HBSE Class 12 Question Paper 2019 Maths

CLASS : 12th (Sr. Secondary) Code No. 4331
Series : SS-M/2019
Roll No. SET : A

xf.kr GRAPH
MATHEMATICS
[ Hindi and English Medium ]
ACADEMIC/OPEN
(Only for Fresh/Re-appear Candidates)
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 20 gSaA
Please make sure that the printed pages in this
question paper are 16 in number and it contains
20 questions.
• iz'u&i= esa nkfgus gkFk dh vksj fn;s x;s dksM uEcj rFkk lsV dks
Nk= mÙkj&iqfLrdk ds eq[;&i`"B ij fy[ksaA
The Code No. and Set on the right side 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.
• mÙkj&iqfLrdk ds chp esa [kkyh iUuk@iUus u NksMsa+A
Don’t leave blank page/pages in your answer-book.

4331/(Set : A) P. T. O.

(2) 4331/(Set : A)
• 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) bl iz'u-i= esa 20 iz'u gSa] tks fd pkj [k.Mksa % v] c]
l vkSj n esa ck¡Vs x, gSa %
[k.M ^v* % bl [k.M esa ,d ç'u gS tks 16 (i-xvi) Hkkxksa
esa gS] ftuesa 6 Hkkx cgqfodYih; gSaA izR;sd
Hkkx 1 vad dk gSA
[k.M ^c* % bl [k.M esa 2 ls 11 rd dqy nl ç'u
gSaA çR;sd ç'u 2 vadksa dk gSA
[k.M ^l* % bl [k.M esa 12 ls 16 rd dqy ik¡p ç'u
gSaA çR;sd ç'u 4 vadksa dk gSA
[k.M ^n* % bl [k.M esa 17 ls 20 rd dqy pkj ç'u
gSAa çR;sd ç'u 6 vadksa dk gSA
(ii) lHkh ç'u vfuok;Z gSaA
(iii) [k.M ^n* ds dqN ç'uksa esa vkarfjd fodYi fn;s x;s gSa]
muesa ls ,d gh iz'u dks pquuk gSA
4331/(Set : A)

(3) 4331/(Set : A)
(iv) fn;s x;s xzkQ-isij dks viuh mÙkj-iqfLrdk ds lkFk vo';
uRFkh djsaA
(v) xzkQ-isij ij viuh mÙkj-iqfLrdk dk Øekad vo'; fy[ksaA
(vi) dSYD;qysVj ds ç;ksx dh vuqefr ugha gSA
General Instructions :
(i) This question paper consists of 20 questions
which are divided into four Sections : A, B,
C and D :
Section 'A' : This Section consists of one
question which is divided into
16 (i-xvi) parts of which 6 parts
of multiple choice type. Each
part carries 1 mark.
Section 'B' : This Section consists of ten
questions from 2 to 11. Each
question carries 2 marks.
Section 'C' : This Section consists of five
questions from 12 to 16. Each
question carries 4 marks.
Section 'D' : This Section consists of four
questions from 17 to 20. Each
question carries 6 marks.
(ii) All questions are compulsory.
(iii) Section 'D' contains some questions where
internal choice have been provided. Choose
one of them.
(iv) You must attach the given graph-paper along
with your answer-book.
(v) You must write your Answer-book Serial No.
on the graph-paper.
(vi) Use of Calculator is not permitted.

4331/(Set : A) P. T. O.

(4) 4331/(Set : A)
[k.M – v
SECTION – A

1. (i) ;fn f : R → R rFkk g : R → R Qyu Øe'k%
f (x) = cos x vkSj g(x ) = 3x 2 }kjk ifjHkkf"kr gSa] rks
fog Kkr dhft,A 1
Find fog, if f : R → R and g : R → R are given
by f (x) = cos x and g(x ) = 3x 2 .

