FUNCTION
DEFINITION:
A function is s relation in which no two different ordered pairs have the same first component.
Definition:
Let A and B be two non-empty sets and f is a relation which associates each element of set A with unique element of set B. This relation is called function and it is denoted by f: A -> B.
Essential requirements for the definition of a function :
A function f : A -> B is defined under the following conditions:
1) Every x ∈ A is associated with some y in B, i.e., a function is defined only when the domain is entirely "used up".
The set B may not be entirely "used up" by the function.
2) The function may associate more than one x to the same y.
3) No element in A should have more than one image in B.
Domain:
The domain of y= f(x) is a set of all real x for which f(x) is defined (real)
Range:
Range of y= f(x) is collection of all outputs {f(x)} corresponding to each real number in the domain.
* FOR FUNCTION f: A -> B;
• Set A is called the domain of the function f and B is called the co-domain of f.
• If a ∈ A, then the elements in B which is assigned to ' a ' is called the image of 'a' and denoted by f(a)
EXAMPLE:
Let A={ a, b, c, d}
B={1, 2,3,4,5}
Relation of A to B is
f={(a,2),(b,3),(c,5),(d,1)}
Then Domain of f= {a,b,c,d}
Co-domain of f= {1,2,3,4,5}
Range of f= {2,3,5,1}
NUMBER OF FUNCTION:
• Number of function from
A to B= nᵐ [ m is the number of elements in setA and n is the number of elements in set B]
TYPES OF FUNCTIONS :
Type Mathematical Form
• Even Function. f(- x)= f(x)
• Odd Function. f(- x)= - f(x)
• Modulus Function (Absolute value Function)
• y= |x|= x, x≥0
-x, x<0
• |x±y| ≤ |x|+ |y|
• |x±y| ≥ | |x| - |y| |
• Greatest Integer function ( Floor function)
[x]=n, n≤x<n+1, n ∈I. [x] indicates
the Integral part of x which is
nearest and smallest integer
to x.
• [x]= x; if x ∈ I
• [x+I] = [x] +I; I is a integer
• [x]+[- x]= - 1, if x ∉ I
0, if x ∈ I
• [x] - [- x] = 2[x] + 1, if x ∉ I
2[x], if x ∈ I
• [x]≥ I => x ≥I, where I is an
integer.
• [x] ≤ I => x < I + 1, where I
is an integer.
• If [x] > n => x≥ n +1; n ∈ I
• If [x] < n => x< n; n ∈ I
• [x] = - [x], if x ∈ I
• [- x] = - [x] - 1, if x ∉ I
•Smallest Integer Function (Ceiling function)
f[x] = [x] ∀ x ∈ I
• [n]= - [n]; where n ∈ I
• [- x] = - [x] +1; where x ∈ R - I
• [x+n] = [x] +n ; where x ∈ R - I
and n ∈ I
• [x]+[- x]= 1, if x ∉ I
0, if x ∈ I
• [x]+[- x]= 2 [x] - 1, if x ∈ I
2 [x], if x ∈ I
• One-one Function (Injective Function).
A Function f: A -> B is said to be one-one Function if different elements of set A have different
images in set B.
• Number of one-one functions
= ⁿ⁽B⁾ P ₙ₍A₎ , if n(B) ≥ n(A) and 0,
if n(B) < n(A).
• Onto Function (Surjective Function)
• A function f: A -> B is said to be onto Function if every every element of B is the f - image of some element of A.
• Range of f= co-domain of f
• Number of onto function from
A to B.
= ⁿᵣ ₌ ₀ ∑ (-1)ⁿ ⁻ ʳ ⁿCᵣ r ᵐ, where
A and B are two sets having m
and n elements respectively.
• n(A)= n(B), then number of
onto function (n(A)) !
• Many - One Function
• A function f: A -> B is said to be Many-One function if two or more
elements in A have the same image in B.
• Number of elements in many one from A to B =
(n(B)ⁿ⁽A⁾ - ⁿ⁽B⁾ P ₙ₍A ₎.
• Into Function
• A function f: A ->B is an into function if there exist an element in B having no pre-image in A.
• One-One Onto Function (Bijective Function).
• A function f : A - > B is a
bijection if it is one-one as
well as onto.
• Number of bijections from
A to B = (n(A)) !
• Inverse Function.
• If a function f: A - > B then
f ⁻¹ : B -> A such that
f ⁻¹(b)= a => f(a)= b.
• f is invertible if it is one-one &
onto.
