Bqsic Concepls
ColculusA: n Introduclion
DifferenlioCl olculus
Inlegrol Colculus
Definife Inlegrol
DifferenlioEl quotions
Answerslo exercises
Concepls& Formuloeo f o Glonce
Solved Problems
AssignmenPl roblems
Answerslo AssignmenPl roblems
Bnsrlleln rnEilrncs
Set Theory:
Sef;
A seti s a welldefinecdo llectioonr ensembloef objectos r elementsE.a che lemenint a seti s unique.
Usuallbyu tn otn ecessarail ys eti s denotebdy a capitalel tteer .g.A , B, .... U,V etc andt hee lements
aree nclosebde tweebnr acket{s } , denotebdy s mallle tter?s, b , ... x, y etc.F ore xample:
A = Seto f alls mallE nglisha lphabets
= {a, b, c, ....,x,y,z} :
B = Seto f all positivein tegerlse sst hano r equatl o 10
= {1,2,3,4 ,5 ,6,7 ,8 ,9 ,1 0}
R = Set of real numbers
={x: -m<x<o}
The elementso f a set can be discrete(e .g.s et of all Englisha lphabetso)r continuou(se .9.s et of real
numbersT).h es etm ayc ontainfi niteo r infiniten umbeor f elementsA. setm ayc ontainn o elementas nd
such a set is called Void set or Null set or empty set and is denoted by 0 (phi). The number of
elementosf a setA is denoteda s n(A)a ndh encen (0)= 0 as it containsn o element.
Union of Sets,'
Uniono f two or mores etsi s the set of all elementsth at belongt o any of theses ets.T he symbolu sedf or
uniono f setsi s 'u'.
i.e. A v B = Union of set A and set 3 = {x: x e A or xeB (or both)i
e.g.l f A={1,2,3,4}andB={2,4,5,6}andC={1,2,6,8} thenAuBvC={1,2,3,4,5,6,8} .
Intersection of Setsi
It is the set of all the elementsw, hicha rec ommonto all the sets.T he symbolu sedf or intersectioonf sets
is'n'i.e. AnB = {x:x e A and xe B}
e.gl fA = {1,2,3,4}a ndB = {2,4,5,6}a ndC = {1,2,6,8it,h enA n B n C= {2I .
Rememberthant( Au B) = n(A)+ n(B)- n( A n B)
Difference of Two Sefs.'
The differenceo f setA to B denoteda s A - B is the seto f thosee lementsth ata re in the set A but not in
thes etB i.e.A - 3 = {x:x e Aandx e B}.
Similar[B - A = {x:x eB andx e A}. Ing eneraAl -B + B-A
e.g.l fA= {a,b ,c ,d }a ndB ={b,c ,e ,f },t henA- B = {?,d }andB -A= {e,0 .
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Subset of a Sef;
A set A is said to be a subset of the set B if each element of the set A is also the element of the set B.
Thesymbouls edis 's' i .e.AsBe (xeA= xeB) .
Each set is a subset of its own set. Also a void set is a subset of any set. lf there is at least one element
in B whichd oesn otb elongto thes etA , thenA is a propers ubseto f setB andi s denoteda sA c B.
e.glfA={a,b,c,d}andB={b,c,d}thenBcAorequivalentlyA:B(i.eAisasupersetofB).
Equatity of Two Sefs.'
SetsAandB ares aidt o bee quailf As B andB sAand wewriteA= B.
Universal Set:
As the namei mpliesi,t is a setw ithc ollectioonf all the elementas ndi s usuallyd enotedb y U. e.g.s eto f
realnumbersRisauniversalsetwhereasasetA={x:x<3}isnotauniversalsetasitdoesnotcontain
the seto f realn umbersx >3. Oncet he univercasle t is knowno, nec and efinet he Complementarsye t
of a seta s thes eto f allt hee lementos f the universasle tw hichd o notb elongto tfat set.e .g.l f A={x:x <
3) then A 1orA " ) = complimentiasreyt o f 4={x:x > 3}. Hencew e cans ayt hatA uA = U i.e.U niono f a
set and its complimentairsy alwayst he Universasle t andA nA = $ i.e.i ntersectioonf the set and its
complimentariys alwaysa voids et.S omeo f the usefulp ropertieos f operationos n setsa rea s follows;
. A (or (A")")=A, A^A" =0 and AuAc =U
I Au$=AandAno=0
r AUU=Uandfu-rU=A
r An(BuC) = (AnB)u(AnC)
r Au (BnC)=(AuB) n (AuC)
De Morgan Laws
. ntg = Ane
r AnB = AUB
lllustration 1.
lf A = {1, 2, 3, 4}, B = {3, 4,
and B- A.
Solution:
A \JB-= n nE ltromD eM organ'Ls aws)
A=U_A ={5,6,7,...,101 ...(1)
E:u_g ={1,2,9,9, 10} ...(2)
AnB ={9,9, 10}
ArlB- - AvB lnomD eM organ'Lsa ws)
= {1,2,5,6,7,8 ,9 ,1 0i (usin(g1 )a nd( 2) )
A- B = {x x e Aand x e Bl = {1,21
B -A = {x: x e B and x e A} = {5, 6, 4.
5, 6, 7] andU = {1,2 , 3, ..., 10},t hene valuateA v B, AoB , A - B
L
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Find A vB, A rB, A- B, B-A, where
i ) A= (7,2,3,4,5,6 7,8,9,101,B ={7,3,5,7,9}
.ii) A=tx:xeN], B = tx:xel]
iii) l={x:xeR]', B= {x:lxl3}
Natural Numbers:
Then umbers1 ,2 , 3, 4... arec alledn aturanl umberst,h eirs eti s denotedb y N.
ThusN={1,2,3,4,5...}
lntegers:
Then umber.s. .-9,- 2, -1,O ,1 ,2 , 3...a rec alledin tegerasn dt hes eti sd enotedb y I orZ .
ThusI (orZ 7= {...- 3, -2, -1,O ,1 ,2,3...}.l ncludinagm ongs eto f integerasr e
r Seto f positivein tegerds enotedb y | * andc onsistso f {1 , 2,3, . . .} (= N)
r Seto f negativein tegersd,e notedb y I - andc onsistso f {..., -3,- 2' -1I
r Seto f non-negatiivnet eger{s0 ,1 ,2,....}c, alleda ss eto f wholen umbers'
r Seto f non-positivinet eger{s. ...,- 3, -2, -1, 0I
Rationat Numbers:
All numbers of the form p/q where p and q are integers and q * 0, are called rational numbers and their
set is denoted by Q.
[^
ThusQ = 1f:p, qel and q+ 0 and HCF of p, q is 1f .
[q - J
It may be noted that every integer is a rational number since it can be written as p/1. lt may also be noted
that eithera ll terminatingo r all recurringd ecimalsa re rationaln umbers. e.g.,p = 0.3 = 0.33333'..
then,10P-P=3+9P =3
1
+ p = 3/9+ p =-, whichi s a rationanl umber.
J
lnational Numbers:
Therea re numbersw hichc an notb e expressedin p/qf ormo r the numbersw hicha re neither ecurring
nor terminatinagr e inationanl umbersT. hesen umbersa re calledi nationanl umbersa nd theirs et is
denotedb y Q" (i.e.c omplementasrye t of Q) e.g. .riZ, 1+./3, n etc. lrrationanl umbersc an not be
expresseads recurrindge cimals.
Reat Numbers.'
Thec ompletese to f rationaal ndi nationanl umbersis thes eto f realn umbersa ndi s denotedb y R. Thus
R = QuQ".
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It may be noted that N c I c Q c R. The real numbers can also be expressed in terms of position of a
pointo n the real line.T he real line is the numberl ine whereint he positioni rf a pointr elativet o the origin
(i.e.0 ) representsa uniquer eal numbera nd vicev ersa.
-3 -2 -1 0 1"12 2 3n
Realn umbelri ne
All the numbersd efineds o far followt he orderp ropertyi. e. if therea re two numbersa and b then
eithera<bora=bora>b.
Expressing a Non-terminating hut Recurring Decimalto Rational Form:
Let a non-terminatinagn d recurringd ecimacl an be expresseda s 0.
non-recurrinpga rth avingm digitsa ndY is the recurringp ait havingn
Now, R = O.XY
= 1O'R = X.V
10'*n R= XY.V
Subtractin(g1 )f rom( 2),w e have
(10'*n-10')R=ff-{
XY-X
>K=
1ot*n -10t
lllustration 2.
Express the rational number as ratio of two
0.12345454545.....
X V such that X is the
\----vJ\----vJ
m digits n digits
digits.
integers whose deaimal expansion is
...(1)
...(2)
Solution:
Letx = 0.1234545454..5..
Then1 03x= 123.45454..5. .
105x= 12345.454.5.. .
Subtractinwge get( 10u- 103x) = 12222
-
*=12222
99000
lllustration 3.
Express 23.6245 as the ratio of two integers.
Solution:
Let R = 23.6245
100R = 2362.45
10000R= 236245.45
9900R =236245-2362
233883
.. ia= -
9900
Intervals:
Intervalsa re basicallys ubsetso f R (seto f realn umbersa) nda re of verym uchi mportancien calculus
as you will get to knows hortly.lf therea ret wo numbersa , b e R sucht hata < b, we can definef our
typeso f intervalsa s follows:
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r Open Interval: (a, b) = {x: a< x< b} i.e. end points are not included.
r ClosedI nterval:[ a, b] = {x: a < x < bi i.e. end pointsa re also includedT. his is possibleo nly
when both a and b are finite.
Semi Open/ Semf Closed/ Half Open/ Half Closed lntervals:
r Open-closedIn terval(: a, bl or la, bl = { x: a < x < b}
Closed- open Interval[:a , b) or [a, b[= { x: a < x < b}
The infinite intervals are defined as follows
r (a,m)={x:x>a},
r [a,co)={x:x>ai
r (-*, b) = ix:x< b)
r (*,bl ={x:x<b}
lntervalsa re particularliym portanitn solvingi nequalitieos r in findingd omainse tc.
tnequatities:
The following are some very useful points to remember:
r a(b=eithera<bor a=b
r a<band b<c=+a<c
r a<b=a+c<b+c VceR
r a ( b = -a > -b i.e. inequalitys ign reversesi f boths idesa re multipliedb y a negativen umber.
r a < b and c < d = a + c < b + d and a -d < b-c.
r a < b= ma < mb if m > 0and ma > mb if m < 0.
r 0<a<b=a'<b'if r>0and a'>b'if r<0
/ 1\
r la+- I > 2Y a> 0 ande qual i tyholdsfora= 1.
\a/
/ 1\
r i a+-l <-2Y a<0andequalityholdsfor a=-1
\a/
lllustration 3.
Find x suchthat3x-3 >x + 5 ana J >
x-1
Solution:
ox-3>x+5= 2x>8=x>4
11
x-1 B
from (1) and (2), we have 4 s x < 9.
Wavy Curue Method:
.....(1)
.....(2)
The
f(x) =
Method of intervals (or wavy
(* - r,')"' (* -
^r)n"
.... (" -
"* )n*
(* - b. )T' (" - br)*' .... (* - oo )''
curve) is used for solving inequalitie'so f the form
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naturanf umbersa nd the numberse 1,a 2,...o, k,b 1,b z,...boa re any realn umberss ucht hata i * b;,
wherei = 1,2,3, . . .k, and j = 1,2,3, . . .p, .
It consistsin thef ollowings tatements:
(i)
(i i)
(i ii)
(iv)
All zerosl of the functionf (x) containedo n the left hand side of the inequalitys houldb e
markedo n the numberli new ithi nked( black)c ircles.
All pointso f discontinuitleso2f the functionf (x) coirtainedo n the left hand side of the
inequalitsyh ouldb e markedo n the numberli new ithu n-inked(w hite)c ircles.
Checkt he valueo f f(x)f or any realn umberg reaterth ant he rightm ostm arkedn umbero n the
numbelri ne.
Fromr ightt o left,b eginninga bovet he numberli ne( ln caseo f valueo f f(x) is positivein step
(iii) othenrvisfero m belowt he numberl ine),a wavy curves houldb e drawnw hich passes
througha ll the markedp ointss o that when passest hrougha simple point3,t he curve
intersectsth e numberl ine,a nd, when passingt hrougha doublep ointat,h e curver emains
locatedo n ones ideo f then umbelri ne.
The appropriatein tervalsa re choseni n accordancew ith the sign of inequality(t hef unction
f(x) is positivew herevetrh e curvei s abovet he numberli ne,i t is negativeif thec urvei s found
belowt hen umbelri ne)T. heiru nionr epresentsh es olUtioonf inequality.
(v)
Remarks:
I The pointw hered enominatoisr zeroo r functiona pproacheisn finityw ill neverb e includedin
the answer.
r Pointso f discontinuitwyi lln everb e includedin the answer.
r lf you are askedt o findt he intervalsw heref (x) is non-negativoer non-positivteh enm aket he
intervalsc losed correspondingto the roots of the numeratora nd let it remain open
correspondintgo the rootso f denominato
tllustration 4.
