Mathematics Problem Solving Techniques of Relation And Functions Class-11 Chapter-2

### Problem Solving Techniques (PST)-1 Function or Not

1. For a Relation to be Function there is only one condition. For a particular value of x , there should be only one values of f(x).

2. In Arrow diagram no element of domain should have more than one image.

3. Finding Function whether a graph represents or not.

Q 3155291164

Let A = { 1,2}, B={3,4} does the relation from A to B defined as { (1,3)(1,4)(2,3)} represents a function.

Solution:

No as 1 has two images 3 and 4.
Q 3135391262

Does the graph represents a function ?

Solution:

No as at x=a , f(a) has two values.
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.

### (PST)-2 Finding Domain of a Function

Basics :

Let  underset ("Function")f : underset("Domain")A -> underset (" range")B

Domain never includes oo or -oo

Domain can be written both in Roster form and set-builder form.

Techniques :

1. Qunatities inside square root should be +ve

if fx) = sqrt(x+5) -> x+5 ge 0 => x ge -5

2. Denominator should not be equal to 0

f(x) = 1/(x-5) -> x ne 5

3. f(x) = log_(x) y quad y >0

x >0 but not equal to 1

x in R^+ - {1}

4. Domain of sum or difference :

e.g., f(x) = sqrt(x-2) + sqrt(x-3)

sqrt(x-2) => x> 2

sqrt(x-3) => x > 3

color{green} ✍️ color{green} mathbf("KEY CONCEPT")

Domain of f(x) should satisfy both conditions

:. Domain of f(x) quad x > 3
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 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 3145191063

The domain of definition of f (x) = sqrt((x+3)/(2-x)(x-5)) is

(A)

(-oo,-3) cup (2,5)

(B)

(-oo ,-3)cup (2,5)

(C)

(-oo ,-3) cup [2,5]

(D)

None of these

Solution:

f(x) = sqrt((x+3)/(2-x) (x-5))

for f(x)  to be defined,

(2-x) (x-5) ne 0

=> x ne 2,5...............(1)

Also, (x+3)/((2-x)(x-5)) ge 0

=> ((x+3)(2-x)(x-5))/((2-x)^2 (x-5)^2) ge 0

=> (x+3) (x-2) (x-5) le 0

=> x in (-oo ,-3] cup (2,5) ..........(2)

From (1) and (2)

x in (-oo ,-3] cup (2,5)
Correct Answer is => (A) (-oo,-3) cup (2,5)
Q 3115180969

The domain of definition of the function f ( x) = log | x | is

(A)

R

(B)

(-oo,0)

(C)

(0,oo)

(D)

R-{0}

Solution:

f(x) = log|x|

For f(x) to be defined,

|x| > 0 , which is always true,

But |x| ne 0

=> x ne 0

Thus , dom (f) = R -{0}
Correct Answer is => (D) R-{0}
Q 3175780666

Let R be a relation in N, defined as R {(x, y) in N xx N : x + 2y = 39} . Find the domain and the range of R.

Solution:

Domain of R = {1, 3, 5, -37}, Range of R= { 1, 2, 3, ..... 19}.
Q 3115880769

Find the domain of definition of the function f(x) given by f(x)= 1/(log_10 (1-x)) + sqrt(x+2)

Solution:

we have f(x) = 1/(log_10 (1-x)) + sqrt (x+2)

Let g(x) = 1/(log_10 (1-x))  and h(x) = sqrt (x+2) Then f (x) = g (x) + h (x)

:. Domain (f) = Domain (g) cap Domain (h)

Now, g(x) = 1/(log_10 (1-x)) is defined for all x for which log_10 (1 - x) is defined and

log_10 (1-x) ne 0 => 1-x > 0 and 1-x ne 1 => x < 1 and x ne 0 => x in (-oo , 0) cup (0,1)

Domain (g) = (-oo, 0) cup (0, 1).

And, h(x) = sqrt(x+2). is defined for all  x satisfying.

x+ 2 ge 0 => x ge -2 => x in (- 2, oo).

:. Domain (h) = [- 2,oo).

