Integration of Safety, Health and Environmental Scania CV AB
JEE MAINS-ENTRANCE EXAM-2019 (12th April-Morning) Solutions MAINS-ENT… · -1+ = + n Cv n Cv Cv nn...
Transcript of JEE MAINS-ENTRANCE EXAM-2019 (12th April-Morning) Solutions MAINS-ENT… · -1+ = + n Cv n Cv Cv nn...
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SOLUTIONS Joint Entrance Exam | IITJEE-2019
12th APRIL 2019 | Morning Session
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Joint Entrance Exam | JEE Mains 2019
PART-A PHYSICS
1.(3)
2.(1)
3.(3)
4.(3) option (3) 5.(3)
2= òI dmr
Þ 20 2sæ öpç ÷è øòb
ar dr r
rÞ
3 3
0( )23-
= s pb aI Þ 2=I mk
Þ3 3
20 0( )2 2 ( )3-
s p = s p -b a b a K Þ
2 2
3+ +
=a b abk
212= p Þ µ
IT HMH T
11 1 cos45 2= ° =
BH B
22 2
3cos302
= ° =BH B \
2 21 2 1
2 1 2
2 30403
æ ö æ ö= = × =ç ÷ ç ÷è øè ø
H T BH T B
\ 1
2
9 6 0.732
= =BB
1 2= +U U U
1 2 1 1 2 2( )+ D = D + Dn n Cv T n Cv T n Cv T
1 1 2 2
1 2
1- +=
+n Cv n CvCv
n n
3 52 32 2 2.15
´ + ´= =
R RR
0, 0, 1= = =A B Y0, 1, 1= = =A B Y1, 0, 0= = =A B Y1, 1, 0= = =A B Y
\
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6.(1)
12 sinq
=utg
22 sin(90 )° - q
=ut
g
1 22 sin 2 sin(90 )q ° - q
× = ×u ut tg g2
1 2 22 2 sin cos´ q q
=ut tg
2
1 22 2sin cosq q
= ×ut tg g2
1 22 sin 2q
= ×ut tg g
1 22
=Rt tg
2pw =
T
=qIT
2 /=
p wqI
2w
=pqI
04µ
= apIBr
0 24µ
= × ppIBr
0 24µ
= × pp
IBr
9 73.8 10 10 22
- - w´ = ´ p´
pqr
9 72
403.8 10 1010 10
- --
p´ = ´ ´
´q
9 2
73.8 10 10 10
40 10
- -
-´ ´ ´
=p´
q
10 73.8 10 1022407
-´ ´=
´q
33.8 10 740 22
-´ ´=
´q
30.030 10-= ´q C
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7.(4)
For 8.(2)
53 10-= ´q C
180D = -cau
180D =acUabc
-D = D +Q U W310 180= +W
130=W J=V IR
=VRI/
=w qRI2 2
2
-=ML TRI T2 3 2[ ]- -=R ML T A
1 22
0
14
=pe
q qFr
1 20 2
1e =
q qF r
2 2
0 2 2-e =A T
MLT L2 2
0 3 2-e =A TML T
1 3 4 20 [ ]- -e = M L T A
2
0 0
1=µ e
C
0 20
1µ =
eC
0 2 2 1 3 4 21
- - -µ =L T M L T A1 2 2
0
- -µ =
M A LT
1 3 4 20
1 2 20
- -
- -e
=µ
M L T AM A LT
2 4 6 40
0
- -e=
µM L T A
2 4 6 40
0
- -µ=
eM L T A
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9.(4)
10.(4) velocity of child (Rightward)
velocity of man (Left ward)
(take Rightward as positive)
Conservation of Linear momentum
11.(3)
At
By diagram
12.(1)
2 3 20
0
- -µ=
eML T A
( )= +W M g a l2a L=w
2 20
1 (1 cos )2
w = - qM L Mgl
202 (1 cos )L gw = - q
0[ 2 (1 cos )]W M g g l= + - q
0[ 2 2 cos ]W M g g g l= + - q
0[3 2cos ]W mg l= - q203 2 12
é ùæ öq= - -ê úç ÷ç ÷ê úè øë û
W mg l
203 2 22
é ùq= - + ´ê ú
ê úë ûW Mg l
20[1 ]W Mg l= + q20[1 ]W Mgl= + q
®CV
®mV
/ 0.7=C MV
( ) 0.7- - =C MV V
0.7= -C mV V
=i tp p
0 50 20(0.7 )= - + -m mV V
0.2=mVsin( )= - +w fy A kx t
cos( )= - - +w w fpV A kx
0, 0= =t x
cos= - w fpV A
= +pV ve
cos = -f ve=f p
stress /strain /
= =D !
