Is f(n) > f(n−1)? (1) n = 8 (2) f(n) = n−1

(1) Since we don't know what the function is we can't comment whether or not f(n)>f(n-1)
Thus, insufficient.

(2) f(n)= n-1
So, f(n-1)= (n-1)-1=n-2
=> f(n)>f(n-1)
Thus, sufficient.

Therefore, the answer is option B.

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The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Since f(n)=n-1 from condition 2), we have f(n)-f(n-1) = (n-1)-((n-1)-1)=n-1-n+2 = 1 > 0 or f(n)>f(n-1).
Condition 2) is sufficient since it yields a unique solution.

Condition 1)
Since we don't know f(7) and f(8) and it doesn't give any information about f(8) and f(7), it is not sufficient.

Therefore, B is the answer.

Asked: Is f(n) > f(n−1)?

(1) n = 8
Since f(n) is unknown
NOT SUFFICIENT

(2) f(n) = n−1
f(n-1) = n-2
n-1>n-2
f(n)>f(n-1)
SUFFICIENT

IMO B

Analyzing the question:
You can think of this question as asking us whether f(n) is a strictly increasing function. The type of functions that can increase then decrease are quadratic or not linear. If f(n) is linear it must be strictly increasing or decreasing.

Statement 1: Tells us nothing about the function. Insufficient.
Statement 2: f(n) is linear, and the slope is 1. Since the slope is greater than 0, f(n) always increases as n increases. Therefore f(n) > f(n-1) for all n. Sufficient.

Ans: B
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while compluting 2nd equation, it is not sufficient when n have negative value. why is it not taken into consideration?

Even if n is negative, S2 would hold true.

Let's take n to be -3.

According to S2:
f(n) = n-1
f (n-1) = n-1-1 = n-2

The question is asking whether n-1>n-2.

Substituting -3 for n
-3-1 > -3-2
-4 > -5

Therefore, S2 is sufficient