HBSE Class 12 Question Paper 2023 Maths

CLASS : 12th (Sr. Secondary) Code No. 231
Series : SS/Annual-2023
Roll No. SET : A

xf.kr
MATHEMATICS
[ Hindi and English Medium ]
ACADEMIC/OPEN
(Only for Fresh/Re-appear/Improvement/Additional 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 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 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.
• 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.

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

(2) 231/(Set : A)
• ijh{kkFkhZ viuk jksy ua0 iz'u&i= ij vo'; fy[ksaA jksy ua0 ds vfrfjDr iz'u&i= ij vU; dqN Hkh u
fy[ksa vkSj oSdfYid iz'uksa ds mÙkjksa ij fdlh izdkj dk fu'kku u yxk,¡A
Candidates must write their Roll No. on the question paper. Except Roll No. do not
write anything on question paper and don't make any mark on answers of objective
type questions.

• d`i;k iz'uksa ds 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 questions, 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 dqy 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 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
[k.M ^l
^l* % 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 ds nks ç'uksa esa vkarfjd fodYi fn;s x;s gSaA vkidks çR;sd esa ls ,d fodYi djuk gSA

231/(Set : A)

(3) 231/(Set : A)
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' : It contains 16 questions from 1 to 16. Each question carries 1
mark.
Section 'B' : It contains 10 questions from 17 to 26. Each question carries 2
marks.
Section 'C' : It contains 5 questions from 27 to 31. Each question carries 4
marks.
Section 'D' : It contains 4 questions from 32 to 35. Each question carries 6
marks.

(iii) Internal choices are given in two questions of Section-D. You have to attempt
one from each.

[k.M – v
SECTION – A

1. ;fn A = R − {3}, B = R − {1} vkSj f (x ) = x − 2 , rks f : A → B gSa % 1
x −3

(A) ,dSdh vkSj vkPNknd (B) u ,dSdh vkSj u vkPNknd

(C) ,dSdh ij vkPNknd ugha (D) cgq,dh vkSj vkPNknd

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

(4) 231/(Set : A)
x −2
If A = R − {3}, B = R − {1} and f (x ) = , then f : A → B is :
x −3

(A) One-one, onto (B) Neither one-one nor onto

(C) One-one, into (D) Many-one, onto

2. tan −1 3 − cot −1 − 3 ( ) dk eq[; eku gS % 1
π
(A) π (B) −
2
(C) 0 (D) 2 3
The principal value of tan −1 3 − cot −1 − 3 is : ( )
π
(A) π (B) −
2
(C) 0 (D) 2 3

 cos α sin α 
3. ;fn A =  , rks A'A gS % 1
− sin α cos α 

(A) I

 cos2 α sin2 α 
(B)  2 
− sin α cos 2 α 

2 cos α 0 
(C)  0
 2 cos α 

(D) 1

231/(Set : A)

(5) 231/(Set : A)
 cos α sin α 
If A =  , then A'A is :
− sin α cos α 

 cos 2 α sin2 α 
(A) I (B)  2 
− sin α cos 2 α 

2 cos α 0 
(C)  0 (D) 1
 2 cos α 

0 1 2
4. ;fn | A | = − 1 0 − 3 , rks |A| dk eku Kkr dhft,A 1
−2 3 0

0 1 2
If | A | = − 1 0 − 3 , then find the value of |A| .
−2 3 0

 5 , x ≥2

5. ;fn f (x ) = ax + b , 2 < x < 10 lHkh x ds eku ds fy, ,d lrr Qyu gS] rks a vkSj b
 21 , x ≥ 10

dk eku gS % 1

(A) a = 1, b = 2

(B) a = 2, b = 1

(C) a = 5, b = 21

(D) a = 21, b = 5

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

(6) 231/(Set : A)
 5 , x ≥2

If f (x ) = ax + b , 2 < x < 10 is continuous for all x, then values of a and b is :
 21 , x ≥ 10


(A) a = 1, b = 2

(B) a = 2, b = 1

(C) a = 5, b = 21

(D) a = 21, b = 5

6. ,d o`Ùk dh f=T;k 3 cm/sec dh nj ls c<+ jgh gSA o`Ùk ds {ks=Qy ds c<+us dh nj Kkr dhft, tc
o`Ùk dh f=T;k 10 cm gSA 1
Radius of a circle is increasing at the rate of 3 cm/sec. Find the rate of change of
area when radius of circle is 10 cm.

