HBSE Class 12 Question Paper 2020 Maths

CLASS : 12th (Sr. Secondary) Code No. 4931
Series : SS-M/2020
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.

4931/(Set : A) P. T. O.

(2) 4931/(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
4931/(Set : A)

(3) 4931/(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.

4931/(Set : A) P. T. O.

(4) 4931/(Set : A)
[k.M – v
SECTION – A

1. (i) ;fn Qyu f : R → R tks f (x) = x3 }kjk ifjHkkf"kr gS]
rks f gS % 1
(A) ,dSdh ij vkPNknd ugha
(B) ,dSdh vkSj vkPNknd
(C) ,dSdh ugha ij vkPNknd
(D) u ,dSdh] u vkPNknd
Let f : R → R is defined as f (x) = x3 then f is :
(A) One-one, into
(B) One-one, onto
(C) Many-one, onto
(D) Many-one, into

(ii) tan−1 x dk eq[; eku gS % 1
 π
(A) 0, 2  (B) [0, π]

 π π
(C) − 2 , 2  (D) buesa ls dksbZ ugha

The principal value of tan−1 x is :
 π
(A) 0, 2  (B) [0, π]

 π π
(C) − 2 , 2  (D) None of these

4931/(Set : A)

(5) 4931/(Set : A)

;fn X + Y = 
5 2
vkSj X − Y = 
3 6
(iii)  , rks
0 9 −2 1
vkO;wg X dk eku gS % 1

 4 4  8 8 
(A)  −1 5  (B)  − 2 10 
   

1 − 2 
(C) 1 4  (D) buesa ls dksbZ ugha
 

5 2  3 6
If X + Y =   and X − Y =   , then
0 9  −2 1
matrix X is :

 4 4  8 8 
(A)  −1 5  (B)  − 2 10 
   

1 − 2 
(C) 1 4  (D) None of these
 

2 4 2x 4
(iv) ;fn lkjf.kd = , rks x dk eku gS % 1
5 1 6 x

(A) 6 (B) ±6

(C) –6 (D) buesa ls dksbZ ugha
2 4 2x 4
If det. = , then the value of x is :
5 1 6 x

(A) 6 (B) ±6
(C) –6 (D) None of these
4931/(Set : A) P. T. O.

(6) 4931/(Set : A)
(v) sec(tan x ) dk x ds lkis{k vodyu dhft,A 1

Differentiate sec(tan x ) with respect to x.

(vi) Qyu f (x ) = x 3 − 3x + 4 dk mPpre gS] tgk¡ x dk
eku gS % 1

(A) –1 (B) 1

(C) 0 (D) buesa ls dksbZ ugha
f (x ) = x 3 − 3x + 4 has a maxima at x is
equal to :

(A) –1 (B) 1

(C) 0 (D) None of these

(vii) Qyu f (x ) = log(sin x ) vUrjky ftlesa fujarj
Ðkleku gS] og gS % 1

 π π 
(A)  0,  (B)  , π
 2 2 

(C) (0, π) (D) buesa ls dksbZ ughas
f (x ) = log(sin x ) is strictly decreasing in
interval :

 π π 
(A)  0,  (B)  , π
 2 2 

(C) (0, π) (D) None of these
4931/(Set : A)

(7) 4931/(Set : A)
tan−1 x
(viii) ∫ dx dk eku Kkr dhft,A 1
1+ x2

tan−1 x
Find the value of ∫ dx .
1+ x2
π /2
(ix) ∫ sin3 x cos 2 x dx dk eku Kkr dhft,A 1
− π /2

π /2
Evaluate ∫ sin3 x cos 2 x dx .
− π /2

3
3 d 2y   dy 
2
 dy
(x) vody lehdj.k x
 2   dx 
+ +x +y = 0
 dx   dx 

dh ?kkr vkSj dksfV Kkr dhft,A 1

Find the degree and order of the differential
3
3
y  2 2
d  dy  dy
equation x +  +x +y = 0.
 dx 
2
 dx  dx
 

dy
(xi) vody lehdj.k (1 + x 2 ) = (1 + y 2 ) dks gy
dx

dhft,A 1

Solve the differential equation :

dy
(1 + x 2 ) = (1 + y 2 )
dx

4931/(Set : A) P. T. O.

(8) 4931/(Set : A)
(xii) ,d FkSys esa 4 lQsn vkSj 6 dkyh xsansa gSaA nks xsansa
izfrLFkkiu ds lkFk ;kn`fPNd fudkyh tkrh gSaA nksuksa xsan
ds dkyh gksus dh izkf;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 both the
balls are black.

