Functions

hemant_malhotra
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  99.70% / XAT 2017  QA : 99.975
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 functionODD 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 originif 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 Insufficientf(x)=x^2+6
f(x)= (x)^2+6=x^2+6=f(x) so even function
g(x)=x^311
g(x)=x^311
(x^3+11)
so neither even nor odd
so f(x)g(x) is neither even nor oddWe 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^sinxfof((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 evenwhich 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 1a) 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 oddso 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 thesef(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)Answer is D
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) noneIf 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 functionf(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)=1x^n
so f(4)=14^3
=63f ( 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=64DOMAIN 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^25x+6)+root(2x+8x^2)
f) log(base x1)) (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=3d) domain of 1/(rootX )
X > 0
here x^2+5x+6 > 0
so x > 2 or x < 3e) ROOT(a) is defined only when a > =0
so root(x^25x+6) is defined when
x^25x+6 > =0
(x2)(x3) > =0
(inf,2]U[3,inf)
now root(2x+8x^2) is defined when 2x+8x^2 > =0
means x^22x+8
(x4)(x+2)
so [2,4] now take intersection of bothf) log(basex1)(x^2+2x+1) (( log(basea)b is defined when a > 0 , b > 0 and a is not equal to 1
x1 > 0
x^2+2x+1 > 0 and x1 = /1
so x=/2
x > 1
(x+1)^2 > 0
which is not true only for x=1
so domain x > 1 and x=/2find domain of root(2x) + root(1+x)
2x > =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 > =0RANGE  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(y1)+x(y1)+y3=0
x real so D > =0
(y1)^24*(y1)(y2) > =0
y1)(y14y+12) > =0
(y1)(3y+11) > =0
(y1)(3y11)
so (1,11/3]Method2: y=1+((2/(x^2+x+1))
now x^2+x+1=x^2+x+1/41/4+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=11/3How many integral values from 1 to 15 can the expression (x^2 +34x 71)/(x^2+2x7) not take?
a) 2
b) 3
c) 4
d) none(x^2 +34x 71)/(x^2+2x7)=y
(1y)x^2+ (342y)x 71+7y=0
b^24ac > =0 for real values therefore
(342y)^24*(1y)(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 valuesA 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
= 462If 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) 7Since 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