If f(x) = x^2 and g(x) = x^(1/2), then g(f(x)) =

g(f(x)) = g(x^2) = √(x^2) = |x|

Answer: C

if the function f(k) is defined as f(k) = -(-1/3k) what is the sum of the first 11 terms for k>0
A) b/w -1/3 and -1/9
B) b/w -1/9 and -1/27
C) b/w 1/27 ad 1/9
D) b/w 1/9 an 1/3
E) 1/3 and 1
ArvindCrackVerbal

Hello Sehaj,
This is a question which tests you on multiple areas. It tests whether you know how to deal with basic functions, it also tests whether you can deal with a sequence and observe a trend/pattern.
As such, the first and foremost thing you have to do in such questions is to find out the first 2 or 3 terms of the sequence using the relationship given.

f(k) = -(\(\frac{-1}{3^k}\)). Note that it’s only the denominator that is raised to a power of k and not the entire fraction.

The equation above can be simplified to look like f(k) = \(\frac{1}{3^k}\) since a negative multiplied with a negative will give us a positive.

k=1 will give us the first term of the sequence; f(1) = \(\frac{1}{3^1}\) = \(\frac{1}{3}\).

k = 2 will give us the second term of the sequence; f(2) = \(\frac{1}{3^2}\) = \(\frac{1}{9}\).

From the first two terms, we know that the sum of the first 11 terms can never be negative. We also know that the sum has to be more than \(\frac{1}{3}\) since we are ADDING only positive values to \(\frac{1}{3}\). This means that the only range that satisfies the above conditions is the one given in answer option E.

Hope that helps!