Mathematics 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.

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?

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}