Let f(x) = (x - l)3 (x +-z)4 (x = s)5 (x + o)
. so/ve thef ottowingin equatiy
- x2 (* _7)t
(i) f(x) > 0
(iii) (x) < 0
Sotution:
We marko n the numberli nez eroso f the function1'., -2,3 and- 6 (withb laekc ircles)a ndt he
pointso f discontinuit0y and 7 (withw hitec ircles)i,s olatet he doublep oints:- 2, and0 , and
draw the curve of signs:
1 The point for which f(x) vanishes ( becomes zero ) called function zeros for e.g. x = a1
2 The points x = bj , j = tl, 2, 3, ..., p] are the point of the discontinuity of the function f(x)
t tf the exponents of a factor is odd then the point catled simple point
o ff the exponent of a factor is even then the point calted doubte point
(it) (x) > 0
(iv) (x) -<.0
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Fromg raphw, e get
(i) x e (-oo, -6)u (1, 3)u(7, m)
(ii) xe G* , -61u {-21u[ 1,3 ]u (7,m )
(iii) xe (-6, -2)u (-2,0) u (0, 1) u (3,7)
(iv) x e t-6, 0) u (0, 1lu [3, 7)
Logarithm:
r Theexpressionlo goa is meaningful fora> 0 andforei ther0<b < 1 orb> 1.
r a - b'o90"
r loo^ b = log" b
log" a
r .1 logr,a - --------p--r ovidedb oth a and b are non-unity.
log" b '
r tft oso>at ,o so,an," "
{;,;rr'^: z] 1.,r
Ittustration5 .
"
(
"i r: t) - y (z + x - y) - z( x.+ y - z),
(i) ,r-ffi bgz thenp rovet haftz v = / f = xvy'.
(ii) Solve for x,
(a) log2x 2log12$-1). (b)
(c) tog, Iogu(Jx + 5 * Ji) = 0.
Solution:
,,,,r",
x(y+z-x) _ y(z+x-y) -z(x+y-z) -n \'/ --- logx logy logz
x(y+z-x) y(z+x-Y) . z(x+Y-z)
:>X= I k ,y= e k andZ= e k
yz(z+x-y) w(x+y-z) xyz
Now,t'zy=e k g * =ek
xy(y-z-x) xy(z:I y\ xyz
SimilarlY,xtl=e k e k -ek
Hence, tfzy = lf = xtfl.
(ii)(a) We can writet he givene xpressiona s
log2x> log2 (x-1) --1r,, > log2x> - log2 (x-1)
,ot t l s'lr)
4los22x = 36 .
Find the interuals where f(x) is positive or negative
iii)
& -1)& - 2). .0
Qx-ilQ+4)
1og.x-1 ,o
(x-I)(x-3)
iv)
=3r*4 ,0
x'-3x+2
(e* -t) xae(i - x)(x - 3)50>0
& -2)& - a)
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? logzx (x -1) > 0 = log2x(x-1 ) > log21
+x(x-1)>1 +x'-x -1>0 =(x-S)(*-1*f l=o 2" 2
By wavy curve method the inequality holds good for
x>- t+Js or x<-1-G
22
1_, ls 1+ Jd i.e (-oo,
;1..r1 zi,-).
(b) 4tosrz, = 36 + 22tost2x - g6
;, 2lo9r4x=z 36 +4x2 = 36 sy2 =9 + x = + 3.
(c) log,l ogu1 fi-** 5'a fi ; = O
= logu(Jx+5*Ji;:t,
=Vx+5+Vx=5
(x+5)-x=5
Dividing(2 ) by (1)
Jx+5-Ji=t
Solving( 1)a nd (3), ZJx + S = O
....(1)
....(2)
.
....(3)
=x= 4.
ii)
iv)
logx =log.y = lo9,
'y-z z-x x-y , provethat{*,y t+xz I'v a 1.
Solve log6(f+Q > Iogl.3Qx). tit) Soive gltogs(x+l)
tf a2 +h2= 23ab, then prove that 619(9 :!) = {flog" + togb ) .
\c)z
=22logrx+3.
Absolute Value:
Let x e R, then the magnitude of.x is called it's absolute value and in general, denoted by lxl and
. | | l-x, x<o
OeTlneaoS lrxr l = II x, x>o
Note that * = 0
""n
be included either with positive values of x or with negative values of x. As we
know all real numbers can be plotted on the real number line, lxl in fact represents the distance of
number'x'fromt he origin,m easureda longt he number-lineT. hus lxl > 0. Secondlya, ny point'x' lying
on the realn umberl inew ill have it'sc oordinatea s (x, 0). Thus it'sd istancef romt he origini s Jf .
Hencelx l= Jx2 .Thusw e cand efinelx la s lxl= Jx2
e.gi fx = -2.5, thelnx l= 2.5, i fx = 3.8t henl xl= f , .3.
Basic Properties of lxl;
' l l " l l = l " l
r lxl>a=x,>aorx<-aif a e R*andx e Rif a e R-
. l " l< a +-a <x< a i f a e R-and no solut ioni f a e R-u{0}
r lx +yl<lxl+lvl
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(i) log172-l x1 l < 3
/r\3 1 +lx-,ltIZJ =a =lx-
=-;r*=$=*=(--,
RSM-91 1-P1-MA- Basic illathematics
' l"-vl>lxl-lvl
Thel ast wop ropertiecas nb ep uti no nec ompacfot rmn amelyl*, l - lyl< l* * yl 5 lxl+l Vl
' l*v=l l xllvl
lxl l " l I pl=H , y+0
tyt lyl
Illustration 6.
Solve for x.
(i) logl2lx - 1l -< 3
Solution:
(ii) lx-11+lx-21<5.
t t'-8 I sinceb aseb < 1
el[g ) --lul-. col.
8lL8 )
( i i ) lx-11+ lx- -2|s5
usinpgr operltxy- 1l+ lx- 21>l2x-31= l2x-3 l< 5
=-5< 2x<8+-1 <x<4.
i) Solve llx- tl +21 <4 ii) Solve cosx + lcosxl 20 for x e [0, 2r]
iii) Solve Uoglxll-<t.
Calculus is the study of the concept of functions and their behavior. These concepts are based on the
theory of real numbers. Broadly, it is classified into two parts: Differential calculus and Integral
Calculus.
Differential Calculus deals with finding the rate of change of one physical quantity with respect to
(written as w.r.t. from now onwards) another when the two are connected by a functional relation. For
example, finding the speed (rate of change of distance with respect to time) when the distance is
given as a function of time. The advantage it offers is that we can get the exact rate of change at any
givent ir, r providedw e knowt he functionadl ependence.
Integral Calculus deals with finding one physical quantity in terms of the other ( i.e, the functional
dependence ) given the rate of change of the first with respect to the second, i.e., it is just the reverse
process of Differential Calculus. For example finding distance as a function of time given, we know
the speed (as a functiono f t, of course)a nd one condition( e.9.,d istancea t a particulatri me). .!.
Before we going for details let us go through the fundamentals of varidbles functions and their types.
Variabte and Constant:
A variableis a quantityth at takeso n variousn umericavl alues.A constanits a quantityw hose
numericavla luer emainsfi xede .g.w hena bodyf allsn eart he surfaceo f earth,it sv elocityin creases
butt he acceleratio(nd uet o gravity)r emainsc onstant.
']llnt6c
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It shouldb e notedt hatw henc onsiderinsgp ecificp hysicapl henomenait, m ayh appenth ato nea nd
the sameq uantityin onep henomenias a constanwt hilei n anotheirt is a variableF. ore xamplet,h e
velocityi n a uniformm otioni s constantw, hereast he velocityin a uniformlya cceleratedm otioni s a
variable.
Quantitiesth at havet he samev alue undera ll circumstanceasr e calleda bsolute constants. For
examplet,h e ratioo f the circumferencoef a circlet o its diameteirs an absolutec onstannt , periodo f
rotationo f eartha bouti tSa xiso r aroundt he sun,u niversagl ravitationaclo nstantp, ermeabilitoyf free
space, etc.
Range of a Variable:
A variable takes on a series of numerical values. The values of the variable are geometrically
depicteda s pointso n a numbers cale.T he set of all numericavl alueso f a variableq uantityi s called
the range of the variable.
e.g. The rangeo f valueso f the variablex = coscrf or all possiblev alueso f a is the interval [-1,1] i.e. -
1<x<1.
Term function is used to define the dependence of one physical quantity on another e.g. volume 'V' of a
sphere of radius 'r' is given by V = * n t . This dependence of 'V' on 'r' would be denoted as
J
V = f (r)a ndw e woulds implys ayt hat' V' is a functiono f 'r'.H ere' f is purelya symbo(l fort hatm attera ny
otherl etterc ouldh aveb eenu sedi n placeo f 'f), and it is simplyu sedt o representth e dependencoef
oneq uantityo n thbo ther.
Definition of Function:
Function can be easily defined with the help of the concept of mapping. Let X and y be any two nonempty
sets. 'iA functionf rom X to Y is a rule or conespondenceth at asslgnst o each elemento f set X,
one and only one elemento f set Y'. Let the conespondenceb e 'f then mathematicallwy e writef : X -+ y
where y = f(x), x eX and y eY. We say that 'y' is the image of 'x' under 'f (or x is the pre image of y).
Two things should always be kept in mind:
I A mapping f: X -+ Y is said to be a function if each elemeht in the set X has it's image in set y. lt
is possible that a few elemedts in the set Y are present which are not the images of any element
in set X.
I Every elementi n set X should have one and only one image. That means it is impossiblet o
have more than one image.f or a specifice lementi n set X. Functionsc an't be multi-valued( A
mappingt hat is multi-valuedis calleda relationf romX to y)
Function Function
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RSM-91 t-P1-MA- Basic Mathematics
(i)
(ii)
(iii)
Let us considesr omeo there xamplesto maket he abovem entionedco nceptsc lear'
Letf:R*-+Rwheref=*.Thiscan'tbeconsideredafunctionaseachxeR-wouldhavetwo
imagesn amelyt Jx . tus itwouldb e a relation.
Letf : [-2,2 ] _+R ,w here* * f= 4. Herey =t J4 *, thatm eansfo re veryx e l-2,2 lwe
wouldh avet wov alueso f y (excepwt henx = t 2). Hencei t doesn ot represenat function.
Letf: R-+Rwherey=x3.HereforeachxeRwewouldhaveauniquevalueofyinthesetR
(asc ubeo f any1wod lstincrte aln umbersa red istinct)H. enceit wouldr epresenat function.
Distinctionb etweena relation and a function can be easilym ade by drawingt he grapho f
y=f(x).
These figures show the graph of two arbitrary curves. ln fig.(a) any line drawn parallel to
y-axisw ouldm eett he curvea t onlyo nep ointT hatm eanse ache lemenot f X wouldh aveo nea ndo nly
onei mageT. husf ig (a)w ouldr epresentht eg rapho f a function'
In fig.(b) certain line (e.g. line L) would meet the curve in more than one points (A, B
and C). Thuse lemenxt o of X wouldh avet hreed istinctim agesT. hust his curvew ill not represenat
function.
Hence if y = f(x) represents a function, lines drawn parallel to y-axis through different points
correspondintog p ointSo f setX shouldm eett hec urvei n onea ndo nlyo nep oint'
Basic Elementary Functions:
(i). Gonstant function:
Y = c where 'c' is a constant, defined for all
real x.
f:R+{c}
defined bY f(x) = c
(ii). Power function: Y = x" (where cr is a constant)
(a)c ri s positivein tegerT. hef unctionis definedin the infiniteln terval- m<x <co.
(b)g is negativein tegerT. hef unctionis definedfo r all valueso f x exceptf or x=0.
Fis. (b)
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(iii). General exponential function: y= a", where a is
' positiven ot equalt o unity.T his functionis defined
fora llv alueso f x.
f:R-+(0,0o)
defined by f(x) = sx
(iv).
(v)-
Logarithmic function: y = log"x, a > 0 but a + 1.
This function is defined for all x>0.
f:(0,m)+R
defined b! f(x) = |n x
Trigonometricf unction:y = sinx,y =cosxd efinedfo r all realx
y = tanx,y = s6cx,d efinedfo r R - (2n+ l7! .
y = cotx,y = coescxd, efinedfo r R-nrr, *i"r" n .,
It must be noted that in all these function the vadable x is expressed in radians.
functionh avea veryi mportanpt ropertyth ati s Periodicity.
Y = stnx
All these
In (x)
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(vi). Algebraic function:
(a)P olynomiaful nctionY: = aoxnt ?tXn-+1 "' + an,w hered o'dl"" ana rer eal
0) andn is a non-negativinet egerc, alledth ep olynomiaolf degreen ' e.g:
constants(a s+
y=ax+b,a+0
y = ax2 + bx +c, a+0
(a linearfu nction)
(a quadraticfu nction)
A polynomiaflu nctioni s definedf or all realv alueso f x'
?6Xn * a.xn
-1 +'....* ?n
(b) RationalF unction\:/ = v '
' bo*t + b.,xt-1+ ....+ b,
e.g:y = a/x (inversev ariation)
The rational function is defined for all values of x except for those where the denominator
becomes zero.
(c) lrrationaflu nction:
2x2 +Ji
e.g.y =
Let y = f(x) be a function.P uttingt he valueso f 'x' in this relationw, e obtaint he corresponding values
of ,y,. Suppose we start putting some values of 'x' in increasing order. The respective values of 'y' that
we obtainm ay turn out to be in increasingo rder,o r in decreasingo rder,o r they may remainc onstant,
or they may even havea mixedt rend,d ependingu pont he type of function.
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14 RSM-91 1-P1-MA-Basic Mathematics
Letustaketwovaluesof x:x1 andx2 (xr < xz). So, yr = f(xl) and y2 = f(x2). Then, the quantity
/v;-v,)
[xz-xr )
-
Let y2 > y1 a !2-Jt is positive = Function is increasing on an average,
xz-xt
. V^ _V,
lf yz< yt = 'z ' I is negative= Functioni s decreasingo n an average.
xz - x't
V^ _V,
lf yz=y .t3 r z t t is zero= Functionis constanot n an average.
xz-xr
As you can see, if xr and x2? rasufficientlfya r apart,t he quantity[ V'
- Vt
] can not give the exact
\xz - xr ,/
ideao f the variationo f y w.r.t.x in the intervallx r,xz]l.t just provideso veraliln formationF.o re xample,
if yz= yt, it doesn ot necessarilmy eant haty is the samef or all x in the interva[lx 1,x 2].T hus,t o obtain
a "sufficientlayc curatein formationw", e havet o choosex 1 ond xe 'lsufficientlcyl ose"t o eacho ther.
