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RAJASTHAN BOARD
QUESTION
PAPER
2024
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RBSE
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Zm_m§H$ Roll No.
Tear Here
Sl.No. :
No. of Questions – 22 SS–15–Mathematics
No. of Printed Pages – 15
Cƒ _mÜ`{_H$ narjm, 2024
SENIOR SECONDARY EXAMINATION, 2024
TEAR HERE TO OPEN THE QUESTION PAPER
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$m{Q>E
For questions having more than one part, the answers to those parts are to
be written together in continuity.
SS–15–Mathematics 5011 [ Turn Over
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5) àíZ nÌ Ho$ {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 16 go 22 _| AmÝV[aH$ {dH$ën {X`o JE h¡&
Q. Nos. 16 to 22 having internal choices.
8) àíZ g§»`m 22 J«m’$ nona na hb H$aZm h¡&
Solve Question number 22 on graph paper.
SS–15–Mathematics 5011
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IÊS> - A
SECTION - A
1) ~hþ{dH$ënr` àíZ …
Multiple Choice Questions :
i) ‘mZ br{OE {H$ g‘wƒ¶ {1, 2, 3, 4} ‘| R = {(1,2), (2,2), (1,1), (4,4), (1,3), (3,3),
(3,2)} Ûmam n[a^m{fV g§~§Y R h¡& {XE JE {dH$ënm| ‘| go ghr CÎma Mw{ZE& [1]
A) R ñdVwë¶ VWm g‘{‘V h¡ {H$ÝVw g§H«$m‘H$ Zht h¡&
~) R ñdVwë¶ VWm g§H«$m‘H$ h¡ {H$ÝVw g‘{‘V Zht h¡&
g) R g‘{‘V VWm g§H«$m‘H$ h¡ {H$ÝVw ñdVwë¶ Zht h¡&
X) R EH$ Vwë¶Vm g§~§Y h¡&
Let R be the relation in the set {1, 2, 3, 4} given by R = {(1,2), (2,2), (1,1),
(4,4), (1,3), (3,3), (3,2)} choose the correct answer in the given options.
A) R is reflexive and symmetric but not transitive.
B) R is reflexive and transitive but not symmetric.
C) R is symmetric and transitive but not reflexive.
D) R is an equivalence relation.
ii) cosec–1(2) H$m ‘w»¶ ‘mZ h¡ :- [1]
π π
A) ~)
2 3
π
g) X) π
6
The principal value of cosec–1(2) is :-
π π
A) B)
2 3
π
C) D) π
6
SS–15–Mathematics 5011 [ Turn Over
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1 2 3 3 −1 3
iii) ¶{X A = VWm B = −1 0 2 h¢, Vmo (2A–B) hmoJm : [1]
2 3 1
1 −5 2 5 6 0
A) 5 6 0 ~) 1 −5 3
−1 5 3 −1 3 5
g) 5 6 0 X) 5 6 0
1 2 3 3 −1 3
If A = and B = −1 0 2 then, (2A–B) will be :
2 3 1
1 −5 2 5 6 0
A) 5 6 0 B) 1 −5 3
−1 5 3 −1 3 5
C) 5 6 0 D) 5 6 0
2 3 x 3
iv) ¶{X 4 5 = hmo, Vmo x H$m ‘mZ h¡ : [1]
2x 5
A) 2 ~) 0
g) 1 X) –1
2 3 x 3
If = ; then the value of x is :
4 5 2x 5
A) 2 B) 0
C) 1 D) –1
SS–15–Mathematics 5011
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dy
v) ¶{X 2x + 8y = sin x, Vmo h¡ : [1]
dx
sin x − 2 cos x − 2
A) ~)
8 8
cos x + 2 cos x + 2
g) X)
2 3
dy
If 2x + 8y = sin x, then is :
dx
sin x − 2 cos x − 2
A) B)
8 8
cos x + 2 cos x + 2
C) D)
2 3
vi) {ZåZ{b{IV ‘| go {H$g A§Vamb ‘| y = x2e–x dY©‘mZ h¡? [1]
A) (1, 0) ~) (2, 0)
g) (2, –∞) X) (0, 2)
In which of the following intervals is y = x2e–x increasing?
