HBSE Class 11 Question Paper 2019 Maths

Code No. 1031
CLASS : 11th (Eleventh) Series : 11-M/2019
Roll No.











xf.kr
MATHEMATICS
[ fgUnh ,oa vaxzsth ek/;e ]
[ Hindi and English Medium ]
(Only for Fresh/School Candidates)

le; : 3 ?k.Vs ] [ iw.kk±d : 80
Time allowed : 3 hours ] [ Maximum Marks : 80
• Ñi;k tk¡p dj ysa fd bl iz'u-i= esa eqfnzr i`"B 16 rFkk iz'u
35 gSaA
Please make sure that the printed pages in this
question paper are 16 in number and it contains
35 questions.
• iz'u-i= esa lcls Åij fn;s x;s dksM uEcj dks Nk= mÙkj-iqfLrdk
ds eq[;-i`"B ij fy[ksaA
The Code No. on the top of the question paper
should be written by the candidate on the front
page of the answer-book.
• Ñi;k iz'u dk mÙkj fy[kuk 'kq: djus ls igys] iz'u dk Øekad
vo'; fy[ksaA
Before beginning to answer a question, its Serial
Number must be written.

1031 P. T. O.

(2) 1031
• mÙkj-iqfLrdk ds chp esa [kkyh iUuk / iUus u NksMsa+A
Don’t leave blank page/pages in your answer-book.
• mÙkj-iqfLrdk ds vfrfjDr dksbZ vU; 'khV ugha feysxhA vr%
vko';drkuqlkj gh fy[ksa vkSj fy[kk mÙkj u dkVsaA
Except answer-book, no extra sheet will be given.
Write to the point and do not strike the written
answer.
• ijh{kkFkhZ viuk jksy ua0 iz'u&i= ij vo'; fy[ksaA
Candidates must write their Roll Number on the
question paper.
• d`i;k iz'uksa dk mÙkj nsus lss iwoZ ;g lqfuf'pr dj ysa fd iz'u-i=
iw.kZ o lgh gS] ijh{kk ds mijkUr bl lEcU/k esa dksbZ Hkh nkok
Lohdkj ugha fd;k tk;sxkA
Before answering the question, ensure that you
have been supplied the correct and complete
question paper, no claim in this regard, will be
entertained after examination.
lkekU; funsZ'k %
(i) lHkh iz'u vfuok;Z gSaA
(ii) bl ç'u-i= esa 35 ç'u gSa] tks fd pkj [k.Mksa % ^v* ^v*]
^c*] ^l* ,oa ^n* esa ck¡Vs x, gSa %
[k.M ^v* % bl [k.M esa ç'u la[;k 1 ls 16 rd
dqy lksyg cgqfodYih; ç'u gSaA çR;sd ç'u
1 vad dk gSA
[k.M ^^cc* % bl [k.M esa ç'u la[;k 17 ls 26 rd
dqy nl ç'u gSAa çR;sd ç'u 2 vadksa dk
gSA
1031

(3) 1031
[k.M ^^ll* % bl [k.M esa ç'u la[;k 27 ls 31 rd
dqy ik¡p ç'u gSaA çR;sd ç'u 4 vadksa dk
gSA
[k.M ^^nn* % bl [k.M esa ç'u la[;k 32 ls 35 rd
dqy pkj ç'u gSaA çR;sd ç'u 6 vadksa dk
gSA
(iii) [k.M ^n* esa nks ç'u esa vkUrfjd fodYi fn;k x;k gSA
vkidks ,d fodYi pquuk gSA
General Instructions :
(i) All questions are compulsory.
(ii) This question paper consists of 35 questions
which are divided into four Sections : 'A', 'B',
'C' and 'D' :
Section 'A' : This Section consists of
sixteen multiple choice
questions from Question Nos. 1
to 16, each of 1 mark.
Section 'B' : This Section contains ten
questions from Question Nos.
17 to 26, each of 2 marks.
Section 'C' : This Section contains five
questions from Question Nos.
27 to 31, each of 4 marks.
Section 'D' : This Section contains four
questions from Question Nos.
32 to 35, each of 6 marks.
(iii) Section 'D' contains two questions in which
internal alternative choices are given. You
have to attempt one alternative.

