Rajasthan Board 12th Model Paper 2026 Maths

RAJASTHAN BOARD

MODEL
PAPER
2026

PRACTICE PAPERS

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iz’u&i= dh ;kstuk& 2026
d{kk & XII
fo"k; & xf.kr
vof/k &3 ?k.Vk 15 feuV iw.kkZad& 80
1- mn~n's ; gsrqvadHkkj&
Ø-l-a mn~n's ; vadHkkj izfr'kr
1- Kku 24 30
2- vocks/k 24 30

3- Kkuksi;ksx 16 20
4- dkS'ky 8 10
5- fo’ys"k.k 8 10
;ksx 80 100

2- iz'uksa ds izdkjokjvadHkkj&
Ø-la- iz'uksa dk izdkj iz'uksa dh vad dqyvad izfr'kr izfr'kr laHkkfor
la[;k izfriz'u ¼vadks dk½ ¼iz'uksa le;
dk½
1- cgqfodYikRed 18 1 18 22.5 33.96 36
2- fjDrLFkku 6 1 6 7.5 11.32 15
3- vfry?kqÙkjkRed 12 1 12 15.0 22.64 42
4- y?kqÙkjkRed 10 2 20 25.0 18.87 40
5- nh?kZmÙkjh; 4 3 12 15.0 7.55 32
6- fuca/kkRed 3 4 12 15.0 5.66 30
;ksx 53 80 100.00 100.00 195
feuV
fodYi ;kstuk % [k.M ^l* ,oa ^n* esga Sa A
3- fo"k; oLrq dk vadHkkj&
Ø-l-a fo"k; oLrq vadHkkj izfr'kr
1 lEcU/k ,oaQyu 3 3.75
2 Áfrykse f=dks.kfefr Qyu 4 5.00
3 vkO;wg 5 6.25
4 lkjf.kd 5 6.25
5 lkarR; ,oa vodyuh;rk 8 10.00
6 vodyt ds vuqÁ;ksx 6 7.50
7 lekdyu 12 15.00
8 lekdyuksa ds vuqÁ;ksx 4 5.00
9 vody lehdj.k 6 7.50
10 lfn'k chtxf.kr 7 8.75
11 f=foeh; T;kfefr 9 11.25
12 jSf[kd Áksxzkeu 4 5.00
13 Ákf;drk 7 8.75
loZ;ksx 80 100

iz'u&i= CY;wfizUV 2026
d{kk &XII fo"k; %&xf.kr le; 3%15 ?kUVs iw.kkZad&80
Ø-la- mÌs'; Kku vocks/k Kkuksi;ksx dkS'ky fo’ys"k.k ;ksx
bdkbZ@

cgqfodYikRed

cgqfodYikRed

cgqfodYikRed

cgqfodYikRed

cgqfodYikRed
vfry?kqÙkjkRe

vfry?kqÙkjkRe

vfry?kqÙkjkRe

vfry?kqÙkjkRe

vfry?kqÙkjkRe
nh?kZmÙkjkRed

nh?kZmÙkjkRed

nh?kZmÙkjkRed

nh?kZmÙkjkRed

nh?kZmÙkjkRed
mibdkbZ

y?kqÙkjkRed

y?kqÙkjkRed

y?kqÙkjkRed

y?kqÙkjkRed

y?kqÙkjkRed
fucU/kkRed

fucU/kkRed

fucU/kkRed

fucU/kkRed

fucU/kkRed
fjDrLFkku

fjDrLFkku

fjDrLFkku

fjDrLFkku

fjDrLFkku
d

d

d

d

d
1 lEcU/k ,oa Qyu 1(1) 2(1)
3(2)

2 Áfrykse f=dks.kfefr 1(1) 1(1) 2(1)
4(3)

Qyu
3 vkO;wg 1(1) 1(1) 1(1) 2(1)
5(4)

4 lkjf.kd 1(1) 1(1) 1(1) 2(1)
5(4)

lkarR; ,oa 8(6)
5 1(1) 2(1) 1(1) 2(1) 1(2)
vodyuh;rk
6 vodyt ds 1(1)
1(2)
1(1) 2(1)
6(5)

vuqÁ;ksx
7 lekdyu 1(1) 2(1) 1(1) 3(1)* 4(1)* 1(1)
12(6)

8 lekdyuksa ds 1(1)
2(1)
1(1) 4(3)

vuqÁ;ksx
9 vody lehdj.k 1(1) 3(1)*
1(1) 1(1) 6(4)

