THE UNIVERSITY OF CALGARY

DEPARTMENT OF MATHEMATICS AND STATISTICS

MATH 265 L01-02 TERM TEST 2 - VERSION 11 WINTER 2017

THURSDAY MARCH 23 TIME: 90 MINUTES

NAMES: LAST & FIRST ID# LAB# LEC#

EXAMINATION RULES

01. This is a closed book examination.

02. Calculators are not permitted.

03. The use of personal electronic or communication devices is prohibited.

04. The exam has nineteen (17) questions

05. Please make sure to show your work in a neat and organized manner for thewritten questions 16 and 17.

06. A University of Calgary Student ID card is required to write the Test. Ifadequate ID is not present, the Student may be asked to complete an Iden-tification Form.

07. Students late in arriving will not be permitted to write the exam thirty ( 30)minutes after the examination has started.

08. No student will be permitted to leave the examination room during the firstthirty ( 30) minutes, nor during the last ten ( 10) minutes of the examination.Students must stop writing and hand in their exam immediately when timeexpires.

09. All inquiries and requests must be addressed to the exam’s Supervisor.

10. Students are strictly cautioned against:

(a) communicating with other students

(b) leaving answer papers exposed to view

(c) attempting to read other students’ examination papers.

11. If a student becomes ill during the course of the examination, he/she mustreport to the Invigilator, hand in the unfinished paper and request that itbe cancelled.

12. Once the examination paper has been handed in for marking, the Studentcannot request that the examination be canceled.

13. Failure to comply with these regulations may result in the rejection of theexamination paper.

Problem Mark

1 - 15

16

17

1

MATH 265 L01-L02 TERM TEST 2 - VERSION 11 WINTER 2017 2

Part I: Consists of 15 multiple choice questions, worth 4 marks each. For eachquestion, clearly circle your choice on this booklet, and record your answer onthe scantron sheet provided. Make sure that you answer all the questions.

01. The derivative of y = tan!

sin(x)"

is given by

A. y′ = sec2!

sin(x)"

cos(x)

B. y′ = sec2!

cos(x)"

C. y′ = sec2(x) cos(x)

D. y′ = cot!

sin(x)"

cos(x)

E. y′ = sec!

sin(x)"

tan!

sin(x)"

cos(x)

02. Let y = ln#

#x2 + 3 x− 4#

#, then y′ is equal to

A. y′ =2 x+ 3

|x2 + 3 x− 4|

B. y′ = ln |2 x+ 3|

C. y′ =1

|x2 + 3 x− 4|

D. y′ =

#

#

#

#

2 x+ 3

x2 + 3 x− 4

#

#

#

#

E. y′ =2 x+ 3

x2 + 3 x− 4

By the chain rule

y'= set ( sine ) ) . ( sine ) )

'

=see ( sink ) ) .

cost )

Recall Clnlxl) 's ¥

Using the chain rule,

we have

I

y'

=

× 2+3×-4. 42+3×-4 )

'

2×+3=

1,2+3×-4

MATH 265 L01-L02 TERM TEST 2 - VERSION 11 WINTER 2017 3

03. If x3 y + ey = sin(x), defines y as a differentiable function of x, then y′ is

A. y′ =sin(x)− 3 x2 y

x3 + ey

B. y′ =cos(x)− 3 x2 y

x3 + ey

C. y′ =cos(x)− x3 − 3 x2 y

ey

D. y′ =cos(x)

3 x2 + ey

E. y′ =cos(x)− x3 − ey

3 x2

04. The derivative of the function y =!

1− 2 x"x, is

A. y′ = −2 x#

1− 2 x$x−1

B. y′ =#

1− 2 x$x

%

ln!

1− 2 x"

+x

1− 2 x

&

C. y′ =#

1− 2 x$x

%

ln!

1− 2 x"

−2 x

1− 2 x

&

D. y′ =#

1− 2 x$x

ln!

