# Thread: Cracking the Math GRE Questions - errors?

1. ## Cracking the Math GRE Questions - errors?

This is from the fourth edition Princeton Review book, Chapter 7 review:

#32 - If $f(x) = \left\{ \begin{array}{lr} \frac x 2 + c \qquad \text{for } 0\le t \le 8 \\ 0 \qquad \text {otherwise} \end{array} \right.$, for what value of c is f(x) the probability density function of a random variable X?

I know that the probability density function f(x) has to satisfy $\int_{-\infty}^{\infty}f(x)dx=1$, which for this function is just $\int_0^8 \frac x 2 + c dx = 8^2 / 4 + 8c=1$, therefore $c=-\frac 7 8$, but that is not an option. The options are 4/3, 2/3, 0, -2/3, and -4/3. The book claims answer -4/3 is correct, but their explanation also seems way off (they have that the integral above is equal to $8^2/4+6c=1$, which still doesn't mean c should be -4/3).

Can someone please either verify that I am correct or let me know where I went wrong?

2. ## Re: Cracking the Math GRE Questions - errors?

Originally Posted by process91
This is from the fourth edition Princeton Review book, Chapter 7 review:

#32 - If $f(x) = \left\{ \begin{array}{lr} \frac x 2 + c \qquad \text{for } 0\le t \le 8 \\ 0 \qquad \text {otherwise} \end{array} \right.$, for what value of c is f(x) the probability density function of a random variable X?

I know that the probability density function f(x) has to satisfy $\int_{-\infty}^{\infty}f(x)dx=1$, which for this function is just $\int_0^8 \frac x 2 + c dx = 8^2 / 4 + 8c=1$, therefore $c=-\frac 7 8$, but that is not an option. The options are 4/3, 2/3, 0, -2/3, and -4/3. The book claims answer -4/3 is correct, but their explanation also seems way off (they have that the integral above is equal to $8^2/4+6c=1$, which still doesn't mean c should be -4/3).

Can someone please either verify that I am correct or let me know where I went wrong?
You are correct with your logic but since $\frac{8^2}{4}=\frac{64}{4}=16$ we actually get that $c=\frac{-15}{8}$

3. ## Re: Cracking the Math GRE Questions - errors?

Whoops, yes - thank you.