HBSE Class 12 Mathematics Question Paper 2018 Set D

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

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 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

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

(2) 3631/(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
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 : D)

(3) 3631/(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) ;fn f (x ) = x + 1 , x ≠ −2 vkSj g(x) = x 2 , rks gof (x) gS % 1
x +2
2 2
 x +1  x + 2
(A)   (B)  
 x + 2  x +1
x2 +1
(C) (D) buesa ls dksbZ ugha
x2 + 2
x +1
If f (x ) = , x ≠ −2 and g(x) = x 2 , then gof (x) is :
x +2
2 2
 x +1  x + 2
(A)   (B)  
 x + 2  x +1
x2 +1
(C) (D) None of these
x2 + 2
 5
(ii) cos sec −1  dk eku gS % 1
 3
5 3
(A) (B)
3 5
4 5
(C) (D)
5 4
 5
The value of cos sec −1  is :
 3
5 3
(A) (B)
3 5
4 5
(C) (D)
5 4

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

(4) 3631/(Set : D)
2 − 1
vkSj B = 
4 3
(iii) ;fn A =    ] rks 2A + B gS % 1
4 2  − 2 1
8 1 6 2
(A)   (B)  
6 5 2 3
 2 4
(C) − 6 − 1 (D) buesa ls dksbZ ugha
 
2 − 1
and B = 
4 3
If A =    ] then 2A + B is :
4 2   − 2 1
8 1 6 2
(A)   (B)  
6 5 2 3
 2 4
(C)   (D) None of these
− 6 − 1
2 3−x
(iv) ;fn = 0, rks x dk eku gS % 1
1 4
(A) 3 (B) −3
(C) 5 (D) −5
2 3−x
If = 0, then value of x is :
1 4
(A) 3 (B) −3
(C) 5 (D) −5
(v) 1 + cot x dk x ds lkis{k vodyt gS % 1
− cos ec 2 x − cos ec 2 x
(A) (B)
1 + cot x 2 1 + cot x
cosec 2 x
(C) (D) buesa ls dksbZ ugha
1 + cot x

The derivative of 1 + cot x w. r. t. x is :
2
− cos ec x − cos ec 2 x
(A) (B)
1 + cot x 2 1 + cot x

3631/(Set : D)

(5) 3631/(Set : D)
2
cosec x
(C) (D) None of these
1 + cot x
(vi) Qyu f(x) = 2 cosx + 3 x dk vf/kdre ;k U;wure eku ds fy, x = …….A 1
π π
(A) (B)
2 4
π π
(C) (D)
3 6
The function f(x) = 2 cosx + 3 x has maxima or minima at x =
……. .
π π
(A) (B)
2 4
π π
(C) (D)
3 6
π
(vii) θ = ij oØ x = a(θ − sinθ), y = a[1 − cosθ] ds yEc dh ço.krk gS % 1
2
(A) 1 (B) −1
(C) 0 (D) 2
The slope of normal to the curve x = a(θ − sinθ), y = a[1 −
π
cosθ] at θ = :
2
(A) 1 (B) −1
(C) 0 (D) 2
x
(viii) ∫ sin 2 dx dk eku gS % 1
2
x sin x
(A) + + c (B) x + sin x + c
2 2
x 1
(C) − sin x + c (D) buesa ls dksbZ ugha
2 2
x
The value of ∫ sin 2 dx is :
2
x sin x
(A) + + c (B) x + sin x + c
2 2
3631/(Set : D) P. T. O.

(6) 3631/(Set : D)
x 1
(C) − sin x + c (D) None of these
2 2
x2 −1
(ix) ∫ dx dk eku gS % 1
x2 + 4
5 x
(A) x+ tan −1 + c
2 2
5 x
(B) x − tan −1 + c
2 2
3 x
(C) x − tan −1 + c
2 2
(D) buesa ls dksbZ ugha
x2 −1
The value of ∫ dx is :
x2 + 4
5 x
(A) x + tan −1 + c
2 2
5 x
(B) x − tan −1 + c
2 2
3 x
(C) x − tan −1 + c
2 2
(D) None of these
3
 2  2
(x) vodyu lehdj.k  d y  +  dy  + sin dy  + 1 = 0 dh dksfV gS % 1
2  dx   dx   dx 
 

(A) 1 (B) 0
(C) 2 (D) buesa ls dksbZ ugha
The order of the differential equation
3
 d 2y  2
  +  dy  + sin dy  + 1 = 0 is :
 dx 2   dx   dx 
 
(A) 1 (B) 0
(C) 2 (D) None of these
dy
(xi) vodyu lehdj.k sin −1 = x dk gy gS % 1
dx

