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Math 250 Skills Assessment Test Page 1
Math 250 Skills Assessment Test
The purpose of this test is purely diagnostic (before beginning your review, it will be helpful to assess both strengths and weaknesses). The test problems cover Calculus I concepts that are essential for second semester calculus. Answers are provided, and each answer has references to relevant review topics either in Math 250 or earlier courses. If anything is unclear, the review material should help.
You may click on the blue words if you wish to jump to an answer or the review topics.
If you would like to print the Skills Assessment so you can work it out on paper, please click Print.
secant line
f (x) = (^1) x
a (^) a + h
Find the slope of the secant line for f (x) =
x between the two points whose x-coordinates are a and (a + h). Answer
- Consider a circle of radius 2 centered at the origin. Find the area of a sector of the circle determined by an angle θ = π 6 radians.^ Answer
- Evaluate the following limits.
a) (^) xlim→ 2 x
2 x^2 + 6x − 20 Answer
b) lim x→ 1 +^ ln(x − 1) Answer
Math 250 Skills Assessment Test Page 2
c) (^) xlim→∞ √^2 x^ + 1 x^2 + 3
Answer
d) (^) xlim→ 0 sin 4x x Answer
- Find y′^ for the following functions.
a) y = 2ex^ + π tan x − 7 cos x Answer
b) y = 3
x +^7 x − sin x + 3 cot x Answer
c) y =^4 3 ln^ x^ −^ sec^ x^ −^ 2 csc^ x^ Answer
d) y = (x^2 + 2x − 1)(ex^ + tan x) Answer
e) y = (4x^2 /^3 − 3
x)(x^3 + 3x − ln x) Answer
f) y = x^2 − 2 x + 1 tan x + cot x Answer
g) y = xex^ + cos x √ (^4) x − 4 Answer
h) y = (x^2 + 3)^1 /^2 − sin(2x) + etan^ x^ Answer
i) y = sin^2 x + sin x^2 Answer
j) y = csc
x^2 + 1 Answer
k) x^2 y + xy^2 = 4 Answer
l) exy^ = ln(x^2 − y) Answer
m) y = xcos^ x^ Answer
- Find the equation of the tangent line to the curve y = 1 + ln x where y = 1 + ln x crosses the x-axis. Answer
Math 250 Skills Assessment Test Page 4
m)
x + 3 x^2 + 6x − 1 dx Answer
- Find the area bounded by the curves y = tan x, x = π 6 , x = π 4 , and y = −1. Answer
- Find the area bounded by the curves y = x and y = 4x^3. Answer
Answers Top of File Math 250 Review Topics
ANSWERS to SKILLS ASSESSMENT
- To find the slope, we need two points. If x = a then y = f (a), and so one point is (a, f (a)). The other point is (a + h, f (a + h)). Then
slope = m = y 2 − y 1 x 2 − x 1
f (a + h) − f (a) a + h − a
f (a + h) − f (a) h
(This quotient has a special form and is called the difference quotient.)
In this problem f (x) =
x, or^ f^ (^ ) =^
( ).^ Thus,
slope = f (a + h) − f (a) h =
a + h
a h =
a − (a + h) a(a + h) h = −h ah(a + h)
a(a + h)
Return to Problem
(see Math 250, Review Topic 1 for help.)
- The area A of a sector of a circle of radius r subtended by a central angle θ is given by the formula A =^1 2 r^2 θ. Thus,
A =^1 2 (2^2 )π 6 = π 3
Return to Problem
(see Math 250, Review Topic 6 for help.)
3 a) Find lim x→ 2 x^2 − 4 2 x^2 + 6x − 20
. This has form
at x = 2, and so we must manipulate. Let’s try factoring.
lim x→ 2 x^2 − 4 2(x^2 + 3x − 10) = lim x→ 2 (x − 2)(x + 2) 2(x − 2)(x + 5) = lim x→ 2 x + 2 2(x + 5)
Return to Problem
(see Math 250, Review Topic 11B
- d) (^) xlim→ 0 sin 4x x = lim x→ 0
sin 4x x = lim x→ 0 4 · sin 4x 4 x
. We know lim x→ 0 sin x x
(See Example 11B.7). This really says lim θ→ 0 sin(θ) (θ) = 1, where θ is the same for the numerator and the denominator. Thus,
xlim→ 0
sin 4x 4 x
This means
xlim→ 0 sin 4x x = lim x→ 0 4 · sin 4x 4 x
Return to Problem
(see Math 250, Review Topic 11B for help.)
- a) y′^ = 2ex^ + π sec^2 x + 7 sin x
Return to Problem
(see Math 250, Review Topic 12B for help.)
- b) Rewrite y as y = 3x^1 /^2 + 7x−^1 − sin x + 3 cot x
y′^ = 3
x−^1 /^2
− 7 x−^2 − cos x − 3 csc^2 x
Return to Problem
(see Math 250, Review Topic 12B for help.)
- c) y′^ =
x − sec x tan x + 2 csc x cot x
Return to Problem
(see Math 250, Review Topic 12B
- d) y′^ =
[
d dx(x
(^2) + 2x − 1)
]
(ex^ + tan x)
- (x^2 + 2x − 1) d dx (ex^ + tan x) = (2x + 2)(ex^ + tan x) + (x^2 + 2x − 1)(ex^ + sec^2 x)
Return to Problem
(see Math 250, Review Topic 12C for help.)