F Ayl
=DamgyA T
2D= =
a pyA Tmg
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13.(1) Case – I (B) (A) Source Observer
Case – II (B) (A) Observer Source
14.(2)
Voltage gain =
Voltage gain
Power gain
Power gain
15.(1)
16.(2)
For
For
1 1500 51500 7.5
-é ù= ê ú-ë ûv v
11 1 1500 7.5 5021500 5
+é ù= »ê ú+ë ûv v Hr
100= WiR3100 10= ´ WoutputR
changeinoutput voltagechange in input voltage
0D=D i
VV
0D ´=D ´C
b i
I RI R
45 10= ´vA
change in output powerchange in input power
=2
0 0 0D ×D D × × D Dæ ö= = = ´ç ÷D ×D D × ´D Dè øC C C C
i b b i b i
V I I R I I RV I I R I b R
23 3
65 10 100 10
100100 10
-
-
æ ö´ ´= ´ç ÷ç ÷´è ø
63
825 10 1010
-
-´
= ´
5 625 10 2.5 10= ´ = ´
1 1 1= +
f u v
1 1 120 5
= +v
20 ( )3-
=v virtual
20 5343
æ ö+ç ÷è ø= =
µreal
appdd
8.8=appd cm
20 50 1000 / 4= ´ =gi Amp
0 2- v
1( ) 2+ =gi R r
1 11900= W =r R0 10- V
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For
17.(2) Total energy
18.(2) Optical path difference
Shift
19.(4)
20.(3)
21.(2) Equivalent resistance
22.(3)
2( ) 10+ =gi R r
2 9900= Wr
2 9900 1900 8000= - = WR0 20- V
3( ) 20+ =gi R r
3 1900= Wr D
3 3 2 10000= - = WR r r
1240 1240108.5 30.4æ ö= +ç ÷è ø
E eV
22113.6 1æ ö´ - =ç ÷
è øZ E
n\ 2
1 138.913.6 4 1 1240108.5 30.4æ öæ ö´ - = ç ÷ç ÷ ´è ø è øn
\ 5=n
( 1)= -µ t
= =lb Dd
\ ( 1)- = ×µ b dtD
Þ ( 1)- =µ lt \1
=-l
µt
0, 1.5= = =ÎI V V\ 1.5Î= V1000 1= =I mA A0»V
Þ 0Î- =Ir Þ 1.5 1 0- ´ =r1.5= Wr
2 21 1 2 21 1,2 2
= =E CV E C V
1 1 0 2 3 0
1 / 3 / 3 / 3= + +
Î Îo
d d dC K A K C A K A
0 1 2 31
1 2 2 3 3 4
3Î æ ö= ç ÷+ +è ø
A k k kCd k k k k k k
( )2 1 2 303
= + +ÎAC k k kd
\ 1 2 31 1
2 2 1 2 3 1 2 3 2 3 1
9( ) ( )
= =+ + + - +
k k kE CE C k k k k k k k k k
2 4 8R R R R R= + + + =2
48
PRe
= =
Þ16 16 4 88
RR´
= Þ = W
0W hv= 34 146.63 10 10 4-= ´ ´ ´ J
206.63 10 4W J= ´ ´20
194 6.63 10 1.661.6 10
eV eV-
-´ ´
= =´
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23.(3) Electrostatic shielding 24.(2)
,
For short dipole at equatorial position
25.(4)
26.(1)
27.(1) by comparing
28.(3) ,
,
0 ˆ= -!P P x
2 2 2 20 0
1 ( ) 14 4
-= +
pe pe+ +M
q qVl d l d 2 2
cos l
l dq =
+0=MV
304MPEd
-=
pe
!!
gain lostQ Q=
1 1 20.5 10 1 50M M L M´ ´ + = ´ ´
2
150 5ML
M= ´ -
0.61 0.61sin . .