• If f(x) = 4x +3 then
f ⁻¹(x)= (x -3)/4
• Equal Functions.
• Two functions f and g are to
be equal if they have the
same domain, range & they
satisfy the condition
f(x)= g(x) ∀ x ∈ domain.
• Composite Function
• Let A, B, C be three non - void sets. f : A -> B, g : B -> C be two Functions, then the Function gof : A -> C defined by
gof(x) = g(f(x)) ∀ x ∈ A called composite function of f and g.
• f : A -> B, g : B -> C & h: C->D, then the composite functions
gof: A -> C & then h: C -> D, then ho(gof) : A -> D is composite function of three functions.
• The composition of functions is not Commutative i.e., fog ≠ gof.
• The composition of functions is associative i.e., f, g, h are three functions such that (fog)oh and fo(goh) exist, then
(fog)oh = fo(goh)
• The composition of two bijections is a bijection.
• The composition of any Function with the identity function is the function itself.
Section I
ILLUSTRATIVE EXAMPLES :
Example . 1:
If f(x) = 3x -2; find (i) f(-1) , (ii) f(2), (iii) f(1/2), (iv) f(x +2), (v) (x²+1)
Solution:
i) We have f(x)= 3x - 2
Replacing x by -1 we get
f(-1)= 3(-1) - 2
= - 3 - 2
= -5
ii) We have f(x)= 3x - 2
Replacing x by 2 we get
f(2)= 3(2) - 2
= 6 - 2
= 4
iii) We have f(x)= 3x - 2
Replacing x by 1/2, we get
f(1/2)= 3(1/2) -2
= 3/2 - 2
= - 1/3
iv) We have f(x)= 3x - 2
Replacing x by (x+2), we get
f(x+2)= 3(x+2) - 2
= 3x + 6 - 2
= 3x +4
v) We have f(x)= 3x - 2
Replacing x by (x² +1), we get
f(x²+1)= 3(x²+1) - 2
= 3x² +3 - 2
= 3x² +1
Example . 2:
If f(x)= (1-x)/(1+x) find f(1/x)
(2009)
Solution:
We have f(x)= (1-x)/(1+x)
Replacing x by 1/x, we get
f(1/x)= (1 - 1/x)/(1+ 1/x)
= {(x -1)/x}/{(x+1)/x
= (x -1)/(x+1)
Example . 3:
If f(x)= (1-x)/(1+x) find f{f(1/x)}
(2013)
Solution:
We have f(x)= (1-x)/(1+x) .........(1)
Replacing x by (1/x), we get
f(1/x)= (1 - 1/x)/(1+ 1/x)
= {(x -1)/x}/{(x+1)/x
= (x -1)/(x+1)
Again replacing x by {(x-1)/(x+1)} in (1) we get,
f{(x -1)/(x+1)}= [1 - (x -1)/(x+1)]/[1+(x -1)/(x+1)]
= (x+1-x+1)/(x+1+x-1)
= 1/x
Example . 4:
If f(x)= (ax+b)/(bx+a) prove f(x). f(1/x) = 1
(1990,2009,2010,2012, 2014)
Solution:
We have f(x)= (ax+b)/(bx+a)
Replacing x by 1/x, we get,
f(1/x)= {a(1/x) +b}/{b(1/x) +a}
= (a + bx)/(b+ ax)
So f(x). f(1/x)= (ax+b)/(bx+a) .
(a + bx)/(b+ ax)
= 1. (Proved)
Example .5 :
If f(x)=x²-x then prove f(h+1)= f(-h) (1997,2007,2009,2010,2013, 2015
Solution:
We have f(x)=x²-x
L.H.S
Replacing x by (h+1) , we get
f(h+1)= (h+1)² - ( h +1)
= h² +1 +2h - h - 1
= h² +h
R. H. S
Replacing x by - h , we get
f(-h) = (-h)² - (-h)
= h² + h
So L. H. S = R. H. S. (Proved)
Example . 6:
If f(x)= e ᵃˣ ⁺ᵇ prove that
eᵇ f(x+y)= f(x) . f(y) (1988)
Solution:
We have f(x)= e ᵃˣ ⁺ᵇ
L. H. S
Replacing x by (x+y), we get
f(x+y)= e ᵃ⁽ˣ⁺ʸ⁾ ⁺ ᵇ
= e ᵃˣ⁺ᵃʸ⁺ ᵇ
and eᵇ f(x+y)= eᵇ(e ᵃˣ⁺ᵃʸ⁺ ᵇ)
= e ᵃˣ⁺ ᵃʸ ⁺ ²ᵇ
R. H. S
Replace x by y, we get
f(y)= e ᵃʸ⁺ ᵇ
And f(x) . f(y) = e ᵃˣ ⁺ᵇ. e ᵃʸ⁺ ᵇ
= e ᵃˣ⁺ ᵃʸ ⁺ ²ᵇ
L. H. S = R. H. S. (Proved)
Example. 7 :
If f(x)= (2x+1)/(2x²+1) and θ(x)= 2f(2x) then find θ(2.5) (1995)
Solution:
We have f(x)= (2x+1)/(2x²+1)
Replacing x by 2x, we get
f(x)= {2(2x) +1}/{2(2x)² +1}
= (4x +1)/(8x² +1)
Therefore, θ(x)= 2f(2x) ={2(4x +1)/(8x² +1).
