HBSE Class 12 Mathematics Question Paper 2017 Set D

CLASS : 12th (Sr. Secondary) Code No. 2031
Series : SS-M/2017
Roll No. SET : D

xf.kr GRAPH
MATHEMATICS
[ Hindi and English Medium ]
ACADEMIC/OPEN
(Only for Fresh Candidates)
(Evening Session)
Time allowed : 3 hours ] [ Maximum Marks : 80
• Ñi;k tk¡p dj ysa fd bl iz'u&i= esa eqfnzr iz'u 20 gSaA
Please make sure that the printed question paper are 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
2031/ (Set : D) P. T. O.

(2) 2031/ (Set : D)
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
carry 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.

2031/ (Set : D)

(3) 2031/ (Set : D)
(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) ekuk f : R → R, f(x) = 3 – 4x }kjk ifjHkkf"kr gS] rks f gS % 1

(A) ,dSdh vkSj vkPNknd
(B) cgq,d vkSj vkPNknd
(C) u rks ,dSdh vkSj u gh vkPNknd
(D) ,dSdh gS] fdUrq vkPNknd ugha gS
Let f : R → R be defined as f(x) = 3 – 4x, then f is :

(A) one-one and onto

(B) many-one and onto

(C) neither one-one nor onto

(D) one-one but not onto

 1 
(ii) cot −1 −  dk eq[; eku gS % 1
 3

π π 2π
(A) (B) (C) (D) π
6 4 3

 1 
The principal value of cot −1 −  is :
 3

π π 2π
(A) (B) (C) (D) π
6 4 3

2031/ (Set : D) P. T. O.

(4) 2031/ (Set : D)
1 3  y 0 5 6
(iii) ;fn 2  + =  gks] rks x vkSj y ds eku gSa % 1
0 x  1 2 1 8

(A) x = 2, y = 3 (B) x = 3, y = 3

(C) x = 4, y = 2 (D) buesa ls dksbZ ugha
1 3  y 0 5 6
If 2  + =  , then the values of x and y are :
0 x  1 2 1 8
(A) x = 2, y = 3 (B) x = 3, y = 3
(C) x = 4, y = 2 (D) None of these
(iv) ;fn A ,d 3 × 3 dksfV dk vkO;wg gks] rks |kA| dk eku gS % 1
2 3
(A) 3k |A| (B) k |A| (C) k |A| (D) k |A|
Let A be a square matrix of order 3. Then |kA| is equal to :
2 3
(A) 3k |A| (B) k |A| (C) k |A| (D) k |A|
kx + 1, ;fn x≤π
(v) ;fn Qyu f (x) f (x ) =  }kjk ifjHkkf"kr x = π ij larr gks]
 sin x , ;fn x >π
rks k dk eku gS % 1
2 1
(A) − (B) − (C) 0 (D) –1
π π
kx + 1, if x≤π
If the function f(x) defined f (x ) = 
by is
 sin x , if x>π
continuous at x = π, then the value of k is :
2 1
(A) − (B) − (C) 0 (D) –1
π π
(vi) o`Ùk ds {ks=Qy ds ifjorZu dh nj bldh f=T;k r ds lkis{k r = 3 lseh ij gS % 1

6 π lseh /ls0 10 lseh /ls0
2 2
(A) (B)

8 π lseh /ls0 4 π lseh /ls0
2 2
(C) (D)

2031/ (Set : D)

