0 ratings0% found this document useful (0 votes) 85 views9 pagesLimits
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content,
claim it here.
Available Formats
Download as PDF or read online on Scribd
R. K. MALIK’S JEE (MAIN & ADV.), MEDICAL + BOARD, NDA, IX & X
Enjoys unparalleled reputation for best results
NEWTON CLASSES in terms of percentage selection
www.newtonclasses.net
LIMITS |JEE ADVANCED PREVIOUS YEAR SOLVED PAPER]
JEE ADVANCED
Single Correct Answer Type Be Gand ope 5
= 4d. does not exist because the left-hand limit is not equal
= J2SSBS them tim #2) is
L to the right-hand limit (IIT-JEE 1998)
\ oo , NERD ctor sin nd 3,6, forerery
4. none ofthese value of 0 then wi
aoe a y= 1,5,=3 y
ce & b=-lb=n @. b= 0, b=? -3n43
{5= en in S2I= 0) 5g (IIT-JEE 1998)
“7 > tig 2MB22— ZEUS is gua 0
"40 (1=c0s 2a)
» a2 b -2
© Wa 4-12 CIT-FEE 1999)
4. none of these ;
(ITER ISS) 9, Forse Rin (2=3) tesquto
ae he
see a. (IEEE 2000)
= (cos? x)
ann be
© r/2 @1 — arrsee200
Teter frye wg ODA)
isequal 10
for[s]20
Iyay=} GI «where [x] denotés the ilar nora deabocke
©, feelel=0
states integer les than or equal to, then lim (3) is METER A,
al b.0 . TL + 0, where mis nonzero
4. none ofthese
(IT-JEE 1985)
a0
5. The value of tim ¥2—_—__~
bt -
2 tie |. The value of lim(sin x)" +(14+-2)"*) = 0, where
(UT-JEE 1991) eee ee ED
ao bel
el 4.2 (IFJEE 2006)
a. exis and it equals 2
Office.: 606 , 6” Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0661-2682623, 9635608812, 8507613968
NEWTONCLASSES.NET
AST37 YEARS]
APTERWISE SOLV
JOUS YEAR C R. K. MALIK’S NEWTON CLASSES
Let gtx) = 0 0 and let p be the left hand derivative of
le Hate If lim go) =,
(IIT-JEE 2008)
A Lim + xIn +b)!" =26 sin? 0, 5>0,
smf 8 (~, n), then the value of is,
(HT. JEE 2011),
GIPIEE 2012)
Let a(@) and Bla) be the roots of the equation
QfiFa DP + (lira -Nr+Tra—H=0 where
a>-L.Then lim, afa) and tim fla) are
L
b-Land=1
2
4. none ofthese
(UEJEE 2012)
Multiple Correct Answers Type
ae -2 =
1. LetL= tim" —* 4
a>0.IfLis finite, then
1
|. areyee 2009
> ars »
Integer Answer Type
1. The largest value ofthe non-negative integer a for which
(IEE Advanced 2014)
2 Let m and n be two positive integers greater than 1. If
then the value of ” is,
(IEE Advanced 2015)
Fill in the Blanks Type
(IT-JEE 1984)
1 im (5) tn
tre «fe
xennnel
otherwise
P+, x40,
and g@)= 4, x
sx
then lim g{f(x)} is =
w(t) .
Ts)
(IE-JEE 1986)
. ABCiran esting nerd in scifi
TIHIAB™= AC an hte side oC en
(r= +)
triangle ABC has perimeter P =
and also
(IT-JEE 1987)
(IIT-JEE 1989)
(HIT-JEE 1990)
(UT-JEE 1996)
(IEE 1997)
‘True/False Type
L 1 lim [fo g()] exists, then both lim f(x) and lim g(x)
exist (OT.JEE 1981)
Subjective Type
av ie - Vix
4. Brute fin FESS (a0)
(UE IEE 1978)
2x
2 F(2) isthe imegral of FSO? 0. Find
lins'eo [vier = 22], rr gee 1979
Office.: 606 , 6" Floor, Hariom Tower, Circular Road, Ranchi, Ph. 0861-2662628, 9696608612, 8607619968,
PREVIOUS YEAR C1 LIMITS [JEE ADVANCED PREVIOUS YEAR SOLVED PAPER]
+ AY sin(a +h}
. 4. Usetheformuta tim
OE-JEE 1980) (HIE-JEE 1982)
$5. Find tim (tan(r7 +2)" (UEJEE 1993)
Answer Key
JEE Advanced Fill in the Blanks Type
Single Correct Answer Type a #Y
1 . ®
&
°.
1B. 4
1. Sa
Multiple Correct Answers Type “ee
1. False
Lae,
Subjective Type
Integer Answer Type °
1 @ 2@ Lak 3. asina +2asina
4.2102
cad
Waa
Ran
ron)
E
d
-
Coe nna aa
606 , 6” Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2662623, 9836608812, 8507613968
NEWTONCLASSES.NET
OLVED PAPERS [LAST 37 YEARS]
ADVANCED PREVIOUS YEAR CHAPTERWI R. K. MALIK'S NEWTON CLASSES
JEE Advanced
Single Correct Answer Type
Bs ai+ 35-2
Viens
iad
sH
2 + p5— 27
in "ak
Alternative Method:
Gay
4.4. The given function is
sins)
tl"
if xeC=,0utL.=)
if x6(0)
Hints and Solutions
Jim, f(3)= im 0=0
lim f00)# tim #0)
a
fin Fd
Lat.» jim EON
LAL.
