Functions - Vikas Saini



  • Q. If f(n) = 14 + 24 + 34 ... n4, then  14 + 34+54 ... (2n – 1)4 be expressed as ?

    (a) f(2n-1) – 16 x f(n)                     (b) f(2n-1) – 8 x f(n)

    (c) f(2n) – 16 x f(n)                         (d) f(2n) – 8 x f(n)

    Solution :-

    Method 1

    Let’s  take n = 2.

    f(n) = f(2) = 17.

    F(2n-1) = f(3) = 98.

    f(2n) = f(4) = 354.

    Value of 14+34 (n=2) = 82.

    (a) negative value

    (b) negative value

    (c)82.

    (d)218.

    Hence option C.

    Method 2

    f(n) = 14+24+34………………….n4.

    f(2n) = 14+24+34………………….(2n)4.

     14+24+34………………….(2n-1)4

    = [14+24+34………………….(2n)4 ] – [(24+44+64………………(2n)4]

    = f(2n) – 16 [14+24+34…………….n4]

    = f(2n) – 16 f(n)

    Hence option (c)

    Q. ‘ f ’ is a real function such that f(x+y) = f(xy) for all real values of x &y. If f(-7) = 7,then the value f(-40)+f(40) .

    Solution :-

    f(0-7) = f(0 x -7)

    f(-7) = f(0) = 7.

    f(0-40) = f(0 x -40)

    f(-40) = f(0) = 7.

    f(0 + 40) = f(0 x 40)

    f(40) = f(0) = 7.

    F(-40)+f(40) = 7 + 7 = 14.

    Q. If f(x) = (sec x + cosec x) (tan x – cot x) and 45 degree < x  < 90 degree,

    Then f(x) lies in which range ?

    Solution :-

    At x = 45 degree

    f(x) = 0.

    At x = 90 degree

    f(x) = infinitive

    Range (0,infinitive)

    Q. Let f(x) = ax2+bx+c, where a,b and c are real numbers. If f(x) attains its maximum value at x = 2, then what is the sum of the roots of f(x)=0 ?

    Solution :-

    f(x) = ax2+bx+c

    differentiate with respect to ‘x’

    f’(x) = 2ax + b

    f’(x) = 0.

    2ax + b = 0.

    x = -b / 2a.

    max value of x given = 2.

    2 = -b / 2a

    -b/a  = 4.

    Sum of roots = -b/a = 4.

    Q. Let f(x) = mx3-100x2+3n, where m and n are positive integers. For how many ordered pairs (m,n) will (x-2) be a factor of f(x) ?

    Solution :-

    If (x-2) is a factor of f(x)

    f(2) = m x 23 – 100 x 22 + 3n.

    0 = 8m+3n – 400.

    8m + 3n = 400.

    Ordered pairs = [400 / LCM (8,3)]

    = 400 / 24

    = 16

    Q. F(n – 1) = 1 / ( 2 – F(n) ) for all natural numbers ‘n’. If F(1) = 2, then what is the value [ F(1)+F(2)………..F(50)] ?

    Solution :-

    F(1) = 2.

    2 – f(n) = 1 / f(n-1)

    2 – f(2) = 1 / f(1).

    f(2) = 2 – ½ = 3/2.

    2 – f(3) = 2/3.

    f(3) = 2 – 2/3 = 4/3.

    2 – f(4) = 3 / 4.

    f (4) = 2 – 3/4 = 5/4.

    Similarly F(50) = 2 – 49/50 = 51/50.

    [f(1)+f(2)……..f(50)]

    = [2+3/2+4/3…………………51/50.]

    = 51.

    Q.  f(x + y) = f(x) + f(y), for all real values of x and y. What is the value of f(2/3) ?

    (a) 2/3 f(1)               (b) 2/3

    (c) 2/3f(0)                (d) 0.

    Solution :-

    f( 2/3 + 1/3) = f(1/3)  +  f(2/3)

    f(1)  = f(1/3) + f(2/3)…….(1)

    f(x + y) = f(x) + f(y)

    let x = y,

    f(x+x) = f(x) + f(x)

    f(2x) = 2 f(x)

    f(2 x 1/3) = 2 f(1/3)

    f(2/3) = 2f(1/3)

    f(1/3) = ½ f(2/3)

    putting this value at equation (1)

    f(1) = 3/2 f(2/3)

    f(2/3)  = 2/3 f(1).