This 'sufficientlyc lose' is the key word here.
To know the rate of change of y w.r.t. x at x = X1, we take x2
very near to x1 (as much as possible), i.e., 'x2 tends to x1' and
then cafcul'rr"Yz-Yt. In the limiting case, we say that x2
xz-xt
nearly coincides with xr and represent it as X2 -) X1. We use the
notation $l 6, Yz -Yt
ds X2 )X1.
oXlx=& xz - xt
'dx'm eanss mallc hangein x ( nearx = xr) and' dy'm eansth ec orrespondincgh angein y.
We call I tn" derivativeo r the differentiaclo efficienotf y w.r.t.x .
dx
(You can understandit plrrysicalblyy takingx as time and y as displacemenotf a body.T hen
denotesth e magnitudeo f velocity).
9I i, atsor epresenteads t'(*) or"
dfl") .
dx dx
Graphicatty:,: l {i.e., f}computed
at x=x1) denotes the slope of the tangent to the curve
oXlx=x,
y=f(x)atx=xr.
We will not here derivet he formulaef or f '(x) of variousf unctionsb, ut we give the resultso f the
derivationhse re.
Basic DifferentiationF ormu lae:
dy
dx
y = constant =
y=X" ;a =COnstan=t +
y= sinx
9=o
dx
9r = o"o-r
dx
9r=
dx "o""
y = tan-1x =
Y = coflx +
y = coseclx
dy
dx
dy
dx
dy
dx
=- 1
1+ x2
*1
=-
1+ x2
-1
l*lJ7-
{"i
I
H,.
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RSM-91 1-P1-MA- Basic Mathematics 15
y = cosx
y=secx
y = cosec x
y = tanx
y = cotx
Y = sin-1x
Y = cos-1x
dV- =-stnx
dx
9I =r""xtanx
dx
dY =-cosecxcotx
dx
9I = ss6'x
dx
dY = - cosec'x
dx
Y = sec -1x
Y=a"
Y=e"
y = log"x
Y=lnx
dx l-lJFl
"'-a*lna
dx
dvy
:-a"
dx
dvl
dy_
ddy
dx
xlna
=!
X
dv1
-j=- - dx J1*
dv -1
-
- -d- -1x=- J1=
Example of Velocity and Acceleration:
Let distancec overedb y a particlem ovingw ith constanta ccelerationa in time t be given by s(t) = u{ +
1at2 alono a same line.
2"
Then velocity at any time t will be the rate of change of s w.r.t. t.
i.e. S= u + at i.e. v(t) = u + at
dt
SomeI mportant Theorems:
The followinga re very importantth eoremsw hichc an be appliedd irectly.
Theorem 1:
lf a functioni s of theform y = kf(x), wherek is a constantt,h en
l|
= f.
#
Theorem 2:
The derivativeo f the sum or differenceo f a finite number of differentiablef unctions is equal to the
sum or drrierenceo f the derivativeso f thesef unctions,
i.e.if y = u (x) + v(x) + w(x), then y'= u'(x) + v'(x) + w'(x).
Theorem 3:
The derivativeo f the producto f two differentiablefu nctionsi s equal to the producto f the derivativeo f
the first function with the second function plus the product of the first function with the derivative of the
secondf unction:i. e.i f y = uv , theny ' = u'v + uv'.
This formula can be extended for the derivatives of the product of any (finite) number of functions.
lllustration 7.
Find the derivativeo f
(i). y=secx+tanx+x1n (ii). Y = x3/2 slnx cosx.
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RSM-91 1 -Pl -MA-Basic Mathematics
Solution:
(i)y=u+v+w
u = du secx'd.- -xl = secxtanx
v = tanx =) V' = seC2X
w=xtl3= *'=!7-2t3
3
y, = Ui+ V,+ W,= SeCXta nX+ SeCl +
I**
(ii) Y = x3/2s inxc osx
n ='xst2,=v sinx,w = cosx
y' = u'v\/ + uv'w + uvw'= gJisin"cosx+x3/2cos2
x-x3l2sin2x.
2
Itlustation 8.
Findt he deivativeo f y = (a + x) d w.r.t.x .
,.1
*""I(a+x)
ox'
=en(a+x+1)
Theorem 4:
lf v = u(x)
. then v' =
' v(x)
lllustration 9:
Differentiawt.er .tx. where' ,= ', !, 2",1x
Solution:
, (r-olfi-(x-4) r1 *-trz
.,-=, r r - l ' , 2 .'.,_
Theorem 5:
lf y - uu, where u and v are functions of x, then y' = vuu-t u' + uu /n u.v'
Illustration 10. i
Differentiate the following w.r.t. x
Solutian:
Usingt heorem3 .
I : t^+ x)ae*
dx ' 'dx
= (a + x)e" + et.1
(i)Y=l
Solution:
(i)Y=l
(ii)y=(srnx)"
dy_
dx
u'v - uv'
v2
1(- Ji 2) 1(Ji 2) _l vx __+. | = _l _+ l . 2x( 2 Jx) 2x\2
^lx)
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RSM-91 1-P1-MA- Basic Mathematics
=
dY = x.xx-1.+1 x" Inx .1 = t' (1+/nx)
dx
(ii)y= (sinx)"'
I = *'(.lnt;*'r l'lsinx )+ (sinx )"'/ns inx 9(z
dx ' dx'
= x2 (sin x ;*'-1 cos x + (sin x)*' In sinx'2x '
Derivative of a Composife Function:
Given a composite function y = f(x), i'e' a function represented by
y= F(u),u = 0 (x) ,or y = FtS(x)1t,h en y'=
dy - dF du
dx du dx
This is calledt he chainr ule.T he rulec an be extendedt o any numbero f cornpositefu nction; e.g.
tfy = f(u(v)t)h, eny ' =1 Y= 31 9g 9Y
dx du dv dx
Illustration 11.
Find the derivative of the
(i) Y = sinf
(iii) Y = sin(nx)3
followingf unctionsw .r.t.x '
(ii)
(iv)
r @x)t
y = cos-' (lnx)
Solution:
(i)Y = sin(x)'L' etu = 12a Y = sinu
* 9I = alrin u) +(x') = cosu.2x = 2x cosx2
dx du' dx'
(ii)y = (/nx )3.L etu = /nx =+Y = u3
-
dY- d(u3o) u =3u21= 3 ( rnx ) , 'dx du dx x x.
(iii)Y = sin(/nx)3
Letu=/nx,v=u3
=y=sinv
- dy - dYd v du - d (sinv) d(u)' 9(tn
- d* ")
=
d" d" d"
-
d" du dx'"'
" '
= cosv. zu' !- 3(ln x)2
"os[(tn
x)3-l
x x ---L' ' - l
(iv)Letu=lnx==cos-tu
=dY=+99 =-LI=-L.
-d*=ffid-
Find the derivative of the following fanctions w.r't' x :
,) Y={
iii) Y = sin3f
v) Y=6x7/2+4xil2-x
ii) y=sin(f-Sx+1)
M v = togtsin(f)I
18 RSM-91 1-P1*MA-Basic Mathematics
Parametric Representationo f a Function and it's Derivatives:
We find the trajectory of a load dropped from an aeroplane moving
horizontallyw ith uniformv elocityv 0 at an altitudey s.
We take the co-ordinate system as shown and assume that the
ioad is dropped at the instant the aeroplane cuts the y-axis. Since
the horizontal translation is uniform, the position of the load at any
ro
timet , isg iveny (x)= vot,y = y, -+
Theset wo equationsa re calledt he parametriecq uationso f the trajectoryb ecauseth e twov ariablesx
andy haveb eene xpressedin termso f thet hirdv ariable't ' (parameter)
i.e.twoequationsx=q(t),y=V(t),wheretassumesvaluesthatlieinagiveninterval(tr,tz).
*ren 9I - dY/dt - (dv/dt) - \t'(t)
dx dx/ dt (dtp/ d t) q'(t)
Illustration 12.
Thef unctiony of x is given,F ind thed erivativeo f y w.r. t. x.
0 x=acost,y=asint. ( i i )
(ii| x=2ln cottandy=tant+cott,
Solution:
x = a(t-sint)y, = a(1-costt) , 0<t <2n.
( i )q = (as int )-' cost =-cot t " dx (a cost)' -sin t
lf we want to compute
l|
at a particulatr, sayt =
f,
, tnen
( gI) =-cotr=-1
'dx)r=ntq 4
(ii)w e naue9 I - [9(1-c ost l]' - a sint
\"/'.- "--- dx [a(t-sint)]' a(1-cost)
(iiis)i nce$= 1 t111 . .(1)
dx dx/dt
dx ^-coseczt -2 4 Nf nrr, ..........._- 't _
dt cot t cost sin t sin 2t
Rlso 9I = sec2t- cosec2-t sin2l - co!2t - - cog2t
o
dt sinztc oszt sinz2 t
dv f-+cos2t)f sin2t ) cos2t
Hence -' = | '-;--- ll ,"' ,- l- ----- = cot2t.
dx \ sin,2t ,[ (-4)J sin2t
2 sin tl2cos tl2
2 sin2tl2
=c ot( v2=) t "nIn- !).
\2 2)
ii)
iii)
,r*=ffi, r=# rinddy/dx.
t fi +f =r -1,x o+ yo= r ' * ! , ,provethfa ty f f=t
If x:cos-1h' Y:sina
#,
showthatdy/dixs independentot'f
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tf
Derivative of an lmplicit Function:
lf x andy are relatedb y the ruleF (x'Y ) = 0 sucht haty cannotb e obtainede ntirelyo r exactlyin ternrs
ofx thenyissaidtobeanimpl ici t funct ionof x'e'g' *'*f =a'+'= tW'
Herew e do notg et a uniquev alueo f y for eachx . Another examplec anb e
"'
* yt - 3x2 y = g
Herea lsoy cannotb e obtainede ntirelyin terms of x. To tino
fr
in such cases start differentiating
thegivenequationasitis(usingruleofcompositefunctions)e'g',forx'+f=a'
We have2 x+ 2^ Y dy =0 = 1u =_" forx2+yt - ,3xry-0
& dx y
- , dv--g ( ^r9r.+y
z* )=o= g =!E*
Wehavezx+3t'ix \ dx )
u--d,
a(V'-x'z)
lfycanbeexpressedentirelyintermsofx,thenyissaidtobeanexp|icitfunctiono{x
Notethateveryexp|icitfunctioncanbewrittenasimplicitfunctionasy-f(x)=0.
Second Derivativeo f a Function:
The secondd erivativeo f y w.r.t.x is the functiono btainedb y differentiatin.o1 f, w'r't' x' It is
represented".
#
ory"orf"(x) e'g'' lf y=xsthen
qI=5xa
"^
d'y = d fgl)=91s*o)=5.4x3 =20x3.
""' d*'
-
ox(oxJ dx\ |
The acceleration 'a' of a particle is the second
dc
t ime) . i .el.f, s =f ( t ) thenv=
fr=f
'(t) anda=
Itlustration1 3.
d2v
Find :-*, where Y = sin2x.
dx'
Solution:
tY
= 2rinxcosx = sin2x
dx
o-Y = 2cos2x.
dx2
derivativeofthedistance's'(givenasafunctionof
d2s=9I=f,,(t).
dtz dt
rina ffit:
(a) Y=cos(x+Y)
fina t+ if y = sn*.
dx'
s! + 2hxf + bf + 2gx + 2fY + c = 0'
tlllltcc
Ltd., FTTDEE Housel2g
RSM-91 1 -Pl-MA-Basic Mathematics
The Behaviour of Functions:
The study of the flight of a shell in empty space yields a formula that gives the dependence of the
range R upon the angle of elevation a and the initial velocity ve i.". n = u6 tl2o
, where g is the
g
acceleration due to gravity. We can now determine at what angle a, the range R will be greatest or
least, and what the condition must be for the range to increase as the angle of elevation s is
increased.
In the oscillationso f a load on a spring (of a railwayc ar or automobile),thed eviationy of the load form
the positiono f equilibriumi s given by y = e-*,1Ac os wt + B sin M) where k, A, B, w (having definite
meaning) are considered constant. This formula gives the information as to what values of the time t
yieldt he maximumo r minimumd eviationa nd for whatv alueso f t, y will increasew ith increasing t.
The following behaviors of a function are important to study:
(i) lncreasea nd Decreaseo f Functions
fncreasing Function: lf y = 1"; and x2 > x1 implies yz>yt for any x belongingt o the interval [a, b],
then y is saidt o be an increasingfu nctiono f x.
f(x) increases in [a, b] e f ,(x) > 0 x in (a, b).
Decreasing Function: lf x2 > xt + Yz < y1 for any x belonging to [a, b], then y is said to be a
decreasing function of x.
f(x) decreases in [a, b] e f'(x) < 0 x in (a, b)
For a constant function, f'(x) = g
For a non-decreasingfu nction,f '(x) > 0
For a non-increasingfu nction,f '(x) < 0
lllustration 14.
Find the interual in which the function is increasing or decreasing
(i) y=sinxinxe(0,2r)
Solution:
(i) Y' = cosx
y'> o r o*r. (o , i l ' [91, r^)
\ 2) \2 )
Hencein creas-in is" [0, i] u(9!, ,*\ \ 2) \2 .)