Now replacing x by 2.5, we get
= [2{4(2.5) +1}]/[8(2.5)²+1}]
= 2(11)/{8(6.25)+1}
= 22/51
Example . 8:
If y= f(x)= (ax+b)/(cx -a) then prove f(y) = x (2005,2008,2012)
Solution:
We have y= (ax+b)/(cx -a) ....(1)
and f(x)= (ax+b)/(cx -a) ....... (2)
Replacing x by y in (2) we get,
f(y)= (ay+b)/(cy - a)
= a.{(ax+b)/ (cx - a) + b}/{c. (ax + b)/(cx - a) - a}. (Putting the value of y from. (1)
=(a²x+ab+bcx-ab)/(acx+bc-acx+a²)
= (a²x + bcx)/(bc +a²)
= x(a² + bc)/(a² +bc)
= x. (Proved)
Example. 9)
If f(x+3)= 3x²-2x +5 then find f(x-1)
(1996, 2012)
Solution:
We have f(x+3)= 3x²-2x +5 ....(1)
Replacing x by (x -4) we get,
f(x - 4+3)= 3(x-4)²- 2(x-4)+5
f(x-1)= 3(x²+16-8x) -2x+8 +5
= 3x²+48 -24x - 2x +13
f(x -1)= 3x²- 26x +61
Alternative method:
We have f(x+3)= 3x²-2x +5 ....(1)
Let (x+3)= y
Then x= y - 3
So, f(y)= 3(y-3)² -2(y-3) +5
f(y)= 3(y²+9-6y) -2y+6+5
f(y)= 3y²+27-18y - 2y +11
f(y) = 3y² - 20y +38
Now replacing x by (x-1), we get
f(x-1)= 3(x-1)²- 20(x-1) +38
= 3(x²+1-2x) - 20x +20 +38
= 3x²+3 - 6x -20x+58
= 3x² - 26x +62
Example . 10)
If f: x -> log₃(x² +2x +3), find f(4).
Solution :
We have f: x -> log₃(x² +2x +3)
Replacing x by 4, we have
f(4)= log₃(4² +2.4 +3)
= log₃27
= 3 log₃3
= 3 . 1
= 3
Example 11:
If y = f(x)= (2x +1)/(3x - m) and f(y) = x, find m.
Solution:
We have
f(y) =x Or, (2y+1)/(3y - m)= x
Or, 2y+1= 3xy - mx
Or, 2y +mx= 3xy -1
Or, y(2+m)= my +1
So, y= (my+1)/(3x -2). .....(1)
Again, it is given that
y= (2x +1)/(3x - m). ..........(2)
Thus, from (1) and (2) we get
(mx +1)/(3x -2)= (2x +1)/(3x - m)
Comparing both sides we get, m=2
Example . 11:
If f(x) = | x | - 3x , find f(-2) (2007)
Solution:
We have f(x) = | x | - 3x
Replacing x by (-2) we get,
f(-2)= | 2 | - 3(-2)
= 2 +6
= 8
Example 12 :. (2008/2017)
If f(x)= | x | - [x], where [x] is the greatest integer not exceeding x, then find the values
i) f(2.5)
ii) f(-2.5).
