 # 5 Function Practice Questions & Explanations

## Question 1:

If f(x) = 100 – x2, is f(a) > f(b) ?

1. a2 > b2
2. a/b > 1

f(x) = 100 - x2

Therefore, f(a) = 100 - a2 and f(b) = 100 - b2

Is f(a) > f(b)?

Or, is 100 - a2 > 100 - b2 ?

Or, is - a2 > - b2 ?

Or, is a2 < b2 ?

1) a2> b2 , clearly its sufficient to tell that a2 is not less than b2 ;SUFFICIENT.

2) a/b > 1

=> (a/b) -1 > 0

=> (a-b)/b > 0, which means either both a-b and b are positive, or both a-b and b are negative.

=> If b>0, a-b>0 means a>b. So a2> b2.

And if b<0, a-b<0 means a2 > b, which means a is bigger negative number than b.(For eg. a=-4, b=-2). So, a2> b2.

So, either way we can say that a2> b2 ; SUFFICIENT.

Alternatively, the statement 2 could also have been analyzed in the following manner.

a/b > 1 → that the magnitude of numerator is more than denominator which would mean a2> b2.

The correct answer is D; each statement alone is sufficient.

## Question 2:

a’ =10/a for all positive prime numbers. And a’ = -a2for all positive non-primes, which of the following is the largest?

(A) 5’ + 6’ + 2’

(B) 7’ – 4’

(C) 4’ – 7’

(D) 4’ + 8’ – 5’

(E) 4’ + 8’ – 2’

Ans:

Since, for all positive prime numbers;

a' = 10/a

And, for all positive non-prime numbers;

a' = -a2

Thus, using these functions to find out the values given in the question, we get;

(A) 5' + 6' + 2' = (10/5 - 6+ 10/2) = (5 - 36 +2) = -29

(B) 7' - 4' = (10/7 - (-42)) = (10/7 + 16) = 122/7

(C) 4' - 7' = ( - 4- 10/7) = ( -16 - 10/7) = -122/7

(D) 4' + 8' - 5' = ( -4- 8- 10/5) = ( -16 -64 - 2) = -82

(E) 4' + 8' - 2' = (-4- 8- 10/2) = (-16 - 64 -5) = -85

Thus, among all the options we can see the largest value is obtained in option B

Alternatively, you could have also analyzed that all the options will generate a negative value except option B. hence that is the largest.

## Question 3:

If [a] represents least integer greater than or equal to a, what is the value of [–3.3] + [–2.9] + [0.1] +  + [1.9] ?

(A) –1

(B) –2

(C) –3

(D) –4

(E) –5

Since [a] represents the least integer greater than or equal to a, this implies that if a is an integer than [a] = a only. But if a is a fraction than, [a] = the least integer greater than a.

Thus, [-3.3]=-3

And,[-2.9]=-2

And,[0.1]=1

And,=1

And,[1.9]=2

Hence, [-3.3]+[-2.9]+[0.1]++[1.9]

=-3-2+1+1+2

=-1

## Question 4:

In which of the following functions, is f(x) = f(1/x), for all values of x greater than 1?

(A) f(x) = 1/x + 1/ (x + 1)

(B) f(x) = |x|

(C) f(x) = 1-x

(D) f(x) = (1 - x2 )2/x2

(E) f(x) = (1 - x2)/x2

(A) f(x) = 1/x + 1/(x + 1) and f(1/x) = 1/(1/x) + 1/(1/x + 1) = x + x/(1 + x)

Thus, f(x) ≠ f(1/x) for all values of x > 1

(B) f(x) = |x| and f(1/x) = |1/x| = 1/(|x|)

Thus, f(x) ≠ f(1/x) for all values of x >1

(C) f(x) = 1 - x and f(1/x) = 1 - 1/x = (x - 1)/x

Thus, f(x)≠f(1/x) for all values of x>1

(D) f(x)=(1 - x2 )2/x2 and f(1/x)=[1 - (1/x)2 ]2/(1/x)2 =[1 - 1/x2 ]2/(1/x2 )=[(x- 1)/x2 ]2/(1/x2 )=(x- 1)2/(x× 1/x2 )=(x- 1)2/x2 =(1 - x2 )2/x2

Thus, f(x) = f(1/x) for all values of x > 1

(E) f(x) = (1 - x2)/x2 and f(1/x) = (1 - (1/x)2)/(1/x)2 = (1 - 1/x2 )/(1/x2 ) = ((x- 1)/x2 )/(1/x2 ) = (x2- 1)/(x2×1/x2 ) = x- 1

Thus, f(x) ≠ f(1/x) for all values of x>1

Thus we can see only for (D); f(x) = f(1/x)

## Question 5:

In a sequence, an + 2 = an + 1 – an + (n + 2) for all n ≥1, where an + 2 represents the (n + 2)th term. If the second and the fourth terms of the sequence are 2 and 5 respectively then what is the value of the 6th term of the sequence?

(A) 5

(B) 7

(C) 8

(D) 9

(E) None of these

For n≥1,

a(n + 2) = a(n + 1) - an + (n + 2) ……Equation(1)

Putting value of n = 2 in the above equation, we get:

a= a- a+ (2 + 2)

In the question, it’s also mentioned that a= 2 , a= 5. Putting in the above equation, we get the value of a= 3

Now by putting value of n = 3 in equation(1), we get:

a5 = a4 - a3 + (3 + 2)

a= 5 - 3 + (3 + 2)

a5 = 7

Now by putting value of n = 4 in equation(1),we get:

a6 = a5 - a4 + (4 + 2)

a6 = 7 - 5 + (4 + 2)

a6 = 8

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