(ii) y' = 6(tanx- 2)(tanx- 3)
+ y' > 0 for tanx > 3 and for tanx < 2
+ x e 1- oot,a n-12u) (tan-13m, ).
(ii) y= 2(tanx)3- 15(tanx)2+ 36(tanx)
(ii) Maxima and Minima of Functions:
A function' f is said to have a maximuma t x=xoi f f(x)<f(xs)x in the immediaten eighborhoodo f x6.
Similarly, a function 'f is said to have a minimum at x = xs if f(x) > f(xo) x in the immediate
neighborhoodo f xe
( r 3n) and decreasi'n" qi"n ' \ 2| '" -'" I
2 ) '
*, , *
23
I
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RSM-91 1-Pl-MA- Basic Mathematics
We haveu sedt he word 'immediateh' ereb ecausea givenf unctionm ay havea ny numbero f 'high'
and 'low' points. lt is just like moving on an uneven surface (which has many 'bumps' and
'depressions')M. athematicallyth, ese bumps are called the points of local maxima and the
depressionasr e calledt he pointso f localm inimaT. he highesto f all the bumpsi s the globalm axima
andt he lowesto f all the bumpsi s the globalm inima.
We stateh eret he preliminarmy ethodso nlyt o findt he maximair ndm inimao f functions:
dv (a). Find i or f(x)
ox
(b). Findt he pointsa t whichi t becomesz ero.T hesep ointsa rec alledc riticapl oints
To findt he pointso f maximaa ndm inimaw e resort o eithero f thef ollowingte sts.
(A) First Derivative Test:
SupposeX = Xoi s a criticalpointi. e.f (xo)= 0.
lf f'(x) changess ign from positivet o negativei n the neighborhoodo f x = xs = lVlaximaa t 'x0"
lf f'(x) changess ign from negativet o positivei n the neighborhoodo f x = xo = Minimaa t'xs'.
(B) SecondD erivativeT est:
tl2v (i) Find * or f "(x)
ox-
(ii) Computeth e valueo f f'(x) at the criticapl oints.
lf it is positive= Minimaa t thev alueso f x
lf it is negative+ Maximaa t thatv alueso f x
lf the function is defined in an interval [a, b], then to find the maximum and minimum values
(i.e.,g lobalm aximuma nd globalm inimumo) f the functionin that intervawl e compareth e valueso f
the function (i.e., y) at all the critical points and also the end points (i.e., f(a) and f(b). Then the
largesat mongt hemg ivest he globalm aximumv aluea nds mallesgt ivest he globalm inimumv alue.
Illustration 15.
Determineth e maximuma nd minimumo f the function
(i). y = f -3x + 3 on thei nterua[t- 3, 3/2].
- x - ! on thein terual0< x <4. Ul). Y=
x+1
utl uteilt.etvcrtv-:x-1+.
Solution:
(i) Fort heg ivenf unctiony,' = 3x2- 3
Fort he criticapl oints3, x2- 3 = 0 = x = t 1
Theny"=6x >0 atx=1
<0 atx=-1
Henceth erei s maximuma tx = - 1 atwhichy = -1 + 3 + 3 = 5
Alsot hereis minimumat -x= 1. Y = 1 -3 + 3 = 1
Nowa tx = - 3,Y = -27 + 9 + 3 = - 15a nda tx = 312,y= 1518
Henceth em inimumva lueo f theg ivenf unctionis -'15a t x = -3 andt hem aximumva luei s 5 at
X = .1. lt shouldb e notedt hatt he valuesa re actuallyth e larges/t smallesvt alueso f the
functionin theg iveni nterval.
21
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RSM-91 1-P1-MA-Basic Mathematics
tu
iii) Find the maximum and minimum values of the function:
V=sin2x-x (t/2<x<td2)
Find the intervals of increase and decrease of fhe function
i) (a) y=4x3-21f+18x+20 (b) y=sinx+cosx,0<x<2n
ii) Find the point of maxima and minima of the foltowing:
(ii)F orc ritica'pld oxints9 I = 0
1.(x+1)-1(x-1)_n ^
(x+1)'
+ No such x, infact
2
' ,
(x+1)2
. vn+1
Jlx ndx=in +1. *c, (' \n+-1)
Jtrt*rt*)0=. t*?;' +c (n+ - 1)
J^:o
(x + 1)'
> 0 in thei ntervaTl. hatm eansis entirelyin creasinign [0,4].
Henceth ereis noc riticaplo initn t hei nterval0< x <4. Atx = 0,v = -1,a nda tx=4,y= 3tS.
Henceth ef unctionis minimuma t x= 0, withy = -1 andi s maximuma tx = 4withy = 375.
A functionF (x)i s calledt he antiderivativoef thef unctionf( x)o n the interva[la , b] if, at all pointso f the
intervalf(x=) P'1*;.
Forexample, theant ider ivatthiveefoufn ct ionf (x)=x' i4s, " . [ f ) '=x2. Thefunct ion*s' -2
3 l.3l 3 -
Y3^v3
and |+1 arealsoant ider ivat ivef (sxo)f= x2.ln facti,* C,whereCisanarbi t raryconstant , is 3'3-
the antiderivativoef x2 . So if a functionf (x) possessesa n anti-derivativFe( x),t hen it possesses
infinitelmy anya ntiderivativeasll, o f themb eingc ontainedin the expressioFn( x)+ C, whereC is a
constant.
lf the functionF (x) is an antiderivativoef f(x), then the
integraol f thef unctionf( x)a nd is denotedb y the symbol
if F'(x) = f(x). lf a functionf (x) is continuouso n an
antiderivatiVTe.h e processo f findingt he antiderivative
differenitn tegrblso f a functiond ifferb y a constant.
Stan d a r d EI e me ntary I ntegr als:
In thef ollowingin tegralsC, standsf or an arbitraryc onstant.
expression F(x) + C is called the indefinite
F(x)OxT. hus, by definition," ff(x)OxF=( x) + 6,
interval [a, b], then this function has an
of a function f(x) is called integration. Two
jcotxd x=l oglsinx l + 6 = - loglcosecxl+c
isecxdx= loglsecx+tanxl+c
t
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11
|.--ql-=1t"n-t"*"
J;0"=loglxl+c ra.+x. a a
ie*dx=e"+c. J tJ-1g*!*=' tan-1 x+c
jsinx dx = -cosx + c
Jcosxd=x s inx+c t+ = sin-1I * " or - "o. - t I * "
L""'* dx= tanx+
"
' "la' - x' a a
Jcosec'xd=x - cotx+c tg = sin-jx +c or-cosx-1+ c
Jtanxdx=- loglcosxl+clo=g lsecx+l c J Jl_ *,
t -g = sec-1x +c or -cosec-1x+c
t xtlx2 -1
1
f^ *112t1 2 xt e.g (r) Jx o*=
1t
*"=5.x--+c
2
e 1 '-2+1 1
(ii) | ;dx =ix-2dx= ^: ; +c=- +c
J x. -2+1 x
The followingp ointsa re to be noted:
. J1 dx = log x + c if x is positive
X
= log (-x) + c if x is negative
becauSse t 'on( -x)*),=- t ,
=
* = J i o-=loslxl+c
. i-.l- dx = sin -tx or - cos -t x. lt does not mean that sin -tx = - cos -t x.
' t J1-x'
Theo nlyl egitimatceo nclusioins t hatt heyd ifferb y somec onstantI.n f act
sin-1x- ( -cosx-1)= sin-1x+ cos-1x = nl2
r lfaisaconstant,then
Jaf(x)dx=a J
tt"lO*
. i t f(x) t g(x)l 6*, = jtlx) dx t j g(x) dx
Illustntion 16.
Evaluate
(i) Ipo* s,* + azf) dx
.-2
vii) l--1 ^ dx
J1+x'
(v) l_ox rsin3x+cos3x
J sin' x cos' x
(i0 ("o"..1-"') dx
(iv) t**
Solution:
(i) i("0 *
",x
+ a2x2)dx
0i )J [cos" . i - " - )
=sinx+zlog lx
= a0 Jdx+ ar Ix dx+ ?zi*' d"= dsX+ ?1
*
.
^r*
.
"
dx =
J"or"
I -e*+c
dx+2F o*- Je"o x
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RSM-9 1 1 -P1-MA-Basic Mathematics
(iii). -2 fx2 +1-1 n( I \ Jf1-r+lx-.- dx=J t --1-+-x ,. dx=J t\f r ---t " lo"= [ o"- l-!! 1+x2 ) J J1+x2
=x-tan-1 x+c
(iv) ;-{6" = f"1
-tr*, o"= l(*,-1*-1-lo,. Jx.+1 J 1+x. J\ 1+xz)
= [x' dx- [ o** [-{- = *t -1
J" J J1+xz
--x+Ian x+c
r"rJ *#t'6"= J.j'tr" , 6x* I;ffi o-
=
Jt"nxsecx dx +
Jcotxcosecxd x = secx-cosecx + c.
Methodso f lntegration:
1. Integrationb y substitution
This methodc onsistso f expressingth e integral
J
ft*l dx, where x is the independent
variablei,n termso f anotheirn tegrawl heres omeo ther,s ay't',i s the independevnat riablex;
andt beingc onnectebdy t her etatioxn = 0 (t). i.e.,
Jrlxpx
=
f t,l{t)l,l'{to) t.
Thism ethodis usefuol nlyw hena relationx = q(t) can be so selectedth att he newi ntegrand
f(x)dxi s of a formw hosei ntegrails known.
lllustration 17.
lntegrate the following w.r.t. x.
(i) e'cos e'
(iii) sec x
sin2 x cosx
1
asinx+bcosx
(i0
(iv)
Solution:
(i) Let e" = t = e*
ff
= f =+d x=e-* .d t
= J" "cose6* x= Je" .cose" .6e1- *= J cosdt t=sint+c= sine " +c
(ii) Letsinx=t + cosxdx=dt
= Jsin2xcosox* =
J,'0,= *
+6= stn3l a6
(iii)J sexcd x= J "*H#T! o"= Jffix, whefr(ex =) s ..x +taxn
= J" " " xdx= toslsecxt+a nx+l 6=to- ls l t "n[ +. * ' ] l . . (4 2)l
(iv) Leta= rcoso, b = rsino
=r=.6'*f
"nO
a=tan-1 b/a
= a sinx +b cosx = r sin( x+cr)
t---
1
dx= f
dx 1t
Jss;nxlgsosx .l rsin(x*o)
= -Jcosec(x+o) dx
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RSM-91 1*P1-MA- Basic Mathematics 25
= l rogl "os"c(x+o) -cot (x|+ +o")= l tos
l""f 1."
where r = Jl\b2 and cr = tan -1 b/a
(v)L etx = a sin0+ dx = a cosOd 0 anda ' -f = a' cos'e
= l 'g= lacosodo|=. r .oe=sin-1I+c J
Ja2 _x, J acos0 J a
Note: t.:--gl-- = 1s;n-r bx + c
t
l"' -(ox + c)2 b a
ad I d" = 1
[
J
'[-2r' . g*. 4 J2 J
-4- x-3
441
dx = 1- t
Jzsrn
'
_,)
J+t 1-
4)
2. Integration by pads
lf u, v be twof unctiono f x, we have
Integratinbgo ths ides,w e get
rdv rdu IJV= lU-oX+ lv-ox
Jdx Jdx
rdv r du
-
IU-dX =UV- IV.-.OX
JdxJdx
Letu=f(x)and*=q(*)
dx
= ft(*),p(*)=or *( x)f rp(x)-d -[ f+ [61*10*d']x
J"'' 'J J\dxJ' .)
Integral of the product of two function = first function x integral of second - integral of
(derivativeo f first x integralo f second)
Note:
r e*1f1x+) f 1x))d x = Je*f(x)d x + Je"f'(x) dx = f(x)e *-ff(*)e'dx + ie*f1x; ox = f(x) e" + c.
Iltustration 18.
-valuate
(i) rt e' dx. (ii) /sin-'x dx.
Solution:
(i) Letf (x)= x, <p(x=) e*
=+ jxe" dx = x.e" - fi.e* dx = xe* - e* + c
(ii)L etf (x)= 51n-t*q,( x)= 1
=+J sin-1dxx - Jsin-11x d. x = sin-1x.-x [!x ox
' '11_* '
= x sin-x1- jx(1-x2) - td' 'x =xs inr l * f i ; * " .
, .2 r 3)- -tx-- |
\ 4)
d,
, (UVJ=
ox'
_dvu+ u- dv
dx dx
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RSM-9 1 1-P1 -MA-Basic Mathematics
Findt he integraol f thef ollowingfu nctions
sin ox cos x
I -cos 2x
1+cos 2x
The difference in the values of an. integral of a function f(x) for two assigned values of the
independenvt ariablex , say a, b, is calledt he definitei ntegral of f(x) over the interval (a, b) and is bo
denotedn y
Jt1xl dx. Thus
Jr1*10,,
= F(b)- F(a),w hereF (x)i s the antid erivativoef f(x). or, we
b
write
Jf(x)6"
=l r1x) l l = r (b) -F(aa) .i s cal ledthleo wer t imi tanbdth eu pper t imi tof integration.
a
Notes:
Oa
f. 1,. r l|f' ,(x)dx= - l f (x) dx JJ
ab
bcb
FP . Jrlx)oxJ=t {xo) x+J r{xo)x
,in"r" . i, *y pointi ns"iooer outsideth ei nterva(la ,b).