3631/(Set : D)

(7) 3631/(Set : D)
1
(A) y = cos x + c (B) y= +c
2
1− x
(C) y = − cos x + c (D) buesa ls dksbZ ugha
dy
Solution of the differential equation sin −1 = x is :
dx
1
(A) y = cos x + c (B) y= +c
1− x2
(C) y = − cos x + c (D) None of these
(xii) ;fn P(A) = 0.5, P(B) = 0.6 vkSj P(A ∪ B) = 0.8, rks P(B/A) gS % 1
1 3
(A) (B)
2 5
1
(C) (D) buesa ls dksbZ ugha
3
If P(A) = 0.5, P(B) = 0.6 and P(A ∪ B) = 0.8, then P(B/A) is :
1 3
(A) (B)
2 5
1
(C) (D) None of these
3
(xiii) ,d vPNh rjg QsaVh xbZ 52 iÙkksa dh xM~Mh ls ,d rk'k dk iÙkk fudkyk x;k vkSj
fQj nwljk iÙkk fudkyk x;k gSA ;fn igyk iÙkk çfrLFkkfir ugha fd;k x;k rks igyk
iÙkk gqdqe rFkk nwljk fpM+h dk gksus dh çkf;drk gS % 1
13 11
(A) (B)
204 204
17
(C) (D) buesa ls dksbZ ugha
204
A card is drawn from a well-shuffled deck of 52 cards and then a
second card is drawn. The probability that the first card is a spade
and the second card is a club if the first card is not replaced is :
13 11
(A) (B)
204 204
17
(C) (D) None of these
204

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

(8) 3631/(Set : D)
(xiv) ;fn A vkSj B nks LorU= ?kVuk,¡ bl çdkj gSa fd P(A ∪ B) = 0.5 vkSj
P(A) = 0.2, rks P(B) gS % 1
3 1
(A) (B)
25 8
3
(C) (D) buesa ls dksbZ ugha
8
If A and B are two independent events such that P(A ∪ B) = 0.5
and P(A) = 0.2, then P(B) is :
3 1
(A) (B)
25 8
3
(C) (D) None of these
8
→ →
(xv) a = 2iˆ + ˆj − 3kˆ vkSj b = iˆ + λˆj + 2kˆ lfn'kksa ds yEc gksus ds fy, λ dk eku gS %

1

(A) 2 (B) 3
(C) 5 (D) 4
→
The value of λ for which the vectors a = 2iˆ + ˆj − 3kˆ and

→
b = iˆ + λˆj + 2kˆ are perpendicular is :

(A) 2 (B) 3
(C) 5 (D) 4
(xvi) 3x + 1 = 6y − 2 = 1 − z js[kk ds fnd~-vuqikr gSa % 1
(A) 2, 1, −6 (B) 1, 2, −3
(C) 6, 1, −2 (D) buesa ls dksbZ ugha
The direction ratios of a line 3x + 1 = 6y − 2 = 1 − z are :
(A) 2, 1, −6 (B) 1, 2, −3
(C) 6, 1, −2 (D) None of these

3631/(Set : D)

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

2. n'kkZb, fd f(x) = 3x + 5, ∀ x ∈ Q ,dSdh gSA 2
Show that f(x) = 3x + 5, for all x ∈ Q is one-one.

3. fl) dhft, % 2
x +y
tan −1 x + tan −1 y = tan −1 , ;fn xy < 1.
1 − xy
Prove that :
x +y
tan −1 x + tan −1 y = tan −1 , if xy < 1.
1 − xy

;fn A = 
3 − 5
4. , rks f(A) Kkr dhft,] tgk¡ f (x ) = x 2 − 5x − 14 2
− 4 2 
 3 − 5
If A =  , then find f(A), where f (x ) = x 2 − 5x − 14 .
− 4 2 
5. f=Hkqt dk {ks=Qy Kkr dhft, ftlds 'kh"kZ (1, −1), (2, 4) vkSj (−3, 5) gSaA 2

Find the area of the triangle whose vertices are (1, −1), (2, 4) and (−3,
5).

