Answer:
-1 3/4
Step-by-step explanation:
As long as 
(or whichever function appears in the denominator) does not approach 0 as 
,
In this case,
so the answer is B.
Please consider the attached file.
We can see that triangle JKM is a right triangle, with right angle at M. Segment KM is 6 units and segment MJ is 3 units. We can also see that KJ is hypotenuse of right triangle.
We will use Pythagoras theorem to solve for KJ as:
Now we will take positive square root on both sides:
Therefore, the length of line segment KJ is 
and option D is the correct choice.
If <em>x</em>² + <em>y</em>² = 1, then <em>y</em> = ±√(1 - <em>x</em>²).
Let <em>f(x)</em> = |<em>x</em>| + |±√(1 - <em>x</em>²)| = |<em>x</em>| + √(1 - <em>x</em>²).
If <em>x</em> < 0, we have |<em>x</em>| = -<em>x</em> ; otherwise, if <em>x</em> ≥ 0, then |<em>x</em>| = <em>x</em>.
• Case 1: suppose <em>x</em> < 0. Then
<em>f(x)</em> = -<em>x</em> + √(1 - <em>x</em>²)
<em>f'(x)</em> = -1 - <em>x</em>/√(1 - <em>x</em>²) = 0 → <em>x</em> = -1/√2 → <em>y</em> = ±1/√2
• Case 2: suppose <em>x</em> ≥ 0. Then
<em>f(x)</em> = <em>x</em> + √(1 - <em>x</em>²)
<em>f'(x)</em> = 1 - <em>x</em>/√(1 - <em>x</em>²) = 0 → <em>x</em> = 1/√2 → <em>y</em> = ±1/√2
In either case, |<em>x</em>| = |<em>y</em>| = 1/√2, so the maximum value of their sum is 2/√2 = √2.