A) (1, 0) B) (2, 0)
C) (2, –∞) D) (0, 2)
sec2 x
vii) dx H$m ‘mZ h¡ - [1]
cosec 2 x
A) sec x – x + c ~) sec x tan x + c
g) tan x + x2 + c X) tan x – x + c
sec2 x
The value of dx
cosec 2 x
A) sec x – x + c B) sec x tan x + c
C) tan x + x2 + c D) tan x – x + c
SS–15–Mathematics 5011 [ Turn Over
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viii) àW‘ MVwWm©e ‘| d¥Îm x2 + y2 = 9 go {Kao joÌ H$m joÌ’$b h¡ : [1]
3π
A) 9π ~)
4
9π
g) X) 3π
4
The area of the region bounded by the circle x2 + y2 = 9 in the first quadrant is :
3π
A) 9π B)
4
9π
C) D) 3π
4
ix) dH«$ y2 = 4x, y - Aj Ed§ aoIm y = 3 go {Kao joÌ H$m joÌ’$b h¡ : [1]
9
A) 2 ~)
4
9 9
g) X)
8 2
Area of the region bounded by the curve y2 = 4x, y - axis and the line y = 3 is :
9
A) 2 B)
4
9 9
C) D)
8 2
4
ds d 2s
x) AdH$bZ g‘rH$aU + 3s 2 = 0 H$s KmV h¡ : [1]
dt dt
A) 1 ~) 2
g) 3 X) 4
4
ds d 2s
The degree of the differential equation + 3 s = 0 is
2
dt dt
A) 1 B) 2
C) 3 D) 4
SS–15–Mathematics 5011
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xi) ¶{X eyݶoVa g{Xe a H$m n[a‘mU 'a' h¡ Am¡a λ EH$ eyݶoVa A{Xe h¡ Vmo λ a EH$ ‘mÌH$ g{Xe
h¡ ¶{X : [1]
A) λ = 1 ~) λ = –1
1
g) a = |λ| X) a=
λ
If a is a nonzero vector of magnitude 'a' and λ a nonzero scalar, then λ a is
unit vector if
A) λ = 1 B) λ = –1
1
C) a = |λ| D) a=
λ
xii) y - Aj Ho$ {XH²$-H$mogmBZ h¡ : [1]
A) 0, 0, 0 ~) 1, 0, 0
g) 0, 1, 0 X) 0, 0, 1
The direction cosine of y - axis is :
A) 0, 0, 0 B) 1, 0, 0
C) 0, 1, 0 D) 0, 0, 1
xiii) Xmo q~XþAm| (–2, 4, –5) Am¡a (1, 2, 3) H$mo {‘bmZo dmbr aoIm H$s {XH²$-H$mogmBZ h¡ : [1]
3 2 8 3 −2 8
A) , , ~) , ,
70 70 70 77 77 77
2 −3 8 8 −2 3
g) , , X) , ,
77 77 77 13 13 13
The direction cosines of the line passing through the two points (–2, 4, –5)
and (1, 2, 3) is :
3 2 8 3 −2 8
A) , , B) , ,
70 70 70 77 77 77
2 −3 8 8 −2 3
C) , , D) , ,
77 77 77 13 13 13
SS–15–Mathematics 5011 [ Turn Over
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xiv) ¶{X P(A) = 0.8, P(B) = 0.5 Am¡a P(B/A) = 0.4 hmo, Vmo P(A∩B) H$m ‘mZ h¡ - [1]
A) 0.32 ~) 0.20
g) 0.40 X) 0.64
If P(A) = 0.8, P(B) = 0.5 and P(B/A) = 0.4, then the value of P(A∩B) is :
A) 0.32 B) 0.20
C) 0.40 D) 0.64
xv) 52 nÎmm| H$s EH$ JS²>S>r ‘| go ¶mÑÀN>¶m {~Zm à{VñWm{nV {H$E JE Xmo nÎmo {ZH$mbo JE, Vmo XmoZm| nÎmm| Ho$
H$mbo a§J H$m hmoZo H$s àm{¶H$Vm h¡ : [1]
26 52
A) ~)
52 102
25 1
g) X)
51 2
Two cards are drawn at random and without replacement from a pack of 52
playing cards, then the probability that both the cards are black is :
26 52
A) B)
52 102
25 1
C) D)
51 2
2) [aº$ ñWmZm| H$s ny{V© H$s{OE :
Fill in the blanks :
i) sin–1 x EH$ Eogm ’$bZ h¡, {OgH$m àm§V ............. h¡& [1]
sin–1 x is a function whose domain is ________.
2π
ii) sin −1 sin H$m ‘mZ ............. h¡& [1]
3
2π
The value of sin −1 sin is _______.
