2IIM Quant Notes  Functions  Part (2/2)

rajesh_balasubramanian
Director, 2IIM Online CAT Preparation  IIT Madras  IIM Bangalore  CAT 100th percentile  CAT 2011, 2012 and 2014.
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 thesefogoh(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
Therefore, Answer – (c)
Answer choice (C).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) * (xy)
c) (x * y)  (x/y)
d) (x+y) * (xy)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
xy = 6/2 = 3
x+y = 6 x 2 = 12
x/y = 6 + 2 = 8a) 3212 = 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 4Find 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, 1log ( 9  x ) ≠ 0
=> log (9 – x) ≠ 1
=> 9x ≠ 10
=> x ≠ 1
And, x + 1 > 0
=> x > 1
Therefore, Answer – (d)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 determinedGiven, 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 = −511512If [X] – Greatest integer less than or equal to x. Find the value of [√1] + [√2] + [√3] + ... + [√100]
615
625
5050
505As 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 = 625Find the value of x for which x[x] = 39
6.244
6.2
6.3
6.5When 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 = 6.5Find the value of x for which x[x] = 15
3.5
5
6.1
None of theseUsing 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.
Answer choice (D)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 thesef(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) (x2) = 0
x = 3, 2
Answer choice (C).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
25f(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
= 36In the previous question if t = 7, for how many values of x, f(x) will be minimum?
1
2
4
8When 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 positivevalued function and f(6) = 81, f(2) = 4, find the value of f(4).
8
18
25
28Taking 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
Answer choice (B).