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10, 0.4.2.0.340, 622,60, yothen
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mays me
amber fteaions =2=2
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R= bedlyoder Twberes eRand-$2225) | snvie-S20
Demin of = [Sexe 5.r eR} *[-5.5] fses
assis
then y= 2-8) +7==3 7a
Wheas=-5 en) =25)-7=17 Deni «
Raneof@,= Say2itiyeR)
su
5 (a Wet fx Ree oh et
pitas «272-2
Wor hen f-i)=CiPst=-t 16
ah omits: Mi @) yb. (Stein s—co83) 8)
hen f3)=(BP = 271928
Then 1)=(7) + 1a 1a
= 1C1.02.0.0.6,28, 0,730,038)
Rangel 018, 750,48)
& rs?
issins-conrs
2s itn scons)?
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ye Rasyt20
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80 fat |
ferz0
Maximum valu of 2)
= minimum vale of ~
Pie,0 |
Maximum vale offi |
21, (by f2e+39,20-T))=208
La 2r+ Iyer and2x- Tray
Solving ite get
Tasty
2”
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Hapa 2
fanensy
eee a)
1 a= (12,3)
arb 242.112.9.0.140.3)
Bram -1.6.248.3)
Bx B= {0,1441,340, 0,63)
ARAM LDC. DZ AE IL
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Siis tener 4eDie,r04
Doms f= 2-1)
La fi=9
pet bogetyeer
>wen=aiey
sy)
is defined, fet 20
se-850
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S-asesa[s 920)
Domain o/= (2.0)
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‘The equation has 3 solutions foe P= 2
‘Number of intgral values ofp is 1
3
Yahe Si 6
32 2333
“The equation has 8 solutions for
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re{ot)
3)
Fors <-3
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foe) = Highes prime factor of =
12-342, 16904
BeBe Ia
Ise3x5,
Range = (3,13. 7,5,2,17)
IEE (Xt) Solutions ranger ef) *)
19. (d) Given, Ax) = cos(logr)
ele sort {}-s00 |
a
= coslogx)-costlog »)~Hfenstog r—log »)
+~ostog x 1ogy) =coslog)-coleg)
He sextog3)-coslog
= cox(og 1) -cos(log y)~costogx)-cosiog 3)
-o
20, 0) Given fe) =toy( 122) a aa)n
Ie
Pox ed
2. (0) ye f= SEF
(0) y= faye SE reo
yoxstyerel
x(y-De ay
yok
at
yeR-(l}
22, (c) Match the option by aking poison the curve.
2 () Use concept of signum funtion.
Pox-300
(6 (e+5)>0
£8 (2-5) 6,20)
red
2.) foy=Xt2,
P-8
#94 0an1
oo)
x#2>0
xe[-2,)UGe)
—
( ad
Ronge=R- {1}
26, (e) yor
y= loge
When > Oandx 1
27. 6) fxr @,x-a) may
132
80. fit.
a fey)=:
28. (B) Simple domain based question.
29.(A) A+1,B>P.C+S,D44,
(+ 9@+ 2009 =-1
xtyehy+204=-1
ORE+y==1y+2004= 1,y=~2005,- 2006
2003, x= +2002 x(R)=2
is not imeger
r=22,y50
£=Oym23n(R)=4
(2s x56
3sys7
=2,3.4,56
yrs.
(3) (GV2.593.518 515406545.)
R>7)
ole!) -{an(a3}(a3}
miR)=3 4. (a) We have, fay = 2s = Vand gin) = 4r-7
Su
6 @)
1)
2 (0)
Anstey
+1 0>2>2"=21>x
Sre(-71)
1. fe) f(x)=' | 73),
2207 BATT
1) sass) fla-y-PZ EP
Weg Ds 2 A PeeDe 2 e708!
2 2m
=Hylanes(25)]
15, (¢) Gienfir+ay.1-«))= 09
Lets ay~nand-
Then x= 22 and yA
Substituting the value of the and yin (9, we ott
3 ;
7c
Le
16. 1 for?
Beret Retes 1)2+720
log te=1)20
og, 1) 034-12 1x22
4-22 0and—2>0
Sreedhar N20
re, UR)
Domain = (1,0). (1,2) 42,70)
11
JEE (X1) Solutions pefth 24xs0
Jl, geet
x-l Oex 20
=e (0,2)
45. Re takyahe-N Axl + fell 2S8S2)
mwerst
nxe2 -1s2<0
Osxel
Be Isxew
=
fons fpteerstiane
feat |
ste} one?) are2]
JP>Oxt of
(2x-20)> e+ of
Qx- 207 Or + aF>0
(2x-2a+2e+0)(2r~2a~-2x-a)>0
asn-af(taroe)
o/(2)-sre9 +
S52 oS
1
K=-2
5
(i) xeRyell=)
(ii) xe FS yy 60%
Gy) reRyeCng] 9
= 0s
-Isyst
(i) y= 4 sin c0s.x= sin x cos.)
