MA 105: MATHEMATICS I

Autumn 2008

MID-SEMESTER EXAMINATION

Day: Monday

Time: 9.30AM - 11.30AM

Date: 8/9/2008

Maximum Marks: 30

Write your Roll Number, D'iu'is'ion and Tutorial Batch on the top page of your ansu)er-

boolc. Complete the 'inder on the top page i,ndicating the page number on which a parti,cular

questi,on 'is answered. If the i,nder is not complete, 2 rnarks will be d,ed,ucted from your

score. Attempt all the quest'ions; each question carries 3 marks.

1. Show that the sequence {o*}n>t is convergent and find its limit where

ar :1 and o 2an

n * l : 4 " " + r

2. (a) State the Mean Value Property of integrals.

(b) Use the Mean Value Property of integrals to prove that any continuous

function f t !a,b] ---+ IR has an anti-derivative.

3. Let "f , [0, 2r] ---+ lR. be defined by /(r) : z * sin r.

(a) Find the critical points of /.

(b) Show that / is strictly increasing over [0,2n'].

4. Let D be the bounded region in the first quadrant of the rg-pIane bounded

by the curves U : 12 and y : 2 - 12. lJsing both the Washer Method and the

Shell Method, find the volume of the solid of revolution obtained by revolving

D about the y-axis.

5. A particle is moving along a plane curve whose polar equation is r : e.0,

where c is a non-zero constant. Let L(c) denote the arc length of the curve

and A(c) denote the area swept out by the position vector of the particle as

d varies from 0 to2n. Compute tr(c) and A(c) in terms of c.

6. (a) Let f : la,b) --.IR be such that / is differentiable on (a,b). If / has a local

minimum at c € (a, b), then show that f '(c) : 0.

(b) State Rolle's Theorem.

7. A function /, continuous on the positive real axis, has the property that

['o f (t) dt: u [" tul dt + r [" y61 a,J t J t J t

for all r ) 0 and all y > 0. If /(1) : 3, find an explicit expression for f(r)

f o r r ) 0 .

8 . F o r r ) 0 , l e t / ( r ) : : t - 1 - l n r , S ( r ) : l n r - I * I l r .

(a) Find the absolute minima of / and g.

(b) Using the observations in (a), prove

1 - 1 5 l n r l - a _ 1r

f o r r ) 0 .

9. Let f (*): $'tor r > 0. Examining the critical points, inflection points and

asymptotes of /, sketch the graph of /.

10. Let f , [0,2] -+ IR be a continuous function satisfying /(0) : f (2).consider

g ( r ) : f ( " + t ) - f ( r ) .

(a) What is the natural domain of g?

(b) Applying the Intermediate value Property to g, what conclusions can you

draw about f?

L. a1:1 and a,,a1 : #fu. Note that

2ana n t L - a n :

f f i - a n :2an- 4a7- an _ a . - 4a7 _ an( I - 4a . )

Now or, ) 0 for all n. So, if. an> | for all n, then an*r- an 10 for all n, and it would follow

that {o,"},21 is a decreasing sequence that is bounded below, and hence {o.}.>t would be

convergent. [1 Mark]

W e p r o v e a . > i f o r a l l n b y i n d u c t i o n . C l e a r l y 7 : a t > ] . A s s u - Q a n ) ] . t n e n

in -2!z- , ) tn Ban ) 4an I r ifr 4a.> 1 iff o., L,4 A n . * r +

which is true. Thus 4,, > j for all n.Let lim,,** an : l. Note that &* > i for all n, implies I > i. Then a'*1 :

yields that I : #, that is, 412:I; hence 1: ] (as I cannot be 0).