(5) 2031/ (Set : D)
The rate of change of the area of a circle w.r.t. its radius r, when r
= 3 cm, is :
(A) 6 π cm2/sec. (B) 10 cm2/sec.
(C) 8 π cm2/sec. (D) 4 π cm2/sec.
x 2 y2
(vii) oØ + = 1 ij og fcUnq ftl ij Li'kZ js[kk x-v{k ds lekUrj gks] gS % 1
25 4
(A) (±5, 0) (B) (±4, 0)
(C) (0, ±2) (D) (0, ±5)
x 2 y2
The point on the curve + = 1 at which the tangent is parallel
25 4
to x-axis, is :
(A) (±5, 0) (B) (±4, 0)
(C) (0, ±2) (D) (0, ±5)
xdx
(viii) ∫ dk eku gS % 1
a4 − x4
1 −1
 x 4 
(A) sin + c (B) sin−1( a 2 − x 2 ) + c
2  4 
a 
1 −1 
 a 2  1 −1 
 x 2 
(C) sin + c (D) sin +c
2  x2  2 a2 
   
xdx
The value of ∫ is :
a4 − x4
1 −1 x 4 
(A) sin + c (B) sin−1( a 2 − x 2 ) + c
2  4 
a 
1 −1 
 a 2  1 −1 
 x 2 
(C) sin + c (D) sin +c
2  x2  2 a2 
   
1

∫ x e dx dk eku gS %
x
(ix) 1
0
1
(A) 1 (B) 0 (C) e (D)
e

2031/ (Set : D) P. T. O.

(6) 2031/ (Set : D)
1
The value of ∫ x e x dx is :
0
1
(A) 1 (B) 0 (C) e (D)
e
(x) oØksa x 2 + y 2 = 2cx ds dqy ds fy, vody lehdj.k gS % 1
dy
(A) 2xy − y2 = 0
dx
dy
(B) 2xy + x 2 − y2 = 0
dx
dy y 2
(C) 2xy + =0
dx 2
dy
(D) + x 2 − y2 = 0
dx
The differential equation for the family of curves x 2 + y 2 = 2cx is :
dy
(A) 2xy − y2 = 0
dx
dy
(B) 2xy + x 2 − y2 = 0
dx
dy y 2
(C) 2xy + =0
dx 2
dy
(D) + x 2 − y2 = 0
dx

(xi) ;fn (1 + x 2 ) dy = −(1 + y 2 ) gks] rks bldk gy gS % 1
dx
−1
(A) tan x + tan−1 y = c
x 
(B) tan−1  = c
y 
1+ x2
(C) log =c
1 + y2

2031/ (Set : D)

(7) 2031/ (Set : D)
1 + x 2
(D) tan−1  =c
1 + y 2 
 
dy
If (1 + x 2 ) = −(1 + y 2 ) then its solution is :
dx
(A) tan−1 x + tan−1 y = c
x 
(B) tan−1  = c
y 
1+ x2
(C) log =c
1 + y2

−1 
2
(D) tan 1 + x  =c
1 + y 2 
 
(xii) lfn'k iˆ − 2 ˆj + kˆ dk lfn'k 4iˆ − 4 ˆj + 7kˆ ij iz{ksi gS % 1
19 5
(A) (B)
2 9
19
(C) (D) 0
9

The projection of vector iˆ − 2 ˆj + kˆ on 4iˆ − 4 ˆj + 7kˆ is :
19 5
(A) (B)
2 9
19
(C) (D) 0
9
x −1 y − 2 z − 3
(xiii) ;fn js[kk,¡ = = rFkk x − 1 = y − 1 = z − 6 ,d nwljs ij
−3 2k 2 3k 1 −5
yfEcr gks] rks k dk eku gS % 1
1 1
(A) − (B) −
7 10
7 10
(C) (D) −
10 7

2031/ (Set : D) P. T. O.

(8) 2031/ (Set : D)
x −1 y − 2 z − 3 x −1 y −1 z − 6
If the lines = = and = = are
−3 2k 2 3k 1 −5
perpendicular to each other, then the value of k is :
1 1
(A) − (B) −
7 10
7 10
(C) (D) −
10 7
(xiv) nks iklksa ds ,dy mNky esa] 8 dk dqy izkIr djus dh izkf;drk gS % 1
1 5
(A) (B)
36 36
7 1
(C) (D)
36 9
In a single throw of two dice, the probability of getting a total of 8
is :
1 5
(A) (B)
36 36
7 1
(C) (D)
36 9
1
(xv) ;fn P(A) = 0 rFkk P(B) = gks] rks P(B/A) dk eku gS % 1
5
1
(A) (B) ifjHkkf"kr ugha
2
(C) 0 (D) 1
1
If P(A) = 0 and P(B) = , then P(B/A) is :
5
1
(A) (B) Not defined
2
(C) 0 (D) 1
(xvi) ;fn ,d U;k¸; flDds dks 8 ckj mNkyk x;k gks] rks 4 fpr izkIr djus dh izkf;drk gS
% 1
35 53
(A) (B)
128 64