- e) (^) y′^ =
x−^1 /^3 −
3 x
− 2 / 3
(x^3 + 3x − ln x)
- (4x^2 /^3 − x^1 /^3 )(3x^2 + 3 −
x
Return to Problem
(see Math 250, Review Topic 12C for help.)
- f) y′^ =
[
d dx (x^2 − 2 x + 1)
]
(tan x + cot x) − (x^2 − 2 x + 1) d dx (tan x + cot x) (tan x + cot x)^2
= (2x^ −^ 2)(tan^ x^ + cot^ x)^ −^ (x
(^2) − 2 x + 1)(sec (^2) x − csc (^2) x) (tan x + cot x)^2
Return to Problem
(see Math 250, Review Topic 12D for help.)
- g) y′^ =
(ex^ + xex^ − sin x)(x^1 /^4 − 4) − (xex^ + cos x)(^1 4 x−^3 /^4 ) (x^1 /^4 − 4)^2
Return to Problem
(see Math 250, Review Topic 12D
- l) Use implicit differentiation.
exy[y + xy′] = 1 x^2 − y (2x − y′)
exyxy′^ +
x^2 − y y′^ = 2 x x^2 − y − yexy
y′^ =
2 x x^2 − y − yexy 1 x^2 − y
Return to Problem
(see Math 250, Review Topic 12F for help.)
- m) Take the natural log of both sides to arrive at
ln y = ln xcos^ x^ = cos x ln x. Now use implicit differentiation 1 y y′^ = − sin x ln x + (cos x) ·
x
Then y′^ = y
− sin x ln x + cos x x
= xcos^ x^
[
− sin x ln x + cos^ x x
]
Return to Problem
(see Math 250, Review Topic 12G
- To find the equation of the tangent line we need the point and slope. The curve crosses the x-axis at f = 0 = 1 + ln x ⇒ ln x = − 1 ⇒ x = e−^1. The point is (e−^1 , 0). To find the slope we need to take the derivative and evaluate the derivative at the point. Thus f ′^ =
x f ′(e−^1 ) =
e−^1 = e. Then y − 0 = e(x − e−^1 ) ⇒ y = ex − 1.
Return to Problem
(see Math 250, Review Topic 12H for help.)
- a) x
3 3
+ C
Return to Problem
(see Math 250, Review Topic 13A for help.)
- b) ex
ln 2
0
= eln 2^ − e^0 = 2 − 1 = 1
Return to Problem
(see Math 250, Review Topic 13A for help.)
- c) sin x
π/ 4
0
= sin π 4 − sin 0 =
Return to Problem
(see Math 250, Review Topic 13A for help.)
- d)
∫ (^) π/ 3
π/ 6
tan^2 x dx =
∫ (^) π/ 3
π/ 6
(sec^2 x − 1)dx = (tan x − x)
π/ 3
π/ 6 = tan π 3
π 3
tan π 6
π 6
π 6
Return to Problem
(see Math 250, Review Topic 13B
- h) Let u = x^2 + 1, du = 2x dx. ∫ (x︸ ︷︷ ︸^2 + 1 u
)^1 /^2 x dx︸︷︷︸ du 2
u^1 /^2 du
=^1
u^3 /^2 3 2
+ C
=^1
(x^2 + 1)^3 /^2 + C
Return to Problem
(see Math 250, Review Topic 13C for help.)
- i) u = x + 1, du = dx.
∫ ︷︸︸︷du dx (x︸ ︷︷ ︸ + 1 u
du u = ln |u| + C = ln(x + 1) + C
Return to Problem
(see Math 250, Review Topic 13C for help.)
- j) u = − tan x, du = − sec^2 x dx. We now need to change the limits of integration; when x = 0, u = − tan 0 = 0; when x = π 4 , u = − tan π 4
u=− 1 ↘
u=0↗
∫ (^) π/ 4
0
e
︷ ︸︸^ u ︷ − tan x (^) sec (^2) x dx ︸ ︷︷ ︸ −du
0
(−eudu)
= −eu
− 1
0
= −e−^1 − (−e^0 ) = 1 − e−^1
Return to Problem
(see Math 250, Review Topic 13C
- k) u = πx, du = π dx,
e
︷︸︸︷^ u πx (^) dx ︸︷︷︸ duπ
=^1
π
eudu
=^1
π eu^ + C = e
πx π
+ C
Return to Problem
(see Math 250, Review Topic 13C for help.)
- l) (^) u = x + 1, du = dx
x = − 1 u = −1 + 1 = 0 x = 0 u = 0 + 1 = 1 u=1↘
u=0↗
− 1 ︸︷︷︸^ x ↓
(x︸ ︷︷ ︸ + 1 u
)^1 /^2 ︸︷︷︸dx du
0
(u − 1)u^1 /^2 du
u−1=x
=
0
(u^3 /^2 − u^1 /^2 )du =
u^5 /^2 −
u^3 /^2
1
0
Return to Problem
(see Math 250, Review Topic 13C
y = 4x^3
y = x
We must first determine where the curves cross. Set x = 4x^3 , then 4 x^3 −x = 0 → x(2x−1)(2x+1) = 0 so x = 0,
. We can use symmetry and find
Area = 2
0
(x − 4 x^3 )dx = 2
x^2 2 − x^4
1 / 2
0
[(^1
) ]
[ 1
]
=^1
Return to Problem
(see Math 250, Review Topic 13D for help.)
Math 250 Web Page Top of File Math 250 Review Topics
