Question

Question #210f3

Answer 1

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)