HBSE Class 12 Mathematics Question Paper 2018 Set C

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

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 : C) P. T. O.

(2) 3631/(Set : C)
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 : C)

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

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

(4) 3631/(Set : C)
lehdj.k X + 
1 5  2 3
(iii) =  esa X dk eku gS % 1
0 4 3 1
 1 2 2 1 
(A)  − 3 3 (B)  
  3 − 3
1 − 2
(C) 3 − 3 (D) buesa ls dksbZ ugha
 
1 5  2 3
In the equation X +  = , X is :
0 4 3 1
 1 2 2 1 
(A)   (B)  
 − 3 3 3 − 3
1 − 2
(C)   (D) None of these
3 − 3
5−x x +1
(iv) ;fn = 0, rks x dk eku gS % 1
2 4
(A) 3 (B) −3
(C) 4 (D) 5
5−x x +1
If = 0, then value of x is :
2 4
(A) 3 (B) −3
(C) 4 (D) 5
(v) cos x 4 dk x ds lkis{k vodyt gS % 1
(A) − sin x 4 (B) 4x 3 sin x 4
(C) − 4x 3 sin x 4 (D) buesa ls dksbZ ugha
The derivative of cos x 4 w. r. t. x is :
(A) − sin x 4 (B) 4x 3 sin x 4
(C) − 4x 3 sin x 4 (D) None of these
(vi) Qyu f(x) = cosx − sinx dk vf/kdre ;k U;wure eku gS x = ……… ij % 1
π 3π
(A) (B)
4 4

3631/(Set : C)

(5) 3631/(Set : C)
π π
(C) (D)
2 3
The function f(x) = cosx − sinx has maxima or minima value at x =
……… .
π 3π
(A) (B)
4 4
π π
(C) (D)
2 3
π
(vii) θ = ij oØ x = a(θ − sinθ), y = a[1 − cos θ] dh Li'kZjs[kk dh ço.krk gS %
2
1
(A) 1 (B) −1
(C) 2 (D) −2
The slope of tangent to the curve x = a(θ − sinθ), y = a[1 − cos θ] at
π
θ= is :
2
(A) 1 (B) −1
(C) 2 (D) −2
x
(viii) ∫ cos 2 dx dk eku gS % 1
2
1
(A) (x − sinx) + c
2
1
(B) (x − cosx) + c
2
1
(C) (x + sinx) + c
2
(D) buesa ls dksbZ ugha
x
The value of ∫ cos 2 dx is :
2
1
(A) (x − sin x) + c
2
1
(B) (x − cos x) + c
2
1
(C) (x + sin x) + c
2

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

(6) 3631/(Set : C)
(D) None of these
x
(ix) ∫ dx dk eku gS % 1
2
x +1
(A) tan −1 x + c (B)
1
2
( )
log 1 + x 2 + c

(C) tan −1 x 2 + c (D) buesa ls dksbZ ugha
x
The value of ∫ dx is :
x2 +1
(A) tan −1 x + c (B)
1
2
(
log 1 + x 2 + c )
(C) tan −1 x 2 + c (D) None of these
2
d y dy
(x) = 1+ vodyu lehdj.k dh ?kkr gS % 1
dx 2 dx

(A) 1 (B) 3
(C) 2 (D) buesa ls dksbZ ugha
d 2y dy
The degree of the differential equation = 1+ is :
dx 2 dx
(A) 1 (B) 3
(C) 2 (D) None of these
dy
(xi) vodyu lehdj.k = x 2 + sin 3x dk gy gS % 1
dx

x3
(A) y= + cos 3x + c
3
x
(B) y= − 3 cos 3x + c
3

x3 1
(C) y= − cos 3x + c
3 3
(D) buesa ls dksbZ ugha

3631/(Set : C)

