# Functions

• We will cover some basic concepts on the topic - Functions

EVEN FUNCTIONS

THOSE FUnCTIONS WHOSE GRAPHS ARE SYMMETRICAL ABOUT Y AXIS ARE KNOWN AS EVEN FUNCTIONS
f(-x)=f(x) then f(X ) will be even
like f(x)=cosx
f(-x)=cos(-x)=cosx=f(x)
so cosx is even function

ODD FUNCTIONS

Those functions which are symmetrical about origin are known as odd functions
and f(-x)=-f(x)
like f(x)=sinx
f(-x)=sin(-x)=-sinx=-f(X )
so sinx is an odd function whose graph is symmetrical about origin

if f(-x) is not equal to f(x) or -f(x) then it is known as asymmetrical graph about origin or y axis

If f(x) = x^2 + 6 and g(x) = x^3 – 11, then h(x) = f(x) – g(x) is a/an
1) Even function
2) Odd function
3) Neither even nor odd function
4) Data Insufficient

f(x)=x^2+6
f(-x)= (-x)^2+6=x^2+6=f(x) so even function

g(x)=x^3-11
g(-x)=-x^3-11
-(x^3+11)
so neither even nor odd
so f(x)-g(x) is neither even nor odd

We have basically 4 types of functions in our cat course

fog(x)= ((f(g(x))

gof(x)=g((f(x))

fof(x)=f((f(x))

gog(x)=g((g(x))

f(x)=sinx
g(x)=2^x
find fog(x),gof(x),fof(x),gog(x)

fog(x)=f((g(x)) = f((2^x))
so replace x by 2^x iin f(x)
sin(2^x)=fog(x)

gof(x) = g((f(x)) = g((sinx))
so replace x by sinx in 2^x
so 2^sinx

fof((x)) = f((f(x)) = sin(sinx))

gog((x) = g((g(x) = 2^(2^x))

f(x) is even and g(x) is even then fog is even
f(x) is odd g(x) is odd then f0g is odd

f(x) is even g(x) is odd then fog is even
f(x) is odd , g(X ) is even then (fog(x)= is even function
so only when both are odd then fog will be odd , if f(x) or g(x) is even then fog will be even

which of the following function's graph is symmetrical abt origin

a) constant
b) sinx+cosx
c) sin((log(x+root(x^2+1))
d) 1+x+x^3
e) more than 1

a) y=0 is symmetrical abt origin but not always possible. y=2 is not possible so rejected

b) sinx+cosx
here sinx is odd

cosx is even
odd + even can't be odd function

c) sin((log(x+root(x^2+1)
here f(x)=sinx
and g(x)=log(x+root(x^2+1)
f(x) and g(x) both odd then fog(x) is odd

d) 1+x+x^3 is neither even nor odd

so a and c are symmterical abt origin

If f(x) = |x – 2| + |x – 3| + |x – 4| and g(x) = f(x + 1), then
1) g(x) is an even function
2) g(x) is an odd function
3) g(x) is neither an odd nor an even function
4) None of these

f(x + 1) = ∣x − 1∣ + ∣x − 2∣ + ∣x − 3∣
Let x = 1, then g(1) = 0 + 1 + 2 = 3
Let x = −1, then g(−1) = 2 + 3 + 4 = 9

so g(x) is not equal to -g(x) or g(-x)
g(x) is neither an even nor an odd function.

Hence, option 3

which of the following functions are odd

a) sin(cosx)
b) sin(2^x)
c) 2^sinx
d) sin(sinx)
e) sin(e^x)
f) e^(sinx)

f(x+y)=f(x)+f(y) for all real x
then f(x) is
a) odd function
b) even
c)neither even nor odd
d) none

If f(x)+f(y)=f(x+y), then f(x) =x*f(1)
now f(x)=x*f(1)
so f(-x)=-x*f(1)=-f(x)
so odd function

f(x) * f(1/x) = f(x) + f(1/x) for all real x, if f(3)=-26 find f(4)

IF f(x)*f(1/x)=f(x)+f(1/x)
then f(x)=1+-x^n

now f(3)=-26
so 1+-(x)^n=-26
so +-(x)^n=-27
so +-(3)^n=-27 (( negative case will be considered bcz in positive case 3^n=-27 not possible ))
so -3^n=-27
so n=3
and f(x)=1-x^n
so f(4)=1-4^3
=-63

f ( x + y ) = f(x) * f(y) then find f(3) if f(1) = 4

if f(x+y)=f(x)*f(y) then f(x)= ((f(1))^x always
Here f(1)=4 = > f(x)=4^x so f(3)=4^3=64