(i) y=sin de
Same range like sin.
-lsyst
(i) y= cousin)
~282sinrs?
Soutsyst
{= (s}-<(l- 0)
vel0.a)
(i) fe) = sink + costs
SG) =1 = 2sin? seo
0<(6indyy <1
bse inant
whe
aes
1+021—4ginzay? 21
isin2ay 21-5
= tsfs1
(bi) fir) = sent + costs
fie) “1 3sin?reostr
say=1-3
sinxy?
a
eee
qsfenst
50 (9 fxyexte~
=(4)
3
zeae
W Juy-2e = 3045
(ty fay=28 +3045
| sore
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(a) DEO DOD
Ayr ay ryr 1-0.
D20
wt
preg
Pau(yr) 20
Pa ay20
(9) faye22 23041
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peo
947 +6 12y-820
Por 120
yek-0-05,3+24)
JEE (XI) Solution (iil) [x] 22
(uy y= re[2)
an (ir) (2x32 Vi
2 [r-J2
2r- 312, 0)
‘5
ae aoe
B-2y)2-4Qr- 1)U r
39+ dgy?—42y—4 82 + By + ABV 820 0) Br-S]>x
ay + ly+120 sete
(9,
pe R-(-7-4¥3,-7 +403) 60)
+6 (3,2)
(9 (> 1 b)22
4€[-2,-4)
(vii) []<-4=> [x] s-5
re(-%,-4)
spfyrtfpas e453 (vy (3157
Y= +s4-6)+8y-3=0 re(-a, +8)
(x) [xJ=- v7
‘no solution
[x)= integer
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x45, 5.5)
fx *€0,1)
sia vf
23 2 rel2n2n+1]
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oh Ad-deesktsess © orneSe-n
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2: +3.xe(0,2)
(923)=-3 6-0)
yesnae
Be 3. xe(-1.0)
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terms will be as follows
1 sw) ft sor), ft
2 1000} [2 ioo0 “ep
= 141+ 1.. + 1(500 terms) = $00
Option (c) is correct Answer.
4. (0) Given fo ~ 120-# 32} + 2-9 +20)=0.
Every modulus function i2 non-negative funeton and
afte non-negative fonctions ada upto get 20 then
individual furetion self equa to zero simakansousy
ia
Fe I2et 32-0 forx=4or8
P= 9+ 20=0 forr= sors,
‘oth the equations are zero atx=4
So, 4s the only solution for this equation,
5 B=987
39< 28 <10and~106—a<-9
$f] =9 and [-23]=~ 10
foe) cos cose
6. (@) Given function i sgn(in(sine))=0
‘we know tat sigur fiction i 0, nly when
= Insina)=0
save!
ame+¥ frm eZ
1 (@) Give funtion i fs) = se —9-+20)=1
[Ase know that siemum function gives 1, only when
0,
B-91+2020
St-4)r-5)>0
Sox € (274) U(5,2)
: cof} se mos
forall © R ,8 6 [-28,2]
4 ops
2 values of is possible
9% (0) fsjearrd
foy=b
AAO) =f) = ab + b= 0
=b=Dora=-I
fiyrar orfisy=b- ax
Case
fes)=ar
FUPUUAM) = ANa)) = 42) = 403 = 9
> 4) = ais notan inter.
\
Case
IUCN =MO- A= fA) =b-4 =9
ab°B
AAMKIO))= AAD 109) =1C10)) = 110)
=10
10 (4-11
Sxebal
and23°=18+9
jtsD<0
2 tes 920
ereB-10)
MW. (@ L yna-2
M. -2ym(24)
ter y-4-2) 0 }
re-tayar?
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ANS ae erent an)
é Domain = (021
a Range = [02]
wy
“fa
ot
w 7
Domain = [02)
; Range= (1.0)
Domain [-20)
Range = (00)
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ity Domsin= [1.1
Range (01)
Sy heap at-b
by =0)ay—b
ere tatery
Precis
so)
und
MeO" Tan
25049 sid
ind
sr fay=(e-2)
Le) —(e-f09?