Solution)
We have, [x]=2 when 2≤ x < 3 and [x]= - 3 when - 3≤ x < - 2
And | x | = x when x ≥ 0
- x when x < 0
Now we have f(x)= | x | - [x]
i) Replacing x by 2.5 we get,
f(2.5)= | 2.5 | - [2.5]
= 2.5 - 2
= 0.5
ii) Replacing x by - 2.5 we get,
f(- 2.5) = | - 2.5 | - [- 2 5]
= 2.5 - (- 3)
= 2.5 +3
= 5.5
Example 13) (2005/2019)
If f(x)= (eˣ -1)/(eˣ+1) and θ(x)= {1+ f(x)}/{1 -f(x)} then prove that θ(x +y)= θ(x) . θ(y)
Solution)
We have,
f(x)= (eˣ -1)/(eˣ+1). ………….(1)
And
θ(x)= {1+ f(x)}/{1 - f(x)}
= {1+ (eˣ -1)/(eˣ+1)}/{1-(eˣ -1)/(eˣ+1). {putting (1)}
= {(eˣ +1+eˣ -1)/(eˣ +1)}/{eˣ+1-eˣ+1)/(eˣ+1)}
= {2eˣ/(eˣ +1)}/{2/eˣ+1)}
= eˣ
so we get θ(x) = eˣ …………(2)
Again θ(x+y)= eˣ⁺ʸ L.H.S
and θ(x) . θ(y)= eˣ . eʸ
= eˣ⁺ʸ R.H.S (proved)
Example 14) (2019)
If f(x)=logₑ{(1+x)/(1-x)} Then prive that f{2x/(1+x²)} = 2 f(x)
Solution)
We have
f(x)=logₑ{(1+x)/(1-x)
L.H.S
f{2x/(1+x²)}= logₑ[{1+ 2x/(1+x²)}/{1 - 2x/(1+x²)}]
= logₑ [{(1+x²+2x)/(1+x²)}/{(1+x²-2x)/(1-x²)}]
= logₑ[(1+x)²/(1-x)²]
= logₑ[(1+x)/(1-x)]²
= 2 logₑ[(1+x)/ (1-x)]
= 2 f(x)
= R.H.S. (Proved)
Example 15) (2018)
If f(x)= {(2x+1)/(2x²+1)} and
g(x)= 2 f(2x) then find g(2.5)
Solution)
We have
f(x)=(2x+1)/(2x²+1)}
f(2x)= (2.2x +1)/(2.(2x)²+1)
= (4x+1)/(8x²+1)
Again g(x)= 2 f(2x)
So, =2{(4x+1)/(8x²+1)} (Ans)
Again g(2.5)= 2{(4(2.5)+1)}/{(8(2.5)²+1)}
=2(10+1)/(50 +1)
= 22/51 (Ans)
EXERCISE 1.1
1) If f(x)= (x²-4)/(x - 2) find
i) f(1)
ii) f(2)
2) If f(x)= 4x² + 2x - 3 find
{f(x+h) - f(x)}/h
3) If f(x)= log₂(x²+3x+4) find
i) f(1)
ii) f(4)
4) 13) If y= f(x) = (x+1)/(x+2) then find a) f(y)
b) f{f(1/x)} (1992,2009)
5) If f(x)= (1-x)/(1+x) find the value of {f(x+h) -f(x)}/h (1998)
6) If f(x)= (x-1)/(x+1) then prove {f(a).f(b)}/{1+f(a)f(b)}=(a-b)/1+ab)
(1993, 1994, 2013)
7) If f(x)=(ax -b)/(bx-a) then prove f(a). f(1/a) - f(b). f(1/b) = 0
(2007)
8)a) If f(x)= (2x² - 3x +4), for what value of x is 2f(x)= f(2x). (2016)
b) If f(x)= 2x² - 5x +4, for what value of x is 2f(x)= f(2x) ? (2005)
c) If f(x)= log x, prove
f(a)+ f(b) = f(ab). (2011)
9) If y = f(x)= (3x +4)/(5x - m) and f(y) = x, find m.
11) f(x)= e ᵖˣ⁺ᵐ , Show that
f(a). f(b). f(c)= f(a+b+c)²ᵐ
12) y= f(x)= (x+1)/(x+2)
Find i) f(y). ii) f{f(1/x)}. (1992)
13) If y = f(x)= (2 - x)/(5+3x) and
z= f(y), express z in terms of x.