× = = ´l l
µ qR P
N A. sin 1.25= =µ qN A
75 10. 0.61 0.241.25
-´= ´ = µR P m
2
2 2tan2 cosgxy x
u= q -
qtan 2q =
1cos5
q =
2 2 92 cos
gu
=q
53
u =
1
1 1 14 2R
= + 141.7 1.73eqR R= + = +
1
1 1 24R+
=4 5.13eqR+
=
143
=R
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29.(4)
30.(3)
9.13eqR =
Bvle =2
21 101 5 101
--´
e = ´ ´ ´
45 10 volt-e = ´
eqIRe =
=eq
VIR
4 45 10 15 109.1 9.13
I- -´ ´
= =
41.648 10I -= ´6164.8 10I -= ´
164.8I A= µ
ˆˆ6 8= - +!S j k
ˆ| |SSS
=!
!
2 2
ˆˆ6 8ˆ(6) ( 8)
j kS -=
+ -
ˆˆ6 8ˆ100j kS -
=
ˆˆ2(3 4 )ˆ10j kS -
=
ˆˆ3 4ˆ5j kS -
=
1 ( )MM l xl
= -
1 2xT M r= w
2( )( )2x
M l xT l x xl
-æ ö= - w +ç ÷è ø
2 2 2( )2x
m l xTl
w -=
x
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PART-B CHEMISTRY
1. (3)
(Yellow ppt)
2.(1) Due to formation of back bonding. is hybridized. Hence it is planar and less basic
than
3.(2)
The distance of corner atom to teat heel a void along diagonal =
Component along any side =
4.(4) ion converted to ion and 5.(1) Probability of finding electron is maximum at region a and c
6.(1)
7.(1)
8.(1)
9.(3)
10.(2)
3 4 3 3 4 3Na PO 3HNO H PO 3NaNO+ ® +
( ) ( )3 4 4 4 3 4 4 32 3H PO 12 NH MoO 21HNO NH PO .12 MoO+ + ®
p dp- p ( )3N SiH 2Sp
( )3 3N CH .
1cos3
q =
34a
3 14 43
´ =a a
4 4 2æ ö- + =ç ÷è ø
a a aa
Cu+ 2Cu + 0Cu .
[ ][ ]3spK s 0.2=
[ ] [ ]24
322
s 24 100.2
3 10
-
-
= ´
= ´
solution
solvent
X X1000m 13.88
X X molar mass of solvent= =
ext.W P v= - D
0.987 9 8.8 Latm= - ´ = -
1Latm 101.3J=8.8L atm 0.9kJ- = -
2(s) 2I I (g)¾¾®
( )2 1 2 1H H Cp T TD = D + D -
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11.(1) Bakelite is a thermosetting polymer.
12.(1)
13. (2) Rate of formation of is equal to twice the rate of dimerisation of 14.(3) Fact 15.(3) 16.(1) Fact 17.(3)
Configuration of galactose is differ around C-4.
18.(4)
19.(4)
20.(3) Oxidising power is directly proportional to SRP. 21.(3) Fact
22.(3) Molar mass of
Molar mass of
Molar mass of 23.(3) RNA is single stranded structure.