1− 2 x"

E. y′ = ln!

1− 2 x"

−2 x

1− 2 x

Differentiate both and keep in

mind that y = y (× )

3×2 .y+x? y't et

. y'

= cost )

(×3teT ) y'

= cos ⇐ ) - 3×2-1

y'

=

cos ⇐ ) -

3×2-1×3+ et

Use Logarithmicdifferentiationluly ) = ln ( c-2×5 '

)

ln ( y ) = X lull -2×1

Differentiate both sides to get

¥ =1.hn ( l - 2× ) t ×

,.ly (1-2×5

= ln ( I - 2× ) - 2×1- 2X

Hence

y'

= y ( luutx ) - EI )

= ( i -2×5 . ( ln ( i -2×1 - EI )

MATH 265 L01-L02 TERM TEST 2 - VERSION 11 WINTER 2017 4

05. The exact value of sin−1

!

sin"16 π

3

#

$

is

A.16 π

3

B.π

3

C.4 π

3

D. −π

3

E. −2 π

3

06. If y = 2 x tan−1%

x&

− ln%

x2 + 1&

, then

A. y′ = 2 tan−1%

x&

+2 x− 1

x2 + 1

B. y′ = 0

C. y′ = 2 tan−1%

x&

D. y′ = 2 tan−1%

x&

+4 x

x2 + 1

E. y′ =2− 2 x

x2 + 1

Reduce the angle

1631=4 It 4¥ ⇒

sin (165) = sin @tt4F)=sin(4§ )

=nn⇐t¥ )

*jf*c¥⇒It follows=nntE )

sin'

@nCfl ) =sii' ( mites ) ) = - Ez

Using product rule and

chain rule

yl=2taidK)t2× ↳+ ,

,I

-

1×2+1 )×2+ ,

= 2 taidx ) +1×

-

2×

x4l x2ti

= 2tai(x )

MATH 265 L01-L02 TERM TEST 2 - VERSION 11 WINTER 2017 5

07. Let f(x) be a differentiable function with inverse f−1(x). Given the table of values

x 1 3 5f(x) 4 3 2f ′(x) 3 1/4 4

!

f−1"′

(3) is equal to

A. 3/4

B. 4

C. 1/3

D. 4/3

E. 1/4

08. Let y =(x+ 1)2 (x+ 3)3

(3− x)3. Using Logarithmic differentiation, y′(0) is equal to

A. −2

B. −4

C. 2

D. 4

E. 1

# IN = pH, ,

= file,

=

Hence (f)/(3) = 4

ln (y ) = 2 lnlxtl ) + 3h ( x + 3) - 3 bn ( 3- × )

Differentiate

¥ = 2¥ ,

+3¥ -3 ÷×t ' '

= ¥,

t It ÷,

y'

= y ( II , + ¥+3 + Ex )

t.eu#jhlEntEtstEt⇒ y

'

@ ) = tj}3- ( ±,

+ Est I ) = 2+1+1=4

MATH 265 L01-L02 TERM TEST 2 - VERSION 11 WINTER 2017 6

09. The degree 3 Taylor polynomial for f(x) =√x at x = 1 is equal to

A. P3(x) = 1 +

1

2(x− 1)−

1

4(x− 1)2 +

3

8(x− 1)3

B. P3(x) = 1 +

1

2(x− 1)−

1

8(x− 1)2 +

1

16(x− 1)3

C. P3(x) = 1 +

1

2x−

1

8x2 +

1

16x3

D. P3(x) = 1 +

1

2x−

1

4x2 +

3

8x3

E. none of the above

10. The electrical resistance R of a wire with unit length and cross-sectional radius x, is

given by the formula R =k

x2, where k > 0 is a constant. If the radius of the wire

decreases by 6%, then the percentage change in the resistance R will approximately

A. decrease by 9%

B. increase by 6%

C. increase by 12%

D. decrease by 6%

E. decrease by 12%

I3X)=fi ) + fly'K . I )

+ tLY÷kt5tt"¥cnP

fx

)=tI=x" 2

⇒ fi )=lftxktzxtn⇒ f 'H=tHence

Izx ) =L + tzlx - i ) - §(× . ,j2+y←,)3 f "x)= - tax"2

⇒ f "n= - ±,

f "tx)=3gx#2 ⇒ f"

d) =3

we have

as a 0Y÷=' 2¥

R

But R'

=- 2¥ It follows

a

,¥~~ -24¥ = - 21¥ . ×g2= -2 AI =

-24%)

KX 2

°

so A,Rz~~ 121°

MATH 265 L01-L02 TERM TEST 2 - VERSION 11 WINTER 2017 7

11. Which of the following best describes the domain of tan−1(x) ?

A.!

0 , π"

B.#

−∞ , −1"

∪!

1 , +∞$

C.!

− 1 , 1"

D.#

− π/2 , π/2$

E.#

−∞ , +∞$

12. If the Mean Value Theorem is applied to f(x) = x3 − x on the interval!

0 , 3"

, which ofthe following lists all the numbers c guaranteed by the theorem?

A. 1 and√3

B. −√3 and

√3

C. −√3 only

D.√3 only

E. 1 only

The domain of tancx )

is the internal ( . as + a)

fcx )=x3 - x is continuous in [ 0,3 ]

and

flex )= 3×2-1exists in ( 0

,3)

By the M .