1031 P. T. O.

(4) 1031
[k.M – v
SECTION – A

1. ;fn X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} ,d lkoZ
leqPp; gS vkSj A = {3, 6, 9, 12} vkSj B = {4, 6, 8, 10,
12} rks (B – A)' gS % 1
(A) {4, 8, 10}
(B) {3, 9}
(C) {1, 2, 3, 5, 6, 7, 9, 11, 12}
(D) {1, 2, 4, 5, 6, 7, 8, 10, 11, 12}
If A = {3, 6, 9, 12}, B = {4, 6, 8, 10, 12} and
X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} is universal
set, then the set (B – A)' is :
(A) {4, 8, 10}
(B) {3, 9}
(C) {1, 2, 3, 5, 6, 7, 9, 11, 12}
(D) {1, 2, 4, 5, 6, 7, 8, 10, 11, 12}

2. ;fn G = {7, 8} vkSj H = {5, 4, 2}, rks G × H ds
mileqPp;ksa dh la[;k gS % 1
(A) 6 (B) 16
(C) 32 (D) 64
If G = {7, 8} and H = {5, 4, 2}, then number of
subsets of G × H is :
(A) 6 (B) 16
(C) 32 (D) 64

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(5) 1031
3. nks o`Ùkksa esa leku yEckbZ ds nks pki dsUnz ij 65° vkSj 110°
dk dks.k cukrs gSa] mu o`Ùkksa dh f=T;kvksa dk vuqikr gS % 1

(A) 22 : 13 (B) 13 : 22

(C) 1:1 (D) buesa ls dksbZ ugha
In two circles, the arcs of same lengths subtend
angles 65° and 110° at the centre. The ratio of
their radii are :

(A) 22 : 13 (B) 13 : 22

(C) 1:1 (D) None of these

4. ;fn sin x = 7 vkSj x f}rh; prqFkkZad esa gS] rks tan x dk
25
eku gS % 1
7 24
(A) (B)
24 7
−7 − 25
(C) (D)
24 24
7
The value of sin x = , x lies in 2nd quadrant,
25
then the value of tan x is :
7 24
(A) (B)
24 7
−7 − 25
(C) (D)
24 24

1031 P. T. O.

(6) 1031
1
5. dk la;qXeh (conjugate) gS % 1
3 + 4i
3 + 4i
(A) 3 + 4i (B)
25
(C) 3 – 4i (D) buesa ls dksbZ ugha
1
The conjugate of is :
3 + 4i
3 + 4i
(A) 3 + 4i (B)
25
(C) 3 – 4i (D) None of these

6. vlfedk 3x − 4 ≥ x + 1 − 1 dk gy gS % 1
2 4
(A) x>1 (B) x≥1
(C) x<1 (D) x≤1
3x − 4 x + 1
The solution of the inequation ≥ −1
2 4
is :
(A) x>1 (B) x≥1
(C) x<1 (D) x≤1

7. ;fn nC9 = nC8 , rks n C17 dk eku gS % 1

(A) 17! (B) 17
(C) 1 (D) buesa ls dksbZ ugha
If n C9 = nC8 , then the value of n C17 is :
(A) 17! (B) 17
(C) 1 (D) None of these
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(7) 1031
8. xq.kksÙkj Js.kh 1 + 2 + 4 + ............. ds igys 5 inksa dk ;ksx
3 9
gS % 1

19 211
(A) (B)
9 81
25
(C) (D) buesa ls dksbZ ugha
3
The sum of first 5 terms of geometric series
2 4
1 + + + ............. is :
3 9
19 211
(A) (B)
9 81
25
(C) (D) None of these
3
9. fcUnq (0, 2) ls xqtjus vkSj x-axis ds lkFk 60° dk dks.k
cukus okyh js[kk dk lehdj.k gS % 1

(A) y = 3x + 2 (B) y = 3x − 2

1 1
(C) y= x +2 (D) y= x −2
3 3

The equation of the line passing through (0, 2)
and making an angle 60° with x-axis is :

(A) y = 3x + 2 (B) y = 3x − 2

1 1
(C) y= x +2 (D) y= x −2
3 3
1031 P. T. O.

