# Quant Boosters - Vikas Saini - Set 3

• If (x-2) is a factor of f(x)
f(2) = m x 2^3 – 100 x 2^2 + 3n.
0 = 8m + 3n – 400.
8m + 3n = 400.
Ordered pairs = [400 / LCM (8,3)]
= 400 / 24
= 16

• Q26) 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)] ?

• f(1) = 2.
2 – f(n) = 1 / f(n-1)
2 – f(2) = 1 / f(1).
f(2) = 2 – 1/2 = 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

• Q27) 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) 2/3 * f(0)
(d) 0

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

• Q28) 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) ?

• Let’s put x = 7.
5f(7) + 4f(11) = 9 (2 * 7 + 1)
5 f(7) + 4 f(11) = 135 ... (1)
Now put x = 11.
5 f(11) + 4f(7) = 9 (2 * 11 + 1)
5 f(11) + 4f(7) = 207 .... (2)
After solving, f(7) = -17

• Q29) F(a,b) = HCF(a,b) / LCM(a,b)
F(1/F(a,b), c) = 1/12.
If a,b,c are distinct positive integers such that any pair of a,b and c are co prime to each other, then what is the sum of a, b and c ?

• 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.

• Q30) 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.

• f(x) = [x]
f(3b-2) = 3b – 2.
3b -2 = a-2
a = 3b.
f(a+2) = a+2.
3b+2 = b+6
2b = 4.
b = 2.
Hence a = 6.
a + b = 8.

61

62

61

61

43

61

63

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