Hence, Domain (f)= (-oo,0) cup (0,1) cap [-2,oo) = [-2,0)cup(0,1)
Q 3145180963

Find the domain of the function f given by f(x) = 1/(sqrt([x]^2 -[x]-6))

Solution:

We have,

f(x) =1/(sqrt([x]^2 -[x]-6))

Clearly, f(x) is defined for all x satisfying

[x]^2 -[x]-6 > 0

=> ([x]- 3)([x]+ 2) > 0

=> x in (-oo,-2) or x in [4,oo)

=> x in(-oo,- 2) cup {4, oo)

Hence, domain (f)= ( -oo,- 2) cup [4, oo).
Q 3115080869

Find the domain of the real function f (x) defined by f (x) =sqrt ((1-|x|)/(2-|x|))

Solution:

We have, f (x)=sqrt ((1-|x|)/(2-|x|))

We observe that f (x) is defined for all x satisfying (1-|x|)/(2-|x|) ge 0

Now, (1-|x|)/(2-|x|) ge 0

=> (|x| -1)/(|x|-2) ge 0

=> | x| le 1 or, | x| > 2

=> x in [-1 ,1]  or x in (-oo ,-2) cup (2,oo)

=> x in (-oo ,-2) cup (2,oo) cup [-1,1]

Hence, domain (f) = (- oo ,-2) cup (2, oo) cup [-1, 1].

### PST-3 Finding Range of a Function

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 3145634563

f(x) = (3)/( 2 - x^2)

Solution:

 f(x) = (3)/( 2 - x^2) x^2 >= 0 => - x^2 <= 0 => 2 - x^2 <= 2

 => 1/(2 -x^2) >= 1/2 => 3/(2 - x^2) >= 3/2 => f(x) >= 3/2

:.  Range  = [ 3/2 , oo)
Q 3185634567

f(x) = | x - 3 |

Solution:

f(x) = | x - 3 | >= 0 => Range = [0, oo)
Q 3115634569

f(x) = 1 - | x - 2 |

Solution:

y = 1 - | x - 2 |

For every x, | x - 2 | >= 0

=> - | x - 2 | <= 0 => 1- | x - 2 | <= 1 => y <= 1
Q 3115734660

f(x) = (|x-4|)/(x -4)

Solution:

 y = (x-4)/(x-4)  or  (- (x-4))/(x-4) , i.e., 1 or -1
Q 3115891760

Find the range of the following functions given by :

f(x) = sqrt(16-x^2)

Solution:

y= sqrt(16-x^2)

y= f(x) = sqrt(16-x^2)

y^2 = 16 -x^2

x= sqrt(16-y^2)

For x to exist & real

16-y^2 ge 0

(y+4) (y-4) le 0

-4 le y le y

y is function of square root , so [-4,0]  isn't in our solution.

So, y in [0,4]
Q 3155891764

f(x) = 1+3 cos 2x

Solution:

Range of 1+3 cos 2x

We know , -1 le cos 2x le 1

-3 le 3 cos 2x le 3

-2 le 1 + 3 cos 2x le 4

### PST-4 Drawing of Function

Q 3115791660

Draw the graph of f(x) = |x-2| + |x-3|

Solution:

f(x) = {tt (( -2x +5 , x < 2), (1, 2 < x < 3) , (2x-5 , x > 3))
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
Q 3175891766

f(x) = { tt (( 3-x text(,) , x > 1), (1 text(,) , x=1),( 2x text(,) , x <1))

Solution:

### PST-5 calculating functional values

Q 3115191069

If 3f(x) + 5 f (1/x) = 1/x=3 for all non-zero x, then f (x) =

(A)

1/14 (3/x + 5x -6)

(B)

1/14 (-3/x +5x -6)

(C)

1/14 (-3/x + 5x +6)

(D)

None of these

Solution:

3 f(x) + 5 f(1/x) = 1/x -3............(1)

Multiplying (1) by 3 :

15 f(1/x) + 9 f(x) = 3/x -9.............(2)

Replacing x by 1/x in (1)

3 f(x) + 5 f(x) = x-3

multiplying by 5

15 f(1/x) + 25 f(x) = 5x -15..............(3)

Solving (2) and (3)

-16 f(x) = 3/x -5x +6

=> f(x) = 1/16 (-3/x) + 5x -6)
Correct Answer is => (D) None of these
Q 3115234160

If f (x) = (1 + x)/(1 - x)  show that  ( f(x) .f(x^2) )/(1 + [ f (x) ]^2) = 1/2

Hots
Solution:

f (x) = (1 + x)/(1 - x) : f (x)^2 = (1 + x^2)/(1 - x^2)

and 1 + [f (x)]^2 = 1 + (1 + x)^2/(1 -x)^2 = ( (1-x)^2 +(1+x)^2)/(1 -x)^2

 = (2(1 + x^2))/(1 -x)^2

 :. ( f(x).f(x^2 ))/(1 +[f (x)]^2) = (1 + x)/(1 - x) xx (1 + x^2)/(1 - x^2) xx (1-x)^2/(2(1 +x^2)) = 1/2