Question #210f3

1 Answer
Nov 6, 2016

Answer:

5. (c)

6. The question is in error. The #3# should be #4# and the answer (d)

Explanation:

Question 5

#v(x) = sqrt(x/(1-x))#

Note that if #x=1# then the denominator of #x/(1-x)# will be zero, so the quotient is undefined. So #1# cannot be part of the domain.

That allows us to immediately eliminate options (d) and (e).

Note that for #x > 1# the numerator will be positive and the denominator negative so #x/(1-x) < 0# and the square root is undefined (at least not Real).

So we can eliminate options (a) and (b).

That only leaves option (c), which is correct:

When #x < 0#, we find that #x/(1-x) < 0#.

When #x in [0, 1)# then #x >= 0# and #1-x > 0# so #x/(1-x) >= 0# and the square root is Real valued.

So #D_v = [0, 1)#

Let:

#y = v(x) = sqrt(x/(1-x))#

Note that by definition of #sqrt(...)# we must have #y >= 0#

Then:

#y^2 = x/(1-x) = (1-(1-x))/(1-x) = 1/(1-x)-1#

Add #1# to both ends to get:

#y^2+1 = 1/(1-x)#

Take the reciprocal of both sides to get:

#1/(y^2+1) = 1-x#

Add #x-1/(y^2+1)# to both sides to get:

#x = 1-1/(y^2+1)#

So for any #y >= 0# there is a value of #x# such that #v(x) = y#

Hence the range of #v(x)# is #[0, oo)#

#color(white)()#
Question 6

#f(x) = sqrt(4-x)#

#g(x) = sqrt(x+3)#

Then:

#D_f = (-oo, 4]#

#D_g = [-3, oo)#

So:

#D_(f+g) = D_f nn D_g = (-oo, 4] nn [-3, oo) = [-3, 4]#

This does not correspond to any of the given options, so there is an error in the question.

graph{sqrt(4-x) + sqrt(x+3) [-4.438, 5.56, -0.64, 4.36]}

#(f+g)(x)# has minima at the ends of the domain and maximum in the centre.

#R_(f+g) = [sqrt(7), sqrt(14)]#

It looks like the error in the question was the #3# instead of a #4# and the correct answer would have been (d)