−1 x
6. (sin x )cos dk x ds lkis{k vodyt Kkr dhft,A 2
−1 x
Find the derivative of (sin x )cos w. r. t. x.

dy
7. Kkr dhft,] ;fn x = a(1 + cos θ), y = a(θ + sin θ).
dx
2
dy
Find , if x = a(1 + cos θ), y = a(θ + sin θ).
dx

8. eku Kkr dhft, % 2

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

( 10 ) 3631/(Set : D)
−1
∫ cot x dx
Evaluate :
−1
∫ cot x dx

9. eku Kkr dhft, % 2
dx
∫ 32 − 2x 2
Evaluate :
dx
∫ 32 − 2x 2
10. vodyu lehdj.k (x 2 + xy ) dy = (x 2 + y 2 ) dx dks gy dhft,A 2

Solve the differential equation :

(x 2 + xy ) dy = (x 2 + y 2 ) dx
11. xf.kr ds ,d ç'u dks rhu Nk=ksa dks fn;k x;k gS ftldks gy djus dh laHkkouk Øe'k% 1 ] 1
2 3
vkSj 1 gSA ç'u dks gy djus dh çkf;drk D;k gS \ 2
4
A problem in Mathematics is given to three students whose chances of
1 1 1
solving it are ] and respectively. What is the probability that
2 3 4
the problem will be solved ?

[k.M – l
SECTION – C

12. fl) dhft, % 4
4 5 16 π
sin −1 + sin −1 + sin −1 =
5 13 65 2
Prove that :

3631/(Set : D)

( 11 ) 3631/(Set : D)
4 5 16 π
sin −1 + sin −1 + sin −1 =
5 13 65 2

n'kkZb, fd Qyu f (x ) = 2 + x , ;fn
 x ≥0
13. , x = 0 ij larr gS ijUrq
2 − x , ;fn x <0
O;qRik| ugha gSA 4
Show that the function :
2 + x , if x ≥ 0
f (x ) =  ,
2 − x , if x < 0
is continuous but not derivable at x = 0.

14. fl) dhft, fd oØ x = y 2 vkSj xy = k yEc ij dkVrh gS] ;fn 8k 2 = 1. 4
Prove that the curve x = y 2 and xy = k cut at right angle, if 8k 2 = 1 .

15. ,d FkSys esa 3 lQsn vkSj 4 yky xsansa gSaA rhu xsansa çfrLFkkiu ds lkFk ,d-,d djds fudkyh
xbZ gSaA yky xsanksa dh la[;k ds fy, çkf;drk caVu Kkr dhft,A4
A bag contains 3 white and 4 red balls. Three balls are drawn one by
one with replacement. Find the probability distribution of the number
of red balls.

16. f=Hkqt dk {ks=Qy Kkr dhft, ftlds 'kh"kZ (1, 2, 4), (3, 1, −2), (4, 3, 1) gSaA
4
Find the area of triangle whose vertices are (1, 2, 4), (3, 1, −2), (4,
3, 1).

[k.M – n
SECTION – D

17. fuEu lehdj.kksa dks vkO;wg fof/k }kjk gy dhft, % 6
x + y + z = 6,
y + 3z = 11,
x − 2y + z = 0.

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

( 12 ) 3631/(Set : D)
Solve the following equations by matrix method :
x + y + z = 6,
y + 3z = 11,
x − 2y + z = 0.

18. fn[kkb, fd oØ y 2 = 4x vkSj x 2 = 4y ,d oxZ x = 0, x = 4, y = 4 vkSj y = 0 ls
f?kjs {ks= dks rhu cjkcj Hkkxksa esa foHkkftr djrk gSA 6

Show that the curves y 2 = 4x and x 2 = 4y divide the area of the
square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts.
vFkok
OR
eku Kkr dhft, %
π
2
sin 2 x
∫ 1 + sin x . cos x dx
0

Evaluate :

π
2
sin 2 x
∫ 1 + sin x . cos x dx
0

19. fcUnq (2, −1, 5) ls js[kk x − 11 = y + 2 = z + 8 ij yEc ds ikn Kkr dhft,A 6
10 −4 − 11

Find the foot of perpendicular from the point (2, −1, 5) on the line
x − 11 y + 2 z + 8
= = .
10 −4 − 11

vFkok
3631/(Set : D)

( 13 ) 3631/(Set : D)
OR

fcUnqvksa (−2, 6, −6), (−3, 10, −9) vkSj (−5, 0, −6) ls xqtjus okys ry dk lehdj.k
Kkr dhft,A
Find the equation of the plane passing through the points (−2, 6, −6),
(−3, 10, −9) and (−5, 0, −6).

20. fuEu L.P.P. dks xzkQ }kjk gy dhft, % 6
U;wure % Z = 5x + 3y
O;ojks/kksa ds vUrxZr %
2x + y ≥ 10,
x + 3y ≥ 15,
x ≤ 10, y ≤ 8, x, y ≥ 0.

Solve graphically the following L. P. P. :
Minimize : Z = 5x + 3y
subject to constraints :
2x + y ≥ 10,
x + 3y ≥ 15,
x ≤ 10, y ≤ 8, x, y ≥ 0.

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