24.(3)
( ) ( )2 2p p I (g) p I (S)C C CD = - 1 10.055 0.031 0.024 cal g k- -= - =
1 12H 24 0.024 50 22.8cal g k- -D = - ´ =
( ) ( )23Fe Phen Fact+
é ùê úë û
4 8C H 2 4C H
2 2Zn 2HCl ZnCl H+ ® +
2 2 2Zn 2NaOH Na ZnO H+ ® +
4alkaline.KMnO3 3 3High tempH C CH CH CH 2CH COOH- = - ¾¾¾¾¾¾¾®
2 2Cl / H O3 2 3 2H C CH CH H C CH CH
| |OH Cl
- = ¾¾¾¾¾® - -
32AB 25 10 kg-= ´
32 2A B 3 10 kg-= ´
3 3A 5 10 and B 10 10- -= ´ = ´
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25.(3) orthosilicates 26.(1) Thermal stability of carbonate of alkaline earth metal increases by increasing size of divalent cation.
27.(2) Square planar complex is formed by removing ligand along axis. So that its splitting pattern follow second option.
28.(1)
29.(1) pKa value is directly proportional to the –I effect. 30.(1) Peptization is a process of converting freshly prepared ppt into colloidal solution when treated with
suitable electrolyte.
Joint Entrance Exam | JEE Mains 2019
PART-C MATHEMATICS
1.(2)
2.(4)
Max value of
3.(2)
4.(1)
5.(3)
6.(4)
( )44SiO ,-
2dz
[ ]4 2 2 3 16 2 3 8= + = Þ + =S a d a d
[ ]6 3 2 5 48= + = -S a d 2 5 16+ = -a d
2 24 12= Þ =d d
22=a
[ ]10 5 44 108 5 64 320= - = - ´ = -S
( ) ( ) 22. 2 0
2¢ = - + - =
-
xf n k n kx xkn x
Þ ( )2 22 2 0- + - =kx x kx x Þ 23 4 0- =kx x
Þ30 or 34
= = =kx x
4=k Þ ( ) 3 3=f x
( )( ) ( )910 21 1 1+ - + +x x x x
( )( ) ( )9 62 3 9 36
9 8 71 1 843 2´ ´
- - = = =´
x x C x
1 18 4 162 2
= = = = = =np npq q p n
( ) ( ) ( ) ( )16
216 16 161 162 0 1 22 2 2
Þ £ = + + = + +Cp x P P P 16 16 16
1 16 120 137 1372 2 2
+ += = = Þ =
k k
1 1 1 1 66 1 17 1 99...3 3 100 3 100 3 100 3 100
é ù é ù é ù é ù é ù- + - - + - - + - - - -ê ú ê ú ê ú ê ú ê úë û ë û ë û ë û ë û
( )67 2 33 67 16 133- - = - - = -
/ 2
0
cotcot cosex
p
+òx
x
/2
0
cos1 cos
=+ò
p x dxx
/2 /2
0 0
11 cos
-
+ò òp p
dx dxx
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7.(4) P is T and is F now is False and r is also False. q is True r is False. So, TTF is truth values of respectively. 8.(1) Consider following cases 10 distinct or 9 distinct or 8 distinct … 0 distinct.
(say) … (i)
… … (ii) Apply
Adding i and ii
9.(4) is the right bisector of the line joining points i.e., y = x
10.(4)
slope of normal
Also, lies on
i.e. or (4, 4) (A little consideration shows that P must lie in 2nd quadrant)
11.(3)
12.(1)
[ ]/2
220
0
sec2 21
= - òp
p px dx
/2
0
tan1 212 22
= -
ppp 1 11 ( 2) , 2
2 2 2= - Þ - Þ = =p p M n
\ 1= -MN(~ )® Ú ÞP q r ~ Úq r ~ Úq r Þ ~ q
\
1 1 1p q r
Þ 1021 21 21
9 8 021 ......+ + =C C C C S
21 2111 12+C C 21 21
20 21+ =C C S -=n nr n rC C