V. T .there exists

c in ( 0,3 ) such that

f- (3) - f @ ) = fk ) (3-0) ( ⇒ 24 -o = ( 3RD 3

( ⇒ 8 = 3 CZ 1

⇐ )c2 =3

⇒ c=- r}

,

-15

only C= -15 belongs to the interval 10,3 )

MATH 265 L01-L02 TERM TEST 2 - VERSION 11 WINTER 2017 8

13. A function f(x) has derivative f ′(x) = x2 (1− x) e−x. Which of the following correctlyclassifies the critical points of f(x) ?

A. f(x) has no local extremum

B. f(x) has a local minimum at x = 0, and a local maximum at x = 1

C. f(x) has a local maximum at x = 0, and a local minimum at x = 1

D. f(x) has a local maximum at x = 0, and no local extremum at x = 1

E. f(x) has a local maximum at x = 1, and no local extremum at x = 0

14. If l’Hopital’s rule is used to evaluate L = limx→0

x− tan−1!

x"

x3, then

A. L = 1

B. L = 1/3

C. L = 0

D. L = 3

E. L = +∞

fix) has critical points x=o and ×=l

Fk ) + 0 + ° -

pylhe first derivativetest

• •

× ° 1 fcx) has a local maximum at

x=l .

It doesn't a local extremum at x=o

This is an indetermination of type °g

Using e' His pilot 's rule

,we have

lim they = HI, .

"5¥÷X→o ×2

Ft= fino -3×2=GIo×¥ ' 'Ez

=fiz ¥+5 's

MATH 265 L01-L02 TERM TEST 2 - VERSION 11 WINTER 2017 9

15. Consider the function f(x) = x2/3!

x2 − 16"

and its derivative f ′(x) =8!

x2 − 4"

3 x1/3.

Which of the following lists all the critical points of f(x) ?

A. −2 , 0 , 2

B. −4 , 0 , 4

C. −2 , 2

D. −4 , 4

E. 0 , 2

since fcx ) is defined in C-as to ),

its critical are the points where

either flex )= 8K¥3×413

is equal to 0, i.e.,

x=±2 or it does not

exist,

i.e.,

x=o .

Hence the critical points are

-2,0 ,2

MATH 265 L01-L02 TERM TEST 2 - VERSION 11 WINTER 2017 10

Part II: questions 16 and 17 below are written questions, worth 6 marks each.Make sure to show all your work and enter your final answers in the box provided.No credit will be given to unsupported answers.

16.!

6 marks"

Find the largest open intervals of increase and decrease of the function

f(x) = 3 x4 + 4 x3 − 12 x2 + 2

Make sure to show your work and enter your answer in the box at the bottom of the page.

f(x), increases in and decreases in

we start by computing f'

* )

flx ) = 12×3+12×2.24 × =12×1×4×-2 ) (2 marks )

= 12 × ( x - 1) ( × + 2)

and fiwlind its sign

flcx ) = o 12×1×-41×+2 ) = 0 ( =) ×= - 2,

0,

1

flex ) - O + O - o +• • • >&marks )

× -20 1

It follows

fx ) T in C-2

,

o ) and (1 ,+ a)

fat into,-2 ) and ( o

,, )

(2marks )

C-2,0 ) & ( t.tn ) C- as - 2) & ( 0,1 )

MATH 265 L01-L02 TERM TEST 2 - VERSION 11 WINTER 2017 11

17.!

6 marks"

Find the coordinates of the point(s) on the graph of the function

f(x) = ln!

2− x− x2"

where the tangent line has slope m =1

2.

Make sure to show your work and enter your answer in the box at the bottom of the page.

The coordinates of the point(s) is(are)

If ( c ,fcc )) is a point on the graph where

the slope of the tangent line is m= I ,we

must have f'

(c) =tz ( 1- mark )

I-

kmowks)

2- x. ×2( 2- ×-×2)1=

- 1- 2×

But flx ) = 2. × -×2

Hence f'e) =z ⇒ LIE = I ⇒

- 2 - 4C = 2 - C - c2 <⇒ c2 . zc -4=0 < ⇒ ( Cti ) ( C- 41=0

( ⇒ c= -1,

4 ( 2 marks )

only c= -1 is in the domain of fcx )

f (4) =ln( 2-4-16 ) can't be evaluated .

( lmark )

The coordinates of the point are ( t.feiD.fi

,lnl2))

( -1,

but))