Rajasthan Board Class 12 Question Paper 2025 Maths

RAJASTHAN BOARD

QUESTION
PAPER
2025

ANNUAL EXAMINATION

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Zm_m§H$ Roll No.

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Sl.No. :

No. of Questions – 20 SS–15–Mathematics
No. of Printed Pages – 19

Cƒ _mÜ`{_H$ narjm, 2025
SENIOR SECONDARY EXAMINATION, 2025

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J{UV
MATHEMATICS

àíZ nÌ H$mo ImobZo Ho$ {bE `hm± \$m‹S>|
g_` : 3 KÊQ>o 15 {_{ZQ>
nyUmªH$ : 80

narjm{W©`m| Ho$ {bE gm_mÝ` {ZX}e …
GENERAL INSTRUCTIONS TO THE EXAMINEES :

1) narjmWu gd©àW_ AnZo àíZ nÌ na Zm_m§H$ A{Zdm`©V… {bI|&
Candidate must write first his/her Roll No. on the question paper
compulsorily.
2) g^r àíZ H$aZo A{Zdm`© h¢&
All the questions are compulsory.
3) àË`oH$ àíZ H$m CÎma Xr JB© CÎma-nwpñVH$m _| hr {bI|&
Write the answer to each question in the given answer-book only.
4) {OZ àíZmo§ _| AmÝV[aH$ IÊS> h¡§, CZ g^r Ho$ CÎma EH$ gmW hr {bI|&
`hm± go H$m{Q>E

For questions having more than one part, the answers to those parts are to
be written together in continuity.

SS–15–Mathematics 7011 [ Turn Over

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5) àíZ nÌ Ho$ {hÝXr d A§J«oOr ê$nmÝVa _o| {H$gr àH$ma H$s Ìw{Q> / AÝVa / {damoYm^mg hmoZo na {hÝXr ^mfm
Ho$ àíZ H$mo hr ghr _mZ|&
If there is any error / difference / contradiction in Hindi & English versions
of the question paper, the question of Hindi version should be treated
valid.

6) àíZ H$m CÎma {bIZo go nyd© àíZ H$m H«$_m§H$ Adí` {bI|&
Write down the serial number of the question before attempting it.

7) àíZ g§»`m 14 go 20 _| AmÝV[aH$ {dH$ën {X`o JE h¡&
Q. Nos. 14 to 20 having internal choices.

8) àíZ g§»`m 20 J«m’$ nona na hb H$aZm h¡&
Solve Question number 20 on graph paper.

SS–15–Mathematics 7011

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IÊS> - A
SECTION - A

1) ~hþ{dH$ënr` àíZ …
Multiple Choice Questions :

i) ‘mZ br{OE {H$ g‘wƒ¶ N ‘| n[a^m{fV  a, b  : a  b  2, b  6 Ûmam àXÎm gå~ÝY R h¡, Vmo R
H$m n[aga hmoJm - [1]
A) 1,2,3 ~) 1, 2,3,4,5
g) 3, 4,5 X) 3,4,5,6
Let R be the relation defined on the set N and given by  a, b  : a  b  2, b  6 ,
then range of R will be -
A) 1,2,3 B) 1, 2,3,4,5
C) 3, 4,5 D) 3,4,5,6
 1
ii) cos 1    H$m ‘w»¶ ‘mZ h¡ - [1]
 2
2 
A) ~)
3 6
 
g) X) 
3 3

 1
The principal value of cos 1    is -
 2
2 
A) B)
3 6
 
C) D) 
3 3

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iii) A   aij  EH$ ñV§^ Amì¶yh hmoJm, ¶{X - [1]
mn