3
SS–15–Mathematics 5011
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3
iii) cos −1 H$m ‘w»¶ ‘mZ ............. h¡& [1]
2
−1
3
The principal value of cos is _______.
2
dy
iv) ¶{X y = cos x hmo, Vmo H$m ‘mZ ............. hmoJm& [1]
dx
dy
If y = cos x , then the value of will be _______.
dx
v) EH$ d¥Îm H$s {ÌÁ¶m r = 3 go‘r na r Ho$ gmnoj joÌ’$b ‘| n[adV©Z H$s Xa ............. h¡&
[1]
The rate of change of the area of a circle with respect to its radius r at r = 3 cm
is __________.
vi) VrZ H$mo{Q> dmbo {H$gr AdH$b g‘rH$aU Ho$ {d{eï> hb ‘| CnpñWV ñdoÀN> AMam| H$s g§»¶m ..............
hmoVr h¡& [1]
The numbers of arbitrary constants present in the particular solution of a
differential equation of third order are _______.
vii) EH$ g{Xe {OgHo$ àma§{^H$ Ed§ A§{V‘ {~ÝXþ g§nmVr hmoVo h¡, .............. H$hbmVm h¡& [1]
A vector whose initial and terminal points coincide, is called _______.
3) A{VbKwÎmamË_H$ àíZ :
Very short answer type questions :
cosθ − sin θ
i) gma{UH$ H$m ‘mZ kmV H$s{OE& [1]
sin θ cosθ
cosθ − sin θ
Find the value of determinant .
sin θ cosθ
ii) gma{UH$m| H$m à¶moJ H$aHo$ (1, 2) Am¡a (3, 6) H$mo {‘bmZo dmbr aoIm H$m g‘rH$aU kmV H$s{OE&[1]
Find equation of line joining (1, 2) and (3, 6) using determinants.
SS–15–Mathematics 5011 [ Turn Over
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iii) EH$ d¥Îm H$s {ÌÁ¶m g‘mZ ê$n go 3cm/s H$s Xa go ~‹T> ahr h¡& kmV H$s{OE {H$ d¥Îm H$m joÌ’$b {H$g
Xa go ~‹T> ahm h¡ O~ {ÌÁ¶m 10 go‘r h¡& [1]
The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the
rate at which the area of the circle is increasing when the radius is 10 cm.
iv) {gÕ H$s{OE {H$ bKwJUH$s¶ ’$bZ (0, ∞) ‘| dY©‘mZ ’$bZ h¡& [1]
Prove that the logarithmic function is increasing on (0, ∞).
( 2 x − 3cos x + e ) dx H$m ‘mZ kmV H$s{OE&
x
v) [1]
Evaluate ( 2 x − 3cos x + e x ) dx .
sin x
vi) 1 + cos x dx H$m ‘mZ kmV H$s{OE& [1]
sin x
Evaluate dx .
1 + cos x
vii) g˶m{nV H$s{OE {H$ ’$bZ y = ex + 1, AdH$b g‘rH$aU y" – y' = 0 H$m hb h¡& [1]
Verify that the function y = ex + 1 is a solution of the differential equation
y" – y' = 0.
viii) Xmo {~ÝXþAm| P(2, 3, 4) Am¡a Q(4, 1, –2) H$mo {‘bmZo dmbo g{Xe H$m ‘ܶ {~ÝXþ kmV H$s{OE& [1]
Find the position vector of the mid point of the vector joining the points
P(2, 3, 4) and Q(4, 1, –2).
ix) g{Xe a = 2iˆ + 3 ˆj + 2kˆ H$m, g{Xe b = iˆ + 2 ˆj + kˆ na àjon kmV H$s{OE& [1]
Find the projection of the vector a = 2iˆ + 3 ˆj + 2kˆ on the vector b = iˆ + 2 ˆj + kˆ .
x) (3a − 5b ) ⋅ ( 2a + 7b ) H$m ‘mZ kmV H$s{OE& [1]
( )(
Evaluate the product 3a − 5b ⋅ 2a + 7b . )
SS–15–Mathematics 5011
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IÊS> - ~
SECTION - B
bKwCÎmar¶ àíZ :
Short answer type questions :
4) {gÕ H$s{OE {H$ g‘wƒ¶ {1, 2, 3} ‘| R = {(1, 2), (2, 1)} Ûmam àXÎm g§~§Y R g‘{‘V h¡ {H$ÝVw Z Vmo
ñdVwë¶ h¡ Am¡a Z g§H«$m‘H$ h¢& [2]
Prove that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is
symmetric but neither reflexive nor transitive.
cosθ sin θ sin θ − cosθ
gab H$s{OE, cosθ + θ
cosθ cosθ sin θ
5) sin . [2]
− sin θ
cosθ sin θ sin θ − cosθ
Simplify cosθ + θ
cosθ cosθ sin θ
sin .