(7) 3631/(Set : C)
dy
Solution of the differential equation = x 2 + sin 3x is :
dx

x3
(A) y= + cos 3x + c
3
x
(B) y= − 3 cos 3x + c
3

x3 1
(C) y= − cos 3x + c
3 3
(D) None of these
(xii) ;fn P(A) = 0.5, P(B) = 0.6 vkSj P(A ∪ B) = 0.8, rks P(A/B) gS % 1
1 3
(A) (B)
2 5
1 1
(C) (D)
3 5
If P(A) = 0.5, P(B) = 0.6 and P(A ∪ B) = 0.8, then P(A/B) is :
1 3
(A) (B)
2 5
1 1
(C) (D)
3 5
(xiii) vPNh rjg QsaVh x;h rk'k dh xM~Mh ls nks iÙks yxkrkj fcuk çfrLFkkiu ds fudkys x;s
gSaA çR;sd ds gqdqe (Spade) dk iÙkk gksus dh çkf;drk gS % 1
1 2
(A) (B)
13 13
1 2
(C) (D)
17 17
The probability of drawing a spade on each of the two consecutive
draws from a well-shuffled pack of cards without replacement is :
1 2
(A) (B)
13 13
1 2
(C) (D)
17 17

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

(8) 3631/(Set : C)
3
(xiv) ;fn A vkSj B nks LorU= ?kVuk,¡ bl çdkj gSa fd P (A ) = vkSj P(B) =
5
1
, rks P(A ∩ B) gS % 1
5

1 3
(A) (B)
25 25

2
(C) (D) buesa ls dksbZ ugha
25
3
If A and B are two independent events such that P(A) = and P(B)
5
1
= , then P(A ∩ B) is :
5
1 3
(A) (B)
25 25
2
(C) (D) None of these
25

→
(xv) ;fn | a→| = → → →
3 , | b | = 2 vkSj a . b = 3, rks a
→
vkSj b ds chp dk dks.k gS
% 1

π π
(A) (B)
2 3
π π
(C) (D)
4 6

→ → → → → →
If | a | = 3 , | b | = 2 and a . b = 3, then angle between a and b
is :

3631/(Set : C)

(9) 3631/(Set : C)
π π
(A) (B)
2 3

π π
(C) (D)
4 6
(xvi) nks fcUnqvksa (−2, 4, −5) vkSj (1, 2, 3) dks feykus okyh js[kk dk fnd~-dksT;k gS % 1
3 2 8
(A) , ,
77 77 77
3 −2 8
(B) , ,
77 77 77
3 2 −8
(C) , ,
77 77 77
(D) buesa ls dksbZ ugha
The direction cosine of a line joining the two points (−2, 4, −5) and
(1, 2, 3) are :
3 2 8
(A) , ,
77 77 77
3 −2 8
(B) , ,
77 77 77
3 2 −8
(C) , ,
77 77 77
(D) None of these

[k.M – c
SECTION – B

2. n'kkZb, fd f : N → N, fn;k gS % 2

x + 1 , ;fn x fo"ke gS ,dSdh gSA
f (x ) = 
x − 1 , ;fn x le gS
Show that f : N → N, given by :
x + 1 , if x is odd
f (x ) =  is one-one.
x − 1 , if x is even

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

( 10 ) 3631/(Set : C)
3. fl) dhft, % 2
π
sec −1 x + cosec −1x =
2
Prove that :
π
sec −1 x + cosec −1x =
2

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

5. f=Hkqt dk {ks=Qy Kkr dhft, ftlds 'kh"kZ (−3, 1), (2, −4) vkSj (5, 1) gSaA
2
Find the area of the triangle whose vertices are (−3, 1), (2, −4) and (5,
1).
−1 x
6. x sin dk x ds lkis{k vodyt Kkr dhft,A 2
−1 x
Find the derivative of x sin w. r. t. x.
dy
7. Kkr dhft,] tcfd θ = π ij x = a(θ + sinθ), y = a(1 − cosθ)A 2
dx 2
dy π
Find , when x = a(θ + sinθ), y = a(1 − cosθ) at θ = .
dx 2
8. eku Kkr dhft, % 2
−1
∫ cos x dx
Evaluate :
−1
∫ cos x dx

9. eku Kkr dhft, % 2
dx
∫ 9 + 25x 2
Evaluate :
dx
∫ 9 + 25x 2 .
3631/(Set : C)

( 11 ) 3631/(Set : C)
dy
10. vodyu lehdj.k xy = x 2 − y 2 dks gy dhft,A 2
dx
Solve the differential equation :
dy
xy = x2 − y2
dx
11. ,d FkSys esa 5 lQsn] 7 yky vkSj 8 dkyh xsansa gSaA çfrLFkkiu ds lkFk ,d-,d djds pkj xsan
fudkyh x;haA de ls de ,d lQsn gksus dh çkf;drk D;k gS \ 2
A bag contains 5 white, 7 red and 8 black balls. Four balls are drawn one by
one with replacement. What is the probability that at least one is white ?

