Functions


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 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 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)

    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) 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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