A) m > 1 ~) m=1
g) n>1 X) n=1

A   aij  is a column matrix, if -
mn

A) m>1 B) m=1
C) n>1 D) n=1

cos  sin 
iv) H$m ‘mZ hmoJm - [1]
sin  cos
A) 0 ~) 1
g) cos 2 X) sin 2

cos  sin 
Value of will be -
sin  cos

A) 0 B) 1
C) cos 2 D) sin 2
v) ¶{X A, 3 × 3 H$mo{Q> H$m EH$ ì¶wËH«$‘Ur¶ dJ© Amì¶yh h¡, Vmo adj A H$m ‘mZ hmoJm - [1]
2 3
A) A ~) A

g) A X) 2A

Let A be a nonsingular square matrix of order 3 × 3. Then adj A is equal to -
2 3
A) A B) A

C) A D) 2A

SS–15–Mathematics 7011

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dy
vi) ¶{X x 2  y 2  2 hmo, Vmo ~am~a hmoJm - [1]
dx

1  2x 2y
A) ~)
2y 1  2x

x y
g)  X) 
y x

dy
If x 2  y 2  2, then is equal to -
dx

1  2x 2y
A) B)
2y 1  2x

x y
C)  D) 
y x

 
vii) {ZåZ{b{IV ‘| go H$m¡Z-gm ’$bZ AÝVamb  0,  ‘| {ZaÝVa ömg‘mZ h¡? [1]
 2

A) sin x ~) cos x

g) tan x X) sin 2x

 
Which of the following function is strictly decreasing function in interval  0,  ?
 2

A) sin x B) cos x

C) tan x D) sin 2x

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 2 x  5 ; ¶{X x3
viii) f  x    , x = 3 na g§VV hmo, Vmo k H$m ‘mZ hmoJm - [1]
 2k ; ¶{X x3
1
A) 1 ~)
6
1
g) 6 X)
2
 2 x  5 ; if x3
f  x   , is continuous at x = 3, then value of k will be -
 2k ; if x3
1
A) 1 B)
6
1
C) 6 D)
2

  sin x  cos x  dx H$m ‘mZ hmoJm -
1
1 1
ix) [1]
0


A) ~) 
2
2
g)
4
X) 2

 
1
Value of  sin 1 x  cos 1 x dx will be -
0


A) B) 
2
2
C) D) 2
4

x) x - Aj, y - Aj, y  cos x, 0  x  go {Kao joÌ H$m joÌ’$b hmoJm - [1]
2
A) 1 ~) 0
g) –1 X) 2

The area bounded by x - axis, y - axis, y  cos x, 0  x  will be -
2
A) 1 B) 0
C) –1 D) 2

SS–15–Mathematics 7011

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dy
xi) AdH$b g‘rH$aU  x  0 H$m x = 0, y = 1 na {d{eîQ> hb hmoJm - [1]
dx

x2 x2
A) y  1  0 ~) y  1
2 2

g) y  2 x2  1  0 X) y  2 x2  1

dy
Particular solution of differential equation  x  0 at x = 0, y = 1 will be -
dx

x2 x2
A) y  1  0 B) y  1
2 2

C) y  2 x2  1  0 D) y  2 x2  1

   
xii) Xmo g{Xem| a Am¡a b Ho$ n[a‘mU H«$‘e… 1 Am¡a 2 h¡ VWm a  b  1 hmo, Vmo BZ g{Xem| Ho$ ~rM H$m
H$moU hmoJm - [1]

 
A) ~)
2 4

 
g) X)
3 6
   
If magnitude of two vectors a and b are 1 and 2 respectively and a  b  1 ,
then angle between these vectors will be -

 
A) B)
2 4

 
C) D)
3 6

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     
xiii) iˆ  kˆ  ˆj  ˆj  kˆ  iˆ  kˆ  iˆ  ˆj H$m ‘mZ h¡ - [1]

A) 1 ~) 0

g) –3 X) –1

    
Value of iˆ  kˆ  ˆj  ˆj  kˆ  iˆ  kˆ  iˆ  ˆj is - 
A) 1 B) 0

C) –3 D) –1


xiv) g{Xe a  iˆ  ˆj  2kˆ Ho$ {XH²$-H$mogmBZ h¡ - [1]

