercise 2.mis Q.No. 7
Solution:

`f, g` are defined for all `x in R`

(i) `(f+g)(x)=f(x)+g(x) = x+ 1 +2x-3`

`=3x-2`

(ii) `(f - g) (x) = f(x)- g (x)`

`=(x+ 1)-(2x-3)=-x+4`

(iii) `(f/g) (x) = (f(x) )/(g(x)) = (x+1)/(2x-3)`

where `x in R` and `x ≠ 3/2`
Q 2665867765

Let `f= {(1, 1), (2, 3), (0, -1), (-1, -3)}` be a function from `Z` to `Z` defined by `f(x) =ax+ b`,
for some integers `a, b`. Determine `a, b`.
Class 11 Exercise 2.mis Q.No. 8
Solution:

We have `f(x) =ax+ b`, for `x = 1 ,f(x) = 1`,

`:. a+ b = 1` ... (i)

For `x=2,f(x)=3`,

`:. 3 = a xx 2 + b` or `2a + b = 3` ... (ii)

subtracting eqn (ii) from (i) `a= 2 , b = -1`
Q 2675867766

Let `R` be a relation from `N ` to `N` defined by

`R = {(a, b) : a, b in N` and `a= b^2}`. Are the following true?

(i) `(a, a) in R` for all `a in N`

(ii) `(a, b) in R`, implies `(b, a) in R`

(iii) `(a, b) in R, (b, c) in R` implies `(a, c) in R`.

Justify your answer in each case.
Class 11 Exercise 2.mis Q.No. 9
Solution:

(i) `a= a^2` is true only, when `a= 0` or `1`. It is not a relation.

(ii) `a= b^2` and `b = a^2` is not true for all `a, b in N`

It is not a relation

(iii) `a = b^2 , b = c^2, :. a =((c)^2)^2 = c^4 => a ≠ c^2`

`:.` It is not a relation.
Q 2655067864

Let `A= { 1, 2, 3, 4}, B = { 1, 5, 9, 11, 15, 16}` and

`f= { ( 1, 5), (2, 9),(3, 1 ), (4, 5), (2, 11)}` Are the following true?

(i) `f` is a relation from `A` to `B`

(ii) `f` is a function from `A` to `B`

Justify you!' answer in each case.
Class 11 Exercise 2.mis Q.No. 10
Solution:

(i) `f` is a subset of `A xx B`.

`:. f` is a relation from `A` to `B`.

(ii) The element `2 in A`, has two images `9` and `11`

`:. f` is not a function from `A` to `B`.
Q 2665067865

Let `f` be the subset of `Z xx Z` defined by `f= { (ab, a+b) : a,b in Z}` .

Is `f` a function from `Z` to `Z`?

Justify your answer.
Class 11 Exercise 2.mis Q.No. 11
Solution:

Let `a= 0, b= 1, :. ab =0, a+b =1 , (0, 1 ) in f`

`a= 0 , b=2 , :. ab= 0, a+ b=2, (0,2) in f`

`:.` For the same element there are different
images.

`:. f` is not a function.
Q 2615067869

Let `A= {9, 10, 11,12, 13,}` and let `f : A -> N` be

defined by `f(n) =` the highest prime factor of `n`.

Find the range of `f`.
Class 11 Exercise 2.mis Q.No. 12
Solution:

For `n = 9; 3` is the highest prime factor of it.

`n = 10 ; 5` is the highest prime factor of it.

`n = 11 ; 11` is the highest prime factor of it.

`n = 12 ; 3` is the highest prime factor of it.

`n = 13 ; 13` is the high''St prime factor of it.

`:.` Range of `f= { 3, 5, 11, 13}`
 
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