HBSE Class 12 Mathematics Question Paper 2018 Set B

CLASS : 12th (Sr. Secondary) Code No. 3631
Series : SS-M/2018
Roll No. SET : B

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.
• 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
3631/(Set : B) P. T. O.

(2) 3631/(Set : B)
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 cgqfodYih; çdkj ds 16 (i-xvi) Hkkxksa esa gSA
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
(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 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.

3631/(Set : B)

(3) 3631/(Set : B)
(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.
[k.M – v
SECTION – A

1. (i) ;fn f(x) = log (1 + x) vkSj g(x) = e x , rks (fog )(x) dk eku gS % 1

(A) log x (B) (
log e x + 1 )
(C) log (1 + x) (D) buesa ls dksbZ ugha
If f(x) = log (1 + x) and g(x) = e x , then value of (fog )(x) is :

(A) log x (B) (
log e x + 1 )
(C) log (1 + x) (D) None of these

 8 
(ii) cos sin −1  dk eku gS % 1
 17 
8 11
(A) (B)
17 17
15
(C) (D) buesa ls dksbZ ugha
17
 8 
The value of cos sin −1  is :
 17 
8 11
(A) (B)
17 17
15
(C) (D) None of these
17

3631/(Set : B) P. T. O.

(4) 3631/(Set : B)
1 4
vkSj B = 
− 1 2
(iii) ;fn A =    ] vkO;wg X bl çdkj gS fd A + B − X = 0]
3 2  0 5
rks X dk eku gS % 1
0 6 2 2 
(A)   (B)  
3 7 3 − 3
0 2 
(C)   (D) buesa ls dksbZ ugha
3 − 3
1 4  − 1 2
If A =   and B =   ] the matrix X such that A + B − X =
3 2  0 5
0, then value of X is :
0 6 2 2 
(A)   (B) 3 − 3
3 7  
0 2 
(C)   (D) None of these
3 − 3
x 12 6 18
(iv) ;fn = , rks x dk eku gS % 1
3 x 2 6
(A) ±4 (B) ± 6
(C) ±8 (D) buesa ls dksbZ ugha
x 12 6 18
If = , then value of x is :
3 x 2 6
(A) ±4 (B) ±6
(C) ±8 (D) None of these
(v) x ds lkis{k tan 3 x dk vodyt gS % 1

(A) 3 tan x sec 2 x (B) tan 2 x sec 2 x

(C) 3 tan2 x sec 2 x (D) buesa ls dksbZ ugha
The derivative of tan 3 x w. r. t. x is :
(A) 3 tan x sec 2 x (B) tan 2 x sec 2 x
(C) 3 tan2 x sec 2 x (D) None of these

3631/(Set : B)

(5) 3631/(Set : B)
(vi) sin 2x ds vf/kdre eku ds fy, x dk eku gS % 1
π π
(A) (B)
2 4
π
(C) (D) buesa ls dksbZ ugha
3
The value of x for which sin 2x attains its maximum, is :
π π
(A) (B)
2 4
π
(C) (D) None of these
3
π
(vii) oØ x = a cos 3 θ, y = a sin 3 θ dh θ = ij vfHkyEc dh ço.krk gS % 1
4
(A) 1 (B) −1
(C) 3 (D) −2
π
The slope of normal to the curve x = a cos 3 θ, y = a sin 3 θ at θ = is
4
:
(A) 1 (B) −1
(C) 3 (D) −2
(viii) ∫ cos 2 x dx dk eku gS % 1
1 1
(A) x + sin 2x + c
2 4
(B) 2 sin x + c
1 1
(C) x − sin 2x + c
2 4
(D) buesa ls dksbZ ugha
The value of ∫ cos 2 x dx is :

1 1
(A) x + sin 2x + c
2 4
(B) 2 sin x + c
1 1
(C) x − sin 2x + c
2 4
(D) None of these
3631/(Set : B) P. T. O.

(6) 3631/(Set : B)
5
3x
(ix) ∫ dx dk eku gS % 1
1 + x 12

(A) ( )
tan −1 x 6 + c

1
(B) tan −1 x 6 + c
2
3
(C) tan −1 x 6 + c
2
(D) buesa ls dksbZ ugha
3x 5
The value of ∫ dx is :
1 + x 12
(A) ( )
tan −1 x 6 + c
1
(B) tan −1 x 6 + c
2
3
(C) tan −1 x 6 + c
2
(D) None of these
2
d 2y  dy 
(x) + 3  + 5 = 0 vodyu lehdj.k dh ?kkr gS % 1
dx 2  dx 