1 1 1 1 1 2
A) , , ~) , ,
4 4 2 6 6 6

1 1
g) , , 2 X) 1, 1, –2
2 2


Direction cosines of vector a  iˆ  ˆj  2kˆ are -

1 1 1 1 1 2
A) , , B) , ,
4 4 2 6 6 6

1 1
C) , , 2 D) 1, 1, –2
2 2

SS–15–Mathematics 7011

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x3 y 4 z 8
xv) {~ÝXþ (1, 2, 3) go OmZo dmbr VWm aoIm   Ho$ g‘mÝVa aoIm H$m H$mVu¶
3 5 6
g‘rH$aU hmoJm - [1]

x 1 y  2 z  3 x 1 y  2 z  3
A)   ~)  
3 5 6 3 5 6

x3 y 4 z 8 x2 y6 z5
g)   X)  
1 2 3 3 5 6
The cartesian equation of the line passing through point (1, 2, 3) and parallel
x3 y 4 z 8
to the line   will be -
3 5 6

x 1 y  2 z  3 x 1 y  2 z  3
A)   B)  
3 5 6 3 5 6

x3 y 4 z 8 x2 y6 z5
C)   D)  
1 2 3 3 5 6
xvi) EH$ {g³Ho$ H$mo VrZ ~ma CN>mbm OmE|, Vmo ݶyZV‘ Xmo ~ma {MV AmZo H$s àm{¶H$Vm hmoJr - [1]
1 3
A) ~)
8 8

1 5
g) X)
2 8
A coin is tossed three times, then the probability to get Head at least two
times will be -
1 3
A) B)
8 8

1 5
C) D)
2 8

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1 1
xvii) EH$ {deof g‘ñ¶m H$mo N>mÌm| A Am¡a B Ûmam ñdV§Ì ê$n go hb H$aZo H$s àm{¶H$VmE± H«$‘e… Am¡a
2 3
h¡& ¶{X XmoZm|, ñdV§Ì ê$n go g‘ñ¶m hb H$a|, Vmo g‘ñ¶m Ho$ hb hmoZo H$s àm{¶H$Vm hmoJr - [1]
1 5
A) ~)
6 6
1 2
g) X)
3 3
Probabilities of solving specific problem independently by students A and B
1 1
are and respectively. If both try to solve the problem independently,
2 3
then the probability that the problem will be solved -
1 5
A) B)
6 6
1 2
C) D)
3 3
4 3
xviii) ¶{X P  A   , P  B   Am¡a A VWm B ñdV§Ì KQ>ZmE± h¡, Vmo P  A  B  H$m ‘mZ hmoJm - [1]
7 7
1 12
A) ~)
7 49
37
g) X) 1
49
4 3
If P  A   , P  B   and A and B are independent events, then value of
7 7
P  A  B  will be -
1 12
A) B)
7 49
37
C) D) 1
49

SS–15–Mathematics 7011

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2) [aº$ ñWmZm| H$s ny{V© H$s{OE : (i go vi)
Fill in the blanks : (i to vi)

 3
i) sin 1   H$m ‘w»¶ ‘mZ ............. h¡& [1]
 2 

 3
________ is the principal value of sin 1  .
 2 

2 1 1 0
ii) ¶{X 1 2  k 0 1  0 hmo, Vmo k = ............. [1]

2 1 1 0
If k  0 , then k = ________.
1 2 0 1

iii) EH$ CËnmX H$s x BH$mB¶m| Ho$ {dH«$¶ go àmßV Hw$b Am¶ ê$n¶m| ‘| R  x   2 x 2  25 x go àXÎm h¡&
O~ x = 10 h¡ Vmo gr‘mÝV Am¶ = ............. [1]
The total revenue in Rupees received from the sale of x units of a product is
given by R  x   2 x 2  25 x . Then marginal revenue = ________ when x = 10.

