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

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`
 
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