(A) 1 (B) 2
(C) 3 (D) 0
2
d 2y
 dy 
The degree of the differential equation + 3  + 5 = 0 is :
2  dx 
dx
(A) 1 (B) 2
(C) 3 (D) 0
dy
(xi) ex = 1 vodyu lehdj.k dk gy gS % 1
dx

3631/(Set : B)

(7) 3631/(Set : B)
−x x
(A) y= e +c (B) y= e +c
(C) y = −e −x
+c (D) buesa ls dksbZ ugha
dy
Solution of the differential equation e x = 1 is :
dx
(A) y = e −x + c (B) y = ex + c
(C) y = − e − x + c (D) None of these
7 9
(xii) ;fn P ( A ) = , P (B ) = vkSj P (A ∩ B ) = 4 , rks P(B/A) gS % 1
13 13 13
4 4
(A) (B)
7 9
7
(C) (D) buesa ls dksbZ ugha
9
7 9 4
If P ( A ) = , P (B ) = and P ( A ∩ B ) = , then P(B/A) is :
13 13 13
4 4
(A) (B)
7 9
7
(C) (D) None of these
9
(xiii) ,d FkSys esa 10 lQsn vkSj 15 dkyh xsansa gSaA nks xsan fcuk cnys yxkrkj fudkyh x;h
gSaA igyh lQsn vkSj nwljh dkyh gksus dh çkf;drk gS % 1
1 1
(A) (B)
3 4
1
(C) (D) buesa ls dksbZ ugha
5
A bag contains 10 white and 15 black balls. Two balls are drawn
in succession without replacement. The probability that first is
white and second is black, is :
1 1
(A) (B)
3 4
1
(C) (D) None of these
5

3631/(Set : B) P. T. O.

(8) 3631/(Set : B)
(xiv) ;fn P(A) = 0.6, P(A ∪ B) = 0.7 vkSj A rFkk B LorU= ?kVuk,¡ gSa] rks P(B) gS %
1
1 1
(A) (B)
2 3
1
(C) (D) buesa ls dksbZ ugha
4
If P(A) = 0.6, P(A ∪ B) = 0.7 and A and B are independent events,
then P(B) is :
1 1
(A) (B)
2 3
1
(C) (D) None of these
4

(xv) ;fn nks lfn'kksa a→ vkSj b→ ds chp dk dks.k 0 gS] rks a→ . b→ dk eku gS % 1

(A) 0 (B) 1
(C) ab (D) −ab
→ → →
If angle between two vectors a and b is 0, then the value of a

→
. b is :
(A) 0 (B) 1
(C) ab (D) −ab
(xvi) funsZ'kkad v{kksa ls leku dks.k cukus okyh js[kk ds fnd~-dksT;k gSa % 1
1 1 1
(A) 1, 1, 1 (B) , ,
2 2 2
1 1 1 1 1 1
(C) ± , ± , ± (D) ± ,± ,±
3 3 3 3 3 3
The direction cosines of a line equally inclined to the coordinate
axis are :
1 1 1
(A) 1, 1, 1 (B) , ,
2 2 2
1 1 1 1 1 1
(C) ± , ± , ± (D) ± ,± ,±
3 3 3 3 3 3

3631/(Set : B)

(9) 3631/(Set : B)
[k.M – c
SECTION – B

n + 1
 , ; fn n fo" ke gS
2. ekuk f (n ) =  2 lHkh ds fy, n ∈ N] n'kkZb, fd f ,dSdh ugha gSA
 n , ; fn n le gS
 2
2
n + 1
 , if n is odd
Let f (n ) =  2 for all n ∈ N, show that f is not one-
 n , if n is even
 2
one.

3. fl) dhft, % 2
π
tan −1 x + cot −1 x = .
2
Prove that :
π
tan −1 x + cot −1 x = .
2

;fn A = 
2 3
4.  vkSj f (x ) = x 2 − 4x + 7, rks f(A) Kkr dhft,A 2
 − 1 2

 2 3
If A =   and f (x ) = x 2 − 4x + 7, then find f(A).
 − 1 2
5. f=Hkqt dk {ks=Qy Kkr dhft, ftlds 'kh"kZ (4, 2), (4, 5) vkSj (−2, 2) gSaA 2
Find the area of the triangle whose vertices are (4, 2), (4, 5) and (−2, 2).

6. x ds lkis{k (sin x )log x dk vodyt Kkr dhft,A 2
Find the derivative of (sin x )log x w. r. t. x.

3631/(Set : B) P. T. O.