 2 9
iv)  

x  3 x   dx  ....................
4
[1]

 2 9
  x  3x  4  dx  __________.
dy
v) AdH$b g‘rH$aU  y  x H$m g‘mH$b JwUm§H$ .............. h¡& [1]
dx
dy
The integrating factor of differential equation  y  x is ________.
dx
vi) g{Xe iˆ  2 ˆj H$m x - Aj na àjon .............. h¡& [1]
The projection of the vector iˆ  2 ˆj on x - axis is ________.

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3) A{V bKwÎmamË_H$ àíZ :
Very short answer type questions :

 1
i) ¶{X A    VWm B   1 1 hmo, Vmo BA kmV H$s{OE& [1]
2
 

 1
If A    and B   1 1 , then find BA.
2

3 2 3
ii)   2 2 3 H$m ‘mZ kmV H$s{OE& [1]
3 2 3

3 2 3
Find the value of   2 2 3 .
3 2 3

iii) AnZr CÎma nwpñVH$m ‘| ’$bZ f  x   x H$m AmboIr¶ {Zê$nU H$s{OE& [1]
Draw a graphical representation of the function f  x   x in your answer book.

1 dy
iv) ¶{X y  e tan x hmo, Vmo kmV H$s{OE& [1]
dx
1 dy
If y  e tan x , then find .
dx
1
v) H$m x Ho$ gmnoj AdH$bZ H$s{OE& [1]
2x  3
1
Differentiate with respect to x.
2x  3
vi) d¥Îm Ho$ joÌ’$b n[adV©Z H$s Xa BgH$s {ÌÁ¶m r Ho$ gmnoj kmV H$s{OE O~{H$ r = 2.5 go‘r h¡& [1]
Find the rate of change of the area of a circle with respect to its radius r when
r = 2.5 cm.

SS–15–Mathematics 7011

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vii) hb H$s{OE  x sin x dx . [1]

Evaluate  x sin x dx .

viii) àW‘ MVwWmªe ‘| d¥Îm x 2  y 2  4 go {Kao joÌ H$m joÌ’$b kmV H$s{OE& [1]

Find the area lying in the first quadrant and bounded by the circle x 2  y 2  4 .

2
d2y  dy 
ix) AdH$b g‘rH$aU xy 2  x    y  0 H$s H$mo{Q> Ed§ KmV kmV H$s{OE& [1]
dx  dx 

2
d2y  dy 
Find the order and degree of the differential equation xy 2  x    y  0 .
dx  dx 
  
x) VrZ g{Xem| a , b Am¡a c Ho$ ¶moJ’$b Ho$ {bE gmhM¶© JwUY‘© {b{IE& [1]
  
Write Associative property for addition of any three vectors a , b and c .

xi) {~ÝXþAm|  3, 2,0  Am¡a 1, 2,5  go hmoH$a OmZo dmbr aoIm H$m g{Xe g‘rH$aU kmV H$s{OE& [1]

Find the vector equation of the line passing through the points  3, 2,0  and
1, 2,5  .

E
xii) ¶{X E Am¡a F Bg àH$ma H$s KQ>ZmE± h¡ {H$ P  F  = 0.3 Am¡a P  E  F  = 0.2 , Vmo P   kmV
F
H$s{OE& [1]

Given that E and F are events such that P  F  = 0.3 and P  E  F  = 0.2 , find

E
P  .
F

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IÊS> - ~
SECTION - B
bKwÎmamË‘H$ àíZ :
Short answer type questions :
4) ¶{X VrZ ’$bZ f, g VWm h g‘wƒ¶ N ‘| n[a^m{fV h¡, Ohm± f  x   2 x, g  y   3 y  4 VWm
h  z   sin z  x, y VWm z  N , {gÕ H$s{OE h   g  f    h  g   f . [2]

If three functions f, g and h are defined in set N, where f  x   2 x, g  y   3 y  4
and h  z   sin z  x, y and z  N , prove that h   g  f    h  g   f .