10 lfn'k chtxf.kr 1(1)
1(1)
1(1) 2(1) 1(1)
1(1) 7(6)

11 f=foeh; T;kfefr 1(1) 1(1) 3(1)* 4(1)*
9(4)

12 jSf[kd Áksxzkeu 4(1)* 4(1)

13 Ákf;drk 1(1) 1(1) 3(1)*
1(2) 7(5)

;ksx 10(10) 7(7) 4(2) 3(1) 2(2) 5(5) 1(1) 6(3) 6(2) 4(1) 2(2) 1(1) 4(4) 6(3) 3(1) 4(4) 4(1) 4(2) 4(1)
80(53)

loZ;ksx 24(20) 24(14) 16(11) 8(5) 8(3) 80(53)

fodYiksa dh ;kstuk %& [k.M ^l* ,oa ^n* esa izR;sd esa ,d vkarfjd fodYi gS uksV%& dks"Bd ds ckgj dh la[;k ^vadksa* dh rFkk vanj dh la[;k ^iz'uksa* ds |ksrd gSA

fo'ks"k %&mDr CY;wfizUV ekWMy iz'ui= dk gS tks iz'uksa ds izdkjksa dks le>us dh lqfo/kk ek= ds fy, gSA ewy iz'ui= dk CY;wfizUV fHkUu gks ldrk gSA

ek/;fed f’k{kk cksMZ jktLFkku] vtesj

mPp ek/;fed ijh{kk & 2026
ekWMy iz’u i=

d{kk&12
fo"k;&xf.kr

le;& 3 ?k.Vs 15 feuV
ERSubject - Mathematics
fo"k; dksM&15
Subject Code - 15

iw . kkZ ad &80

ijh{kkfFkZ;ksa ds fy, lkekU; funsZ’k %
BS
Genral Instruction to the Examinees :

1. ijh{kkFkhZ loZiFz ke vius iz’ui= ij ukekad vfuok;Zr% fy[ksaA
Candidate must write first his/her Roll No. on the question paper compulsorily.
2. lHkh iz’u djus vfuok;Z gaSA
All the questions are compulsory.
3. izR;sd iz’u dk mŸkj nh xbZ mŸkjiqfLrdk esa gh fy[ksaA
Write the answer to each question in the given answer book only.
4. ftu iz’uksa esa vkUrfjd [k.M gSa] mu lHkh ds mŸkj ,d lkFk gh fy[ksaA
For questions having more than one part the answers to those parts are to be

written together in continuity.

1

5. iz’u dk mŸkj fy[kus ls iwoZ iz’u dk Øekad vo’; fy[ksaA
Write down the serial number of the question before attempting it.
6. iz’u i= ds fgUnh o vaxt
sz h :ikUrj esa fdlh izdkj dh =qfV@varj@fojks/kkHkkl gksus
ij fgUnh Hkk"kk ds iz’u dks gh lgh ekusAa
If there is any error/difference/contradiction in Hindi & English versions of the

question paper, the question of Hindi version should be treated valid.
7. iz’u Øekad 14 ls 20 esa vkUrfjd fodYi gSA

ER
There are internal choices in QuestionNo. 14 to 20.
BS

2

[k.M&v
SECTION- A

1- cgqfodYih; iz’u ¼i ls xviii½
Multiple Choice Question (i to xviii)
(i) ;fn A  {1, 2,3} gks rks vo;o ¼1] 2½ okys rqY;rk laca/kksa dh la[;k gS&
¼v½ 1 ¼c½ 2
¼l½ 3 ¼n½ 4 ¼1½
If A  {1, 2,3} then the number of equivalence relations with element

(ii)
(1, 2) is-
(a) 1
(c) 3
;fn sin 1 x  y rks
ER
¼v½ 0  y  
(b) 2
(d) 4


¼c½   y 
2


2

¼l½ 0  y   ¼n½   y  ¼1½
2 2
BS
If sin 1 x  y then
 
(a) 0  y   (b)   y
2 2
 
(c) 0  y   (d)   y
2 2
0  1 3 5
(iii) ;fn A  0 2 rFkk B
0 
rks dk eku gksxk&
 0
0 0 1 0
¼v½ 0 0 
¼c½ 0 1
 