D
I Geometricallyt,h e definite integal jftt"l dx denotes the area bounded by the curve
y =f (x) , thea xis ofxand thetwoordinatesx = a and x = b.
e.g : Work done a on body as it is displacedf rom the position x=a to x=b is equalt o Area
undert he force- displacemengt raphb oundedb y x=a & x=b.
h
i.e.,W =
Jf .O*; whereF ist hef orcea pptiedo n theb ody
lllustratioin 9 .
Findt he areao f the curveb oundedb yy =
Solution:
JJ
Area= l' ydx= lI et,- *d- x
1-J1
/ a.- r3
=tf_ et r* )" =_e-6^_ e sq.units.
\2)1 z
e'*,x=landx=3.
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RSM-91 1-P1-MA- Basic Mathematics 27
lllustration 20.
Evaluate
^lVn (iil tI
U
1
F
(i) | x'^e '^^. - dx .
I
n
=+^ - /[ "^.+r1 .)'1, ! tla=,+ - +.; i1"'.0*
srn'x
- ^- *oX
(1 - x' )o"
Solution:
(i) Integratingb y partsw e have
t"'"'*=0[ ."- '+lJ Sr*{o*=}*+"o'-" "
d l. 2)o d 2 o
1 f"r*,' tl e, 1l e2 -1
=_r-
2\ 2 )o 2l-2 2) 4
( i i )L etx=s in0+dx= cosOd 0 and0 = sin-1x;w henx =0,s in0=0
Whenx =112s.i n0=n/4
'm -1 n/ra
ecosede ''f ^
d (r- *'\"'' j cos'e d
nl4
=1et"no-) i/a J t.taneoe
0
=f,-{-'onc ose=)(f ,-{ross ec{(= X-,onA= }-}r"sz'
(lntegratinbgy parts)
An equationc onsistingo f the independenvt ariable,t he dependentv ariablea nd the derivativeso f
dependenvt ariablew ith respectt o independenvt ariablei s calleda DifferentiaEl quation.
For example:
dy . - -:-... q. a2x=oi
d"*ycosx=t'n"' o*,
(2x + 3v ,! *h^' + av) = o \-- ,rdX VI
r ^t312 l. (ov)' I o'v l'.[*J| ="e
The ordero f a differentiael quationi s the ordero f the highestd erivativeo ccurringi n it. A solutiono r
integralo f a differentiali s a relationshipb etweent he dependenta nd independentv ariable,n ot
involving the differential coefficients such that this relation and the derivatives obtained from it
satisfies the given differential equation. This also implies that a differential equation can be derived
fr:omi ts solutionb y the processo f differentiationa nd elimination.
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I
RSM-9 1 1-P1 -MA-Basic Mathematics
Illustration 21.
Showthaty=e-x(Acosx
-d.2*+v: . r2+d2vy=0.
dx' dx
+ Bsinx) is a sorution of the differentiat equation
Solution:
I =
"*(-
Asinx+ Bcosx)- e-*(Acosx'+B sinx)
dx
v vrvve^.seilr^fl
* =e-' (-Acosx- Bs inx)- e-*( -Acosx+ Bsinx)- 9I
dx.
vv""\/
dx
d'v 2dv
=
--*+2y=0.
dx' dx
How fo so/ye the differential equation:
we will considert he firsto rderl ineard ifferentiael quation.I n ordert o solvea differentiael quationw e
need to separatet he variabless o that we can integrate.T he generalf orm of the differential equation
shallb e or shallb e broughtt o as f(x) Ox+ rp(y)d y = 0
or f(x) + <p1yd; yldx = 0 . . (1)
Integratingw ith respectt o x, we get
Jrt*n*. JotulfloX= Go ff (x)ox , J,p(v)=O cv w hicihs s otutioonf( 1)
lllustratian 22.
Solve (xf +fl dx + gf + fl dy = e.
Solution:
We havex (1+y2)dx+ y(1+x2)dy= 0
o, ;l;o* 1-J-6y - g
1+ x' 1+ y'
sothJa#to -* ffiov="
1^1-
> -log(xz + 1)+
Tlog(1
+ y2) = c "
or (x2+1)( f*1)=e'" = a, is the requireds olutionc ontaininga rbitraryc onstant,a,.
Solye the following differential equations:
#.Y
= xex
rf tfl#+ (f + xf) = o
Eliminate a and bfrom the relation y= a sinx + b cosx +x sinx.
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RSM-91 1-P1-MA- Basic Mathematics 29
(iii)
ffisntn$T 0r [tnGrst
Exercise 1.
Exercise2 .
(i)
( i i )
( i i i )
( i )
( i i i )
( i i )
Exercise3 .
Exercise 4.
(i)
(iii)
Exercise 5.
(i)
(iii)
(v)
Exercise 6.
( i )
Exercise 7.
(iXa)
( i i )
Exercise 8.
(iXa)
(b)
(ii)
xe(-4, 1)w(2,512)
x e (1, e) u (3, .o)
x e (1.8)
21x5t2+ 103t2-1
2xll(x2+ i ) In a l
11+xzy11+x2+xa1
dy _ t(2-t3)
dx 1-213
-sin(x + y)
1+s in(x+ y)
-d-2-v; =n-2e "nv
dx'
(ii) xe(413,1)u(2,m)
(iv) x e (-co, 1l u (2, 3l
(iii) x=1
Au B = {1,2,3,4,5,6, 7, B, 9, 10} ; AnB = {1, 3, 5, 7, 9}
A- B = {2, 4,6,8, JO}' B -A = 0
A uB=l ;AnB=N; A-B=0,8-A=set ofn on posi t ivein tegers.
AUB=R, AnB=B; A-B =R-{x:lxl<3}=R-[-3,-g]; B-A =O
x e [-1, 3]
t r.l tr I x€ l-e, -- jul-, el
L et Le I
[
-l
( i i ) xe [3r fo,i ) . I r ,
(ii)
(iv)
(vi)
2")
(1/x). sec2 (ln(x)
(sinx)* iln (sinx) + x cotxl
(x)n*, ex
nl-l In-
\n/ n
(b) -
ax+hy+g
hx+by+f
(1 \ r 1\
Decreasiningl ; ,31, Increasinign l -m, - j u (3,co)
\z / \ z)
lncreasin[g 0 ,r /4). (4,2"1, decreasii"n fg + ,+'l \4 .l - \4 4)
Maxima Minima
(a)x=1 x=3
(b)x=al2 x=0anda
Max. = latx=-1. min. - -n atx- n
2222
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RSM-9 1 1-P1 -MA-Basic Mathematics
Exercise 9.
'
(i)
(iii)
Exercise 10.
(i)
(iii)
log l(1+x3;+; 6
-x' cosx + 2x sinx + 2 cosx + c
xy=e'(x-1)*c
logl!=X+Y
cy xy
sinsx
-._+C
5
tanx-x+C
1"-1u=k
22"
t!*v = 2cosx
dxz
(ii)
(iv)
(i i)
(iv)
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1.
2.
4.
5.
6.
7.
8.
12.
16.
RSM-91 1-P1-MA- Basic Mathematics
Set is a well-defined collection or ensemble of objects'
/ , t \
I a+l l>z v a > 0 ande qual i thyo ldsfo ra = 1.
\a/
/ 1\
la* - l<-Z v a < 0 ande qual i tyholdsfoar= -1.
\ a/
wavy curve method is used to solve following type of inequalities:
A(xl
Let say p(x) =
6#
where A(x), B(x) are any two polynomials,t hen P(x)>0,P (x)>g,P (x)<g'
P(x) < 0.
loo^b=log"b wherec>0.
log" a
ar>ar>O if b> 1
f f logoa>. ,l ogoar=0.Qt<az
if 0<b<1
;4 = glo9" x
Function is used to define the dependence of one physical quantity on other in which every
element is domain x has unique image in y if function is written in therefore y = f(x).
lf y = Ltt, then y'= u'v i uv'
t rv= u ! *1t,h eny '= dY = u'v-uv'
"' v(x)'-"-' dx v2
lf y = (x) is a function, then it will be increasing for any interval (xr, xz) where X1, X2 2re critical
pointsobtainedbyput t in*g =Oor f ' (x)=0i f andonlyi ff oreveryvalueofx sin(x1,x2) , f ' (x)
-dx
>0.
lf y = 11";i s a function,t hen it will be decreasingf or any interval( xr, xz) where X1,X 2? f€
criticat points obtained by putting
l|
= o or f'(x) = 0 if and only if for every value of xq in
(Xr, xz), f'(x) < 0.
y = f(x) will posses maxima at any point xs if f'(xs) = 0 and f"(xe) < 0.
y = f(x) will posses minima at any point xo if f (xo) = 0 and f"(x0) > 0.
lntegrationis inversep rocesso f differentiatioin.e . it is sometimesc alleda nti derivativeo f a
function.
.r v (x) du{ x) u (x)dv (x)
lff (x)= u(x)v(xth),e nfr f(x)
= -t *1;
. Jdu(x) n
- l
-
fu(x)v(x)d=xu (x)[ v(x)dx-l l " " t^t lv(x)oxlo x
J-','-' \ / .'J \ / JL dx J I
Thisr ulei s calledp roducrtu leo f integration'
9.
10.
11.
13.
14.
15.
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17.
18.
19.
20.
RSM*91 1-P1-MA-Basic Mathematics
b
Jt(x)ox
= F(b)- F(a), whereF (x)i s antiderivatiovef f (x).
;"
l-. / .
lf(x)dx: - lf(x)dx
J
ab
ocb
f-, l. I
Jf (x)dx =
Jf
(x)dx +
Jf
(x)dx,w herec isa nyp ointin sideo ro utsidteh ei nterva(at ,b ).
aac
An equationc onsistingo f the independenvt ariable,t he dependenvt ariable and the derivative
of dependenvt ariablew .r.t.i ndependenvt ariablei s calleda differential equationl ike -x-d. :v +V=0.
dX
tllllcc UA. HtfEt nou".,Ze-A, frtu S.rrL S"*rpif
RSM-91 1-P1-MA- Basic Mathematics 33
$0lltEPlrn Bo utrttlills
Problem 1.
Let f(x) =
Subjective :
(x-s)(x +2)(x+5)
. Find intervals where f(x) is positive or negative.
(x + r)(x -t)
Solution:
As can be seen, f(x) > 0 V x > 7 as all the braces
are posi t iveF. or 3 <x<7, the quant i tyx - 7 < 0
an rest of the terms are positive, hence f(x) < 0 for
(3, 7). Similarlyfo r otheri ntervalsT. hus
f(x) > 0 v x e (-5,-2)u (-1, 3) u (7,.o)
and f(x)< 0 V xe (-oo ,-5)u(-2, -1) u (3, 7).
Probtem 2.
Solve logo.t(x - 1) < logo.onx( * 1) .
Solution:
logo. (x- 1)< logoon(-x 1) = logo. . (x- t ) . l iogo.s(x- 1)
= logo.(ux - 1)< logo.o(nx - 1)
L
= logor (x- 1)< logo.on-( x1 ) = (x- 1) ' > x- 1
x' -3x + 2 > 0+ (x -l) (x-2) > 0
= xe ( -4, 1) u(2, oc)
Also for log function to define
(x-1)>0=x>1
From (l) and (ii) x e (2, a:.)
Problem 3.
lf I x -11 + | x +11 = 2, find x.
Solution:
Case(i)lfx<-1
=(-x-1)*1-x=2
>-2x=2=x=-1
Gase (ii) lf -1< x < 1
=X+1 +1-X=2
+2=2=-1 <x<1
Case (iii) lfx ) 1 = x +1 + x- 1 =2
=>x=1
Thusf rom( l),( ii),a nd (iii)- 1 < x < 1
Alternate solution:
I x +11 + | x-11=2=l(x+ 1) + (1 -x) |
. . ( i )
. . . ( i i )
.. ( i )
; , , ,
. . . . ( i l )
. . . . ( i i i )
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311 RSM-9 1 1-P1-MA-Basic Mathematics
I
l
(x + 1) and (1 - x) should have same sign
(x + 1) (x- 1)> 0 + (x + 1)(x-1)< 0
-1 <x<1.
Problem 1.
t y = r(i - x11ay 1 -(a - b)tan,
E
thenp rovteh at
#
=
Eo
Solution:
Letx=acos't+bsin2t
= a -x =1a -b)sin2t and x -b = 1a -b)cos2t
=> y = (a -b)sint cost - (a -b) t
Differentiatin(gi) w.r.t.t $ = Z1U- a)cost sint
dt
Differentiatin(gii )w .r.t.t
dv
fr
= t " -b) (cost2-s in2t ) - (a- b) = 2(b- a)sin2t
From(i ii)a nd( iv)! I = t"n, =
\H
Problem 5.
Evaluatefx (togx S2d x.
Solution:
| =
Jx(bxs) ,d x- (toxs) 2J *0"- J{{n"o) ,J xdx)dx
=t 'troo*-) l'z nsx.lt* = f;Qoo-n J xrooxdx
Now Jxtooxdx
= tosx Jxox- f{nox)' Jxdx)dx
=f ;nnI^:-+ r.= f ;,ont^it-* ={,on-*** ",
= f;,on*-**"'
Prohlem 6.
. . . ( i )
. . ( i i )
. . . (iii)
...(iv)
Evaluate sin2x cos2x dx
sin2 xcos2 xdx
J t t r 12 0 (sln" x + cos" x,|
Put tan3x + '1 = t
n/4J0
Solution:
fr
(sin3 x * cos")'
"'f o,n' xsec2 x
I j-.:::jd* (divideb yc osox)
d (tan'x+1)-
t4J
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RSM-91 1-P1-MA- Basic Mathematics 35
tan2x sec2x o" =
*
Therefore
^';o!" n' *t9"1i o" =
d (t + tan' xJ
Problem 7.