1 1 1 
5) {gÕ H$s{OE tan  tan 1  . [2]
2 3 4

1 1 1 
Prove that tan  tan 1  .
2 3 4

1
6) EH$ Eogo 2 × 2 Amì¶yh  aij  H$s aMZm H$s{OE, {OgHo$ Ad¶d aij  i  2 j Ûmam àXÎm hmo& [2]
2
1
Construct a 2 × 2 matrix  aij  , whose elements are given by aij  i2j .
2

7) EH$ {Ì^wO H$m joÌ’$b kmV H$s{OE, {OgHo$ erf©  3,0  ,  4,2  Am¡a  5,1 h¡& [2]

Find the area of the triangle, whose vertices are  3,0  ,  4,2  and  5,1 .

8) f  x   1  x  1  x  Ûmam àXÎm ’$bZ H$m AdH$bZ kmV H$s{OE Am¡a Bg àH$ma f  1 kmV
H$s{OE& [2]
Find the derivative of the function given by f  x   1  x  1  x  and hence find
f  1 .

SS–15–Mathematics 7011

15
d2y
9) ¶{X y  A sin x  Bcos x hmo, Vmo {gÕ H$s{OE {H$ 2  y  0 . [2]
dx

d2y
If y  A sin x  Bcos x , then prove that  y  0.
dx 2

10) Eogr Xmo YZ g§»¶mE± kmV H$s{OE {OZH$m ¶moJ 15 h¡ Am¡a {OZHo$ dJm] H$m ¶moJ ݶyZV‘ hmo& [2]
Find two positive numbers whose sum is 15 and the sum of whose squares is
minimum.

 3  2x  x dx kmV H$s{OE&
2
11) [2]

Find  3  2x  x 2 dx .

12) g‘mH$bZ Ho$ AZwà¶moJ Ûmam d¥Îm x 2  y 2  16 H$m joÌ’$b kmV H$s{OE& [2]

Find the area of circle x 2  y 2  16 using applications of integrals.


13) Xmo g{Xem| a Am¡a b Ho$ n[a‘mU kmV H$s{OE, ¶{X BZHo$ n[a‘mU g‘mZ h¡ Am¡a BZHo$ ~rM H$m H$moU 60° d
1
BZHo$ A{Xe JwUZ’$b H$m ‘mZ h¡& [2]
2

Find the magnitude of two vectors a and b , having the same magnitude and such
1
that the angle between them is 60° and value of their scalar product is .
2

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IÊS> - g
SECTION - C
XrK© CÎmamË‘H$ àíZ :
Long answer type questions :

2x  9
14) kmV H$s{OE  dx . [3]
x 2  8 x  25

2x  9
Find  dx .
x  8 x  25
2

AWdm/OR
sin x
kmV H$s{OE  sin  x  a  dx .

sin x
Find  sin x  a dx .
 

15) AdH$b g‘rH$aU y dx  x dy  2 y dy  0 H$m ì¶mnH$ hb kmV H$s{OE&
2
[3]

Find the general solution of the differential equation y dx  x dy  2 y 2 dy  0 .
AWdm/OR

d2y dy
{gÕ H$s{OE y  e 2 x
AdH$b g‘rH$aU 2  4  4 y  0 H$m EH$ hb h¡&
dx dx

Prove that y  e 2 x is a solution of differential equation :

d2y dy
2
 4  4y  0
dx dx

SS–15–Mathematics 7011

17
16) EH$ {Ì^wO H$s ^wOmAm| Ho$ {XH²$-H$mogmBZ kmV H$s{OE ¶{X {Ì^wO Ho$ erf© {~ÝXþ A(3, 5, –4), B(–1, 1, 2)
Am¡a C(–5, –5, –2) h¢& [3]

Find the direction cosines of the sides of the triangle whose vertices are A(3, 5, –4),
B(–1, 1, 2) and C(–5, –5, –2).