− sin θ
5 −1 2 1 2 1 5 −1
6) Xem©BE {H$ 6 3 4 ≠ 3 4 6 7 . [2]
7
5 −1 2 1 2 1 5 −1
Show that 6 7 3 4 ≠ 3 4 6 7 .
1 2
7) Amì¶yh 3 4 H$m ghI§S>O kmV H$s{OE& [2]
1 2
Find the adjoint of matrix .
3 4
dy
8) ¶{X sin2x + cos2y = 1 hmo, Vmo kmV H$s{OE& [2]
dx
dy
If sin2x + cos2y = 1, then find .
dx
SS–15–Mathematics 5011 [ Turn Over
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9) log ( cos ⋅ e x ) H$m x Ho$ gmnoj AdH$bZ H$s{OE& [2]
Differentiate log ( cos ⋅ e x ) with respect to x.
4 dy
10) ¶{X x = 4t , y = h¡ Vmo kmV H$s{OE& [2]
t dx
dy 4
Find , if x = 4t , y = .
dx t
11) {gÕ H$s{OE {H$ R ‘| {X¶m J¶m ’$bZ f(x) = x3 – 3x2 + 3x – 100 dY©‘mZ h¡& [2]
Prove that the function given by f(x) = x3 – 3x2 + 3x – 100 is increasing in R.
sin x cos x dx H$m ‘mZ kmV H$s{OE&
3 3
12) [2]
Evaluate sin 3 x cos3 x dx .
13) d¥Îm x2 + y2 = a2 go {Kao joÌ H$m joÌ’$b kmV H$s{OE& [2]
Find the area enclosed by the circle x2 + y2 = a2.
14) EH$ g‘m§Va MVw^w©O H$m joÌ’$b kmV H$s{OE, {OgH$s g§b¾ ^wOmE± g{Xe a = iˆ − ˆj + 3kˆ Am¡a
b = 2iˆ − 7 ˆj + kˆ Ûmam {ZYm©[aV h¡& [2]
Find the area of the parallelogram whose adjacent sides are determined by the
vectors a = iˆ − ˆj + 3kˆ and b = 2iˆ − 7 ˆj + kˆ .
15) EH$ ݶm¶ nm§go H$mo CN>mbm J¶m h¡& KQ>ZmAm| E = {1, 3, 5}, F = {2, 3} Am¡a G = {2, 3, 4, 5} Ho$
{b¶o P(E/F) Am¡a P(F/E) kmV H$s{OE& [2]
A fair die has been tossed. Find P(E/F) and P(F/E) for the events E = {1, 3, 5},
F = {2, 3} and G = {2, 3, 4, 5}.
SS–15–Mathematics 5011
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IÊS> - g
SECTION - C
XrK©CÎmar¶ àíZ :
Long answer type questions :
x2
16) x + a dx H$m ‘mZ kmV H$s{OE&
6 6 [3]
x2
Evaluate dx .
x6 + a6
AWdm/OR
x
( x + 1)( x + 2 ) dx H$m ‘mZ kmV H$s{OE& [3]
x
Evaluate x + 1 x + 2 dx .
( )( )
dy
17) AdH$b g‘rH$aU x + 2 y = x 2 ( x ≠ 0 ) H$m ì¶mnH$ hb kmV H$s{OE& [3]
dx
dy
Find the general solution of the differential equation x + 2 y = x2 ( x ≠ 0) .
dx
AWdm/OR
AdH$b g‘rH$aU (e + e ) dy – (e – e–x) dx = 0 H$m ì¶mnH$ hb kmV H$s{OE&
x –x x
[3]
Find the general solution of the differential equation (ex + e–x) dy – (ex – e–x) dx = 0.
( ) (
18) {XE JE aoIm-¶w½‘ r = 3iˆ + 2 ˆj − 4kˆ + λ iˆ + 2 ˆj + 2kˆ Am¡a )
( ) (
r = 5iˆ − 2 ˆj + μ 3iˆ + 2 ˆj + 6kˆ ) Ho$ ‘ܶ H$moU kmV H$s{OE& [3]
Find the angle between the pair of lines given by r = ( 3iˆ + 2 ˆj − 4kˆ ) + λ ( iˆ + 2 ˆj + 2kˆ )
and r = ( 5iˆ − 2 ˆj ) + μ ( 3iˆ + 2 ˆj + 6kˆ ) .