(1\ 7 '
lffor non-zerox , a (x) + bfl' | =' - 5 wherea zb, thenr ina [rQ)ax'
\x/ x J '
1
Solution:
Here af (x)+bf [1)=l-u
\X/ X
Put x = I in (t), we get
X
/.1 \
af l- I+bf (x) = X-5
\x/
From (1) and (2), we get
r (x)=^ ' ^(9-n" l - -1 a'-b'\X .) a+D
2 2, 2
HencTer ( * )o=* ^l ^ [ [9-o*16" - - l fox
,' ' ' a'-b' {\x. ) a+D t
= , .(^,n*-0"')'- 5, r*f = ^f ^("rnz-1-s(a-n))
a2_b2 t- 2 ), a+b.
11
a2_b2\ 2 ' )
=- :--:l^iln2-s)*4-ol .'.l r,*,0=* ^l ^[ "( u,2-s).+l
a2_brl-,"'
-I 2J i" ar-brL' 2J
Probtem 8.
Solve3 e'tanYd x + (1 -e') sec2Yd Y= g-
Solution:
Given3 e'tanydx+ (1 -e") sec2yd y = 6
3e* sec' v 3e* , sec2Y -,-. = -- dx-- 'dV+--- -ox=- -oY
1-e" tany 1-e^ tany
3e"
+ - dx= 2cosec2dy y.
1- e*
Integratinbgo ths ides,w e get 3log(e"-1 ) = - log( cosec2 y * cot 2y)* c.
Probtem 9.
Find thea reab oundedb y the curves7 a= f and x = 3 - 2f '
Solution:
The verticeso f the parabolax = y2a nd x = 3 - 2y2a rer espectivelOy (0,0 ) and M(3,0 ).
Thusg ivenc urvesin tersecatt (1,1 ).
;i# *
. .(1)
...(2)
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-.
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artt-
36 RSM-91 1-P1-MA-Basic Mathematics
Problem 10.
Findt he areab oundedb yt he curvesy = tanx ,y =sinx and
Solution:
and (1, -1). So that the required area = area
(MPOOM)= 2area (MPOM)
/a a
[ ' . - \ =z l[ J- io"."f" ,l?/-#, 0" I \.d ir t )
fr
4
Requireddr €€=r fttanx )Ox-
0
pl4 =lnlsecxllA/o+cos
"lo
= rn(J2).[+-''] \ / \"./2 )
(t- J2) .t 4 : t ;-' *)nz=)((t-Jr\J|+rn2) square
Jz22\\/t
units.
=z ll*',-,,1t ?'ul( '- " )"]';
= ,('-- ol- !-F -23t2)
(3 ) 3J2\
=
t. #rJ,
=+=4 s quaurnei ts.
n
X=- -4
4
forO < x
fr
4
lsin x dx .
J
0
lf.''
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RSM-91 1-P1-MA- Basic Mathematics
Objective :
Problem 1.
Differentiatioonf sin-11:! 's'1'1'J v i5
2
(A)
1. *
-2 (c)
Ax
Solution:
(G). Let Ji = t and Y = 51n-r
11
or! = si1'1-1
#
=sin-1(c os20)
Probtem 2. l
The teastp ositiveva lueo f lx + /l + lx - 3l
(A) 1
(c) 3
Solution:
(D)' We know that lx * Yl < lxl + lYl
= lx+ 1l + lx-31 = lx+ 1l + 13-xl
>lx+1+3-xl
>4.
Problem 3.
(t-*2) av
tf v=tnl+lthen:? is
'
\1+x') ox
(A) -4xt,
\"/
1-xa
-4v @ ::+ 1-x
Solution:
= y= ,i"-'[,i.(i-rt=))t -r, =L-2*n'ts ot haXt = #
:*
B) - - : ./x (7 + x)
(D) none of these
(putt=tan0)
is equalto
(B) 2
(D) 4
-4 (B)
1- -xtt
t, - '-'7-x4
(G).y = In (1 - x'; - In 11 + x2)
-
dy = -2*
-
2x - -2x(1+x2 +.1-x2) =
=O*,
=
d*
-
T-7
*x' 1-xa 1-xa
Problem 4.
The absolutem aximumo f y = v3 - 3x + 2 in 0 <x <2' is
(A) 4
(c) 2
(B) 6
(D) 0
"o
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=:rilfa.'
38 RSM-91 1-P1-MA-Basic Mathematics
Solution:
+ absolutem aximum= 4.
Problem 5.
ff [ert') o, = 4 + C, thent (x)i s J5
(A)4tnx
(C) tn (x)
Sotution:
(A). We have [sr(") dx = f- + c J5
. +et(")=Xa=
"ln1x4;
+f(x)=4ln(x)'
Problem 6.
The number tan f is
(A) an integer
(C) an irrationanl umber
Solution:
Problem 7.
(A). E=3x2-3
dx
'*=o=x=t1
ox
+x=l,sincexe[0,21
6)!=o ox
rc) dt! =o
dx'
(B) (n xf
(D) none of these
(B) a rationalnumber
(D) a prime number
Let tan 5o be a rational number then tan 1g" = jlgn-s"
_ is also rational
= tan ,Oo_ 3tan10o- tan310o i^ ^,^^ ^ __
t _ gt"n, 10o
Is also a rationawl hichi s not true
=+t an 5oi s an irrationanl umber.
The differential equation of all non-vertical tines on a plane is
@4) =o dy
@{1+ =s
dy'
Solution:
(c). Theg enerael quatioonf non-verticrainl esin a planeis ax + by = .r,b + 0
=a+o * =o= I = * + 4=0. dx dx b dx2
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RSM-911 -P1-MA- BasicM athematics 39
Problem 8.
Solution:
tr/4
Thevatueof | $ar i " j Jcos3 xcos(x-a)
@2lJcos"+sin"-fosa]
Q 2lJcos"+s'n".Jsln"]
Solution:
sincr dx
Problem 9.
The number log27 is
(A) an integer
(C) an irrational number
Solution:
(C). Let logr7 =P = 2P =7
-
Problem 11.
Problem 10.
r l-1::9:l!- dx = f(x) + c then f(x) is
JCOTX-nnX
A) -'ot4'
I
,",
cos4x
8
Solution:
,", -t'ntut'
(D) none of these
J - will alwaytsru ef orx belongtso
lx-11
(B) (1' a)
(D) (-n' a)
t4(A). J
rl4 =J
= 2[Gos. , + sr r ' " tanx)o'o= 2[Jcos" + sin" - Jcf i ]
(B) a rationalnumber
(D) a prime number.
(A). r 1+cos4x r1+2cos22x_ 1 , '
I dx - l----------- ' ox
Jcotx-tanx ,.?tj_slnx
stn x cos x
12cos22x cosxsinxlx = I isin+xdx= -cos4X
*c
J cos2x 2.1 8
rhee quat|i "o |, . l*l =
(A) 0
(c) (-1,1)
(A()B, H). e|r- e'. l*l =* =1 ".*l
@ 2lJcosa+ s in". Jcos"]
(D) none of these
J 11FF
-(""t-."""
+ sinxsincr)
sincr.sec" x 'A+
---qx = l#, where t = cos ct + sin cr tan x
Jcoscr,+sinatanx "Vt
Now,i f P is integero r rationalthenL HS is evena nd RHS odd. Hencen ot possible.
Hence logr T is irrational.
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RSM-9 1 1-P1-MA-Basic Mathematics
sincelxl . ll-x- r l| =|1 " . *xl l 'ist rueonlvi r
- * ) 0 +xe{0}u( 1, co)
Problem 72.
Thef unetio'nf ( x) =(+a) wittb e increasining thein terval |'x)
(A) (1,2e) @) @, e)
(c) (2, 2e) D) (g,y) '\4 4)
Solution:
(B)(,D ).r 1*1[=!9 uI], *, o
(x )
+f ,(x)i=;i["i' ] *,',",=(.::g=[9]
for functionto be increasingf '(x) > 0
/a\ :+ rogl9]>o
-3t1
+ x<e:+xe(o,e).
\X/ X
Problem 13.
tf y = tog2{ tog2-xd}x,t hen! is equatto
6) tx? t.osg2ee x
' ' log.(2x)x
Solution:
(A), (B).y = tog2(tog2x)
dv1 1
d"
=
log, *
tlog' ex-log'e =
Read the following write up carefully:
Letx e R be any realn umbers ucht hatx liesb etweena nyt wo consecutivien tegerss ay (l - 1) and I
!.".,| - 1 < x < l, thenw e cana lwaysfi ndt hisu niquein tegelr.
Letu s callt his I as superi ntegravl alueo f x.
We denotei t symbolicailays (x).
Fore xamplei:f x = 2.63,t hen( x)= 3, if x = -2.63,t hen( x)= _2.
Now answer the following questions:
Problem 14.
Ther angeo f functiony - < x >
ffx e R-r.s
x
B' )1xlog, xlog.2
(D) none of these
xfog, xlog" 2log"2 xlog" xlog" 2
(B) [- 1, 1]
Q) n, 2l
log, e
xlog" x
(A) t0, 1l
(c) t0,2l
Solution:
(A) lfx e R-, then x < <x> < 0
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RSM-91 1-P1 -MA- Basic Mathematics 41
(A)
(c)
Solution:
(C) m (x) = cos 10x + cos (-gx)
m (x) = cos 10x + cos 9x
(n)9n1on1^1
m"[t+_ Jt= c--o-s 4_+ c-o-s-_ 4 = 7*u: _ J, - J,
tr m (x) = cos < o,u x * cos <-r2> x, then the vatue
"t ^(X). ^(i)
,t
(B)-1
1 (D) -1--T2
= r f i )=.o"9+cos5n = o+( - l )5= -1 =+
\2) 2
1
.12
1
-1+
.1-2
.'.0<<x><1
x
.'.ye[0, 1].
Problem 15.
'fll *' [l) =l ,-*,'a ) .
\4) \2) \. J2l
l;\ t; )'"
Problem 16.
tf m(x) = ti'{3'
"
t -t'} for1 <x s2, thenth ev alueo r
^'l
(o)r ; l;
(c)i
Sotution:
(B)for 1<x<2
<x>=2
+ m(x)=.ini?-"']
[3 )
+ m'(x=)
""r[?-"')t-z*) = t'[
Problem 17.
t: @ -l;
@-)3
,E):''E*,'F3=
Statemen-t 1: lf f(x) and g(x) are monotonocallyin creasingf unctionsi n domain D, then
t(x) - g(x) is a/so increasing.
because
Statemen-t2 : For two function,f (x) andg (x) increasingin domainD , f(x) + g(x) is always
increasing.
(A) Statemen-t 1 is True, Statement- 2 is True; Statement-2is a correct explanationf or
Statement-1
(B) Statemen-t 1 is True,S tatemen-t2 is True;S tatement-2is IVOI a correcte xplanationfo r
Statement-1
(C) Statemen-t1 is True,S tatemen-t2 is False
(D) Statemen-t1 is False,S tatemen-t2 is True
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RSM-91 1-P1-MA-Basic Mathematics
Solution:
Let h(x) = f(x) + g(x)
+ h'(x) = f'(x) + g'(x) since f'(x) and g'(x) are both greater than zero hence h'(x) > 0 V x.
Hence h(x) is increasing.
Let H(x) = f(x) - g(x)
+ H'(x) = f'(x) - g'(x) since f'(x) and g'(x) are both positive but f'(x) may be greater than or
lesst han g'(x).H enceH (x)c an be increasingo r decreasing.
Problem 18.
Solution:
(A) (q) (B) (q) (c) (p, q, r, s)
(A). LHS = 2*too"{vt z)rtos^(ztx)=toe^(xtv)
(D) (p)
+ LH S = 2 (groo" x
;(1o9"
Y-1o9z")
1"rog" r ;(los"
z-log" x)
("o0" z ;(loo"
x-losu v)
+LHS=2a"=2
(B),
(c). l l lx-21- 21-11=l
x<2
=+l l-xl- 1l= 1
= llxl- 1l = 1
x<0
l-x- t1= 1
x+1=t1
x= - 2,4
x>0
lx- 1;= 1
x-1=+1
x= 2,0
x>2
+ llx - 41- t1= 1
x<4
l-x+31 =1
x-3=+1
x= 4,2
= x = 2,4,6
x>4
lx-s1= 1
x-5=+1
x= 6,4
(D) . f f ty)=1* )+f (Y) - . , . . ' ,^^ ^'
\ 2 )
-Z- + f(x)i s a constanftu nction
.'.9 r(*)= o .
OX
l],ahtrrti* II
(A) lf x, y, z> 0 and a> 0, *1, then
(rtog" v-tos" z)(y,on" z-too. x)(rtoou x-tog. y
) =
(p) o
(Bl Then umber.osf otutionso f ltogazlxl=l srnxi s/are Vx i0 (q) 2
(C) lllx - 2l - 2l - 1l = 1, then x can be 0 4
(D)
''(ry)-f
Q)lf0 ), tnen ftr(x) = (s) 6
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RSM-91 1-P1-MA- Basic Mathematics 43
1.
3.
6.
Subjective :
Level- I
Solve for x, where
(i) x'- 3x + 2> 0
(iii) e" -7e* + 10 = 0
2. Find the derivative of the following functions w'r.t. x :
(i) y=x4+3x2-6
(iii) y=12x2-3)2
(V) e* -1 V =-
' e" +1
*' -1 ,o
X
y=(1 + 4x3111+2x2'1
y=/n(x3-sinx)
(x + 1\2
v=-
' ("*2)3(x+3)a
y = x sin-1x
( i i )
x+1
(vii) Y=sin-'JZ
(ii)
(iv)
(vi)
(viii)
, when x and y are given bY
x=acos't,y=bsin3t
x = (voc oscr)t , y = (vos incr)t - gt212w, here ve,c L? rtd g are constants.