0 0 0 1
¼l½ 1 1 
¼n½ 1 0 
¼1½
 

3

0  1 3 5
If A    and B   then value of AB is-
0 2 0 0 

0 0 1 0
(a)  (b) 
0 0  0 1

0 0 0 1
(c)  (d) 
1 1  1 0 

x 2 6 2
(iv) ;fn 18 x  18 6 gks rks x cjkcj gS”

If
x 2 6 2

¼v½
¼l½

18 x 18 6
(a)
(c)
6
&6

ER
then the value of x is -

2
6
-6
3
¼c½  6
¼n½

(b)  6
(d) 0
0 ¼1½

(v) vkO;wg A  1 4 
dk lg[k.Mt Kkr dhft,&

BS
 4  3  2  3 
¼v½ 1 2 
¼c½ 1  4 
  
 4 3 4 1
¼l½  1  2  ¼n½ 3 2 
¼1½
  
2 3
Final adjoint of matrix A  
1 4 

 4  3  2  3 
(a)  (b) 
 1 2  
 1  4 
 4 3 4 1
(c)   (d) 
 1  2 3 2 

4

(vi) x ds lkis{k sin 2 x dk vodyt gS&
¼v½ cos 2 x ¼c½  cos 2 x
¼l½ sin 2x ¼n½ cos 2x ¼1½
The derirative of sin 2 x with respect to x is-
(a) cos 2 x (b)  cos 2 x
(c) sin 2x (d) cos 2x
1
(vii) [ x( x  1)  1] 3 , 0  x  1 dk mPpre eku gS&
1
 1 3 1
¼v½  
 3
¼l½ 1

 1 3
(a)  
 3
1
ER ¼c½

¼n½ 0
1
2

Maximum value of [ x( x  1)  1] 3 , 0  x  1 is -

(b)
1
2
¼1½

(c) 1 (d) 0
dy
BS
(viii) ;fn x  y   rks dk eku gS &
dx
¼v½ 1 ¼c½ -1
¼l½ 2 ¼n½ -2 ¼1½
dy
If x  y   then value of is -
dx
(a) 1 (b) -1
(c) 2 (d) -2
(ix)  cos 2x dx dk eku gS &
sin2x sin 2 x
¼v½  C ¼c½ C
2 2
¼l½  sin 2x  C ¼n½ sin 2x  C ¼1½

5

Value of  cos 2x dx is -
sin2x sin 2 x
(a)  C (b) C
2 2
(c)  sin 2x  C (d) sin 2x  C
(x) izFke prqikZ’k esa o`Ÿk x 2  y 2  4 ,oa js[kkvksa x = 0, x = 2 ls f?kjs {ks= dk
{ks=Qy gS &

¼v½  ¼c½
2
 
¼l½

(a) 

(c)

3

3
ER ¼n½

(b)

(d)
4
Area surrounded by circle x 2  y 2  4 and lines x = 0, x = 2 in first
quadrant is -

2

4
¼1½

dy
vody lehdj.k  y  cos x dk lekdy xq.kd gksxk &
BS
(xi)
dx
¼v½ sin x ¼c½ cos x
¼l½ e x ¼n½ e  x ¼1½
dy
Integrating factor of differential question  y  cos x is -
dx
(a) sin x (b) cos x
(c) e x (d) e  x
        
(xii) i .( j  k )  j .(i  k )  k .(i  j ) dk eku gS &
¼v½ 1 ¼c½ &1
¼l½ 3 ¼n½ 0 ¼1½
        
Value of i .( j  k )  j .(i  k )  k .(i  j ) is -
(a) 1 (b) -1
(c) 3 (d) 0

6


(xiii) ;fn a ,d ek=d lfn’k gS vkSj ( x  a ).( x  a )  8 rks x gS &
¼v½ 2 2 ¼c½ &9
¼l½ 3 ¼n½ 9 ¼1½
    
If a is a unit vector and ( x  a ).( x  a )  8 then x is -
(a) 2 2 (b) -9
(c) 3 (d) 9
        
(xiv) ;fn a  i  7 j  7 k rFkk b  3i  2 j  2k rks a  b dk eku gS &
¼v½ 19 2 ¼c½ 2 19

(xv)
 
¼l½ 19


(c) 19
ER
  
¼n½ buesa ls dksbZ ugha
   
If a  i  7 j  7 k and b  3i  2 j  2k then value of a  b is -

(a) 19 2 (b) 2 19
(d) None of these
nks js[kkvksa ds ijLij yEcor gksus dk izfrca/k gS &
¼v½ a1a2  b1b2  c1c2  0 ¼c½ a1a2  b1b2  c1c2  1
¼1½