2. f :10,2] - - -+ IR is cont inuous, / (0) : f (2) ,g( r ) : f ( "+t ) - f ( * ) .( a ) Thena tu ra ldoma ino f g i s [0 ,1 ] (as f ( "+ t ) - f ( " ) wou ldno tmakesense fo re i t he r r (0

4 a n * 1 4 a n l 7 4 a " l I

1a n + t )

4

[1 Mark],2"\, fot all n

4An-l L

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or r > 1 but makes sense for z e [0, 1])Now

g(0) : / (1) - / (0) and e(1) : f (2) - f ( t ) : / (0) - / (1) ( us ing / (0) : f (2D

Note that e(0) : e(1) i f f /(0) : /(1).The IVP can be aplied to g as 9 is continuous on [0, ]].

[1 Mark]

l-rt ^t , M.rkl

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[] MarklIf 9(0) I gG) then 9(0) and g(1) are non-zero numbers with opposite signs and, by the IVP

appl ied to g, there exists c € (0,1) (open inerval!) such that g(c):0, that is , / (c_1_1) : / (") .

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3. (u) Since / has a local minimum at c € (4,, b), there exists some d > 0 such that /(r) 2 f (")

f o r a l l r e ( c - d , c f d ) c ( a , b ) .Since / ' (c) exists, f ' (") : f ' *(") : f -(") .Thus i f we show that / i (c) ) 0 and f -(") < 0, then

it will follow that //(c) :0. t1 Mark]

Now, / i (c ) : l im7, -6* J@+h)-Jk) . For h e (0 ,d ) ,c +h € (c ,c l d ) and f (c+h) - / ( " ) > 0 ,

tnus fst9:l1d ) 0 for all such h and hence f*(") 20. [] Mark]

N e x t , / 1 ( c ) : J 3 p _ . F o r h € ( - d , O ) ' c * h e ( c - d , c ) a n d f ( c + h ) - / ( " ) ) 0 ,

tnus flstp ( 0 for all such h and hence f _(c) ! 0. l lMarkl

continuous on [a, b],(b) Rolle's Theorem: Let f : la,bl ------+ IR be a function such that / is

differentiable on (a, b) , and f(o) : f(b).

/ Then there exists some c e (a,b) such that f ' (c):0.

/ i

[1 Mark]

4. Consider f t I-n,2nl ------+ lR, /(r) : lrl * sin(r).( a ) N o t e t h a t , f o r r € f - r , 0 ) , f ' ( * ) : - 1 + c o s ( z ) a n d , f o r r e [ 0 , 2 n ) , f , ( r ) : 1 * c o s ( z ) .S i n c e / ' ( r ) d o e s n o t e x i s t a t r : 0 a n d s i n c e f ' ( r ) : 0 f o r r : z r : t h e c r i t i c a l p o i n t s o f f a r e0 and n.. [1 Mark]( b ) C o n s i d e r t h e i n t e r v a l [ 0 , r ] . F o r a n y r l , r . 2 € [ 0 , 2 ] , w i t h r l L r 2 , t h e r e e x i s t s c € ( q , 1 2 ) C(0,n) such that f (r2) - f ("r): f '(c)(rz - 11) (This follows from Lagrange's MVT). sincef ' ( " )>0 , f ( rz )> f ( r t ) . Hence/ iss t r i c t l y inc reas ingon [0 ,2 r ' ] . Byas imi la rargument , / i sstrictly increasing on [zr,2n-]. Conclusion: / is strictly increasing on [0,2n]. [2 Marks]

s . f ( " ) : t r - 2 - l n r + f ' ( r ) - 1 - I + f ' ( r ) : 0 i f r : L . A I s o , / " ( 1 ) : # ] , : r : 1 ) 0 .Thus / has alocal minimum at r :1, and since / is def i rred on (0,oo),r :*1- iJactual ly apoint of absolute minimum for /. [1 Marklg ( r ) : l n r - 1 + : I * s , @ ) : i - # : * ( r - * ) + g , ( r ) : 0 i f r : r . A r s o ,g"(7) : -