 1
(ii) sin−1 −  dk eku gS % 1
 2
π π
(A) − (B) −
3 6
π
(C) − (D) buesa ls dksbZ ugha
4
 1
The value of sin−1 −  is :
 2
π π
(A) − (B) −
3 6
π
(C) − (D) None of these
4
(iii) ,d ,sls 3 × 2 vkO;wg dh jpuk dhft,] ftlds vo;o
1
aij = | i − 3 j | }kjk çnÙk gSaA 1
2
Construct a 3 × 2 matrix whose elements
1
are given by aij = | i − 3 j |.
2
4331/(Set : A)

(5) 4331/(Set : A)
3 x 3 2
(iv) = ds fy, x dk eku gS % 1
x 1 4 1

(A) ±2 3 (B) ±3 3

(C) ±2 2 (D) buesa ls dksbZ ugha
3 x 3 2
The value of x for which =
x 1 4 1
is :

(A) ±2 3 (B) ±3 3

(C) ±2 2 (D) None of these

(v) ( ) dk vodyu dhft,A
x ds lkis{k sin x 2 + 5 1

Differentiate sin(x + 5) w. r. t. x.
2

(vi) o`Ùk ds {ks=Qy ds ifjorZu dh nj bldh f=T;k r ds
lkis{k tcfd r = 3 cm, gS % 1

(A) 6π cm2 /sec (B) 4π cm2 /sec

(C) 5π cm2 /sec (D) buesa ls dksbZ ugha
The rate of change of the area of a circle
with respect to its radius r when r = 3 cm
is :

(A) 6π cm2 /sec (B) 4π cm2 /sec

(C) 5π cm2 /sec (D) None of these

4331/(Set : A) P. T. O.

(6) 4331/(Set : A)
(vii) x = 2 ij oØ y = x 3 − x dh Li'kZjs[kk dh ço.krk Kkr
dhft,A 1
Find the slope of tangent to the curve
y = x 3 − x at x = 2.

(viii) ∫
(
sin tan−1 x ) dx dk eku Kkr dhft,A 1
2
1+ x

Find the value of ∫
(
sin tan−1 x ) dx .
2
1+ x
1
(ix) ∫ sin5 x cos 4 x dx dk eku gS % 1
−1

(A) 1 (B) −1
(C) 0 (D) buesa ls dksbZ ugha
1
The value of ∫ sin5 x cos 4 x dx is :
−1

(A) 1 (B) −1
(C) 0 (D) None of these

d 2y dy
(x) 2x 2 2
−3 + y = 0, vodyu lehdj.k dh
dx dx

dksfV gS % 1

(A) 2 (B) 0

(C) 1 (D) buesa ls dksbZ ugha
4331/(Set : A)

(7) 4331/(Set : A)
The order of the differential equation
2
2d y dy
2x 2
−3 + y = 0, is :
dx dx

(A) 2 (B) 0

(C) 1 (D) None of these

2
d 3y  d 2y 
(xi) + 2  − dy + y = 0, vodyu lehdj.k
dx 3  2 dx
 dx 
dh ?kkr gS % 1

(A) 3 (B) 2

(C) 1 (D) buesa ls dksbZ ugha
The degree of the differential equation
2
d 3y  d 2y  dy
+ 2 2  − + y = 0, is :
dx 3   dx
 dx 

(A) 3 (B) 2

(C) 1 (D) None of these

(xii) ;fn P (E ) = 0.6, P (F ) = 0.3 rFkk P (E ∩ F ) = 0.2,

rks P (E / F ) Kkr dhft,A 1

If P (E ) = 0.6, P (F ) = 0.3 and P (E ∩ F ) = 0.2,
then find P (E / F ) .

4331/(Set : A) P. T. O.

(8) 4331/(Set : A)
(xiii) ;fn P (A ) = 0.3, P (B ) = 0.6 rFkk A vkSj B LorU=
?kVuk,¡ gSa, rks P (A vkSj B ) dk eku Kkr dhft,A 1

If P (A ) = 0.3, P (B ) = 0.6 and A and B are
independent events, then find the value of
P (A and B ).