( ( t \ ) x =aIcos+tt oo[tan7JvJ =,a s int
4. Find the intervals of decrease and increase of the functions:
(i) 2x2 - In x (ii) x'e-*
5. Find the point of maxima and minima of the following:
(i) y=2e"+e-'
lnt6grateth ef ollowingfu nctions:
. (i) x1t3 + x1t2 + x3t5
(iii) sect x cosec'x
cos2x
(v)
cos- x sln- x
(ii) Y=x2(a-x)2
Find9I
dx
(i)
(ii)
(ii)i
(iv)
(ii)
(iv)
(vi)
x2 -2x+3
sin2 x
X'
5cosx-3sin
^' * -c+os'x
3cosx+4
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J.-<q---
M
RSM-91 1-Pl-MA-Basic Mathematics
T. sotu"
"fl('"s"')'*tos-, ,,- lf.
8. Evaluate:
(i) fi2" * sx2)ox (ii) ilo" 0 lx
1n/
(iii)
I^o ( -x )3 dx (iv)
'f z * s9 in " .."
o d cos'x
9. Calculateth e areao f the regionb oundedb y the curvey (y _ 1) = x and y_axis.
10. Formt he differentiael quationo f the following.
Iii, *y =
( i i i ) ?"'+ be-*+ x2 (ii) y = ae^+ be-" t'=4a(x+b)
Fill in the btanks (11 - 13):
11. Thei ntervalosf x, sucht hatt og2fl x + ll + lx _ 4l + t) < 3 is
12. The intervals of increase of the function sinx + cosx (0 < x < 2n) are
13. Theareab etweenx-axisantdh ec urvey =xr f romx=0tox=3 is sq. units.
True / False (,l4 - i5}:
14. Therea re 4 possibles olutionso f the equationx 2- 3lxl + 2 = g.
15. The maximum value of y = s cos x - ] cos Sx occurs whe Tt ..
3
)n *=
6
, then the value of a is 5.
I ,*,,.,
FlllICC ka., Htrlet aou".,Zg-n, Xri;
RSM-91 1-P1-MA- Basic Mathematics 45
(i)
(iii)
(v)
3.
6.
7.
Level- ll
1.
2. Findt he derivativeo f the followingf unctionsw .r.t.x :
x2 +2
v=-
' x2 _ x-2
sinx
' 1+cosx
cosx 1. . x
v=-+-lntan-
' 2sinzx 2 2
Solve for x, where
(i) In3 x2< 16e e(x+ 2)2
(iii)
' " " - ' . [ - ! )
(ii) x2+Slxl-6=0
(ii)
(iv)
\ / =- 2x2- 1 t r----;
x{1 + x'
, 1+x2
v' = /fl ------ ^ 1-x'
vz
Thecurve'2y=+ dividesthecur*v'e* f =Sintwopar tsF' indtheareaosfm al lerpar t '
4. Findt he differenceb etweenth e greatesat ndt he leastv alueo f the functionf (x) = sin x + sin
Zx, lsin 3x in [0, ru]
3
5. Integrateth e followingfu nctions:
(i)
(iii)
(v)
e* sin(bx + d)
x2 -1
^'n
x6 -1
Vsec" x secxtanx
xa +x2 +1
n 2\x' +1)
dv- = lOO.. X
dx
(ii)
(iv)
x2
t^
{1-xo
(x+r)(x+losx)2
(ii)
(iv)
x2 +1
Findt he areae nclosedb y the curvey = e*, x-axisa nd the two ordinatesx = -1 and x = 1'
Find the
(i)
(iii)
(v)
8. Solvet he followingd ifferentiael quations:
(i)
dy xy2 +x
dx x2y +y
integrals of the following functions:
esrn x w
1
r
VX+x
cosx cos2x cos3x
2x
(ii)
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46 RSM-91 1 -P1-MA-Basic Mathematics
11.
12.
(iii) cosecx logydy + x2y2dx= 0 (iv)
(v)
"(r!*
zu) = *u9I
\. dx -) 'dx
$ =
""*t
+x2el
OX
L lf A = {(x,y) : y = e*, x € R}, B = {(x, y) : y = e-', x e R}, C = {(x, y) :y = x, x e R},
D = {(x, y) I y = 1lx, x e R - {0}}, then find
(i) AnB
(iii) CnD
10. Solve
(i) 7x-3>(x+1)2>x+3
(z\/,
(ii) ros.Io*11*tosrilsrlr*ll=r.
\ x)
s(,/c,(
10)_..
(ii) AnC
13.
14.
Thesolutiono f the integralequatio(nx t *yr)dy=xydxisy=y(x). lf y(1)= 1 andy (xo)= then find the value s of xs.
l fy=1* , , " t , * , " .1" . ,provethatgI=yf " l * " , l X-cr (x-crXx-cz) dx
"1c.,_x
cr_xl'
The maximuma nd minimumv alueso f f(x) = 2x3- gxz+ 12x- 5 V x e [0, 3].
Q"logu "
logno
"loga
5 - 3log,o(x/10-) glogloxo+ togo2.
SOlve fOf x.
5
_ , evtvg tut A.
fff = p1*1is a potynomoiafdt egre3e, t henp roveth at4 l urq) = p(") dtP!"). ' dx(" dxz ) dx3
15.
CaI'JCG Ltd,, FIITJEE House,2g-L Katu Sn i, S.*
RSM-91 1-P1-MA- Basic Mathematics 47
Objective:
Level- |
1' The set of all integersx sucht hat lx - 3l t z is equalt o
(A) t1, 2,3,4, 5\
(c) {2, 3, 4}
2. The valueso f x for whichl ogrtzx > logtrgx is equalt o
(B) {1, 2,3, 4)
(c){-4, -3,-2}
(B) (1, "o)
(D) none of these
(B) (1, 0) u (2, 3l
(D) none of these
(B) (2, - 1)
(D) (1, - 2)
(B)a> J2
(D)a>1
(B) 1
(D) none of these
(B)t an-1(e ')* c
(D) none ofthese
(B) 4/3
(D) 16/3
4. n, =
T##
anol|
=a x+ b ,t hen(a ,b )i se quatol
(A) (0, 1)
(c) (0, 1l
3. The intervailn which
x2 -2x--3 is negative,
xt -2x
(A) (- 1, 0) u (2, 3)
(c) t- 1, 0) u (2, 3l
(A) (- 1, 2)
(c) (- 2, 1)
(A) 3/4
(c) 3
7. i-?-!L is equatro
Jg^+e-^
(A) 2 tan-1 (e*) * C
(B) tan-1( et * ) *C
(A)2t3
(c) B/3
5. Thevalueof 'a'forwhichthefunctionf (x) =sinx- cosx-ax+ b'decreasesforx e Ris
given by
(A)as JZ
(C)a<1
6. lf x liesi n [0, 1],t hent he minimumv alueo f x2+ x + 1 is
B. Theareaboundedb ythecurvey=xlx| ,x-axisandtheordinatex=-1 ,x= l isgivenb y
9. Thesolut iono(f1 - V)
" I +(1 +x)Y=0is
OX
(A)ln(xy)+x+y=C (B)ln(xy)+x- y=C
(C) In (xy)- x-Y = C (D)noneo f these
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FRSM-
91 1-Pl-MA-Basic Mathematics
10. lf y = s cos (log x) + b sin (tog x), then
"
(A) d2v dv x- --*+ x:', * V = 0
dx' dx
(c) xr4**9I-v=o
dx' dx
(
log,log, I\
(A) 0
(c) n
11.
(B) 1
(D) -n
12. lf a2 +b2 +c2 = 1, then ab + bc + ac lies in the interval
,""',l[;-,'1]' ]
er[ -r],]
(D) none of these
, ' , [ * ]
,',[#)
(B) sin-1(2")+r<
(D) none of these
(B)B-n=6
(D) none of these
(B) 1
(D) none of these
13. lf y = {x +.,/x+ J". t"- is equalto
,^,[#)
r' o' \[z1v -1))
Thea ntiderivatiovef L w.r.tx is
,lt - +'
(A) log, e.sin-l(2" )+ t
tnen !I
dx
14.
16.
f| is eouarto
17.
I
witnr especxt is
(C) cos-1(z ')tog,e + I(
15. lf fortwo sets A and B, A u B = A n B = A, then we have
(A)A-B--0
(C)A=B
lf y = 1nl xlt hen
(A)1x, ' o
X
(c)1*, * o
X
The differential
equalto
(A) 0
(c) x
1e)a,x*o
lxl
(D) none of these
coefficienot f cos-1( 1Z"o"r- 5sinx )
r'--- 13 J'whereo<x<
(Bx) 'o?- - t=v=o
dxz dx
(D) none of these
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RSM-91 1-P1-MA- Basic Mathematics 49
19.
ln nl t - . - t rs.
14 2)
18.
lf(x) dx
0
The function f (x) = 5;n+ x + cos4 x in
(A) decreasing
(C) nothing can be said
fV l + sin2xd x (0< x < nl 2) ise quatlo
lnlrinx+cosx+C
(C)cosx-sinx+C
20. lf f (x) = lsin xl + lcos xl, then
(A) 0
(c)2
21. lfe * + eY- ex+Y, t hen
(A) sv-"
(C)-gv-x
is equalto
(c) (1, 2)
25. The value sf i-srli-dx is equalto
J coso'' x
(A)Jsinx+ c
(C) 2Gecx +c
(A)0<t<5
(c)t>5
(B) increasing
(D) none of these
(B)sinx-cosx+C
(D) none of these
is equalto
(B) 1
(D) 4
dy
dx
(B) sx-v
(D) -s"-v
22.
23.
24.
lf f(x) = cot-1(Jcos2x)
/*\
then f'l ;lisequalto
\o/
(A)
(c)
r e-" (1- x)
l-----+dx is equalto
J cos' (xe-* )
(A) tan(xe-*)+c
(C) tan2 (xe-- )+ c
The function 2x3 -gxz +12x+4 is decreasing
(A) (-*, 1) and (2, oo)
.l
E
E
!5
/ \ - t (B) -lcos(xe ^)l +c
\ - I
(D) none of these.
in the interval(s)
(B) (-*, 1) only
(D) (2, m) only
(B) Jcosx +c
(D) Jcosecx + c
(B)0<t<3
(D)t>3
2
{3
- - 2
{J
(B)
(D)
r3
A particlem ovesa longa straighlti nea ccordi"n3gto the law s =!-gt'+9t+17, wheres is in
meters and t in seconds. Velocity decreases when
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l,-<- --
50 RSM-g11 -P1 :,,riA-BasicM athematics
28- lf y=s;6x+bx2 has its extremevaluesatx=-1 and x= 2, then p=(a, b) is
(A)( 2,- 1) F' )' \ (.z , -1.] 2)
/ r\
(c)l-2,
, ) (D)nonoef t hese
r ^,213 Letf(x) =14-x'| , then f has a
(A) a local minima at x = 0
(C) a local maxima atx = -2
fcos 2x
ox ts equat to
., COSX
(A)2 sinx+ logl(sec-x tanx)+l 6
(C)2 sinx+ logl (secx+ tanx)+l c
lfy = 6es-r( rnx), thent hev atueo r $ is
dx
29.
30.
31.
27' rrv=JF.#,then2xl|+v isequalto
(A) 2x
s-?Ji
(A)
(c)
dv
;i
= Vcosx,t heny ise quatlo
(A) ce"os"
(c)e''n"
fJ------ is equatro
Jr/1+x+r/x
2 ^,^ ) ^, (A)
;(1
+x)''" -'-x2t3 +c
t"c2t' 9r r+ x\3t*29 xt " +c 2
f$o* is equatto
J VCOS X
(n)$+c
VSIN X
g)!+c
Vtanx
(B) 2Ji
(D) none of these
(B) a local maxima atx = 2
(D) none of these
(B)2 sinx- logl (secx- tanx)l+ 6
(D)2 sinx- log(f secx+ tanx)l+ c
-___ 1
x{1- (ln x)'
,|
- l
ffi
F)-+ x.{1- (ln x)'
(D) ---i-----T
x{1+ (ln x)'
32.
(B) e"ot"
(D) ces'n"
34.
t'2g'l2 9r r+ x )2/*3 9 xt / ' * c
fOl? U+x)3t2-2
^ztz
' '3' * " 3
(a) $+c vcosx
('D ) --?-: + c
(sinx )"''
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RSM-91 1-P1-MA- Basic Mathematics 51
35. rhev atuoer Jef"i #.x
is equalto
(A)#r.c
tcy -Ji- *
"
36. Thev atueo f flogE1gldx is equatlo
J (logx)'
(n) -tl+c
(losx )-
(c) ;l-+c logx
27. [-Ig:-d" is equato
J(x+1)'
(A- .2) e^ ;rJ
aX (c) ---: " (x + 1)"
1s-1$ +c
(D) none of these
(B)- I++c
(losx)-
1ol19 9r*"
oX (B) --: ^ ' ' (x+1)'
AX (' D) -: x+1
38. J"-{o*'+8x+3)
.(A) (2x + 1)2 e" + k
(C) (ax+2)e" +k
dx equals
39.
r cos 2x'+x+1
1;4,n ,**2*o"
=
(A) log(x2 + sin2x + 2x)+ c
1^ (C)
7
log (x'+sin 2x+2x)+c
fg"1l)"- dx is equarto
J (zx + a)'
(A) e"
' +c 2x+1
(C) e'(2x + 3)+ c
(B) (x+1)2e"+k
((D) none of these
(B) -log(x2 + sin2x+ 2x) + c
(D) none of these
(B) e" (2x + 1)+ c
(o) 2f-3+c
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52
41.
