Pre-Calculus Test 1

Instructor: Prof. A. Mingarelli
Jan. 30, 1997

NOTATION: Pi = 3.14159... is our usual Greek number

1 a) [5] Evaluate the following quantities exactly (i.e., not as numbers found by a calculator):

a)sin (Pi/12)

Solution:1. a) [sqrt(3)-1]/[2sqrt(2)] ... ,Write Pi/12 = Pi/3 - Pi/4, use the sin-difference formula and simplify. (see Assignment 1 for a similar question along with Sample Test 1, question 1.)

1 b) [5] Solve the following equation for x: 2 cos2(x) + sin(x) = 2 where x lies between "0" and "Pi".

Solution: Pi/6 and 5Pi/6 ...By inspection we see that x = 0 and x = Pi are solutions but they are not the ones we want, right? (because we want 0 < x < Pi for our solutions). Use of the trig. identity cos2(x) = 1 - sin2(x), we get, after some simplification and rearrangement, -2 sin2(x) + sin(x) = 0. Now, factor out the sin-term here...You'll find that sin(x) (1 - 2 sin(x) ) = 0, i.e., sin(x) = 1/2 (or sin(x) = 0, which we know gives nothing new). Finally, one solution of sin(x) = 1/2 is Pi/6 for x in the given interval (the other is 5Pi/6) (see Assignment 1 for a similar question along with Sample Test 1, question 2.) ).

2. [10] lim x -> o {sin(3x) - sin(4x)}/12x

Solution: -1/12 ...Rewrite {sin(3x) - sin(4x)}/12x as (1/4)sin(3x)/3x - (1/3)sin(4x)/4x. As x -> 0 we know that sin(mx)/mx -> 1 for any non-zero number,m, right? Thus, the last expression converges to 1/4 - 1/3 = -1/12 (see Sample Test 1, question 3, for a similar question.)

3. [10] Find lim n -> infinity {n/(n+1)}2

Solution: 1 ...Rewrite {n/(n+1)}2 as {1 - 1/(n+1)}2 and use a limit theorem: Since the last display converges to 1 it follows that the former display also converges to 1 (see Sample Test 1, question 4, for a similar question.)

4. Let f(x) = x-1, for x < -1; f(x) = 1 for -1 <= x <= 1; and f(x) = x for x > 1.

a) [3+2] Where is f discontinuous? Why?

Solution: f is discontinuous at x = 1. ... Why? Because the left- and right-hand limits there are different. Actually, limx -> -1- f(x) = limx -> -1- (x-1) = -2 while limx -> -1+ f(x) = limx -> -1- (1) = 1. The result follows.

b) [3+2] Find limx -> 1+ f(x) and limx -> 1- f(x)

Solution: 1 and 1 ... Since f(x) = x for x > 1, it follows that limx -> 1+ f(x) = limx -> 1+ x = 1. Furthermore, since f(x) = 1 for x < 1, and x close to 1, it follows that limx -> 1- f(x) = limx -> 1+ 1 = 1. That's all (see Sample Test 1, question 5, for a similar question.)

Total: 40 points