7. ijoy; y 2 = 4ax dh Li'kZjs[kk dh ço.krk fcUnq (at 2 , 2at ) ij gS % 1

2
(A) t (B)
t

1
(C) (D) buesa ls dksbZ ugha
t

The slope of the tangent to the curve y 2 = 4ax at the point (at 2 , 2at ) is :

2
(A) t (B)
t

1
(C) (D) None of these
t

231/(Set : A)

(7) 231/(Set : A)
2
sec x
8. ∫ cos ec 2x dx dk eku gS % 1

(A) − tan2 x + c

(B) − cot 2 x + c

(C) tan x − x + c

(D) buesa ls dksbZ ugha

sec 2 x
∫ cos ec 2x dx is equal to :

(A) − tan2 x + c

(B) − cot 2 x + c

(C) tan x − x + c

(D) None of these

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

(8) 231/(Set : A)
9. eku Kkr dhft, % 1

∫ x cos 2x dx
Evaluate :

∫ x cos 2x dx
10. y = a sin(x + b) ls çnÙk oØksa ds ifjokj dh vody lehdj.k gS % 1

d 2y d 2y
(A) =y (B) a = by
dx 2 dx 2

d 2y d 2y
(C) b = ay (D) +y = 0
dx 2 dx 2

The differential equation representing the family of curve y = a sin(x + b) is :

d 2y
(A) =y
dx 2

d 2y
(B) a = by
dx 2

d 2y
(C) b = ay
dx 2

d 2y
(D) +y =0
dx 2

231/(Set : A)

(9) 231/(Set : A)
11. vody lehdj.k dy = y tan x dk O;kid gy fudkysaA 1
dx

dy
Find general solution of the differential equation = y tan x .
dx

12. ;fn P (E ) = 3 , P (F ) = 3 , E vkSj F Lora= ?kVuk,¡ gSa] rks P(E ∪ F) dk eku gS % 1
5 10

9 9
(A) (B)
10 50

36
(C) (D) buesa ls dksbZ ugha
50

3 3
If P (E ) = , P (F ) = , E and F are independent, then P(E ∪ F) is :
5 10

9 9
(A) (B)
10 50

36
(C) (D) None of these
50

13. ,d ikls dks 6 ckj Qsadk tkrk gSA 4 ckj fo"ke la[;k vkus dh çkf;drk Kkr dhft,A 1

A die is thrown 6 times. Find the probability of getting odd number 4 times.

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

( 10 ) 231/(Set : A)
14. ,d ;kn`fPNd pj ds çlj.k Kkr djus ds fy, lw= fyf[k,A 1
Write the formula for finding variance of a random variable.

15. ;fn ,d bdkbZ lfn'k iˆ ds lkFk π , ĵ ds lkFk π vkSj k̂ ds lkFk θ U;wudks.k cukrk gS] rks θ dk
3 4
eku gS % 1

π
(A)
6

π
(B)
4

π
(C)
3

π
(D)
2

π π
If a unit vector makes with iˆ , with ĵ and an acute angle θ with k̂ , then
3 4
θ is :
π
(A)
6
π
(B)
4
π
(C)
3
π
(D)
2

16. leryksa 2x + y + 3z = 2 vkSj x − 2y = 5 ds chp dk dks.k Kkr dhft,A 1

Find the angle between the planes 2x + y + 3z = 2 and x − 2y = 5.

