GMAT Functions Questions With Solutions

f(x+1) + f(x-1) = f(x). So, f(x+2) + f(x) = f(x+1). Adding both the equations, f(x+2) + f(x-1) =0. Substituting x = 48, we get f(50)+f(47) = 0

$$f(1)^2 = f(1) => f(1) = 1$$

$$f(2)*(f(1/2) = f(1) => 4x = 1$$

So, $$f(1/2) = 1/4$$

The minimum value is obtained when 2x+1 = 3-4x => 6x = 2 => x = 1/3
So, f((x) = 2*1/3 + 1 = 5/3

Consider the first term:

$$\sqrt{1+1/1^2+1/2^2}$$ = $$\sqrt{9/4}$$ = 3/2

Second term: $$\sqrt{1+1/2^2+1/3^2}$$ = $$\sqrt{49/36}$$ = 7/6

First term + Second term = 3/2 + 7/6 = (9+7)/6 = 16/6 = 8/3 = 3 - 1/3

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Required sum = 2008 - 1/2008

It has been given that f(x) =  max(2x+3,6-x).

When x is greater than 1, the value of 2x+3 will be greater than 6-x. As x increases above 1, the value of 2x+3 will increase and the value of 6-x will decrease. Therefore, for x>1, f(x) = 2x+3.

When x is less than 1, the value of 6-x will be greater than 2x+3. As 'x' decreases below 1, the value of 6-x will increase and the value of 2x+3 will decrease. Therefore, for x<1, f(x) = 6-x.

f(x) = 2x+3 is x > 1 and f(x) = 6-x is x<1.

As we can see, the value of f(x) will increase both after 1 and before 1. Therefore, the least value of f(x) must occur at x=1.

Hence, the minimum value of f(x) is when x=1 and equals 5.

$$f(x, y, z) = xyz - (x + y + z)$$

For xyz to be maximum, all the three numbers should be equal to 12.

For xyz - (x + y + z) to be maximum two them should be equal to -12 and the other should be equal to -12.

Since the absolute values of them are not equal, they have to be as close as possible and x+y+z should be negative.

-12, -11 and 10 is the closest such combination

$$f(x, y, z) = (-12)(-11)(10) - (-12 - 11 + 10) = 1320 + 13 = 1333$$

For n, the number of elements in set S is $$^nC_2$$. Tho friends have 3 numbers in total and 1 common element. The number of common friends is the set formed by the non-common elements of the friends plus the number of elements in the set which have the common element other than the two friends = 1 + n-1 - 2 = n-2.

$$a_n$$ + $$b_n$$ for even n = $$p*b_{n-1}$$ + $$q*b_{n-1}$$
= $$(p+q)*b_{n-1}$$
$$b_{n-1}$$ = $$q*a_{n-2}$$ = $$qp*b_{n-3}$$
= $$q^2*p*a_{n-4}$$ = $$q^2p^2*b_{n-5}$$
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= $$(qp)^{n/2-1}*b_1$$ = $$(qp)^{n/2-1}*q$$
So, $$a_n + b_n$$ = $$q(pq)^{(n/2) -1}(p+q)$$

f(3) = 9a + 3b + c = 0 f(5) = 25a + 5b + c

f(2) = 4a + 2b + c

f(5) = -3f(2) => 25a + 5b + c = -12a -6b -3c

=> 37a + 11b + 4c = 0 --> (1)

4(9a + 3b + c) = 36a + 12b + 4c = 0 --> (2)

From (1) and (2), a - b = 0 => a = b

=> c = -12a

The equation is, therefore, $$ax^2 + ax - 12a = 0 => x^2 + x - 12 = 0$$

=> -4 is a root of the equation.

f(3) = 9a + 3b + c = 0 f(5) = 25a + 5b + c

f(2) = 4a + 2b + c

f(5) = -3f(2) => 25a + 5b + c = -12a -6b -3c

=> 37a + 11b + 4c = 0 --> (1)

4(9a + 3b + c) = 36a + 12b + 4c = 0 --> (2)

From (1) and (2), a - b = 0 => a = b

=> c = -12a

The equation is, therefore, $$ax^2 + ax - 12a = 0 => x^2 + x - 12 = 0$$

a + b + c = a + a - 12a = -10a.

But the value of a is not given. Therefore, the value cannot be determined.

The number of enemies of any elements is all the elements in the set formed using the numbers other than these two = $$^{n-2}C_2$$ = $$1/2 * (n^2 - 5n + 6)$$.

$$S = \frac {a*w}{t + p*w}$$ or $$S = \frac {a}{t/w + p}$$

So, as w increases, the denominator decreases and hence $$S$$ increases.