    Option (a)

    Q.  5f(x) + 4f( 4x+5 / x – 4)  = 9(2x+1), here x is real number and not equal to 4.

    What is the value of f(7) ?

    Solution :-

    Let’s put x = 7.

    5f(7) + 4f(11) = 9 (2 x 7 +1)

    5 f(7) + 4 f(11) = 135………….(1)

    Now put x = 11.

    5 f(11) + 4f(7)  = 9 (2 x 11 +1)

    5 f(11) + 4f(7) = 207………..(2)

    After solving

    f (7)  = -17.

    Q. F(a,b) = HCF(a,b) / LCM(a,b)

    F(1 / F(a,b) , c)  = 1/12.

    a,b,c are distinct positive integers such that any pair of a,b,c are co prime to each other, then what is the sum of a,b and c ?

    Solution :-

    1 / F(a,b) = LCM(a,b) / HCF (a,b)

    Any two co-prime number possible for value of a & here.

    F ( any value of a & b, c ) = HCF (a,b,c) / LCM (a,b,c).

    HCF(a,b,c) / LCM (a,b,c) = 1/12.

    LCM (a,b,c) = 12.

    Only possible triplet here is (1,2,4)

    a+b+c = 1+2+4 = 8.

    Q. Let f(x) = [x]. If ‘a’ and ‘b’ are two real numbers such that f(3b – 2) = a -2 and f(a+2) = b +6, then find the sum of a and b.

    Solution :-

    Suppose :-  f(x) = [x]

    f(3b-2) = 3b – 2.

    3b -2 = a-2

    a = 3b.

    f(a+2) = a+2.

    3b+2 = b+6………(2)

    2b = 4.

    b = 2.

    Hence a = 6.

    a + b = 8.

    Q. If f(x) = 4x / (4x + 2)

    Then find the value of f(1/100)+f(2/100) ... f(99/100).

    Solution :-

    f (x) = 4x / 4x + 2.

    f ( 1 – x ) = 4(1 – x) / 4(1 – x ) + 2.

    = 4 / 4 + 2.4x

    = 2 / 2 + 4x.

    f(x) + f(1 – x) = [4x  / 4x + 2] + [ 2 / 4x + 2 ]

    = 1.

    f(1 / 100) + f(1 – 1/100) = f(1) + f(99/100) = 1.

    f (2/100) + f( 1 – 2/100) = f(2) + f(98/100) = 1.

    Similarly f(49/100) + f(50/100) = 1.

    Hence f(1/100) + f(2/100) + f(3/100)………………………..f(99/100)

    = 49.

    Q.  The minimum value of ax2+bx+c is 7/8 at x = 5/4. Find the value of expression at x = 5, if the value of the expression at x = 1 is 1.

    Solution :-

    Let f(x) = ax2+bx+c

    f’(x) = 2ax + b.

    2ax + b = 0.

    x = -b / 2a

    At x = 5/4

    b = -5, a = 2.

    At x = 1,

    a(1)2 +b(1)+c = 1.

    a+b+c = 1.

    2 – 5 +c = 1.

    C =4.

    f(x) = 2x2-5x+4.

    f(5) = 2(5)2 – 5(5)+4 = 29.

    Q.  If f(x) = x4 + x3 + x2 + x + 1, where x is a positive integer greater than 1.What will be the remainder if f(x5) is divided by f(x) ?

    Solution :-

     f(x) = x4 + x3 + x2 + x + 1.

    f(x5) = x20 + x15 + x10 + x5 + 1.

    Let’s take x = 2.

    f (x) = f(2) = 31.

    f (x5) = f(32) = 220 + 215 + 210 + 25+1.

    f(32) mod  f(2) =  220 + 215 + 210 + 25+1  mod 31

    = (25)4 + (25)3+(25)2 +(25 ) + 1 mod 31

    = 1+1+1+1+1

    = 5.

    Q . If x is real, then find  smallest value of the expression 3x2 – 4x +7.

    Solution :-

    Suppose f(x) = 3x2-4x+7

    f’(x) = 6x – 4

    6x – 4 = 0

    x = 2/3.

    f(2/3) = 17/3.

     


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