There is no need to simplify, just plug in the point (2, 1):
- B Start by plugging (0, 2) into y = ax^2 + bx + c. Then, c = 2. Next, take the derivative of y = ax^2 +
bx + c: = 2ax + b. Given that y = x + 2 is normal to y = ax^2 + bx + c, the slope of y = ax^2
+ bx + c, or , is − 3. Thus, by plugging x = 0 into = 2 ax + b, b = −3. Finally, plug the
point (2, 1) and the values of c and b into y = ax^2 + bx + c. Therefore, a = .
- D This question is testing whether you know your derivatives of trigonometric functions. If you
do, this is an easy problem.
The derivative of sec x is sec x tan x, and the derivative of csc x is −csc x cot x. That makes the
derivative here sec x tan x − csc x cot x.
- C If we take the limit as x goes to 0 from the right, we get an indeterminate form , so let’s use
L’Hôpital’s Rule. We take the derivative of the numerator and the denominator and we get
= = . Now, when we take the limit
we get = 1.
- C You should recognize this integral as one of the inverse trigonometric integrals.
Step 1: As you should recall, = tan−1(x) + C. The 4 is no big deal. Just multiply the
integral by 4 to get 4tan−1(x). Then we just have to evaluate the limits of integration.