¼l½ a1a2  b1b2  c1c2  1 ¼n½ a1a2 b1b2 c1c2  0 ¼1½
BS
The condition that two lines are perpendicular to each other is -
(a) a1a2  b1b2  c1c2  0 (b) a1a2  b1b2  c1c2  1
(c) a1a2  b1b2  c1c2  1 (d) a1a2 b1b2 c1c2  0
3 1
(xvi) ;fn P  A   , P  B   vkSj A o B Lora= ?kVuk,a gS rks P  A  B  gS &
5 5
3 3
¼v½ ¼c½
25 5
3 3
¼l½ ¼n½ ¼1½
35 15
3 1
If P  A   , P  B   and A, B are independent events then P  A  B  is -
5 5
3 3
(a) (b)
25 5

7

3 3
(c) (d)
35 15
(xvii) rhu iklksa dks ,d lkFk mNkyus ij dqy ifj.kkeksa dh la[;k gS &
¼v½ 6 ¼c½ 36
¼l½ 216 ¼n½ 12 ¼1½
The total number of outcomes when three dices are thrown simultaneously is -
(a) 6 (b) 36
(c) 216 (d) 12
(xviii) nks ?kVuk,a A o B ijLij Lora= gkax
s h ;fn &

ER
¼v½ P ( A) P( B )  P( A  B) ¼c½ P ( A) P( B )  P( A  B)
¼l½ P ( A)  P ( B )  P ( A  B ) ¼n½ P ( A)  P ( B)  1
Two events A and B will be mutually independent if -
(a) P ( A) P ( B )  P ( A  B ) (b) P ( A) P ( B )  P ( A  B )
(c) P ( A)  P ( B )  P ( A  B ) (d) P  A   P  B  1
¼1½

2- fjDr LFkkuksa dh iwfrZ dhft, ¼i ls vi½ &
Fill in the blanks ¼i to vi½ -
BS
(i) ;fn f ( x)  4 x  3 rks f (1)  ...................... gSA ¼1½
If f ( x)  4 x  3 then f ( 1)  .........................
x 2 6 2
(ii) ;fn 18 x  18 6 rks x dk eku ...................... gSA ¼1½

x 2 6 2
If  then value of x is ......................
18 x 18 6
(iii) ,d o`Ÿk dh f=T;k r  6 cm ij {ks=Qy esa r ds lkis{k ifjorZu dh
nj ------------------ gSA ¼1½
The rate of change of the area of a circle with respect to its radius at
r  6 cm ...................
1
tan 1 x
(iv) 0 1  x 2 dx dk eku ----------------------------- gSA ¼1½

8

1
tan 1 x
Value of  1  x 2 dx is ...................
0

d2y  dy 
(v) vody lehdj.k 2  sin    0 dh ?kkr --------------------------- gSA ¼1½
dx  dx 
d2y  dy 
Degree of differential equation 2  sin    0 is ....................
dx  dx 

(vi) lekUrj prqHkqZt dh Hkqtk,a a  3iˆ  ˆj  4kˆ vkSj b  iˆ  ˆj  kˆ gks rks {ks=Qy
------------------------ gSA ¼1½
 

3-

(i)
  
ER
If sides of parallelogram are a  3iˆ  ˆj  4kˆ and b  iˆ  ˆj  kˆ then the area
is ...............

vfr y?kwŸkjkRed iz’u ¼i ls vi½ &
Very Short Answer Type Questions ( i to vi) -
1 3 

 1  3
;fn A   2 4  , B   2  4  rks BA dk eku Kkr dhft,A ¼1½

1 3   1  3
BS
If A    , B  then find the value of BA.
2 4  2  4 
2 3
(ii) eSfVªDl A  1 4
dk lg[k.Mt vkO;wg (adj A) Kkr dhft,A ¼1½

2 3
If matrix A   then find adjoint matrix (adj A).
1 4
(iii) fl} dhft, fd Qyu f ( x)  cos x ] (0,  ) esa gzkleku gSA ¼1½
Prove that function f ( x )  cos x is decreasing in (0,  ) .

(iv) cos x  dk x ds lkis{k vodyu dhft,A ¼1½

Find derivative with respect to x for cos  x  .

dy
(v) ;fn y  2 sin x rks Kkr dhft,A ¼1½
dx

9

dy
If y  2 sin x then find .
dx
(vi) ,d mRikn dh x bdkbZ;ksa ds foØ; ls izkIr dqy vk; :i;ksa eas
R  x   3x 2  36 x  5 ls iznŸk gS] tc x = 15 gS rks lhekar vk; Kkr
dhft,A ¼1½
The total revenue in rupees obtained from the sale of x units of a product is
given by R  x   3x 2  36 x  5 . Find the marginal cost when x = 15.