2031/ (Set : D)

(9) 2031/ (Set : D)
105
(C) (D) buesa ls dksbZ ugha
128
If a fair coin is tossed 8 times, the probability of getting 4 heads is
:
35 53
(A) (B)
128 64
105
(C) (D) None of these
128
[k.M – c
SECTION – B
2. ;fn f (x ) = 4x + 3 , x ≠ 2 gks] rks fn[kkb, fd fof (x) =
6x − 4 3
2
x gS] x ≠ ds fy,A 2
3
4x + 3 2 2
If f (x ) = , x ≠ , show that fof (x) = x for all x ≠ .
6x − 4 3 3
3. fn[kkb, fd % 2
1 2 3
tan−1 + tan−1 = tan−1
2 11 4
Show that :
1 2 3
tan−1 + tan−1 = tan−1
2 11 4
 3 − 1
4. izkjfEHkd :ikarj.k dk iz;ksx djds vkO;wg   dk O;qRØe Kkr dhft,A 2
− 4 2 
Using elementary transformations, find the inverse of the matrix
 3 − 1
− 4 2  .
 
3 1 −2
5. lkjf.kd 0 0 −1 dk eku Kkr dhft,A 2
3 −5 0

2031/ (Set : D) P. T. O.

( 10 ) 2031/ (Set : D)
3 1 −2
Evaluate the determinant 0 0 −1 .
3 −5 0

dx
6. eku Kkr dhft, % ∫ 2
2
9x − 12x + 8
dx
Evaluate : ∫ 2
9x − 12x + 8
π
2
sin4 x
7. ∫ sin4 x + cos4 x dx dk eku Kkr dhft, 2
0

π
2
sin4 x
Evaluate : ∫ 4
x + cos 4 x
dx
0 sin

8. a vkSj b dks foyqIr djrs gq, xy = ae x + be − x ds vuqlkj vody lehdj.k Kkr
dhft,A 2

Find the differential equation corresponding to xy = ae x + be − x by
eliminating a and b.

9. fn[kkb, fd x + y = tan−1 y vody lehdj.k y 2 dy + y 2 + 1 = 0 dk gy gSA 2
dx

Show that x + y = tan−1 y is a solution of the differential equation
dy
y2 + y2 + 1 = 0 .
dx

10. tan2 x dks sec2 (x 2 ) ds lkis{k vodyu dhft,A 2

Differentiate tan2 x w.r.t. sec 2 (x 2 ) .

2031/ (Set : D)

( 11 ) 2031/ (Set : D)
11. ,d vufHkur ikls dks nks ckj mNkyk x;kA igyh mNky ij 4, 5 vkSj 6 dk izkIr gksuk rFkk
nwljh mNky ij 1, 2, 3 ;k 4 ds izkIr gksus dh izkf;drk Kkr dhft,A 2
An unbiased die is tossed twice. Find the probability of getting 4, 5 or 6
on the first toss and 1, 2, 3 or 4 on the second toss.
[k.M – l
SECTION – C

12. fln~/k dhft, % tan−1( x ) = 1 cos−11− x  , x ∈ [0,1]. 4
2 1 + x 
1 1 − x 
Prove that : tan−1( x ) = cos−1  , x ∈ [0,1].
2 1 + x 

13. ;fn x y = y x gks] rks dy fudkfy;sA 4
dx
dy
Find if x y = y x .
dx

14. vUrjky [0, 2] esa Qyu 3x 4 − 2x 3 − 6x 2 + 6x + 1 ds fujis{k mPpre rFkk fujis{k
fuEure eku Kkr dhft,A 4
Find the absolute maximum and absolute minimum values of the function
3x 4 − 2x 3 − 6x 2 + 6x + 1 in the interval [0, 2].

