Math Module Preliminary Test Solutions

Transcription

1 SSEA Summer 207 Mat Module Preliminar Test Solutions. [3 points] Find all values of tat satisf =. Solution: = ( ) = ( ) = ( ) =. Tis means ( ) is positive. Tat is, 0, wic implies. 2. [6 points] Find all values of tat satisf 2 < 0. Solution: It migt be tempting to write 2 < and cancel on bot sides to get <. But tis is not correct since we do not know if is positive or negative wic could result in a cange in direction of te inequalit. One wa to solve tis problem is to use completion of squares as sown below: 2 < < 0 ( ) 2 < 2 4 ( ) 2 < < 2. Now we can use te absolute value propert to get: 2 < 2 < < < < <.

2 SSEA Mat Module Preliminar Test Solutions, Page 2 of 7 Jul 3, [4 points] A ra of ligt comes in along te line + = from te second quadrant and reflects off te -ais as sown in te following figure. Te angle of incidence is equal to te angle of reflection. Write an equation for te line along wic te reflected ligt travels. 4. [6 points] Te grap of function f is sown below. Solution: Recall tat = m + b is te slope-intercept equation of te line wit slope m and -intercept b. Te incoming ra as equation = +, wic means its slope is and intercept is. Because of te geometr, te outgoing ra as te same magnitude of slope as te incoming one but as te opposite sign. So te outgoing ra as slope. Also, b etending te outgoing ra all te wa toward te ais, we see tat it as a -intercept of. Terefore, te outgoing ra as te equation =. f() Matc te functions wit te graps given below. (a) = f( ), (C), curve is reflected about te -ais. (b) = f(), (A) curve is reflected about te -ais. (c) = f() +, (D) curve is sifted unit in te positive -direction. (d) = f( + ), (B) curve is sifted unit in te negative -direction.

3 SSEA Mat Module Preliminar Test Solutions, Page 3 of 7 Jul 3, 207 A B C D 5. [5 points] Let f() = a + b and g() = c + d. Wat condition must be satisfied b te constants a, b, c, d in order tat (f g)() = (g f)() for all? Recall tat (f g)() is te composite function f(g()). Solution: (f g)() = f(g()) = f(c + d) = a(c + d) + b and (g f)() = g(f()) = g(a + b) = c(a + b) + d. Terefore, (f g)() = (g f)() implies (ac) + (ad + b) 0 = (ca) + (cb + d) 0. Comparing te coefficiets of 0 and, we get ad + b = cb + d.

4 SSEA Mat Module Preliminar Test Solutions, Page 4 of 7 Jul 3, [6 points] Compute te following limits (a) lim Solution: B plugging in te limiting value of, we get: lim = = (b) lim Solution: Dividing te numerator and denominator b 2, we get: lim = [6 points] Te slope of te tangent line to te curve = f() at te point ( 0, f( 0 )) is te number m 0 f( 0 + ) f( 0 ) (provided te limit eists). Using tis, find te slope of te tangent line to te following curve at te point (0, 0). ( ) 2 sin, 0 f() = 0, = 0. Solution: Using te definition of te slope at 0 = 0,, we get: f( 0 + ) f( 0 ) m 0 f() f(0) 0 ( 2 sin ) sin = 0, ) (

5 SSEA Mat Module Preliminar Test Solutions, Page 5 of 7 Jul 3, 207 b te squeeze teorem, since: ( ) sin ( ) sin ( ) lim lim sin lim ( ) 0 lim sin [6 points] Te graps in te following figure sow te position s, velocit v = ds/dt, and acceleration a = d 2 s/dt 2 of a bod moving along a coordinate line as functions of time t. Wic grap is wic? Solution: Note tat curve A crosses te -ais at points were curve C flattens. Tis means curve A is te derivative of curve C. Using a similar logic, we see tat curve B is te derivative of curve A. So C represents displacement, A represents velocit, and B represents acceleration.

6 SSEA Mat Module Preliminar Test Solutions, Page 6 of 7 Jul 3, [2 points] Find d/d for eac of te following. (a) = 3 3( 2 + π 2 ). Solution: d d = d ( π 2) = d ( ) 3 d ( ) 3 2 d ( ) 3π 2 = d d d d (b) = 2 sin ( ). Solution: We will use bot product and cain rules ere as follows: d d = d ( ( )) 2 sin d = d d ( ) ( ) d ( ( )) 2 sin + 2 sin, (product rule) d = 2 2 sin ( ) + 2 cos ( ) 2, (cain rule) = sin ( ) + cos ( ). (c) 2 2 =. Solution: We will use implicit differentiation and product rule as follows: d d (2 2 ) = d d () d d (2 ) d d (2 ) = 0, (product rule) d = 0, (implicit differentiation) d d d =. 0. [6 points] Witout using a calculator, estimate te value of (.0002) 50. Clearl sow our steps. Hint: Use te approimation f() f( 0 ) + f ( 0 )( 0 ) for 0. To identif f() and 0, write (.0002) 50 as ( ) 50.

7 SSEA Mat Module Preliminar Test Solutions, Page 7 of 7 Jul 3, 207 Solution: We will use linearization for tis problem. If function f is differentiable at = 0, ten te approimating function: L() = f(a) + f (a)( a) is called te linearization of f at = 0. In oter words, a complicated function f beaves like te straigt line L() in te close vicinit of = 0. B writing (.0002) 50 as ( ) 50, we identif f() as ( + ) 50 wit close to 0 = 0. So, let us consider te linearization of f() = ( + ) 50 at = 0: L() = f(0) + f (0)( 0) = ( + 0) ( + 0) 49 ( 0) = So f() L() near = 0, and tis implies f(0.0002) = ( ) 50 L(0.0002) = =.0.. [0 points] Te accompaning figure sows a portion of te grap of a twice differentiable function = f(). At eac of te five labeled points, classif and as positive, negative, or zero. P Q R S T Solution: Recall tat te first derivative is negative wen te function decreases, positive wen it increases, and zero wen it flattens out at a point. Te second derivative is positive wen te curve is concave up, negative wen it is concave down, and zero wen concavit canges at a point.