14) If f(x +2)= 5x² -3x +2, find the value of f(x -2). (2015)
15) If f(2x -1)= (3x -1)/(x+1) find
i) f{ f(4)}
ii) f(1 - 3x)
16) If f(x)= (3x +1)/(x -1),
find f(2-x)
17) Given f(x)= x+2 and g(x)= (x²-4)/(x-2), when f(x)= g(x) ? Give reasons for your answer. (2016)
18) f(x)= (x²-5x+6)/(x²-8x+12) Show f(2) is not defined and also find f(-5). (1988)
Answer)
1) i) 3, ii) unidentified
2) 2(4x +2h +1)
3) i) 3 ii) 5
4) i) (y+1)/(y+2).ii) (3x+2)/(4x +3)
5) - 2
7) (7x +8)/(12x +31)
8) a) ± 1. b) ±1
9) 3
12) i) (y+1)/(y+2) ii) (2x+3)/(3x+5)
13) (8+7x)/(31+12x)
14) 5x² - 43x +94
15) i) 51/49 ii) (4-9x)/(4-3x)
16) (7-3x)/(1-x)
17) f(x)= g(x), when x≠ 2
18) 8/11
Section II
Domain
----------
Tips for finding the domain of a function :
1. Algebraic Functions :
(i) Denominator should be non-zero.
(ii) Expression under the even root should be non-negative.
2. Logarithmic function:
Log ₓy is defined when x > 0,
y > 0 and y ≠ 1.
3. Exponential Function:
a ˣ and e ˣ are defined for all real values of x, where a > 0
and e > 0.
ILLUSTRATIVE EXAMPLES :
Example. 1 :
Find the domain of the defination
(x²-5x+6)/(x²-8x+12). (2002)
Solution:
Let f(x)= (x²-5x+6)/(x²-8x+12)
Clearly, f(x) will be undefined if x²-8x+12=0
i.e., If x² - 6x - 2x +12=0
i.e., If x(x - 6) - 2(x - 6)=0
i.e., if (x - 6)(x - 2)=0
i.e., if x=6 or,.x= 2
Therefore, the domain of defination of f(x) is - ∞ < x< ∞ but x≠ 6 and x≠2
Example .2)
Find the domain of defination
√(x² +x -12). (2001, 2013)
Solution:
Let f(x)= √(x² +x -12)
Now, (x² +x -12)
= x² +4x - 3x - 12
= x(x +4) - 3(x + 4)
= (x+4)(x - 3)
So f(x)= √{(x+4)(x - 3)}
Clearly, (x+4)(x+3) ≥0 when x≤-4 or, x≥3
And clearly
(x+4)(x+3) < 0 when - 4 < x < 3
Hence, it follows that f(x) is real when x≤4 or, x ≥3 and it is imaginary when - 4 < x < 3 .
Therefore, the domain of defination of f(x) is :
x ≤ 4 , x ≥3 or - ∞ < x ≤ -4, 3≤x<∞
Example .3:
Find the domain of defination
(x+2)/√(x² - 9x)
Solution
Let f(x) = (x+2)/√(x² - 9x)
= (x+2)/√{x(x - 9)}
Clearly, the function f(x) will be defined when x(x - 9)>0
i.e., when x< 0 or, x> 9. Therefore, the domain of defination of f(x) is:
x< 0, x > 9 or - ∞<x<0, 9< x< ∞
Example - 4:
Find the domain of defination
Log (x² - 7x +12). (2000,2008)
Let f(x) = Log (x² - 7x +12)
Clearly, f(x) is defined when
(x² - 7x +12) > 0 or, (x-3)(x-4)>0
It is readily seen that the condition is satisfied when x < 3 or, x > 4.
Therefore, the required domain of defination of f (x) is,
- ∞ <x <3 or, 4 < x< ∞
Exampe 5) (2018)
Find tne domain of the definition of (x²+x+5)/(x²-6x+8)
Solution)
Clearly x² - 6x +8=
Or, x²- 4x -2x +8=0
Or, x(x-4) - 2(x-4)=0
Or, (x-4)(x-2)=0
either x=4 , x=2
So Domain of the function -α≺x ≺α
excluded x=2 and x= 4 (Ans)
EXCERCISE 1.2
Find the domain of defination of the following function :
1) (2x²-7x +6)/(3x² -7x+2). (2015)
2) (x -1)/(x² - 5x +6)
3) √(x² - 7x +12)
4) √(x² -1)
5) x/√(x - 1). (2016)
6) (x+2)/√(x²- x -2). (1994)
7) (4x -5)/√(x² - 7x +12). (1998/19)
8) 5/√{(x+1)(x -3)}. (2004)
9) √(x -3) + √(7 - x)
10) (3x - 7)/√(4 - x²) (2011)
11) x/(x² -9) (2017)
Answers)
1) ( - ∞, 1/3)∪(1/3, 2)∪ (2,∞)
2) R - {2, 3}
3) x ≤3 and x ≥ 4
4) x ≤ - 1 and x ≥ 1
5) x > 1 or 1 < x < ∞
6) - ∞ ≤x < -1, 2 ≤ x < ∞
7) - ∞ < x < 3, 4 < x < ∞
8) - ∞ < x < -1, 3 < x ∞
9) 3 ≤ x ≤ 7
10) -2 < x < 2
or - ∞ <x< -2,2< x <∞
Section III
Range
--------
Method for finding the range of a function y = f(x).