(xiii) A vkSj B nks Lora= ?kVuk,¡ gSaA ;fn P (A ) = 0.3 vkSj
P (B ) = 0.4] rks P (A/B ) dk eku Kkr dhft,A 1

A and B are independent event such that
P (A ) = 0.3 and P (B ) = 0.4, find the P (A/B ).

(xiv) ,d ;kn`PN;k pj X dk izkf;drk caVu fuEufyf[kr gS % 1

X 0 1 2 3 4 5 6 7
P(X) 0 k 2k 2k 3k k2 2k2 7k2 + k

k dk eku Kkr dhft,A

A random variable X has the following
probability distribution :

X 0 1 2 3 4 5 6 7
P(X) 0 k 2k 2k 3k k2 2k2 7k2 + k

Find k.

4931/(Set : A)

(9) 4931/(Set : A)

(xv) lfn'kksa a→ = 2iˆ + 2 ˆj − 5kˆ vkSj b→ = ˆj − kˆ ds ;ksx dh
fn'kk esa bdkbZ lfn'k (unit vector) Kkr dhft,A 1

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

(xvi) ml js[kk dk lfn'k lehdj.k Kkr dhft, tks fcUnq
iˆ + 2 ˆj + 3kˆ ls xqtjrh gS vkSj 3iˆ + 2 ˆj − 2kˆ lfn'k
dh fn'kk esa gksA 1

Write the equation of line passing through
the point with position vector iˆ + 2 ˆj + 3kˆ
and in the direction 3iˆ + 2 ˆj − 2kˆ in vector
form.

[k.M – c
SECTION – B

1
2. ;fn f : R → R] f (x ) = (3 − x ) 3 3
}kjk iznf'kZr gS] rks
fof (x) Kkr dhft,A 2
1
3 3
If f : R → R be given by f (x ) = (3 − x ) , find fof (x).

3. fl) dhft, fd cos −1 4 + cos −1 12 = cos −1 33 2
5 13 65
4 12 33
Prove that cos −1 + cos −1 = cos −1
5 13 65

4931/(Set : A) P. T. O.

( 10 ) 4931/(Set : A)
− 2
4. ;fn A =  4  vkSj B = [1 3 − 6], rks (AB )' Kkr
 5 
dhft,A 2

− 2
If A =  4  and B = [1 3 − 6], find (AB )' .
 5 

y +k y y
5. fl) dhft, y y +k y = (3y + k ) k 2 2
y y y +k

y +k y y
Prove that y y +k y = (3y + k ) k 2
y y y +k

6. Kkr dhft, fd fuEufyf[kr Qyu x = 2 ij lrr gS ;k ugha % 2
f (x ) = x 3 − 3, x ≤ 2

= x 2 + 1, x > 2

Find out whether the following function is
continuous or not at x = 2 :
f (x ) = x 3 − 3, x ≤ 2

= x 2 + 1, x > 2

4931/(Set : A)

( 11 ) 4931/(Set : A)
7. ;fn x = a(cos θ + θ sin θ) 2
y = a (sin θ − θ cos θ) ,

rks θ = π ij dy dk eku Kkr dhft,A
4 dx
If x = a (cos θ + θ sin θ)
y = a (sin θ − θ cos θ) ,
dy π
then find , at θ = .
dx 4

x1 1 
8. ∫ e  x − x 2  dx dk eku Kkr dhft,A 2

1 1 
Evaluate ∫ e x  − 2  dx .
x x 

π /2
sin3 x
9. ∫ sin3 x + cos3 x dx dk eku Kkr dhft,A 2
0
π /2
sin3 x
Evaluate ∫ sin3 x + cos3 x dx.
0

10. vody lehdj.k x dy + 2y = x 2 , x ≠ 0 dk lkekU; gy
dx
Kkr dhft,A 2

Find the general solution of the differential
dy
equation x + 2y = x 2 , x ≠ 0.
dx
4931/(Set : A) P. T. O.

( 12 ) 4931/(Set : A)
11. ,d ikls dks 6 ckj Qsadk tkrk gSA le la[;k vkuk lQyrk gSA
4 lQyrk vkus dh izkf;drk Kkr dhft,A 2

A dice is thrown 6 times. If getting an even
number is success, find probability of getting 4
successes.