DOMAIN OF y=f(x) means value of x that will give real values of f(x) or y

Find domain of following
a) x^2+2x+3
b) 1/(x^2+3x+2))
c) root(x^2+5x+6)
d) 1/root(x^2+5x+6))
e) root(x^2-5x+6)+root(2x+8-x^2)
f) log(base x-1)) (x^2+2x+1))

a) domain of any polynomial always all real numbers so x^2+2x+3 's domain=R

b) domain of 1/(X ) is all real number except zero
so x^2+3x+2 should not be equal to zero
so x should not be equal to -1,-2
so R~{-1,-2}

c)) domain of root (X ) is X > =0
so root(x^2+5x+6)
x^2+5x+6 > =0
so (x+2)(x+3) > =0
so x > =-2 or x=-3

d) domain of 1/(rootX )
X > 0
here x^2+5x+6 > 0
so x > -2 or x < -3

e) ROOT(a) is defined only when a > =0
so root(x^2-5x+6) is defined when
x^2-5x+6 > =0
(x-2)(x-3) > =0
(-inf,2]U[3,inf)
now root(2x+8-x^2) is defined when 2x+8-x^2 > =0
means x^2-2x+8
(x-4)(x+2)
so [-2,4] now take intersection of both

f) log(basex-1)(x^2+2x+1) (( log(basea)b is defined when a > 0 , b > 0 and a is not equal to 1
x-1 > 0
x^2+2x+1 > 0 and x-1 = /1
so x=/2
x > 1
(x+1)^2 > 0
which is not true only for x=-1
so domain x > 1 and x=/2

find domain of root(2-x) + root(1+x)

2-x > =0
so x < =2

1+x > =0
so x > =-1
so draw real line and take intersection
-1 < = x < =2 is domain of the function

domain of root(X ) is X > =0

RANGE - MIN AND MAX VALUE OF f(x) or y for which x is real

Find range of the function x^2+x+3/(x^2+x+1))

method1-

y=x^2+x+3/(x^2+x+1))
yx^2+yx+y=x^2+x+3
x^2(y-1)+x(y-1)+y-3=0
x real so D > =0
(y-1)^2-4*(y-1)(y-2) > =0
y-1)(y-1-4y+12) > =0
(y-1)(-3y+11) > =0
(y-1)(3y-11)
so (1,11/3]

Method-2: y=1+((2/(x^2+x+1))
now x^2+x+1=x^2+x+1/4-1/4+1
(x+1/2)^2+3/4
so min value is 3/4 and maax infinity
so y=1+2/inf
y=1
and y=1+2/(3/4))
y=1+8/3=11/3

How many integral values from 1 to 15 can the expression (x^2 +34x -71)/(x^2+2x-7) not take?

a) 2
b) 3
c) 4
d) none

(x^2 +34x -71)/(x^2+2x-7)=y
(1-y)x^2+ (34-2y)x -71+7y=0

b^2-4ac > =0 for real values therefore
(34-2y)^2-4*(1-y)(-71+7y) > =0
now on simplify
y^2- 14y +45 > =0
that is y < =5 and y > =9 so y can't take values
6,7,8 so three integral values

A function V(a, b) is defined for positive integers a, b and satisfies V(a, a) = a, V(a, b) = V(b, a), V(a, a+b) = (1 + a/b) V(a, b). The value represented by V(66, 14) is ?
(a) 364
(b) 231
(c) 455
(d) 472
(e) none   ( CAT 2009 )

here we know value of V(a,a+b)
so 1st term should be less than 2nd term, but in V(66,14) 66 > 14
so change this we know that V(a,b)=V(b,a)
so V(66,14)=V(14,66)
now proceed
V(66,14 ) = V(14,66= (33/26)*V(14,52)
= (33/26) * (26/19)* V(14, 38 )
= (33/19) *(19/12)* V(14, 24)
= (33/12) *(12/5)* V(14,10)
= (33/5)* (7/2)*V(10, 4)
=(231/10)* (5/3) *V(4, 6)
= (77/2)* 3 *V(4, 2)
= (231/2)*2* V(2, 2)
= 231*2
= 462

If f(x − 1) + f(x + 1) = f(x) and f(2) = 6, f(0) = 1, then what is the value of f(50) ?
1) −7
2) 6
3) 1
4) 7

Since we know both f(0) and f(2), we can find f(1).
f(1) = f(0) + f(2) = 7
f(2) = f(1) + f(3)

f(3) = f(2) – f(1)
= 6 – 7
= −1
Also, f(3) = f(2) + f(4)
f(4) = f(3) – f(2) = −7
Continuing in a similar way, we can find out
f(0) = 1
f(1) = 7
f(2) = 6
f(3) = −1
f(4) = −7
f(5) = −6
f(6) = 1
f(7) = 7 and so on
After every 6 integral values of x, f(x) repeats itself.
f(6) = f(12) = f(18 ) = f(24) = f(30) = f(36) = f(42) = f(48 ) = 1
f(49) = 7
f(50) = 6

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