math module preliminary test solutions - stanford …acharyan/pdf/prelimtestsol.pdf · math module...

��

��

SSEA

Summer 2017

Math Module Preliminary Test Solutions

1. [3 points] Find all values of x that satisfy |x− 1|= 1− x.

Solution:

|x− 1|= 1− x⇒ |−(1− x)|= 1− x⇒ |−1||(1− x)|= 1− x⇒ |(1− x)|= 1− x.

This means (1− x) is positive. That is, 1− x ≥ 0, which implies x ≤ 1.

2. [6 points] Find all values of x that satisfy x2 − x < 0.

Solution: It might be tempting to write x2 < x and cancel x on both sides to getx < 1. But this is not correct since we do not know if x is positive or negative whichcould result in a change in direction of the inequality. One way to solve this problemis to use completion of squares as shown below:

x2 − x < 0

⇒ x2 − x+1

4− 1

4< 0

⇒(x− 1

2

)2

<1

4

⇒√(

x− 1

2

)2

<

√1

4

⇒∣∣∣∣x− 1

2

∣∣∣∣ < 1

2.

Now we can use the absolute value property to get:

⇒ −1

2< x− 1

2<

1

2

⇒ −1

2+

1

2< x− 1

2+

1

2<

1

2+

1

2⇒ 0 < x < 1.

SSEA Math Module Preliminary Test Solutions, of 7 July 31, 2017

3. [4 points] A ray of light comes in along the line x+ y = 1 from the second quadrant andreflects off the x-axis as shown in the following figure. The angle of incidence is equalto the angle of reflection. Write an equation for the line along which the reflected lighttravels.

Solution: Recall that y = mx+ b is theslope-intercept equation of the line withslope m and y-intercept b. The incom-ing ray has equation y = −x + 1, whichmeans its slope is −1 and y−intercept is1. Because of the geometry, the outgoingray has the same magnitude of slope asthe incoming one but has the oppositesign. So the outgoing ray has slope 1.Also, by extending the outgoing ray allthe way toward the y−axis, we see thatit has a y-intercept of −1. Therefore, theoutgoing ray has the equation y = x−1.

4. [6 points] The graph of function f is shown below.

1

x

y

−1

f(x)

Match the functions with the graphs given below.

(a) y = f(−x), (C), curve is reflected about the y-axis.

(b) y = −f(x), (A) curve is reflected about the x-axis.

(c) y = f(x) + 1, (D) curve is shifted 1 unit in the positive y-direction.

(d) y = f(x+ 1), (B) curve is shifted 1 unit in the negative x-direction.

Sudarsan N.S. Acharya [email protected]


1

x

y

−1

1

x

y

−1

1

x

y

−1

1

x

y

−1

A

D

B

C

5. [5 points] Let f(x) = ax+ b and g(x) = cx+d. What condition must be satisfied by theconstants a, b, c, d in order that (f ◦ g)(x) = (g ◦ f)(x) for all x? Recall that (f ◦ g)(x)is the composite function f(g(x)).

Solution: (f ◦ g)(x) = f(g(x)) = f(cx + d) = a(cx + d) + b and (g ◦ f)(x) =g(f(x)) = g(ax + b) = c(ax + b) + d. Therefore, (f ◦ g)(x) = (g ◦ f)(x) implies(ac)x1 + (ad+ b)x0 = (ca)x1 + (cb+ d)x0. Comparing the coefficiets of x0 and x1, weget ad+ b = cb+ d.



6. [6 points] Compute the following limits.

(a) limx→1x4 + x2 − 1

x2 + 5.

Solution: By plugging in the limiting value of x, we get:

limx→1

x4 + x2 − 1

x2 + 5=

14 + 12 − 1

12 + 5=

1

6.

(b) limx→∞5x2 + 8x− 3

3x2 + 2.

Solution: Dividing the numerator and denominator by x2, we get:

limx→∞

5x2 + 8x− 3

3x2 + 2= lim

x→∞

5 + 8x− 3

x2

3 + 2x2

=5

3.