AWdm/OR

Xem©BE {H$ {~ÝXþAm| (1, 2, 3), (3, 4, 5) go hmoH$a OmZo dmbr aoIm, {~ÝXþAm| (–1, 2, 4), (2, –1, 4) go
OmZo dmbr aoIm na bå~ h¡&

Show that the line through the points (1, 2, 3), (3, 4, 5) is perpendicular to the line
through the points (–1, 2, 4), (2, –1, 4).

17) ¶{X A Am¡a B ñdV§Ì KQ>ZmE± h¡ Vmo {gÕ H$s{OE {H$ A Am¡a B ‘| go H$‘ go H$‘ EH$ Ho$ hmoZo H$s
àm{¶H$Vm  1  P  A  P  B  hmoJr& [3]

If A and B are two independent events, then prove that the probability of occurrence
of at least one of A and B is given by  1  P  A  P  B  .

AWdm/OR

52 Vmem| H$s JS²>S>r go EH$ nÎmm Imo OmVm h¡& eof nÎmm| go Xmo nÎm| {ZH$mbo OmVo h¡ Omo BªQ> Ho$ h¡& Imo J¶o nÎmo Ho$
BªQ> H$m hmoZo H$s àm{¶H$Vm kmV H$s{OE&

A card from a pack of 52 cards is lost. From remaining cards of the pack, two
cards are drawn and they are found to be both diamonds. Find the probability of
the lost card being a diamond.

SS–15–Mathematics 7011 [ Turn Over

18
IÊS> - X
SECTION - D
{Z~§YmË‘H$ àíZ :
Essay type questions :

3 dx
18) 1 x 2 x  1 H$m ‘mZ kmV H$s{OE& [4]
 
dx
3
Find the value of 1 x 2 x  1 .
 
AWdm/OR
1
  x  a  x  b  dx H$m hb H$s{OE&
1
Solve  dx .
 x  a  x  b 

x  8 y  19 z  10
19) {~ÝXþ (1, 2, –4) go OmZo dmbr Am¡ a Xmo Z m| ao I mAm|   Am¡ a
3 16 7
x  15 y  29 z  5
  Ho$ bå~ aoIm H$m g{Xe g‘rH$aU kmV H$s{OE& [4]
3 8 5
Find the vector equation of the line passing through the point (1, 2, –4) and perpendicular
x  8 y  19 z  10 x  15 y  29 z  5
to the both lines :   and   .
3 16 7 3 8 5
AWdm/OR
{ZåZ{b{IV aoIm ¶w½‘ Ho$ ~rM H$s ݶyZV‘ Xÿar kmV H$s{OE :
x y z x 5 y 2 z 3
  Am¡a   .
2 2 1 4 1 8
Find shortest distance between the following pair of lines :
x y z x5 y 2 z 3
  and   .
2 2 1 4 1 8

SS–15–Mathematics 7011

19
20) AmboIr¶ {d{Y Ûmam CÔoí¶ ’$bZ Z  3 x  2 y H$m A{YH$V‘ ‘mZ {ZåZ{b{IV ì¶damoYm| Ho$ A§VJ©V kmV
H$s{OE : [4]

x  y  2  0, x  2 y  7

VWm x  0, y  0

Determine graphically the maximum value of the objective function Z  3 x  2 y
subject to the following constraints :

x  y  2  0, x  2 y  7

and x  0, y0

AWdm/OR

AmboIr¶ {d{Y Ûmam CÔoí¶ ’$bZ Z  x  3 y H$m ݶyZV‘ ‘mZ {ZåZ{b{IV ì¶damoYm| Ho$ AÝVJ©V kmV
H$s{OE :

x  y, x  y  4, x  2 y  8

VWm x  0, y  0

Determine graphically the minimum value of the objective function Z  x  3 y
subject to the following constraints :

x  y , x  y  4, x  2 y  8

and x  0, y0



SS–15–Mathematics 7011

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Study Materials
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