231/(Set : A)

( 11 ) 231/(Set : A)
[k.M – c
SECTION – B

1
17. ;fn f : R → R, f (x ) = (3 − x 3 )3 ] rks fof(x) Kkr dhft,A f −1(x ) Hkh Kkr djsaA 2

1
( )
If f : R → R, f (x ) = 3 − x 3 3 ] then find fof(x). Hence find f −1(x ) .

18. n'kkZb, fd % 2

x x
tan −1 = sin −1
a2 − x2 a

tgk¡ |x| < a.

Show that :

x x
tan −1 = sin −1
a2 − x2 a

where |x| < a.
2 1
2 −3 4
19. ;fn A =  vkSj B = 4 5 , rks AB Kkr dhft,A 2
− 1 2 3
2 3

2 1
2 −3 4
If A =   and B = 4 5 , then find AB.

− 1 2 3
2 3

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

( 12 ) 231/(Set : A)
20. fl) dhft, % 2

1 bc a (b + c )
1 ca b(c + a ) = 0
1 ab c (a + b )

Prove that :

1 bc a (b + c )
1 ca b(c + a ) = 0
1 ab c (a + b )

21. ;fn f (x ) = x cos x ] rks f ' (x ) Kkr dhft,A 2

If f (x ) = x cos x ] find f ' (x ) .

 2x  dy
22. ;fn y = cos −1  , − 1 < x < 1 , rks Kkr dhft,A 2
1 + x 2  dx

 2x  dy
If y = cos −1  , − 1 < x < 1 , then find .
2 dx
1 + x 

23. eku Kkr dhft, % 2

2x + 3
∫ x 2 + 3x + 2 dx
Evaluate :

2x + 3
∫ x 2 + 3x + 2 dx
231/(Set : A)

( 13 ) 231/(Set : A)
24. eku Kkr dhft, % 2
π /2
sin 4 x
∫ sin 4 x + cos 4 x
dx
0
Evaluate :
π /2
sin 4 x
∫ sin 4 x + cos 4 x
dx
0

25. vody lehdj.k dy + y sec x = tan x ,  0 ≤ x < π  dks gy dhft,A 2
dx  2
dy  π
Solve the differential equation + y sec x = tan x ,  0 ≤ x <  .
dx  2

26. ,d U;k¸; ikls dks ,d ckj Qsadus ls vkus okyh la[;kvksa dk ek/; Kkr dhft,A 2
Find mean of the number obtained on a throw of an unbiased die.

[k.M – l

SECTION – C

27. lehdj.k 2 tan−1(cos x ) = tan−1(2 cosec x ) dks gy dhft,A 4

Solve the equation 2 tan −1(cos x ) = tan −1(2 cosec x ) .

28. Qyu f (x ) = x 2 − 4x − 3 ij ySxzkat e/;eku çes; dk varjky [1, 4] ij lR;kiu dhft,A 4

Verify Lagrange mean value theorem for the function f (x ) = x 2 − 4x − 3 on the
interval [1, 4].

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

( 14 ) 231/(Set : A)
29. 20 cm f=T;k okys o`Ùk ds varxZr cuus okys lHkh vk;rksa esa oxZ dk {ks=Qy mPpre gksrk gSA ml oxZ
dk {ks=Qy Hkh Kkr dhft,A 4

Prove that among all rectangles inscribed in a circle of radius 20 cm, square has
the maximum area. Also find the area of that square.

30. rhu flDdksa esa] ,d nksuksa rjQ fpr (Head) okyk flDdk gS] nwljs flDds esa 60% ckj fpr vkrk gS
vkSj rhljk U;k¸; (Unbiased) flDdk gSA ;fn ,d flDdk ;kn`PN;k pquk tkrk gS vkSj mls mNkyus ij
fpr (Head) vkrk gS] rks mlds nksuksa rjQ fpr okyk flDdk gksus dh çkf;drk Kkr dhft,A 4

There are three coins. One is two headed coin (having head on both sides),
another is a biased coin that comes up head 60% of the time and third is
unbiased coin. One of the three coins is chosen at random and tossed. If head
occurs what is the probability that it is two headed coin.