AWdm/OR
SS–15–Mathematics 5011 [ Turn Over
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Xem©BE {H$ {~ÝXþAm| (1, –1, 2), (3, 4, –2) go hmoH$a OmZo dmbr aoIm {~ÝXþAm| (0, 3, 2) Am¡a (3, 5, 6) go
OmZo dmbr aoIm na b§~ h¡& [3]
Show that the line through the point (1, –1, 2), (3, 4, –2) is perpendicular to the line
through the point (0, 3, 2) and (3, 5, 6).
19) EH$ n[adma ‘| Xmo ~ƒo h¢& ¶{X ¶h kmV hmo {H$ ~ƒm| ‘| go H$‘ go H$‘ EH$ ~ƒm b‹S>H$m h¡, Vmo XmoZm| ~ƒm| Ho$
b‹S>H$m hmoZo H$s ³¶m àm{¶H$Vm h¡? [3]
A family has two children. What is the probability that both the children are boys
given that at least one of them is a boy?
AWdm/OR
EH$ nmgo H$mo EH$ ~ma CN>mbm OmVm h¡& KQ>Zm "nmgo na àmá g§»¶m 3 H$m And˶© h¡' H$mo E go Am¡a "nmgo na
àmá g§»¶m g‘ h¡' H$mo F go {Zê${nV {H$¶m OmE Vmo ~VmE± ³¶m KQ>ZmE± E Am¡a F ñdV§Ì h¡? [3]
A die is thrown. If E is the event 'the number appearing is a multiple of 3' and F be
the event 'the number appearing is even' then find whether E and F are independent?
IÊS> - X
SECTION - D
{Z~§YmË‘H$ àíZ :
Essay type questions :
20) (1 − 4x − x )dx H$m ‘mZ kmV H$s{OE&
2
[4]
Evaluate (1 − 4x − x )dx .
2
AWdm/OR
1
5x x5 + 1 dx H$m ‘mZ kmV H$s{OE&
4
[4]
−1
1
Evaluate 5 x 4 x5 + 1 dx .
−1
SS–15–Mathematics 5011
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21) aoImAm| l1 Am¡a l2 Ho$ ~rM H$s ݶyZV‘ Xÿar kmV H$s{OE {OZHo$ g{Xe g‘rH$aU h¡ : [4]
(
r = iˆ + ˆj + λ 2iˆ − ˆj + kˆ )
(
r = 2iˆ + ˆj − kˆ + μ 3iˆ − 5 ˆj + 2kˆ )
Find the shortest distance between the lines l1 and l2 whose vector equations are
(
r = iˆ + ˆj + λ 2iˆ − ˆj + kˆ )
(
r = 2iˆ + ˆj − kˆ + μ 3iˆ − 5 ˆj + 2kˆ )
AWdm/OR
{~ÝXþ, {OgH$s pñW{V g{Xe 2iˆ − ˆj + 4kˆ go JwOaZo d g{Xe iˆ + 2 ˆj − kˆ H$s {Xem ‘| OmZo dmbr aoIm H$m
g{Xe Am¡a H$mVu¶ ê$nm| ‘| g‘rH$aU kmV H$s{OE& [4]
Find the equation of the line in vector and in Cartesian form that passes through
the point with position vector 2iˆ − ˆj + 4kˆ and is in the direction iˆ + 2 ˆj − kˆ .
22) {ZåZ{b{IV ì¶damoYm| Ho$ A§VJ©V Z = 4x + y H$m AmboIr¶ {d{Y go A{YH$V‘rH$aU H$s{OE& [4]
x + y < 50, 3x + y < 90, x > 0, y > 0
Maximize Z = 4x + y subject to constraints x + y < 50, 3x + y < 90, x > 0, y > 0
by using graphical method.
AWdm/OR
{ZåZ{b{IV ì¶damoYm| Ho$ AÝVJ©V Z = 3x + 2y H$m AmboIr¶ {d{Y go A{YH$V‘rH$aU H$s{OE& [4]
x + 2y < 10, 3x + y < 15, x > 0, y > 0.
Maximize Z = 3x + 2y subject to constraints x + 2y < 10, 3x + y < 15, x > 0, y > 0
by using graphical method.
SS–15–Mathematics 5011
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