(e(e-r'J)
a6ax
Osas3
50,8 nota inetion
but as unique image for ec lesa of deen
Lr EKy
OF echt)
Lo & Is dfined when $4020
or ~4< 11 frthisimeral fis fined
Deminis <1
2) so= eR
Value off) sea when
Veg, (t—1)20and 250
log, (\=8)2 Land rt
log (1-2) 20g, and re
eteretandecto rs ld ret O fara = fase a fey sisal HY
2620) i
one
{a
a P|
Ta sdera) ons
=o bent
nae | peri
ae
Range =(-3,0) et
pescos
paar
Deain= (0,2)
| Range = (0.1)
o
Domain = (02)
Range = (10) DPM 38=0,rE!
acre!
P-12er 35 =0
5.7
F124 35=-1
= 12e+36=0
x= 6 not possible ab xis an integer
only 2 solutions,
a tn rastijedear4iFal
Clearly domain of f(x) is [
Now, »'
f [5 .
(x=2f§}~6amay an
Hence range offi (5.V8)
7 1
Forsto be rea,
(29-34) 24(y-1)(21~1y) [+ Diserininane 20}
3 y'+289-3y2—7y" -71+ By
By! -112y +3600
ay ly +4520
=(y-9)(y-5)20
3(y-9)(9-5)20
sy sSory29
_yeannot lie berween $ and 9.
0) Do this by graph transformation
Or
a 1
LO)= 5 > Vos=t
al
=fnd=* t
ania ve
2 Do this by graph transformation
or
Abed = [40D]
=Aip20
2) Do this by graph transformation
or
Range off = [-1. 11
> logy, Gila)+2) = [ogy loa,,3]
18,0008)
00+ 2f 2) 00+ 30
Forx=7, (1) + 2/1) = 10+ 30= 100
For 11, 3/(11) + 2/3) = 140,
A) £0) -
=a gg AOE
19. (0004)
GPa x+2 (x
(x) + Slap
20.(30)
2x3
21, (0) Given fy) = log y = fly) = tog (05), then
foe (4) =n + log( ty) = log 1 = 0.
2, (1+x)cosy=2?
Seay
(Past Year Questions)
JEE MAIN
1. (©) His given,
AQ) = sin 0 (sin 0+ sin 38)
= (sin + sin 0 — sin @sin 0
sin @~ sin? 9) sin = sin?
= sin? Bos! 8 = Qin 8 cos
which is true fora 8.
@ faarea = fas aa 6 oy Casek: 2091
2h-Vid-x- bro
ar-hir nd = Vee
velba > red
0
i ais rom 190 sorage= (0 ou f
ees). bgt) Cosel: xel2)
3 (0 Gies, fa) REED 5 MELE 2
(Fedee2) G2 AVF-H+ 1-605 +620
Former +30 eNT=0 > r2160
art ti) t
and rain + He +2) 20 socal
are
ioe 2sin* x08? x
ee £
4. @ Mo Rb6 Cae, B
Give, fee
Tegabetistonas:
i
re
=
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8. (Frome gen eco get
: fot res
pare ate ee")
Wiit-terclsoeteyel :
POrE roo?) 60"-1
BE 52 06f01<1
seaf{S)aae 4
an | 89 Seton fa)
Forborsinotfi)4-<0->1082
-Mubiply eq. (i) by #i& suberact from eq, (/) and -1>0
Frm Vy cae etn,
FeChOu(=)
sal
stetsar
From Eas () and), wegetthe domain of fix)
v.22)
So, commis exacly wo denen.
JEE (Xt) Solutions then mat
rine
and
Epinned ser eeetdroae?
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£4, 4
a) 9
py ate o{!)nievE.
1
a(t} -are-tone
orm
8 OMS ~O
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Adding union (9 24)
BenfO-ny= 1
aay)
44)
AB)"
14) YM + (8H) 230
90-2920
G0 aF or Nas?
etos oe pias!
intra or sin? r=?
4
* Toul seh = 4
18. fads a)= Ve + IHF, domo.)
Fie)~sie)~ A I=, doowia 1}
A880 I=3-V, dri 0,11
0)
So,coomnon domi (0,1)
JEE ADVANCED
&
16. (a) Sines, y=
wen
17. ©) Fordomsinafy
InesQinrat and x+250
webreO mad 2
B 2exctentaigd
m1 (2)-s0-4 ais
3 6620-10 a} AB
0
are biee
= ft tt tT he
Hence the solon set's,
em DOCH UII.)
ence (9) a) te comet options Guid
1, (a) Sioce. din off) nd fate D, and Dy Ths ey
oman of fs) + AGI 8D, Dy Hence, sive Sea
aor : : (|
30,[1900) . ae ttlith
oi * 2
AI+fd-9 9 et
Be +4e 4x10
Tetr-5=0
pacha 1s iat
4 4
So, oun of eal oots= 4