(xiv) ,d FkSys esa 4 lQsn vkSj 6 dkyh xsansa gSaA nks xsan
çfrLFkkiu ds lkFk ;kn`fPNd fudkyh x;h gSaA nksuksa xsan
dkyh gksus dh çkf;drk Kkr dhft,A 1

A bag contains 4 white and 6 black balls.
Two balls are drawn at random with
replacement. Find the probability that both
balls are black.

→
(xv) a = 2iˆ + 2 ˆj − 5kˆ vkSj b→ = 2iˆ + ˆj + 3kˆ ds ;ksxQy
ds vuqfn'k ek=d lfn'k Kkr dhft,A 1

Find the unit vector of the sum of the
→ →
vectors a = 2iˆ + 2 ˆj − 5kˆ and b = 2iˆ + ˆj + 3kˆ .

(xvi) ;fn ,d js[kk ds fnd~ vuqikr 2, −1, −2 gSa] rks js[kk ds
fnd~-dksT;k Kkr dhft,A 1

If direction ratio's of a line are 2, −1, −2,
then find the direction cosines of the line.

4331/(Set : A)

(9) 4331/(Set : A)
[k.M – c
SECTION – B

2. eku yhft, fd N esa ,d f}vk/kkjh lafØ;k ∗, a ∗ b = a rFkk
b dk L. C. M. }kjk ifjHkkf"kr gSA 5 ∗ 7 Kkr dhft,A 2

Let ∗ be the binary operation on N given by a ∗ b
= L. C. M. of a and b. Find 5 ∗ 7.

 3π 
3. sin −1  sin  dk eq[; eku Kkr dhft,A 2
 5 

 3π 
Find the principal value of sin −1  sin .
 5 

ekuk A = 
2 4  1 3
4. , B = , 2A + B dk eku Kkr
3 2  − 2 5
dhft,A 2

2 4  1 3
Let A =  , B= , find 2A + B.
3 2  − 2 5

;fn A = 
1 2
5. , rks |2A| = 4|A| n'kkZb,A 2
4 2
1 2
If A =  , then show that |2A| = 4|A|.
4 2

4331/(Set : A) P. T. O.

( 10 ) 4331/(Set : A)

6. lehdj.k 2x + 3y = sin y ls dy Kkr dhft,A 2
dx

dy
Find from the equation 2x + 3y = sin y.
dx

7. ;fn x = 2 at 2 , y = at 4 , rks dy Kkr dhft,A 2
dx

dy
If x = 2 at 2 , y = at 4 , then find .
dx

8. eku Kkr dhft, % 2

∫ x . sin x dx
Evaluate :

∫ x . sin x dx
9. eku Kkr dhft, % 2

3
1
∫ x dx
2

Evaluate :

3
1
∫ x dx
2

4331/(Set : A)

( 11 ) 4331/(Set : A)
x y
10. a rFkk b dks foyqIr djrs gq, oØ + = 1 dks fu:fir
a b
djus okys vodyu lehdj.k Kkr dhft,A 2

Find differential equation corresponding to
x y
+ = 1 , by eliminating a and b.
a b

11. ,d U;k¸; flDds dks 10 ckj mNkyk x;k gSA Bhd N% fpr vkus
ds fy, çkf;drk Kkr dhft,A 2

A fair coin is tossed 10 times. Find the
probability of exactly 6 heads.

[k.M – l
SECTION – C

12. fuEu dks ljyre :i esa O;Dr dhft, % 4

1 + x2 −1
tan−1 ,x ≠0
x
Write the simplest form of the following :

1 + x2 −1
tan−1 ,x ≠0
x

4331/(Set : A) P. T. O.

