Answers and Explanations

MISCELLANEOUS TOPICS FOLLOW-UP TEST

1 . E

This    complex-looking question    becomes easily  manageable  once    you spend   a   few moments

asking  yourself    some    questions   about   the make-believe    definition  given   in  the question

stem.   When    would   an  integer not fall    within  the domain  of  a   function?   When    it  would   render

the function    undefined.  Noticing    that    in  each    choice  the function    contains    a   fraction,   you

would   then    ask,    what    would   make    a   fraction    undefined?  A   value   of  zero    in  the denominator.

So  in  each    choice, get the forbidden   values  by  setting the denominator equal   to  zero.

In  (A),    x^2     –   4   ≠   0,  so  x   ≠   ±2. The forbidden   values, then,   are ±2, and the panvoid of  the function

is  2   +   (–2)    =   0.  In  (B) x^3     –   x   ≠   0,  so  x(x^2 – 1)  ≠   0,  meaning that    x(x +   1) (x   –   1)  ≠   0,  and therefore

that    x   =   0,  x   =   –1, and x   =   1 fall  outside the domain  of  the function.   The panvoid of  the

function,   then,   is  0   +   (–1)    +   1   =   0.  Notice  that    the function    in  (C) is  not in  simplest    form:

If  (x  +   3)(x    –   3)  ≠   0,  then    x   ≠   ±3, and the panvoid of  the function    is  –3  +   3   =   0.  In  (D),    x   ≠   0,  so  the

panvoid of  the function    is  0.  The answer  must    be  (E):    If  x^2     –   x   ≠   0,  then    x(x –   1)  ≠   0,  so  x   ≠   0   and

x   ≠   1.  The panvoid of  (E) is  therefore   0   +   1   =   1.

2 . B

Think   about   the general form    of  a   quadratic   equation:   ax^2    +   bx  +   c   =    0 .    If  the roots   of  the

equation    are the complex numbers 4   ±   i,  then    in  the quadratic   formula  ,  the

real    part    of  4   ±   i   equals   ,  and the imaginary   part    of  4   ±   i   equals   .  That    is, 

and  .  You can express each    of  these   equations   a   bit more    simply, since   the a-