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### Transcript of JEE MAINS-ENTRANCE EXAM-2019 (12th April-Morning) Solutions · Joint Entrance Exam | JEE Mains 2019...

Vidyamandir Classes

VMC | JEE Mains-2019 1 Solutions | Morning Session

SOLUTIONS Joint Entrance Exam | IITJEE-2019

12th APRIL 2019 | Morning Session

Vidyamandir Classes

VMC | JEE Mains-2019 2 Solutions | Morning Session

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

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

\

Vidyamandir Classes

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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

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

Vidyamandir Classes

VMC | JEE Mains-2019 5 Solutions | Morning Session

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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VMC | JEE Mains-2019 6 Solutions | Morning Session

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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VMC | JEE Mains-2019 7 Solutions | Morning Session

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-

-´ ´

= =´

Vidyamandir Classes

VMC | JEE Mains-2019 8 Solutions | Morning Session

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

Vidyamandir Classes

VMC | JEE Mains-2019 9 Solutions | Morning Session

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

Vidyamandir Classes

VMC | JEE Mains-2019 10 Solutions | Morning Session

Joint Entrance Exam | JEE Mains 2019

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 -

Vidyamandir Classes

VMC | JEE Mains-2019 11 Solutions | Morning Session

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- -= ´ = ´

Vidyamandir Classes

VMC | JEE Mains-2019 12 Solutions | Morning Session

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

Vidyamandir Classes

VMC | JEE Mains-2019 13 Solutions | Morning Session

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

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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VMC | JEE Mains-2019 14 Solutions | Morning Session

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

Vidyamandir Classes

VMC | JEE Mains-2019 15 Solutions | Morning Session

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

Vidyamandir Classes

VMC | JEE Mains-2019 16 Solutions | Morning Session

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

Vidyamandir Classes

VMC | JEE Mains-2019 17 Solutions | Morning Session

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