$ + #1,:, : 1 ) 0 . Tlius 9 has a Iocal il inirnrr- at tr : 1, and since g isdefined on (0,oo), r: 1 is actually a point of absolute minimum for g. [1 Mark]S i n c e / ( 1 ) - 1 - 1 - l n 1 : 0 , u s i n g t h e r e s u l t i n ( a ) o n e h a s f ( " ) > / ( 1 ) : 0 f o r a i l r ) 0 ,that is, r - I ) lnr for all r > 0. [] Mark]S ince 9(1) : ln1- 1+ 1 :0 , us ing the resu i t in (a ) one has 9( r ) > 9 (1) :0 fo r a l i r > 0 ,that is, lnr ) 1- j for all r > 0. [] Mark]

6. f(r): Y@ r 0) =+ /(r) : t# : t=y. Thus, f, > 0 for r 1ei so that /is.strictiy increasing over (0, si], and f' < 0 for r > e* so that / is strictly decreasing over[e ] ,oo) . C lear ly r :e t i s theon lycr i t i ca lpo in to f f , in fac t ,apo in to f loca lmax imumof / .[1Mark]

Further, f" (r) :- ?r3 -zr2 (t -zlnt)

Thus, f" < 0 if. r < ei so thatf is strictly concave (that is,strictly concave downward) over(0, e3;, and, f" > 0 i f

" > ,3 ,o

that / is strictly convex (thatis, strictly concave upward) over(e8, oo). Clearly, r : eE is theonly point of inflection of f .

[1 Mark]Now, lim"*6 /(r) : -oo andlim"-* f (r) : 0 so that the y-axis is a vertical asymptote of/ and the r-axis is a horizontalasymptote of /. [] Mark]Correct graph [] Mark]

, . r : e e + # : " " * .

: Io'" J er.\ cre%o : ,tr * * fr'n "* o, : ry k*llo"

:ry@'^"-r).A(c):I lo'" , '0,: ;

I ,^ e2dde

- r lor*f'"

4c t - r o

: l y " n " - t ] .4 C -

8.(a) Appealing to the Leibnitz rule or the FTC, one can differentiateftu fr faI f ( t )d t :y I f ( t ) td t+* I f ( t )d t

J r J t J rwith respect to r to obtain f ("ily : af @) + Il f ft)dt.

P u t r : l t o o b t a i n r a

f (y)a: af G) + J, f Q)dt.

Thus

f @): f (t)a + I l f (Dat

aand

f , tu): l f ( t )+ f (y))y- l t (J)a+ I i f ( t )at l _af @)- I l f @at

y2 a2which is continuous. This shows that / is continuously differentiable.

b) f(a)y :Ba * I i f(Dat + f '@)a + f@): 3* f( i l + f '@) : i+ f (y): 3lny * c, where c is some constant.S i n c e / ( 1 ) : 3 , c : 3 ;

thus /(y) : 3lng * 3 for y > 0.

9,(a) Mean Value Property of integrals: lf. f : la,b] ------+ IR is a continuousthere exists some c e [a,b] such that Il f @)a, : f (c)(b - a).

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[] uart<l

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function, then

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(b) Let f , la,b] ------ IR be continuous. Define F(r) : I: f (t)dt. Note that the integralexists since / is continuous. Then, for h, suchthat r +he [a,b], one hasF(r + h) - F(r) _ I:t' f (t)dt - I: f (t\dt pr*h

h h t:

J, f (t)dt ' [1 Mark]

By the MVp of integrals, there exists some c between r and,r * D such that

f,'*o f(t)dt : f (c)(r + h - r) : f (c)h,

thusF ( r + h ) _ F ( r )

- : f ( c ) .If h tends to zero (so that c also tends to 0), then /(c) tends to f(r)as / is continuous. Hence

,.^ F(r + h) - F(r) _ rt-\

h - O h - J \ r ' )

showing that F,(r): r. This tr, is an anti_derivative of /.

L0. Washer Method:

sh.n M\j;i::: : * f' {/r)'dv + * l,' rt/z - at,da : *(t * *, :,.

[1 Mark]

[2 Marks]

[1 Mark]voiume :n

I ' r [ (2- r )2pr- rzldr : nn

Jo @ _ rz)d,r _ n"(; _ I,

: *