( 12 ) 4331/(Set : A)
13. k dk eku Kkr dhft, rkfd Qyu %

k cos x
 , ; fn x ≠ π
f (x ) =  π − 2x 2] x = π ij larr gksA 4
π 2
 3 , ; fn x =
 2

Find the value of k, so that the function :
k cos x π
 , if x ≠
f (x ) =  π − 2x 2
π
 3 , if x =
 2
π
is continuous at x = .
2

14. vUrjky Kkr dhft, ftuesa f (x) = 2 x 3 − 3 x 2 − 36 x + 7
ls çnÙk Qyu f fujUrj o/kZeku gSA 4

Find the interval in which the function

f (x) = 2 x 3 − 3 x 2 − 36 x + 7 is strictly increasing.

15. rk'k ds 52 iÙkksa dh ,d lqfefJr xM~Mh ls nks iÙks mÙkjksÙkj
çfrLFkkiuk ds lkFk fudkys tkrs gSaA bDdksa dh la[;k dk
çkf;drk caVu Kkr dhft,A 4

4331/(Set : A)

( 13 ) 4331/(Set : A)
Two cards are drawn with replacement from a
well shuffled pack of 52 cards. Write the
probability distribution of the number of aces
obtained.

→ → → →
16. lfn'k a + b vkSj a − b esa ls çR;sd ds yEcor ek=d

lfn'k Kkr dhft, tgk¡ a→ = iˆ + ˆj + kˆ , b→ = iˆ + 2 ˆj + 3kˆ gSA 4

Find a unit vector perpendicular to each vector
→ → → → →
a +b and a − b, where a = iˆ + ˆj + kˆ,

→
b = iˆ + 2 ˆj + 3kˆ .

[k.M – n

SECTION – D

17. fuEu lehdj.kksa dks vkO;wg fof/k }kjk gy dhft, % 6

x − y + z = 4,

2x + y − 3z = 0,

x + y + z = 2.

4331/(Set : A) P. T. O.

( 14 ) 4331/(Set : A)
Solve the following equations by a matrix method :

x − y + z = 4,

2x + y − 3z = 0,

x + y + z = 2.

x2 y2
18. nh?kZo`Ùk + = 1 ls f?kjs {ks= dk {ks=Qy Kkr dhft,A 6
a2 b2

Find the area enclosed by the ellipse :
x2 y2
+ =1
a 2 b2
vFkok

OR

eku Kkr dhft, % 6

π /2
cos5 x
∫ sin5 x + cos5 x
dx
0

Evaluate :

π /2
cos5 x
∫ sin5 x + cos5 x
dx
0

4331/(Set : A)

( 15 ) 4331/(Set : A)
→ →
19. r = iˆ + ˆj + λ(2iˆ − ˆj + kˆ ) vkSj r = 2iˆ + ˆj − kˆ + µ

(3iˆ − 5 ˆj + 2kˆ ) js[kkvksa ds chp dh U;wure nwjh Kkr dhft,A 6

Find the shortest distance between the lines :
→
r = iˆ + ˆj + λ(2iˆ − ˆj + kˆ ) and

→
r = 2iˆ + ˆj − kˆ + µ (3iˆ − 5 ˆj + 2kˆ )

vFkok
OR

lery dk lehdj.k Kkr dhft,] tks fcUnqvksa (1, 1, −1),
(6, 4, −5) vkSj (−4, −2, 3) ls xqtjrk gSA 6

Find the equation of plane passing through the
points (1, 1, −1), (6, 4, −5) and (−4, −2, 3).

20. vkys[k }kjk fuEu jSf[kd çksxzkeu leL;k dks gy dhft, % 6

vf/kdre % Z = 250x + 75y
O;ojks/kksa ds vUrxZr %
5x + y ≤ 100,

x + y ≤ 60,

x ≥ 0, y ≥ 0.

4331/(Set : A) P. T. O.

( 16 ) 4331/(Set : A)
Solve the following linear programming problem
by graphical method :

Maximize : Z = 250x + 75y

subject to the constraints :

5x + y ≤ 100,

x + y ≤ 60,

x ≥ 0, y ≥ 0.

s

4331/(Set : A)