(vii)  x  x  2  dx dk eku Kkr dhft,A ¼1½

(viii)

(ix)

(x)
Find value of  x

ER
 x  2 dx .
ijoy; y 2  4ax dh fu;rk dk lehdj.k fyf[k,A
Find equation of directrix for parabola y 2  4ax .
ijoy; y 2  4ax dh vody lehdj.k Kkr djksA
Find the differential equation of parabola y 2  4ax .
lfn’k a  ˆi  ˆj  kˆ dk ifjek.k Kkr dhft,A
¼1½

¼1½

¼1½

Compute the magnitude of vector a  ˆi  ˆj  kˆ .
BS
(xi) lfn’k 7iˆ  ˆj  8kˆ dk lfn’k ˆi  3jˆ  7kˆ ij iz{ksi Kkr djksA ¼1½
Find the projection of vector ˆi  3jˆ  7kˆ on the vector 7iˆ  ˆj  8kˆ .

1  A
(xii) ;fn P  A   , P( B)  0 rc P  B  dk eku Kkr djksA ¼1½
2  
1  A
If P  A   , P ( B )  0 then find value of P   .
2 B

[k.M & c
Section - B

4- fl) dhft, fd f  x   3x  4 }kjk iznŸk Qyu f :R  R ,dSdh gSA ¼2½
Show that the function f :R  R given by f  x   3x  4 is injective.

10

cos x 3 
5- tan 1 ,  x  dks ljyre :i esa O;Dr dhft,A ¼2½
1  cos x 2 2
1 cos x 3 
Write the function tan ,  x  in simplest form .
1  cos x 2 2

8 0  2 2
   
6- ;fn A 4 2 , B  4 2 rFkk 2 A  3x  5B gks] rks x dk eku Kkr
3 6  5 1

dhft,A ¼2½

7-
If

of x .
8 0

3 6 

1 0 1
 
ER
 2  2
A   4  2  , B   4 2 
 
 5 1 
and 2 A  3x  5B then find the value

;fn A  0 1 2 gks rks fn[kkb, fd 3 A  27 A
0 0 4
¼2½

1 0 1
BS
A  0 1 2
If , then prove that 3 A  27 A .
0 0 4

1  1  dy
8- ;fn y  sec  2 x 2  1  rks Kkr dhft,A ¼2½
  dx

1  1  dy
If y  sec  2  then Find .
 2x 1  dx
9- log  cos e x  dk x ds lkis{k vodyt Kkr dhft,A ¼2½
Find derivative of log  cos e x  with respect to x .

10- vUrjky Kkr dhft,] ftlesa f  x   4 x3  6 x 2  72 x  3 }kjk iznŸk Qyu
f o/kZeku gSA ¼2½
Find the intervals in which function f  x   4 x3  6 x 2  72 x  3 is mcreasing .

11

d 3
11- ;fn f  x   4 x 3  4 tgka f  2   0 gS rks f  x  Kkr dhft,A ¼2½
dx x
d 3
If f  x   4 x3  4 where f  2   0 then find f  x  .
dx x
12- nks ijoy;ksa y  x 2 ,oa y 2  x ls f?kjs {ks= dk {ks=Qy Kkr dhft,A ¼2½
Find the area of the region bounded by two parabolas y  x 2 and y 2  x .
  
13- ;fn a  b  5iˆ  ˆj  3kˆ vkSj b  ˆi  3ˆj  5kˆ gks rks n’kkZb, fd lfn’k a  b

vkSj a  b yEcor gSA ¼2½
   

ER
If a  5iˆ  ˆj  3kˆ and b  ˆi  3jˆ  5kˆ then prove that vectors a  b
 
and a  b are perpendicular .

dx
[k.M&l
SECTION- C

14- Kkr dhft,  5x  2 x
2 ¼3½
BS
dx
Evaluate  2
5x  2 x
vFkok@OR
5x  2
Kkr dhft,  1  2 x  3x dx
2 ¼3½
5x  2
Evaluate  dx
1  2 x  3x 2
15- vody lehdj.k  tan 1 y  x  dy  1  y 2  dx dk gy Kkr dhft,A ¼3½
Solve the differential equation  tan y  x  dy  1  y  dx
1 2

vFkok@OR
dy 1
vody lehdj.k 1  x 2   2 xy  dk fof’k"V gy Kkr dhft,
dx 1  x2
;fn y = 2, x = 1 ¼3½

