# Question #210f3

##### 1 Answer

#### Answer:

**5**. (c)

**6**. The question is in error. The

#### Explanation:

**Question 5**

Note that if

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

Note that for

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

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

When

When

So

Let:

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

Note that by definition of

Then:

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

Add

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

Take the reciprocal of both sides to get:

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

Add

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

So for any

Hence the range of

**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]}

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

It looks like the error in the question was the