CAT 2017 – Slot 2 – Quantitative Ability – If f(ab) = f(a)f(b) for all positive integers a and b
Q. 1: If f(ab) = f(a)f(b) for all positive integers a and b, then the largest possible value of f(l) is
Let us take the case when a=b=1
So, f(1) = f(1) f(1)
f(1) = [f(1)]^2
f(1)[f(1)-1] = 0
f(1) = 1
So, the maximum value of f(1) = 1
Answer: 1
Q. 2: Let f(x) = 2x-5and g(x) = 7-2x. Then |f(x) + g(x)| = |f(x)| + |g(x)| if and only if
A) 5/2 < x < 7/2
B) x ≤ 5/2 or x ≥ 7/2
C) x < 5/2 or x ≥ 7/2
D) 5/2 ≤ x ≤ 7/2
|f(x) + g(x)| = |f(x)| + |g(x)|
Putting value of f(x) and g(x), we get,
|2x-5| + |7-2x| = 2
1st Case: When x<=5/2
-2x + 5 +7 – 2x = 2
=> x=5/2
2nd Case: 5/2 < x < 7/2
On solving, we get, 2=2, which satisfies the condition
3rd Case: x ≥ 7/2
2x-5 – 7+2x = 2
x=7/2
So, the answer should be 5/2<= x <= 7/2
Option (D)
Q. 3: An infinite geometric progression a1, a2, a3,… has the property that an = 3(a(n+ l) + a(n+2) +….) for every n ≥ 1. If the sum a1 + a2 + a3 +……. = 32, then a5 is
A) 1/32
B) 2/32
C) 3/32
D) 4/32
For any n ≥ 1, an = 3 (a(n+1) + a(n+2) + ……..)
So, a1 = 3 (a2 + a3 + ……) or r = 1/4 and
a1 + a2 + a3 +………… = 4a1/3 = 32
So, a1 = 24
GP = 24, 6, 1.5,…….
a5 = 1.5/16 = 3/32
Option (C)
Q. 4: 
A) 25/151
B) 1/2
C) 1/4
D) 111/55
Correct Answer:- A
Explanation:-
All the terms like 1/5, 1/8,… 1/299 will cancel out
Checkout Other Questions of CAT 2017 Slot 2 Paper:
Verbal Ability : | Q.01- Q.06 | Q.07- Q.12 | Q.13- Q.18 | Q.19- Q.21 | Q.22- Q.24 | Q.25- Q.29 | Q.30 – Q.34 |
Logical Reasoning : | Q.01- Q.04 | Q.05- Q.08 | Q.09- Q.12 | Q.13- Q.16 | Q.17- Q.20 | Q.21- Q.24 | Q.25 – Q.28 | Q.29 – Q.32 |
Quantitative Aptitude: | Q.01- Q.05 | Q.06- Q.10 | Q.11- Q.15 | Q.16- Q.20 | Q.21- Q.25 | Q.26- Q.30 | Q.31 – Q.34 |
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