[k.M – l
SECTION – C
12. fl) dhft, % 4
 1 + sin x + 1 − sin x  x
cot −1  = .
 1 + sin x − 1 − sin x  2
Prove that :
 1 + sin x + 1 − sin x  x
cot −1  = .
 1 + sin x − 1 − sin x  2

 2 −1 ; fn
13. n'kkZb, fd f (x ) = x x ≥1
] x = 1 ij larr gS ijUrq O;qRik| ugha gSA 4
 1 − x ; fn
x <1
x 2 − 1 if x ≥ 1
Show that f (x ) =  is continuous but not derivable at x
 1 − x if x < 1
= 1.

x2 y2
14. n'kkZb, fd (x 0 , y 0 ) ij vfrijoy; − =1 dh Li'kZjs[kk dk lehdj.k
a2 b2
xx 0 yy 0
− = 1 gSA 4
a2 b2

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

( 12 ) 3631/(Set : C)
x2 y2
Show that the equation of tangent to the hyperbola − = 1 at
a2 b2
xx 0 yy 0
(x 0 , y 0 ) is − = 1.
a2 b2

15. ,d FkSys esa 4 lQsn vkSj 6 yky xsansa gSaA FkSys ls 4 xsansa ;kn`PN;k fudkyh x;h gSaA lQsn xsanksa
dh la[;k dk çkf;drk caVu Kkr dhft,A 4
An urn contains 4 white and 6 red balls. Four balls are drawn at
random from the urn. Find the probability distribution of the number
of white balls.
→ →
16. a = iˆ + 2 ˆj − kˆ vkSj b = 2iˆ + 3 ˆj + kˆ ij vfHkyEc ek=d lfn'k Kkr dhft,A 4
→
Find a unit vector perpendicular to the a = iˆ + 2 ˆj − kˆ and
→
b = 2iˆ + 3 ˆj + kˆ .
[k.M – n
SECTION – D
17. fuEu lehdj.kksa dks vkO;wg fof/k }kjk gy dhft, % 6
2x − y + z = − 1,
−x + 2y − z = 4,
x − y + 2z = −3.
Solve the following equations by matrix method :
2x − y + z = − 1,
−x + 2y − z = 4,
x − y + 2z = −3.

18. ijoy; y 2 = 4ax vkSj x 2 = 4ay, a > 0 ds chp ds {ks= dk {ks=Qy Kkr dhft,A 6
Find the area of region included between the parabola y 2 = 4ax and
x 2 = 4ay, a > 0 .
vFkok
OR
eku Kkr dhft, %
3631/(Set : C)

( 13 ) 3631/(Set : C)
π /2
x
∫ sin x + cos x dx
0

Evaluate :
π /2
x
∫ sin x + cos x dx
0

19. fcUnq (0, 2, 7) ls js[kk x + 1 = y − 1 = z − 3 ij vfHkyEc dk ikn (foot) Kkr dhft,A
−1 3 −2
6

Find the foot of perpendicular from the point (0, 2, 7) to the line
x +1 y −1 z − 3
= = .
−1 3 −2
vFkok
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 = 2x + 3y
O;ojks/kksa ds vUrxZr %
x + y ≤ 4,
3x + y ≥ 4,
x + 5y ≥ 4,
x ≤ 3, y ≤ 3, x, y ≥ 0.
Solve graphically the following L. P. P. :
Minimize : Z = 2x + 3y
subject to constraints :
x + y ≤ 4,
3x + y ≥ 4,

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

( 14 ) 3631/(Set : C)
x + 5y ≥ 4,
x ≤ 3, y ≤ 3, x, y ≥ 0.

s

3631/(Set : C)