7. [6 points] The slope of the tangent line to the curve y = f(x) at the point (x0, f(x0)) isthe number

m = limh→0

f(x0 + h)− f(x0)

h(provided the limit exists).

Using this, find the slope of the tangent line to the following curve at the point (0, 0).

f(x) =

x2 sin

(1

x

), x 6= 0

0, x = 0.

Solution: Using the definition of the slope at x0 = 0,, we get:

m = limh→0

f(x0 + h)− f(x0)

h

= limh→0

f(h)− f(0)

h

= limh→0

h2 sin(1h

)− 0

h

= limh→0

h sin

(1

h

)= 0,



by the squeeze theorem, since:

−1 ≤ sin

(1

h

)≤ 1

⇒ −h ≤ h sin

(1

h

)≤ h

⇒ limh→0−h ≤ lim

h→0h sin

(1

h

)≤ lim

h→0h

⇒ 0 ≤ limh→0

h sin

(1

h

)≤ 0.

8. [6 points] The graphs in the following figure show the position s, velocity v = ds/dt,and acceleration a = d2s/dt2 of a body moving along a coordinate line as functions oftime t. Which graph is which?

Solution: Note that curve A crosses the x-axis at points where curve C flattens.This means curve A is the derivative of curve C. Using a similar logic, we see thatcurve B is the derivative of curve A. So C represents displacement, A representsvelocity, and B represents acceleration.



9. [12 points] Find dy/dx for each of the following.

(a) y = x3 − 3(x2 + π2).

Solution:

dy

dx=

d

dx

(x3 − 3x2 − 3π2

)=

d

dx

(x3)− d

dx

(3x2)− d

dx

(3π2)

= 3x2 − 6x.

(b) y = 2√x sin

(√x).

Solution: We will use both product and chain rules here as follows:

dy

dx=

d

dx

(2√x sin

(√x))

=d

dx

(2√x)

sin(√

x)

+ 2√xd

dx

(sin(√

x)), (product rule)

= 21

2√x

sin(√

x)

+ 2√x cos

(√x) 1

2√x, (chain rule)

=sin (√x)√

x+ cos

(√x).

(c) x2y2 = 1.

Solution: We will use implicit differentiation and product rule as follows:

d

dx(x2y2) =

d

dx(1)

⇒ d

dx(x2)y2 + x2

d

dx(y2) = 0, (product rule)

⇒ 2xy2 + x2 2ydy

dx= 0, (implicit differentiation)

⇒ dy

dx= −y

x.

10. [6 points] Without using a calculator, estimate the value of (1.0002)50. Clearly showyour steps.

Hint: Use the approximation f(x) ≈ f(x0) + f ′(x0)(x−x0) for x ≈ x0. To identify f(x)and x0, write (1.0002)50 as (1 + 0.0002)50.



Solution: We will use linearization for this problem. If function f is differentiableat x = x0, then the approximating function:

L(x) = f(a) + f ′(a)(x− a)

is called the linearization of f at x = x0. In other words, a complicated function fbehaves like the straight line L(x) in the close vicinity of x = x0.

By writing (1.0002)50 as (1 + 0.0002)50, we identify f(x) as (1 + x)50 with x close tox0 = 0. So, let us consider the linearization of f(x) = (1 + x)50 at x = 0:

L(x) = f(0) + f ′(0)(x− 0)

= (1 + 0)50 + 50(1 + 0)49(x− 0)

= 1 + 50x.

So f(x) ≈ L(x) near x = 0, and this implies f(0.0002) = (1+0.0002)50 ≈ L(0.0002) =1 + 50× 0.0002 = 1.01.

11. [10 points] The accompanying figure shows a portion of the graph of a twice differen-tiable function y = f(x). At each of the five labeled points, classify y′ and y′′ as positive,negative, or zero.

P Q R S Ty′ − + + 0 −y′′ + 0 − − −

Solution: Recall that the first derivative is negative when the function decreases,positive when it increases, and zero when it flattens out at a point. The secondderivative is positive when the curve is concave up, negative when it is concavedown, and zero when concavity changes at a point.


math module preliminary test solutions - stanford …acharyan/pdf/prelimtestsol.pdf · math module...

Documents