# Quant Boosters - Vikas Saini - Set 3

• 4x = k/2.
2y = k/4.
z = k/8.

(k/2) + (k/4) + (k/8) = 35.
=> 4k + 2k + k/8 = 35.
=> 7 k/8 = 35
=> k = 40.
x = 5, y = 5, z=5.

Max value = 5^7/8

• Q21) If f(n) = 1^4 + 2^4 + 3^4 ... n^4, then 1^4 + 3^4 + 5^4 ... (2n – 1)^4 can 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)

• 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 1^4 + 3^4 (n=2) = 82.
(a) negative value
(b) negative value
(c) 82.
(d) 218.
Hence option C.

Method 2
f(n) = 1^4 + 2^4 + 3^4 ... + n^4.
f(2n) = 1^4 + 2^4 + 3^4 ... + (2n)^4.
1^4 + 2^4 + 3^4 + ... (2n-1)^4
= [1^4 + 2^4 + 3^4 ... (2n)^4 ] – [(2^4 + 4^4 + 6^4 ... (2n)^4]
= f(2n) – 16 [1^4 + 2^4 + 3^4 ... n^4]
= f(2n) – 16 f(n)
Hence option (c)

• Q22) 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) .

• 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.

• Q23) If f(x) = (sec x + cosec x) (tan x – cot x) and 45 degree < x < 90 degree, then f(x) lies in which range ?

• At x = 45 degree
f(x) = 0.
At x = 90 degree
f(x) = infinitive
Range (0,infinitive)

• Q24) Let f(x) = ax^2 + 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 ?

• f(x) = ax^2 + 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.

• Q25) Let f(x) = mx^3 - 100x^2 + 3n, where m and n are positive integers. For how many ordered pairs (m,n) will (x-2) be a factor of f(x) ?

• 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.

61

63

64

44

61

34

62