2125 2=202=S
1- = -z i z ( ) ( )0, and 1,0i
2 23 4 12 8 6+ = Þ = -dyx y y xdx
68
= -dy xdx y
Þ4 2( )3
- = = -dx y givendy x
2 3= -y x
,x y 2 23 4 12+ =x y2
223 4 123
æ ö + =ç ÷è ø
y y
2 2 2 2 94 12 36 16 364
+ = Þ = Þ =y y y y
3 31 1,2 2
æ ö= ± Þ = ±ç ÷è ø
! !y x
31,2
æ ö-ç ÷è øp
Þ2
2 5 25 5 55 252 4 2æ ö+ = + =ç ÷è ø
1 12 3sin sin53
- --
1 1 1 1 15 4 5 4 12 3 5 4 56 56cos cos cos cos sin13 5 13 5 13 5 13 5 65 2 65
- - - - -é ù æ ö é ù= - = × + × - < = = -ç ÷ê ú ê úë û è ø ë û
p
2 2 4+ =x y
2 0+ =dx ydydt dt
( )1 25 25 / sec.3 3
= - = - = -
xdxdy dt cmdt y
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13.(4)
14.(2)
15.(3)
Area of PAB
Length of chord
16.(4)
is required vector
i.e.,
17.(2)
Differentiating both sides w.r.t x
2 2 2 2
01
4
- æ ö= - -ç ÷ç ÷
è øò
yA y dy
2 2 22 3
02 12
-é ù
= - -ê úê úë û
y yA y2
12 12
é ù= - -ê ú
ê úë û
y yy
( ) ( ) ( )242 2 1 1 2 1 2 112
é ù= - - - - -ê úë û
( ) ( )12 2 1 2 2 2 1 2 23
é ù= - - - + -ê úë û
( ) 2 22 2 1 2 2 13
é ù= - - - +ê ú
ë û
( ) 2 2 22 2 1 1 2 2 13 3 3
é ù æ ö= - - = - = +ç ÷ê ú ç ÷
ë û è ø
4 52 23 3æ ö= -ç ÷è ø
2 3 2 54 8 2 45 1 3 1 18 2 4 12 2
é ù é ù+ê ú ê ú- - é ù é ùë û ë û= = =ê ú ê ú- -ë û ë û
A
2 3 2 50 2 0 15 1 3 1 12 0 1 02 2
é ù é ù-ê ú ê ú - -- - é ù é ùë û ë û= = =ê ú ê ú
ë û ë ûB
2 4 0 1 4 2 4 24 1 1 0 1 4 1 4
- + - -é ù é ù é ù é ù= = =ê ú ê ú ê ú ê ú- - - - -ë û ë û ë û ë û
AB
1 12 2
= × = ´ ×PA PB h AB Þ 12 5 13´ = ´h
6013
=h \12013
( ) ( )! !
( ) ! ( ) ( )4 4 0 16 16 82 0 4
+ ´ - = l = - + -
!" "
!
i j ka b a b i j k ! !( )2 2= l - -!i j k
!(2 2 ) 3 12 4l - - l = Þ l =!i j k \ ! !8 8 4- -!i j k
! !( )4 2 2- -!i j k
( ) ( )( )
3
04 2= -ò
f xt dt x g x
( ) ( ) ( )34 ( ) 2 ( ) ( )¢ ¢´ = - +f x f x x g x g x
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18.(4)
Then solution will exist only if
or
are 5 solutions is
19.(4)
(put )
20.(4)
IF
Solution is x.IF = (Put )
And apply by parts and replacing t we get
Given
( )2
1lim 4 6 9 2 1848®
´ ´ = Þx
4 21 sin cos 3+ =xn x4 4sin sin 3= -x x
4 2sin 0 and sin 3 0= =x x
=x nx 13 = Îpx m m n z
. ., 0, , 2 , , 2= p p - p - pi e x5 5,2 2-é ùê úë û
p p
3
42 1-
+òxx x
3
4 42 1
= -+ +ò òx dx
x x x x ( )2 2
3 232 3 1 .3 . 1
= - ´+ +
ò òx x dx
x x xx x( ) ( )32 1log
3 3 1= -
+òdtx H
t t
3 =x t
( )32 1 1 1log 13 3 1
æ ö+ - -ç ÷+è øòx dtt t
( ) ( )32 1log 1 log log 13 3
+ - - + +é ùë ûx t t c ( )3
33
2 1log 1 log3 3 1
æ ö= + - +ç ÷ç ÷+è ø
xx cx
( )3233
1 log 13 1
= + ++
xx cx
33
31 1log3
æ ö++ç ÷ç ÷
è ø
x cx
31log + xx
2 1 0æ ö
+ - =ç ÷è ø
y dx x dyy
2 1 0+ - =dxy xdy y
2 31
+ =dx xdy y y
21 1
-ò= =
dyy ye e
\1
31 .