31. ml f=Hkqt dk {ks=Qy Kkr dhft, ftlds 'kh"kZ fcUnq A(1, 1, 1), B(1, 2, 3) vkSj C(2, 3, 1) gksA 4
Find the area of the triangle having the vertices A(1, 1, 1), B(1, 2, 3) and
C(2, 3, 1).
[k.M
[k.M – n
SECTION – D

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

2x + 3y + 3z = 5

x − 2y + z = −4

3x − y − 2z = 3

231/(Set : A)

( 15 ) 231/(Set : A)
Solve the following system of equations by matrix method :

2x + 3y + 3z = 5

x − 2y + z = −4

3x − y − 2z = 3

vFkok
OR

fl) dhft, %
1 + a 2 − b2 2ab − 2b
2 2
2ab 1−a +b 2a = (1 + a 2 + b 2 )3
2b − 2a 1 − a 2 − b2

Prove that :

1 + a 2 − b2 2ab − 2b
2 2
2ab 1−a +b 2a = (1 + a 2 + b 2 )3
2b − 2a 1 − a 2 − b2

33. o`Ùkksa x 2 + y 2 = 4 vkSj (x − 2)2 + y 2 = 4 ds chp ds {ks= dk {ks=Qy Kkr dhft,A 6

Find the area of the region enclosed between the circles x 2 + y 2 = 4 and
(x − 2)2 + y 2 = 4 .

vFkok
OR
231/(Set : A) P. T. O.

( 16 ) 231/(Set : A)
fuEufyf[kr {ks= dk {ks=Qy Kkr dhft, %
{(x, y) : 0 ≤ y ≤ ( x 2 +1), 0 ≤ y ≤ (x + 1), 0 ≤ x ≤ 2 }
Find the area of the region
{(x, y) : 0 ≤ y ≤ ( x 2 +1), 0 ≤ y ≤ (x + 1), 0 ≤ x ≤ 2 }

34. js[kkvksa x + 1 = y + 1 = z + 1 vkSj x − 3 = y − 5 = z − 7 ds chp dh U;wure nwjh Kkr dhft,A 6
7 −6 1 1 −2 1
x +1 y +1 z +1
Find the shortest distance between the lines = = and
7 −6 1
x −3 y −5 z −7
= = .
1 −2 1

35. ,d HkksT; inkFkZ esa 80 ek=d foVkfeu A vkSj 100 ek=d [kfut gksus pkfg,A nks HkksT; F1 vkSj F2
miyC/k gSaA HkksT; F1 esa 3 ek=d foVkfeu A vkSj 4 ek=d [kfut gSA HkksT; F2 esa 6 ek=d foVkfeu
A vkSj 3 ek=d [kfut gSAa HkksT; F1 ds ,d ek=d dh dher #0 4 vkSj F2 ds ,d ek=d dh dher
#0 6 gSA bl leL;k dks jSf[kd çksxzkeu leL;k ds :i esa O;Dr dhft,A nksuksa HkksT; dks feykdj ,slk
HkksT; cuk;sa tks fuEure ewY; dk gks vkSj foVkfeu vkSj [kfut dh vko';drkvksa dh iwfrZ djrk gSA 6

A diet is to contain at least 80 units of Vitamin A and 100 units of minerals.
Two foods F1 and F2 are available. Food F1 costs Rs. 4 per unit and F2 costs
Rs. 6 per unit. One unit of F1 contains 3 units of Vitamin A and 4 units of
minerals. One unit of food F2 contains 6 units of Vitamin A and 3 units of
minerals. Formulate the linear programming problems. Find the minimum cost
for diet that consists of mixture of two foods which meets the minimal
nutritional requirement.

S
231/(Set : A)