(8) 1031
10. fcUnq (–1, 1) dh js[kk 12x − 5y = 9 ls nwjh gS % 1
(A) –26 (B) 8
(C) 2 (D) 0
The distance of the point (–1, 1) from the line
12x − 5y = 9 is :
(A) –26 (B) 8
(C) 2 (D) 0

x 3 − 2x 2
11. lim dk eku gS % 1
x →2 x 2 − 5 x + 6
(A) 0 (B) 4
(C) −4 (D) buesa ls dksbZ ugha
3 2
x − 2x
The value of lim is :
x →2 x 2 − 5 x + 6
(A) 0 (B) 4
(C) −4 (D) None of these

12. 3 cot x + 5 cosec x dk x ds lkis{k vodyt gS % 1
(A) 3 cosec 2 x + 5 cosec x cot x
(B) 3 cot2 x + 5 cosec 2 x
(C) − 3 cosec 2 x − 5 cosec x cot x
(D) buesa ls dksbZ ugha
The derivative of 3 cot x + 5 cosec x w.r.t. x is :
(A) 3 cosec 2 x + 5 cosec x cot x
(B) 3 cot2 x + 5 cosec 2 x
(C) − 3 cosec 2 x − 5 cosec x cot x
(D) None of these

1031

(9) 1031
1 dy
13. ;fn y = 2
, rks gS % 1
ax + bx + c dx

2ax + b 1
(A) 2 2
(B)
(ax + bx + c ) 2ax + b

− (2ax + b )
(C) 0 (D)
(ax 2 + bx + c )2

1 dy
If y = 2
, then is :
ax + bx + c dx

2ax + b 1
(A) 2 2
(B)
(ax + bx + c ) 2ax + b

− (2ax + b )
(C) 0 (D)
(ax 2 + bx + c )2

| x |
14. ;fn f (x ) =  x x ≠0
] rks lim f (x ) dk eku gS % 1
 0 x =0 x →0

(A) 0 (B) 1

(C) –1 (D) vfLrRo esa ugha
| x |
 x ≠0
If f (x ) =  x , then lim f (x ) is :
 0 x =0 x →0

(A) 0 (B) 1

(C) –1 (D) Does not exist

1031 P. T. O.

( 10 ) 1031
15. vk¡dM+ksa 14, 17, 18, 19, 20, 22, 23, 27 dk ek/; ds
lkis{k ek/; fopyu gS % 1

(A) 20 (B) 0

(C) 3 (D) 14
The mean deviation about mean of the following
data 14, 17, 18, 19, 20, 22, 23, 27 is :

(A) 20 (B) 0

(C) 3 (D) 14

16. ;fn ,d flDds dks rhu ckj mNkyk tk;s] rks 1 fpr vkSj 2
iV vkus dh izkf;drk gS % 1

1 1
(A) (B)
2 4

1 3
(C) (D)
8 8

If a coin is tossed thrice, then the probability of
getting 1 Head and 2 tails is :

1 1
(A) (B)
2 4

1 3
(C) (D)
8 8
1031

( 11 ) 1031
[k.M – c
SECTION – B

17. 500 dkj j[kus okyksa esa 400 ds ikl dkj A vkSj 200 ds
ikl dkj B gSA fdrus dkj ekfydksa ds ikl nksuksa izdkj A vkSj
B dh dkjsa gSa \ 2
Out of 500 car owners, 400 owned car A and
200 owned car B. How many car owners have
both car A and B ?

18. fl) dhft, fd % 2

π  π  π  π 
cos − x  cos − y  − sin − x  sin − y  = sin(x + y )
4  4  4  4 
Prove that :
π  π  π  π 
cos − x  cos − y  − sin − x  sin − y  = sin(x + y )
4  4  4  4 

19. lehdj.k cos x = − 1 dk O;kid gy Kkr dhft,A 2
2
Find the general solution of the equation
1
cos x = − .
2
6
 1
20.  2x 2 −  ds izlkj esa e/; in Kkr dhft,A 2
 3

Find the middle term in the expansion of
6
 2 1
 2x −  .
 3

1031 P. T. O.

( 12 ) 1031
21. A.P. 25, 22, 19, ……… ds dqN inksa dk ;ksx 116 gSA
ml A.P. esa fdrus in gSa \ 2
The sum of a certain number of terms of A.P.
25, 22, 19, ……… is 116. Find the number of
terms.
2
y2
22. vfrijoy; x − = 1 dh mRdsUnzrk Kkr dhft,A 2
16 9
Find the eccentricity of the hyperbola
x 2 y2
− = 1.
16 9

23. js[kkvksa x − 2y + 2 = 0 vkSj x + 3y + 4 = 0 ds chp dk
dks.k Kkr dhft,A 2
Find the angle between the lines x − 2y + 2 = 0
and x + 3y + 4 = 0 .

24. ;fn y = (7x + 6 tan x )x 5 , rks dy Kkr dhft,A 2
dx
dy
If y = (7x + 6 tan x )x 5 , then find .
dx

25. nks ckjackjrk caVuksa ds fopyu xq.kkad (C.V.) 30 vkSj 50 gSaA
;fn muds izeki fopyu Øe'k% 12 vkSj 15 gSa rks muds
lekarj ek/; Kkr dhft,A 2

If coefficient of variation of two distributions are
30 and 50 and their standard deviations are 12
and 15 respectively. Find their arithmetic means.