42.
RSM-91 t-Pt-MA-Basic lttathematics
Statement-1: lim sin(1/x) -,1
x +o 1lx
because
Statemen-t 2: lim^ i .
slnx -1
x+0 X
(A) Statemen-t 1 is True, Statemen-t 2 is True; Statement-2is a correcte xplanationfo r Statement-1
(B) Statemen-t1 is True,S tatemen-t2 is True;S tatement-i2s NOTa correcte xplanatiofno r Statement-1
(C) Statemen-t1 is True,S tatemen-t2 is False
(D)S tatemen-1t is FalseS, tatemen-2t is True
Statemen-t1 : Areau ndert he curvey = x3a ndx -axisf orx e t-2,zlis zero.
because
z
Statemen-Zt :
Jx3Ox
= O.
-2
(A) Statemen-t 1 is True, Statemen-t 2 is True; Statement-2is a correcte xplanationfo r Statement-1
(B) Statemen-t1 is True,S tatemen-t2 is True;S tatement-i2s NoT a correcte xplanatiofno r Statement-1
(C)S tatemen-t1 is True,S tatemen-t2 is False
(D)S tatemen-1t is FalseS, tatemen-2t is True
jLllg3 Ltd., FIIDEE House,zg-+ Kalu Sarar) Saruapriya whar, New Dethi _r7o otq rn ffi
53
2
-3
1
-6
6.
7.
RSM*91 1-P1-MA- Basic Mathematlcs
Level- ll
1. lf fsec2xdx = f(g(x))+ c , wherec is constanto f integration, thenw hicho f the followingm ay
J
3. Area enclosed between the parabbla Y = x2 and * = f is equal to
be true
(A)domainoff=R-{0}
(C) f(x) = 1'
2. 6 [--A-=mtan-1[nt"nI)+c, then
J5+4cosx \ 2)
(A)m=1
1 (C) n=-
3
1 (A)
5
(C) area of region {(x, y) : x2 < y < lxl 3
4. Let f(x) = J1- x2 , then
(A)d omaino ff (x)i st -1, 1l
(C)d omaina ndr angea res ame
{-r
5. Let(x)= !-x,then
1+x'
(A) domain of f(x) is R - {-1}
(C) fof(x) = x
(e) zfx-x2;dx
0
(D) none of these
(B)r angeis [0, 1]
(D) none of these
(B) range of f(x) = R - {-1}
(D) none of these
(B) B=
(D) B --
(B) range of g^= R
(D) g(x) = sec'x
2
(B) m =;
(D)m+n=1
(B) min. at x = 30o
(D) max. at x = sin-1*
(B) t0, oo)
("DL3)[ 1 9..") )
ffi [----9I- = Atan-1x+Btan-1]+c, then
J1 x2+ 1)(x2+ 4) 2
(A) A =;
(c) A=-1
Thefunctio1n + 2sinx+ 3cos2(xo =
-.?),t
(A) min. at x = 90'
(C) max. at x = sin-11 J5
B. The intervals is which function f(x) - x3 + 5x2 - 1 is strictly increasing isiare
n,[--,-T]
(c()- .",+l
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54 RSM-91 1-P1-MA-Basic Mathematics
9. Thevalueso f k< 0forwhicht hefunctionf(x=) xz+ kx+ 1 isstricflyin creasingin 11,2l
( )-2
(c) 1
10. A particlem ovesi n a line accordingt o the equations =
constantst in (sec)a nd s = distancec overedb y line
(A) velocity after 2r"" i= 2" * b
S
(C) accelerationis proportionaltos -3
(B)-1
(D) 3
where a, b and c are
(B) accelerationis independenot f s
(D) none of these
1.
2.
3.
11. Aparticlei s movinga longa linea ccordingtothela ws =tan-1t+ at2+ bt+ cwherea, b, care
constants.l t is seen that'at t = 1, s is 3.5 m and velocity= 3 m/seca nd accelerationis 1.5
m/sec2,t hen
(A)a=1 1B)b=2
(C)a+b+c= Y;
12. Thec riticapl artso f functionsf( x)= 2"o"2x - cos4xi n (0,n ) is/are
(D) b =;
13. 'The functionf (x) = (x' - 2x + 2) e* + 3 is increasingfo r
(A) (0, m)
(c) (-1, 1)
(^);
(cT)
4
(C) local maxima at x = l
e
( ' ) ;
@+
(B) x e(-co, oo)
(D) none of these
14. A particlem ovesi n a straightl inea ccordingto x = t1l2t,h en
(A) accelerationi s negativeV x
(B) accelerationis directlyp roportionatl o cube of velocity
(C) accelerationis equalt o magnitudeo f velocitya t t = 3
(D) none of these
15. The function x' V x > O will have
(A) local minima at x = 1
e
(B)m aximumva lueo f functionis j.
e1/e
(D)m inimumva lueo f functioni. *e""
lfy=21on*+bx'+xhasitsextremumvalueatx=-1 andx=2,thenfindthevalueofa-2b.
l f f (x)=x3+6x2+px+2, i f thelargestpossibleintervailn whichf (x) isadecreasingfunct ion
is (- 3, - 1), then find the value of p.
Findtheintegralvalueofxforwhichthefunctionf(x)=xe-6x2+9x+Sisdecreasing.
at2 +Zbt+e
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55
RSM-91 1 -P't-MA- Basic Mathematics
t
Comprehension:
Comprehension - |
lf f(x) = lllx- 1l* lx- 2l-21-31.
1. Numbeor f solutionosf f(x)= 1
(A) 4
(c) 6
2. Numbeor f solutionosf f(x)=
(A) 4
(c)6
(B)2
(D) 0
(B) 2
(D) 0
3' Area bound by the curve y = f(x) and x-axis is
(A) 5 sq. units (P) 7 sC' units
(c1 s sq. units (D) 11 sq' units
Comprehension ll:
Read the following write up carefully:
Considert he integralso f the form:
'
:
I"-
(f(x) + f '(x))dx '
By product rule considering eY(x) as first integral and e" f (x) as second one, we get
t=e" f (x) -J " " t '1*10*J*" * t '1* ;O*=exf (x)+c.
HenceJ " - { t t * ) * f ' (x) )dx:e* f (x)+c'
Now answer the following questions:
r -(1+sinxcosx) .
4. le" | ___________|= ls_ equat ro
J \ cos'x )
(A) e" sec'x + c
(C) e.log lsecxl + 6
- { 1 1 ).
c. ldx is equalto
J\ In x (tn x). /
(A)I n(lnx)+c
(c) ll+c tnx
(B) e" tanx + c
(D) none of these
(B)x+lnx+c
(D) none of these
6. tf
Je*(tanx-togcosx)dx
= F(x),w hereF (x)= (x) loglsecxlwitFh( 0)= 0, thenr angeo f f(x)i s
(A) (-m, m)
(c) RComprehension
lll:
Read the following write up carefully:
It has beeno bservedth ati ntegralos f the type Jf'(x)f(x)Ox
givesf '(x)dx= dt thust akings implifietdo rm Jt(t1O' t
Now answer the following questions:
can be evaluated putting f(x) = l which
(B) R-
(D) R * t1)
JllltGG Ltd.,FrrTJEE House,2g-+ K"lu s"oL s"*"p'iy" wht', Ilt 573942
56 RSM-9'l 1 -P1 -MA-Basic Mathematics
7. {5#J."
(n)a,r'n( *':t)*.
'12 l.,lzx 1 ,",f,"n[fri])."
8. I
sl-t{dx
is equatro . t1x"+ x+1) .
(A)(xu + x + J) + s
1cy .-{ x-+x+1
L
J"-t, * togux )dxi s equato
(A)x*+s
(C) log"x2 + 6
(Bs) in-lr#.l."
(V2xl
(D) x2 *1*"
x'
(B)
(D)
4 ___;__ t c
(x"+x+1)
I
t
_Ia
x++x-5+1
(B) logux + c
(D) none of these
Matcht heC olumn
M4ch the intervaol f monotoni
incrqases in (- m, .o
(B) y = log(x. . f i .1) (q) increaseisn (0,3 )
(r) increasesin (- oo,0 ) and Oecre-seJin 10, .o;
1n (D)
4x" -9x' + 6x
(s) increases
'"
(;, r) ano decreases in (- oo, Q) r,r
(''
;)
u (1' m)
1+x+ x2 - 1_ x+ x2
x)=sinx+cosx
(o" - t ) (D) f(x)= xl "- -'
I
\a^ +1/
(s) f(0) = 1 and odd function
ffi
RSM-91 1- P1 -MA- Basic Mathematics 57
[]ts wtnsT 0ls slG1 tlr tll lTP n0B Httls
t .
2.
Subjective :
Level- |
(i)
(iii)
(i)
(iii)
(v)
(vi)
(vii)
3. (i) -9tant
a
( -
-,
1lu [2, co) (ii)
In2, In5
4x3 + 6x (ii)
8x (2x2-3) (iv)
2e* | 1e*+112
(x+1)2 | z 3 4 I
S +tt[ + 3f Lx + 1 (x+2) (x + 3)l
1
-_
tl1-x'-2x
t-1, 1l
4x (10x3+ 3x +1)
(3x'- cos x)/ (x3 - sinx)
( * .-.)
| _-: + srn 'x I
[Jt-*' )
tan o -9! seco
v0
t4 +212 -1
A
(viii)
(ii)
(iv)
4. (i)
(i i)
(iii) tan t
Maxima
(i)
(ii) x= al2
(i) 3
\ ' /438 *+re* 2 r t rz *9xu,u * "
(iii) -2cot2x+C
(v). -2 cosec2x+C
lt 1^l
\ta' 4' 'J
(i) 12
(iii) 1t25o
1
- sq. units
o
( i ) xY"+ 2Y'- xY+ x' -2=O
(iii) YY" + Y'2 = 0
[-3,4]
9
F
. [t I r t f
lncreasins'"
LZ,*1,
decrasins't
Lo';j
Increasingin (0, 2), decreasingin (-oo0, lu [2,m)
6.
Minima
*=
1rn1
22
x=0,a
( i i ) - -1+1 1xxtxo^- ^*C
(iv) 5sinx+ 3cosx+ 2tanx+ C
(vi) -(3cosecx+4cotx) + Q
(ii) log 2
(iv) 3^12-1
7.
8.
(ii)
9.
10
12.
14.
11.
13.
15.
Y" =y
[0. 1-l,[!l. 2^)
L 4) L4 . l
T
nnJrle M., Hnee nou"e2g', 6573942
58 RSM-91 1-P1-MA-Basic Mathematics
Level- ll
1. ( i )
(i ii)
2.
x e [-1, 1]
[r*G ) l-. ml
12 )
_x2 -8x+2
(ii) + 1,+2
(ii)
(iv)
4.
(i)
(',,)
(v)
,(".|)sq units
(i)
(ii)
(iv)
oX
; i fsin(bx +d ) - bcos(b*x o ) ]u"
(1+b'J'
2. .Et.
u
(sec x)"'' + c
_x+3_ ta1n.' x+-c1
+c
(i ii)
(v)
6.
"-
1
e
./{.st 2in-'x I
7. (i) ++c ( i i )
(iii) 2tnl1 +Jil*c (iv)
(,_v., '' sin6 x sin2 x sin4 x x
u
*
a
*
to
+4+c
8. (i) (f * t) = c(x2+ 1) (ii)
( lnv+t ) . . (lll) l= x'cos x-2(xsinx+cosx)+c \v)'
(iv) e"+"-r***" (v)
3
e(i). {(0, 1)} (ii).
(iii). {(1, 1) , (-1, -1)}
10.(i). xe(1,4) (ii). *u (r-J5, o)r(r*.,6, *)
11. Jg
" 13.
14. x = 100
/
^
\Z
I\x/ ' -x-2] '
1
1+cosx
-cosecx cot2x
4x2 +1
/^
xV1 + x'
4x
1- xa
3. difference= 2
5.
I
l.{2
Ssrn
'x" +c
1t . r
''3
6 (" * Inlxl/ + c
x-2tan-lx+c
x-tan-1x* {*"
3
yln-I-xtnI+c
ee
(x'y)=key/"+c
0
-5,4
J']llcc rtd.' FrrrJEE House,2g-+ Kala garail saruapriya uihar, Ivew Delhi -tto ot6, ph 26srgg4, 26g54102, Fax 2dgr3g42
RSM-91 1*P1-MA- Basic Mathematics 59
Objective:
Level- l
Level- ll
1.
4.
7.
10.
13.
1.
4.
7.
10.
13.
16.
19.
22.
25.
28.
31.
34.
37.
40.
c
B
A
A
cc
B
cc
B
B
B
D
D
2.
5.
L
11.
14.
17.
20.
23.
26.
29.
32.
35.
38.
41.
2.
5.
8.
11.
14.
A
B
A
D
B
B
D
A
B
D
D
D
D
D
3.A
6.A
9.8
12. C
15. C
18. B
21. C
24. C
27. B
30. D
33. D
36. C
39. C
42. D
3. A,B,C
6. A,D
9. A,B
12. A, B, C
15. A, D
A,B
A,B
A,D
A,B
A,B,C
B,C
A,B
A,B
A,C,D
A,B
3.C
6.8
9.4
2.C
5.C
B,D
1.A
4.B
7.A
(c) (s) (D) (s)
(c) (p) (D) (q)
1. (A) (0 (B) (P)
2. (A) (s) (B) (r)
NumericBaal sed:
Nlatch the Column
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