-
ò yey
1
.-
= = -ò tyxe t e dt1
- = ty
1 1 1- - -1
= + + +y y yxe e e cy
11 1
æ ö= + +ç ÷è ø
yx cey
1 1= =x y Þ1-
=Cc
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Now at
21.(3)
22.(3) There are two equilateral and total number of triangles =
is required probability
23.(2) Volume of 11 pipe
(to be minimum)
min if
24.(3)
Sum of
25.(4) Let
26.(3)
m = 3
123 1 3 12
2 2
æ öç ÷= = + = -ç ÷è ø
y x ee e
1 1a b
+-a -b
( ) ( )( )( )1 1 21 2 1 1
a -b +b -a a +b- ab=
- -b -a -b+ab
( )2 22529375 375
25 2 375 271375 375
--
= =-- -
112
=
DS 63C
\ 63
2 110
=C
1 10 1 ( ) ( )0 1
lé ù = l = lë û
la b c f say
3( ) 1 1= l -l +f x
\ 2( ) 3 1 0¢ = l - =f x
13
l = ±
( ) 163
¢¢ = l > l = +f x at \13
l =
5 2 10 2 1
3 1
aé ùê ú= ê úê úa -ë û
B
( ) ( ) ( )| | 5 1 3 2 1 2= - - - a -a + - aB 225 2 2= - + a - a \ 21| | 1 0
2 2 25= + =
a - a -A
Þ 2. ., 2 2 24 0a - a - =i e:a 2 1a + =d
( )( ) ( )21 2sin sin 2 sin 12
= × + - +y x x x( ) ( ) ( )cos 2 cos 2 2 1 cos 2 22 2 2
+ - +æ ö= - - ç ÷
è ø
x x [ ]1 1 cos22
= - -
2sin 1= -y
2 21
1 8- =
x y
2 38- =mm
3 113
= +
= -
y m
x
3 113
= +
= -
y m
x
811
= +e
3=e
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27.(2)
at (0, 1)
28.(4)
29.(1)
X = 14
Variance =
30.(3)
13 10 531 8 433
+= =
-
4 + =e xy e
1- -= =
+ydy ydx ee x
( ) ( )( )( )
12
2 2
1é ùæ ö+ + - +ç ÷ê úè øë û= -+
y y
y
dye x y e ydxd y
dx e x
( ) ( )
( )24
1 11 1é ùé ùæ ö æ ö- - - +ê úç ÷ ç ÷ê úè ø è øë ûë û= -
+
e ee e
e x21
=e
2 5 1 2 13 2 1- + -
= = =-
x r
2 3 , 1 2 ,1+ - + -r r r
( ) ( ) ( )2 2 3 3 1 2 1 13 0+ + - + - - + =r r r
13 13 04 6 3 6 1 13 0
+ =+ - + - + + =!"""""#"""""$
rr r r
1= -r ( 1, 3, 2)- -
( ) ( ) ( )3 2 3 1 2 4 1 16+ + - + + - =r r r
6 9 1 2 4 4 16+ - + + - =!"""#"""$r r r
7 7=r ( )5,1,0
1=r2 2 26 4 2 36 16 4 56= + + = + + =PQ 2 14=
1 2 3 4 114
+ + +=
x x x x 5 10.......16
6+
=x x
1 2 10..... 44 96 140+ + = + =x x x
( )2 2
4S
-x x ( )22000 14
10= - 200 196= - 4=
( )3pæ öf =ç ÷è ø
hofog x tan3pæ ö= ç ÷
è øhof ( )3= hof ( )1/ 43= h 1 3
1 3-
=+
tan4 3p pæ ö= -ç ÷è ø
tan12pæ ö= -ç ÷
è ø
11tan12pæ ö= ç ÷
è ø