1031

( 13 ) 1031
26. ,d FkSys esa 2 lQsn vkSj 3 yky xsan gSaA 2 xsan ;kn`PN;k
fudkyh tkrh gSaA 1 lQsn vkSj 1 yky xsan vkus dh izkf;drk
Kkr dhft,A 2
A bag contains 2 white and 3 red balls. 2 balls
are selected at random. Find the probability of
getting 1 white and 1 red ball.

[k.M – l
SECTION – C

27. fl) dhft, fd % 4

sin 5x − 2 sin 3x + sin x
= tan x
cos 5x − cos x
Prove that :
sin 5x − 2 sin 3x + sin x
= tan x
cos 5x − cos x

28. xf.krh; izsj.k ds fl)kUr ls fl) dhft, % 4
n (n + 1)(n + 2)
1 × 2 + 2 × 3 + 3 × 4 + …… n(n + 1) =
3
Prove by the principle of mathematical induction :
n (n + 1)(n + 2)
1 × 2 + 2 × 3 + 3 × 4 + …… n(n + 1) =
3

29. ,d A.P. ds n inksa dk ;ksx 3n 2 + 5n gS ;fn mldk mok¡
in 164 gS] rks m dk eku Kkr dhft,A 4

If sum of n terms of A.P. is 3n 2 + 5n and its mth
term is 164. Find the value of m.
1031 P. T. O.

( 14 ) 1031
30. P (2, –3, 4) vkSj Q (8, 0, 1) dks feykus okyh js[kk ij ,d
fcUnq R ftldk x-coordinate 4 gS fdlh vuqikr esa foHkkftr
djrk gSA fcUnq R ds funsZ'kkad Kkr dhft,A og vuqikr Hkh Kkr
dhft, ftlesa R, PQ dks foHkkftr djrk gSA 4
A point R on line PQ with x-coordinate 4 divides
the line joining P (2, –3, 4) and Q (8, 0, 1). Find
the coordinate of point R. Also find the ratio in
which R divides PQ.
4x + 5 sin x
31. dk x ds lkis{k vodyt dhft,A 4
x + 7 cos x
4x + 5 sin x
Differentiate w.r.t. x.
x + 7 cos x
[k.M – n
SECTION – D
32. fl) dhft, % 6
x −y
(cos x − cos y )2 + (sin x − sin y )2 = 4 sin2  
 2 
Prove that :
x −y
(cos x − cos y )2 + (sin x − sin y )2 = 4 sin2  
 2 
vFkok
OR

lehdj.k sec2 2x = 1 − tan 2x dk eq[; gy vkSj O;kid
gy Kkr dhft,A
Find the general solution and principal solution
of the equation sec 2 2x = 1 − tan 2x .

1031

( 15 ) 1031
33. ( 3 + 2 )4 + ( 3 − 2 )4 dk eku Kkr dhft,A 6

Evaluate ( 3 + 2 )4 + ( 3 − 2 )4

vFkok
OR

n
 1 
;fn  2 4 + 11  ds izlkj esa izkjaHk ls 5osa vkSj var ls 5osa
 
 34 

inksa dk vuqikr 6 : 1 gS] rks n dk eku Kkr dhft,A

If the ratio of 5th term from the beginning and
5th term from end in the expansion of
n
 1 
 4 1 
 2 + 1  is 6 : 1 . Find the value of n.
 34 

2
y2
34. nh?kZo`Ùk x + = 1 ds ukfHk vkSj 'kh"kZ ds funsZ'kkad] mRdsUnzrk
25 4
vkSj ukfHkyac dh yEckbZ Kkr dhft,A 6

Find the coordinates of foci, vertices, eccentricity
and length of latus rectum of the ellipse

x 2 y2
+ = 1.
25 4

1031 P. T. O.

( 16 ) 1031
35. fuEufyf[kr ckjackjrk caVu dk ek/; vkSj izeki fopyu
(S.D.) Kkr dhft, % 6

oxZ-vUrjky 70-75 75-80 80-85 85-90 90-95 95-100 100-105
ckjackjrk 3 4 7 6 5 3 2

Find mean and standard deviation of the following
frequency distribution :
Class-Interval 70-75 75-80 80-85 85-90 90-95 95-100 100-105

Frequency 3 4 7 6 5 3 2

S

1031