# 2IIM Quant Notes - Functions - Part (2/2)

• The value of fogoh(9) could be, if
f(x) = 1/x
g(x) = 1/(x−2)
h(x) = √x
a) 3
b) 1/3
c) -5
d) None of these

fogoh(9) means f(g(h(9)))
Start by solving h(9) = √9 = 3, -3
Taking 3, g(3) = 1/(3−2) = 1 => f(1) = 1
Taking -3, g(3) = 1(−3−2) = −1/5 => f(-1/5) = -5

For this question, assume the following operators:
A * B = A^2 - B^2
A - B = A/B
A + B = A * B
A / B = A + B
Which of the following expression would yield the result as x subtracted by y?
a) (x * y)-(x+5)
b) (x/y) * (x-y)
c) (x * y) - (x/y)
d) (x+y) * (x-y)

Never solve each expression. Solve by assigning values of x and y
Let x = 6, y = 2 Therefore, required x – y = 4
x * y = 6^2 – 2^2 = 32
x-y = 6/2 = 3
x+y = 6 x 2 = 12
x/y = 6 + 2 = 8

a) 32-12 = 32/12 not equal to 4
b) 8 * 3 = 8^2 - 3^2 = 55 not equal to 4
c) 32 - 8 = 32/8 = 4 (Our answer)
d) 12 * 3 = 12^2 - 3^2 not equal to 4

Find the domain of: { 1/(1−log(9−x)) } + √(x+1)
(-∞,9)
[-1,9)
[-1,9) excluding 0
(-1,9)

In the expression,
9 - x > 0
=> x < 9
Also, 1-log ( 9 - x ) ≠ 0
=> log (9 – x) ≠ 1
=> 9-x ≠ 10
=> x ≠ -1
And, x + 1 > 0
=> x > -1

If f(x).f(1/x) = -(f(x) + f(1/x)) and f(5) = 124. Find the value of f(1/8)
511
511/512
−511/512
Can’t be determined

Given, f(5) = 124 = 5^3 – 1. So, f(x) could be x^3 – 1. However, the given relationship in the question must be satisfied.
Given, f(x).f(1/x) = -(f(x) + f(1/x))
=> f(1/x) = −f(x)/{(f(x)+1)}
=> f(1/5) = -f(5)/(f(5)+1) = −124/125 = 1/125 − 1 = = (1/5)^3 – 1
Therefore, f(x) = x^3 – 1
=> f(1/8) = (1/8)^3 – 1 = −511512

If [X] – Greatest integer less than or equal to x. Find the value of [√1] + [√2] + [√3] + ... + [√100]
615
625
5050
505

As can be seen,
Thus, the series becomes:
3 x 1 + 5 x 2 + .... + 19 x 9 + 10
=> ∑(2n+1)n + 10 (where, n = 1 to 9)
=> ∑ 2n^2 + ∑n + 10
=> 2 x n (n+1) * (2n+1)/6 + n * (n+1)/2 + 10
=> 2 x 9 x 10 x 19/6 + 9x10/2 + 10
=> 570 + 45 + 10 = 625

Find the value of x for which x[x] = 39
6.244
6.2
6.3
6.5

When x = 7,
x[x] = 7 x 7 = 49
When x = 6,
x[x] = 6 x 6 = 36
Therefore, x must lie in between 6 and 7 => [x] = 6
=> x = 45/[x] = 39/6 = 6.5

Find the value of x for which x[x] = 15
3.5
5
6.1
None of these

Using similar approach as previous question, [x] = 3
=> x = 15/3 => x = 5 which is not possible since 3 f(x).g(x) = 1 which implies that f(x) and g(x) are essentially inverse of one another. So, one just have to look for an option which has equal number of fs and gs on both side of the equation.

If f(x)=(x+6)/(x+2). Find the value of x for which f(x) = $f^{-1}$(x)?
-3
2
Both A and B
None of these

f(x) = $f^{-1}$(x) when f(x) = x
(x+6)/(x+2) = x
x + 6 = x^2 + 2x
x^2 + x – 6 = 0
(x+3) (x-2) = 0
x = -3, 2

If f(x) = |x| + |x+3| + |x+6| + .... + |x+3t|, where x is an integer and t is a positive integer, find the minimum value of f(x) when t = 6
63
36
30
25

f(x) = |x| + |x+3| + |x+6| + ... + |x+18|
Minimum value of this function will occur when x = -9 i.e. the middle term is at its minimum which is 0.
Therefore, f(-9) = 9 + 6 + 3 + 0 + 3 + 6 + 9
= 2 x 18
= 36

In the previous question if t = 7, for how many values of x, f(x) will be minimum?
1
2
4
8

When t = 7,
f(x) = |x| + |x+3| + |x+6| + ……………………………..+ |x+21|
This expression has two middle terms: |x+9|, |x+12|
The value of f(x) will be minimized when the sum of the two middle terms are minimized
|x+9|+ |x+12| should be minimum
This happens when -12 ≤ x ≤ -9, Note that for x = -12, -11, -10, -9 the value of the sum of the middle terms = 3. Therefore, for all 4 values of x, f(x) will have minimum value.

If f(x)/f(x−1)=(x−2)/(x+1), for all x ≥ 0 and f(x) is a positive-valued function and f(6) = 81, f(2) = 4, find the value of f(4).
8
18
25
28

Taking x = 5,
f(5)/f(4)=(3)/(6) ……………(1)
Taking x = 4,
f(4)/f(3)=(2)/(5)
=> f(5) = f(2) * (3)/(4)
Putting f(5) in equation (1),
We get (f(4))^2 = f(2)f(6) = 81 x 4
f(4) = 9 x 2 = 18