= Slim
ig!
LHL #REL, Therefor, lim f(x) doesnot exis.
- Putting 8 0, we set by
Mi Shae
sin.
2078 § 9 cin"
= by +B, sin 8b sin? B+. + sin“
Taking limit as 9» 0, we obtain
sinnd
Yim ne 5, bon
im BEZA—281S hg
stas2e—2uta = [205 ane
a asinte ain
Office.: 606 , 6” Floor, Hariom Tower, Circular Road, Ranchi, Ph.; 0861-2662628, 9696608612, 8607619968,
NEWTONCLASSES.NET
LAST 37 YEARS]
APTERWISE SOLVE
PREVIOUS YEAR CI LIMITS [JEE ADVANCED PREVIOUS YEAR SOLVED PAPER]
Wee betel axel
p= left hand derivative of te~ at
> lim ge=-1
1 ate (i— a)
lim g(t A)=—1
tim |
MBL tog cor"
w)
oy telecon)"
tim 2h (Applying L’Hosptal’s Rule)
ost tn S23) “ec
Hn = >(*)ia(*)
ih ds itm
sin(esins) (xia
+ lin a 5 ml + xl +52)!" 2 in?
IMeose—e") iene
oss)(l+c0sxy(cosxe*)
Lis finite nonzero, Then n = 3 (38 for n= 1,2, L= 0, and for
n=4,L=o9)
APTERWISE SOLVE
(an) mx tani] sinne
12.4. Given tim '=0, where a is non-rer0
awe
ornl(a—nn— 1) =0
PREVIOUS YEAR CI
16 in| (eins)! +(e
= lim ina)"
1
(Using Hospital’ rue des.
[Using Hosp 1 jad
= 2wsaetI=0
606 , 6” Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2662623, 9836608812, 8507613968
EWTONCLASSES.NET R. K. MALIK'S
NEWTON CLASSES
Multiple Correct Answers Type
Numerator if = and then
‘Alternative Method:
Integer Answer Type
1.0) tim [raetsine— Dalle
BY Ca inte 4
(sinx=1)
site
eoxie)-))
2°" st egal
2n-m
Fill in the Blanks Type
mx (es)
1 tim (1=s) an = tim
Alternative Method:
Tim (1—sytan
sins, cemmne!
om A laere
e+ x20
and 600=44,
5. 2
(0°) = sind) = 60") =
Him et FG) = a((0) = esd) = (0) = (0) #1 = 1
Hence, im g(/(4)} = 1
Office.: 606 , 6” Floor, Hariom Tower, Circular Road, Ranchi, Ph.; 0861-2662628, 9696608612, 8607619968,
NEWTONCLASSES.NET
LAST 37 YEARS]
APTERWISE SOLVE
PREVIOUS YEAR C! LIMITS [JEE ADVANCED PREVIOUS YEAR SOLVED PAPER]
eh) af fee -20)
tin | =
(Gray 1 Alternative Method:
7
|. In AABC, AB = AC, AD 1 BC (Dis the midpoint of BC),
Let r= radius of cicumeircle
nf! +se)"
2c=2-F
samaota ace bx 8c 4D = hark
2
nf (+20)
42h
[vse (2)
=
Office.: 606 , 6" Floor, Hariom Tower, Circular Road, Ranchi-1, Ph. 0861-2662623, 9636608612, 8507613968,
EWTONCLASSES.NET
5
z
&
IOUS YEAR C R. K. MALIK'S NEWTON CLASSES
True/False Type
1, Fae, Conse fx) = ESI gs) = 5
A=© Then, lim 0)
(0) exits, bt tim f(x) and lim g(x) do nt eis im 28 x~sin2 (a)
‘Therefore, satomen is false.
(Applying Hospital's Rae)
Subjective Type
rae jm— RATAN Et (Applying L’Hospital’s Rule)
fae = (Applying L Hospital's Rule)
wanenanat: = lim SATE (Applying L’Hospital’s Rule)
(Sera ~ Sa) (Varas + Ja) (fSaT +25)
)
(ars 2h) (ave +25) (ante + Ds) sina hyena
(fomo) t
@, tim @[sin(a + h)—sina] + 2ahsin(a +h) +H? sin(a +h)
*
RS [LAST 37 YEARS]
faces +2 2221 + ti 2asin(a +
2 3 favan + Vix)
erate 1 We
(jar2a-vlia) 3 Mia dina ne
+ limhsin(a+)
Alternative Method: Alternative Method:
tim Met 2s VBE sim (2 AY sina + h)— a? sin
Faeroe \ *
a+ Wana +h)
1
(Applying L’Hospiat's Rua)
=2asinabcooa
1) Set
Jira
tim 2 TD
My Tee-1
PREVIOUS YEAR C!
= tim =" tim F +1)
20412202
Office.: 606 , 6” Floor, Hariom Tower, Circular Road, Ranchi, Ph.; 0861-2662628, 9696608612, 8607619968,
NEWTONCLASSES.NET LIMITS [JEE ADVANCED PREVIOUS YEAR SOLVED PAPER]
Alternative Method:
ssl)
606 , 6” Floor, Hariom Tower, Circular Road, Ranchi-1, Ph.: 0651-2662623, 9836608812, 8507613968
NEWTONCLASSES.NET