[k.M – l
SECTION – C

12. lehdj.k tan−1 1 − x = 1 tan−1 x , x > 0 dks gy dhft,A 4
1+ x 2
1− x 1
Solve the equation tan−1 = tan−1 x , x > 0.
1+ x 2

13. ;fn y = (sin x )sin x , 0 < x < π, rks dy Kkr dhft,A 4
dx
dy
If y = (sin x )sin x , 0 < x < π, find .
dx

14. fcUnq t = π 4 ij oØ x = a sin3 t , y = a cos3 t dh Li'kZ
js[kk dk lehdj.k Kkr dhft,A 4

Find the equation of tangent to the curve

x = a sin3 t , y = a cos 3 t at point t = π .
4

4931/(Set : A)

( 13 ) 4931/(Set : A)
15. ,d f=Hkqt ABC ds 'kh"kksZa ds fLFkfr lfn'k (position vector)
A(iˆ + ˆj + 2kˆ ), B (2iˆ + 3 ˆj + 5kˆ ) vkSj C (iˆ + 5 ˆj + 5kˆ ) gS]
rks ∆ABC dk {ks=Qy Kkr dhft,A 4
The vertices of a triangle ABC are given by
position vector A(iˆ + ˆj + 2kˆ ), B (2iˆ + 3 ˆj + 5kˆ ) and
C (iˆ + 5 ˆj + 5kˆ ) . Find its area.

16. fdlh fof'k"V leL;k dks A, B vkSj C }kjk Lora= :i ls gy
djus dh izkf;drk,¡ Øe'k% 1 , 1 vkSj 1 gSaA ;fn rhuksa
2 3 4
Lora= :i ls gy djrs gSa] rks leL;k gy gksus dh izkf;drk
Kkr dhft,A 4

Probability of solving
specific problem
1 1 1
independently by A, B and C are , and . If
2 3 4
they all try the problem independently, find the
probability that problem is solved.

[k.M – n
SECTION – D

17. fuEufyf[kr lehdj.kksa dks vkO;wg fof/k ls gy dhft, % 6

2x + 3y + 3z = 5,

x – 2y + z = – 4,

3x – y – 2z = 3.

4931/(Set : A) P. T. O.

( 14 ) 4931/(Set : A)
Solve the following system of equation by Matrix
method :

2x + 3y + 3z = 5,

x – 2y + z = – 4,

3x – y – 2z = 3.

18. o`Ùk x 2 + y 2 = 4 ls js[kk x + y = 2 }kjk dkVs x;s y?kq {ks=
dk {ks=Qy Kkr dhft,A 6

Find the area of smaller part of the circle
x 2 + y 2 = 4 cut-off by the line x + y = 2 .

vFkok
OR

fl) dhft, fd oØ y 2 = 4x vkSj x 2 = 4y, x = 0, y = 0
x = 4 vkSj y = 4 }kjk cus oxZ dks rhu cjkcj Hkkxksa esa ck¡Vrs

gSaA 6

Prove that the curves y 2 = 4x and x 2 = 4y divide
the area of the square bounded by x = 0, y = 0,
x = 4 and y = 4 in three equal parts.
4931/(Set : A)

( 15 ) 4931/(Set : A)
19. ml lery dk lehdj.k Kkr dhft, tks leryksa

→ →
r .(2iˆ + 2 ˆj − 3kˆ ) = 7 vkSj r .(2iˆ + 5 ˆj + 3kˆ ) = 9 ds izfrPNsn

ls xqtjrk gS vkSj (2, 1, 3) fcUnq ls Hkh xqtjrk gSA 6

Find the equation of the plane passing through the

→
intersection of the planes r .(2iˆ + 2 ˆj − 3kˆ ) = 7 and

→
r .(2iˆ + 5 ˆj + 3kˆ ) = 9 and through the point (2, 1, 3).

vFkok
OR

js[kkvksa → →
r = (iˆ + 2 ˆj + kˆ ) + λ(iˆ − ˆj + kˆ ) vkSj r = (2iˆ − ˆj + kˆ )

+ µ(2iˆ + ˆj + 2kˆ ) ds chp dh fuEure nwjh (S.D.) Kkr

dhft,A 6

Find the shortest distance between the lines

→ →
r = (iˆ + 2 ˆj + kˆ ) + λ(iˆ − ˆj + kˆ ) and r = (2iˆ − ˆj + kˆ )

+ µ(2iˆ + ˆj + 2kˆ ) .

4931/(Set : A) P. T. O.

( 16 ) 4931/(Set : A)
20. fuEu jSf[kd izksxzkeu leL;k (L.P.P.) dks xzkQh; fof/k }kjk gy
dhft, % 6

U;wure % Z = 18x + 10y
O;ojks/kksa ds vUrxZr %
4x + y ≥ 20,

2x + 3y ≥ 30,

x, y ≥ 0.

Solve the linear programming problem by
graphic method

Minimize : Z = 18x + 10y under the constraints :

4x + y ≥ 20,

2x + 3y ≥ 30,

x, y ≥ 0.

s
4931/(Set : A)