15. A vkSj B ckjh-ckjh ls ,d flDds dks mNkyrs gSa] tc rd fd muesa ls dksbZ ,d fpr dks
izkIr djrk gS vkSj [ksy thr ysrk gSA muds thrus dh Øe'k% izkf;drk Kkr dhft,A 4
A and B throw a coin alternatively till one of them gets a head and wins
the game. Find their respective probabilities of winning.

16. fln~/k dhft, fd fcUnq ftudh fLFkfr lfn'k a = 2iˆ − ˆj + kˆ , b = iˆ − 3 ˆj − 5kˆ rFkk
c = 3iˆ − 4 ˆj − 4kˆ }kjk iznÙk gS] ,d ledks.k f=Hkqt cukrs gSaA 4

Prove that the points whose position vectors are given by a = 2iˆ − ˆj + kˆ ,
b = iˆ − 3 ˆj − 5kˆ and c = 3iˆ − 4 ˆj − 4kˆ form a right-angled triangle.

2031/ (Set : D) P. T. O.

( 12 ) 2031/ (Set : D)
[k.M – n
SECTION – D

17. fuEu lehdj.kksa dks vkO;wg-fof/k }kjk gy dhft, % 6
3x + y + 2z = 3,
2x − 3y − z = −3 rFkk
x + 2y + z = 4 .
Solve the following system of equations by matrix method :
3x + y + 2z = 3,
2x − 3y − z = −3 and
x + 2y + z = 4 .

18. js[kk y = x + 2 rFkk oØ y = 1 x 2 + 2 ds chp f?kjs gq, {ks= dk {ks=Qy fudkfy,A 6
3
Find the area enclosed between the straight line y = x + 2 and the curve
1
y = x2 + 2 .
3
vFkok
OR

oØksa y = x 2 + 5 rFkk y = x 3 vkSj js[kk,¡ x = 1 rFkk x = 2 ds chp f?kjs gq, {ks= dk
{ks=Qy Kkr dhft,A
Find the area between the curves y = x 2 + 5 and y = x 3 and the lines x
= 1 and x = 2.

19. js[kkvksa r = iˆ + 2 ˆj + kˆ + λ(iˆ − ˆj + kˆ ) rFkk r = 2iˆ − ˆj − kˆ + µ(2iˆ + ˆj + 2kˆ ) ds chp
U;wure nwjh 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 ) .

2031/ (Set : D)

( 13 ) 2031/ (Set : D)
vFkok
OR

leryksa r .(2iˆ − 3 ˆj + 4kˆ) = 1 rFkk r .(iˆ − ˆj ) + 4 = 0 ds izfrPNsnu ls xqtjrs gq, rFkk
r .(2iˆ − ˆj + kˆ ) + 8 = 0 ij yfEcr lery dk lehdj.k Kkr dhft,A

Find the equation of the plane passing through the intersection of
ˆ ˆ ˆ ˆ ˆ
planes r .(2i − 3 j + 4k ) = 1 and r .(i − j ) + 4 = 0 and perpendicular to
r .(2iˆ − ˆj + kˆ ) + 8 = 0 .

20. fuEu jSf[kd izksxzkeu leL;k dks xzkQh; fof/k ls gy dhft, % 6
O;ojks/kksa − x + 3y ≤ 10 ; x + y ≤ 6 ; x – y ≤ 2 rFkk x, y ≥ 0 ds vUrxZr
z = x + 2y dk vf/kdrehdj.k dhft,A
Solve the following linear programming problem graphically :
Maximize z = x + 2y subject to the constraints − x + 3y ≤ 10 ; x + y ≤ 6 ;
x – y ≤ 2 and x, y ≥ 0.

s

2031/ (Set : D) P. T. O.