( 10 ) 3631/(Set : B)
dy
7. Kkr dhft,] tcfd x = cos 2θ + 2 cos θ, y = sin 2θ − 2 sin θ. 2
dx
dy
Find , when x = cos 2θ + 2 cos θ, y = sin 2θ − 2 sin θ.
dx

8. eku Kkr dhft, % 2

−1
∫ sin x dx
Evaluate :
−1
∫ sin x dx

9. eku Kkr dhft, % 2

dx
∫ 1 − 4x 2
Evaluate :

dx
∫ 1 − 4x 2
10. vodyu lehdj.k (x 2 + y 2 ) dx + 2xy dy = 0 dks gy dhft,A 2
Solve the differential equation :
(x 2 + y 2 ) dx + 2xy dy = 0
11. ,d FkSys esa 3 yky vkSj 5 dkyh xsansa gSa vkSj nwljs FkSys esa 6 yky vkSj 4 dkyh xsansa gSAa çR;sd
FkSys ls ,d xsan fudkyh xbZ gSA çkf;drk Kkr dhft, fd nksuksa dkyh gSaA 2
A bag contains 3 red and 5 black balls and a second bag contains 6 red
and 4 black balls. A ball is drawn from each bag. Find the probability
that both are black.

[k.M – l
SECTION – C

3631/(Set : B)

( 11 ) 3631/(Set : B)
12. fl) dhft, % 4
 + 1− x2  π 1
2
−1  1 + x  = + cos −1 x 2
tan
 1+ x2 − 1− x2  4 2
 
Prove that :
 1+ x2 + 1− x2  π 1
tan −1   = + cos −1 x 2
 1+ x2 − 1− x2  4 2
 

13. n'kkZb, fd Qyu f(x) = |x − 1|+ |x + 1|, lHkh ds fy, x ∈ R, x = −1 ij
vodyuh; ugha gSaA 4
Show that the function f(x) = |x − 1|+ |x + 1|, for all x ∈ R, is not
differentiable at x = −1.

π
14. θ = ij oØ x = 1 − cos θ, y = θ − sin θ dh Li'kZjs[kk Kkr dhft,A 4
4
Find the equation of tangent to the curve x = 1 − cos θ, y = θ
π
− sin θ at θ = .
4
15. nks iklksa dks nks ckj mNkyus ij ;ksx 9 vkus ds fy, çkf;drk caVu Kkr dhft,A 4
Find the probability distribution of the number of times a total of 9
appears in two throws of two dice.

16. ;fn → → →
a = iˆ + ˆj + kˆ vkSj b = ˆj − kˆ ] rks ,d lfn'k c Kkr dhft, tks bl çdkj gS

%→ → → → →
a × c = b vkSj a . c = 3. 4
→ → → → → →
If a = iˆ + ˆj + kˆ and b = ˆj − kˆ ] find a vector c such that a × c = b
→ →
and a . c = 3.

[k.M – n
SECTION – D

3631/(Set : B) P. T. O.

( 12 ) 3631/(Set : B)
17. fuEu lehdj.kksa dks vkO;wg fof/k }kjk gy dhft, % 6
8x + 4y + 3z = 19,
2x + y + z = 5,
x + 2y + 2z = 7.
Solve the following equations by matrix method :
8x + 4y + 3z = 19,
2x + y + z = 5,
x + 2y + 2z = 7.

18. oØ x 2 = 4y vkSj js[kk x = 4y − 2 ls f?kjs {ks= dk fp= cukb, vkSj mldk {ks=Qy Kkr
dhft,A 6
Draw a sketch of the region bounded by the curve x 2 = 4y and the line
x = 4y − 2 and determine its area.
vFkok
OR

eku Kkr dhft, %
π
x
∫ 4 − cos 2 x dx
0

Evaluate :
π
x
∫ 4 − cos 2 x dx
0

19. js[kk x = y − 1 = z − 2 esa fcUnq (1, 6, 3) dk çfrfcEc Kkr dhft,A 6
1 2 3

x y −1 z − 2
Find the image of the point (1, 6, 3) in the line = = .
1 2 3

vFkok

3631/(Set : B)

( 13 ) 3631/(Set : B)
OR

fcUnqvksa (0, 1, 1), (1, 1, 2) vkSj (−1, 2, −2) ls xqtjus okys ry dk lehdj.k Kkr
dhft,A

Find the equation of the plane passing through the points (0, 1, 1), (1,
1, 2) and (−1, 2, −2).

20. fuEu L.P.P. dks xzkQh; fof/k }kjk gy dhft, % 6
U;wure % Z = −3x + 4y
O;ojks/kksa ds vUrxZr %
x + 2y ≤ 8,
3x + 2y ≤ 12,
x, y ≥ 0.

Solve graphically the following L. P. P. :
Minimize : Z = −3x + 4y
subject to constraints :
x + 2y ≤ 8,
3x + 2y ≤ 12,
x, y ≥ 0.

s
3631/(Set : B) P. T. O.