12

dy 1
Find specific solution of differential equation 1  x 
2
 2 xy 
dx 1  x2
at y = 2, x = 1.
16- n’kkZb, fd fcUnqvksa (4, 7, 8) (2, 3, 4) ls gksdj tkus okyh js[kk fcUnqvksa
(-1, -2, 1), (1, 2, 5)ls tkus okyh js[kk ds lekarj gSA ¼3½
Show that the line passing through the points (4, 7, 8) (2, 3, 4) is parallel to
the passing through the points (-1, -2, 1), (1, 2, 5)
vFkok@OR
1  x 7 y  14 z  3
p dk eku Kkr dhft, rkfd js[kk,a   vkSj

7  7x y  5 6  z
3p

1

5
ER
ijLij yac gSA

Find value of p when given lines

7  7x y  5 6  z
3p

1

5
3

are perpendicular..
3


3p
3p

1  x 7 y  14 z  3

2
2

and
¼3½

17- ,d ikls dks Qsda us ds ijh{k.k ij fopkj dhft,A ;fn ikls ij izdV la[;k 3
BS
dk xq.kt gS rks ikls dks iqu% Qsd
a s vkSj ;fn dksbZ vU; la[;k izdV gks rks ,d
flDds dks mNkysAa ?kVuk ^U;wure ,d ij la[;k 3 izdV gksuk* fn;k x;k gS rks
?kVuk ^flDds ij izdV gksuk* dh lizfrca/k izkf;drk Kkr dhft,A ¼3½
Consider the experiment of throwing a die, if a multiple of 3 comes up, throw
the die again and if any other number comes, toss a coin. Find the conditional
probality of the event ‘the coin shows a tail’, given that ‘at least one die shows
a 3.’
vFkok@OR
,d fo’ks"k leL;k dks A vkSj B ds Lora= :i ls gy djus dh izkf;drk,a Øe’k%
1 1
,oa gSA ;fn nksuksa Lora= :i ls leL;k gy djus dk iz;kl djrs gSa rks
2 3
izkf;drk Kkr dhft, fd leL;k gy gks tk,xhA ¼3½

13

1 1
Probality of solving specific problem independently by A and B are and
2 3
respectively. If both try to solve the problem independently. Find the probability
that the problem is solved.

[k.M & n
Section - D

x sin x
18-  1  cos x dx dk eku Kkr dhft,A
0
2 ¼4½

1


1
3
x4

x sin x
Evalute  1  cos 2 x dx
0

1

 x  x  dx
3 3
ER
dk eku Kkr dhft,A
vFkok@OR

¼4½

1
1
 x  x  dx
3 3
BS
Evalute  x4
1
3

19- nks js[kkvksa ds e/; dks.k Kkr dhft, tcfd js[kk,a
      
r  3i  2 j  4k   i  2 j  2k  
     

r  5i  2 j   3i  2 j  6k  ¼4½
Find the angle between two lines when lines are
      
r  3i  2 j  4k   i  2 j  2k  
     

r  5i  2 j   3i  2 j  6k 
vFkok@ OR
js[kkvksa l1 vkSj l2 ds chp U;wure nwjh Kkr dhft, ftuds lfn’k lehdj.k gS&
            
 
r  i  j   2i  j  k rFkk r  2i  j  k   3i  5 j  2k  ¼4½

14

Find shortest distance between lines l1 and l2. The vector equations of lines
            
  
are r  i  j   2i  j  k and r  2i  j  k   3i  5 j  2k 
20- fuEu O;ojks/kksa ds vUrxZr z  4 x  y dk vf/kdre Kkr dhft, &
x  y  50
3 x  y  20 ¼4½
x0 , y0
Maximise z  4 x  y subject to the constraints -
x  y  50

x0 , y0

ER
3 x  y  20

vFkok@ OR
fuEu vojks/kksa ds vUrxZr z  3x  5 y dk vf/kdre Kkr dhft, &
3 x  5 y  15
5 x  2 y  10
x0 , y0
¼4½

Maximise z  3 x  5 y subject to the constraints -
BS
3 x  5 y  15
5 x  2 y  10
x0 , y0

TTT

15

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