Step 1. Find the domain of the
function y = f(x).
Step 2. Solve the equation y = f(x)
to find x in terms of y. Find
the real values of y for
which x is real.
The set of values of y so obtained makes up the range of f.
* NOTE
To find the real values we have to use the formula b² - 4ac = 0 where a, b, c are the constant from quadratic equation ax²+ bx +c = 0
ILLUSTRATIVE EXAMPLES :
Example 1)
Find the Range of the function
x/(1+x²). (2006)
Solution :
Let y = x/(1+x²)
Or, x²y + y = x. ..........(1)
Or, x²y - x + y =0
Or, x = {1±√(1 - 4y²)}/2y
Since x is finite and real, we have,
y≠ 0 and 1 - 4y² ≥ 0
Or, (1 -2y)(1+2y) ≥0
Or, - 1/2 ≤ y ≤ 1/2
Also, from (1) we see that y= 0 when x=0.
Therefore, the required range of the given function is
-1/2 ≤ y ≤ 1/2.
Example . 2)
Find the range of √(3x² - 4x +5)
Solution:
Let y =√(3x² - 4x +5)
=> y² = (3x² - 4x +5)
=> 3x² - 4x + (5 - y²) =0
=> For x to be real, b² - 4ac = 0
Do (-4)² - 4.3.(5-y²) ≥0
=> y ≥ √(11/3)
So Range of y = [√(11/3 , ∞}
EXERCISE 1.3
Find the Range of the following:
1) 2x/(4 +x²). (2003)
2) √(x - x²)
3) x²/(1 + x²)
4) (3x² - 2x +12)/(x² +2x +4)
5) (x² -4)/(x - 2)
Answer)
1) - 1/2 ≤ y < 0 and 0 < y ≤ 1/2
2) - 1/2 ≤ y ≤ 1/2
3) [0, 1)
4) [ 7/3, 5]
5) - ∞ < y < 4
Section IV
Odd and Even Function
--------------------------------
A function f(x) is said to be an odd function if f(- x)= - f(x). Then, It is odd function.
A function f(x) is said to be an even function if f(- x)= f(x). Then it is even Function.
Example 1). (2001)
Prove that log[√(1+x²) - x} is an odd funtion of.x.
Solution)
Let f(x)= log{√(1+x²) - x}
f(-x)= log[√{1+(-x)²} - (-x)]
= log{√(1+x²) +x}
Rationalising
=log [{√(1+x²) +x}{√(1+x²) -x}/√(1+x²) -x}]
= log[1/{√(1+x²) -x}]
= log 1 - log {√(1+x²) -x}
= 0 - f(x)
= - f(x)
Hence, f(x) is an odd function of x
Example 2) show that 3x² + 5x⁴ -3 is a even function.
Solution
Let f(x)= 3x² + 5x⁴ -3
then, f(-x)= 3(-x)² + 5(-x)⁴ -3
= 3x² + 5x⁴ -3
= f(x)
Hence, f(x) is an even function of x.
Example 3) (2018)
Show that the function
x{(3ˣ-1)/(3ˣ+1)} is even
Solution)
Let f(x)= x{(3ˣ-1)/(3ˣ+1)}
f(-x)= -x{(3⁻ˣ-1)/(3⁻ˣ+1)}
= -x{(1/3ˣ -1)/(1/3ˣ+1)}
= -x{(1 - 3ˣ)/(1+ 3ˣ)
= -x{-(3ˣ-1)/(3ˣ+1)}
= x{(3ˣ-1)/(3ˣ+1)
= f(x)
Hence, f(x) is an even function of x.
Exercise
1) Show that x +x³ is an odd function.
2) If f(x)= {(1+eˣ)/(1 - eˣ)} , then show that f(x) is an odd function.
(2006)
3) If f(x)= {(eˣ+1)/( eˣ- 1)} , then show that f(x) is an even function.
4) If 5ˣ - 5 ⁻ˣ , then show it is even function.
5) log{(1 -x)/(1+x)} show it is even function
6) log{√(1+x²) +x} show that it is even function.
No comments:
Post a Comment