HBSE Class 12 Question Paper 2018 Maths

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

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 : A) P. T. O.

(2) 3631/(Set : A)
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 : A)

(3) 3631/(Set : A)
(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 (gof )(x) dk eku gS % 1

(A) e 1+ x (B) 1+x

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

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

 3
(ii) sin cos −1  dk eku gS % 1
 5

4 3
(A) (B)
5 5
2
(C) (D) buesa ls dksbZ ugha
5

 3
The value of sin cos −1  is :
 5

4 3
(A) (B)
5 5
2
(C) (D) None of these
5

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

(4) 3631/(Set : A)
4 2 3 
vkSj B = 
1 3 7
(iii) ;fn A =    ] rks 2A + B gS % 1
1 5 7 0 4 1 
9 7 13 9 13 7 
(A)   (B)  
2 14 15 2 14 15
7 9 13
(C) 2 14 (D) buesa ls dksbZ ugha
 15 
4 2 3 1 3 7
If A =   and B =   ] then 2A + B is :
1 5 7 0 4 1 
9 7 13 9 13 7 
(A)  (B)
2 14 15 2 14 15
 
7 9 13
(C)  (D) None of these
2 14 15 
3 −4
(iv) ;fn = 3, rks m dk eku gS % 1
m 5
(A) 3 (B) 4
(C) −3 (D) buesa ls dksbZ ugha
3 −4
If = 3, then value of m is :
m 5
(A) 3 (B) 4
(C) −3 (D) None of these
(v) sin x 3 dk x ds lkis{k vodyt gS % 1

(A) cos x 3 (B) 3x 2 cos x 3

(C) 3x 2 cos x (D) buesa ls dksbZ ugha
Derivative of sin x 3 w. r. t. x is :
(A) cos x 3 (B) 3x 2 cos x 3
(C) 3x 2 cos x (D) None of these

3631/(Set : A)

(5) 3631/(Set : A)
(vi) Qyu f(x) = sin3x + 4 dk vf/kdre vkSj fuEure eku Øe'k% gS % 1
(A) 5 vkSj 3 (B) 6 vkSj 4
(C) 4 vkSj 3 (D) buesa ls dksbZ ugha
The maximum and minimum value of function f(x) = sin3x + 4 are
respectively :
(A) 5 and 3 (B) 6 and 4
(C) 4 and 3 (D) None of these
π
(vii) oØ x = a cos 3 θ, y = a sin 3 θ dh θ = ij Li'kZ- js[kk dh ço.krk gS % 1
4
(A) 1 (B) 2
(C) −1 (D) buesa ls dksbZ ugha
π
The slope of tangent to the curve x = a cos 3 θ, y = a sin 3 θ at θ = is
4
:
(A) 1 (B) 2
(C) −1 (D) None of these

(viii) ∫ tan 2 x dx dk eku gS % 1

(A) tan x − x + c (B) cot x − x + c
(C) sec x − x + c (D) buesa ls dksbZ ugha
The value of ∫ tan 2 x dx is :
(A) tan x − x + c (B) cot x − x + c
(C) sec x − x + c (D) None of these
3x
(ix) ∫ dx dk eku gS % 1
1 + 2x 4
(A)
3
2
(
tan −1 2x 2 + c)
(B)
3
2 2
(
tan −1 2x 2 + c )
3631/(Set : A) P. T. O.

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

(A) 2 (B) 3
(C) 1 (D) buesa ls dksbZ ugha
2
d 2y  dy  dy
The degree of the differential equation xy + x  −y =0
dx 2  dx  dx
is :
(A) 2 (B) 3
(C) 1 (D) None of these
dy
(xi) = tan 2 x vodyu lehdj.k dk gy gS % 1
dx
(A) y = tan x − x + c
(B) y = cot x − x + c
(C) y = sec x − x + c
(D) buesa ls dksbZ ugha
dy
Solution of the differential equation = tan 2 x is :
dx
3631/(Set : A)

(7) 3631/(Set : A)
(A) y = tan x − x + c
(B) y = cot x − x + c
(C) y = sec x − x + c
(D) None of these
7 9
(xii) ;fn P ( A ) = , P (B ) = vkSj P (A ∩ B ) = 4 , rks P(A/B) gS % 1
13 13 13
2 5
(A) (B)
9 9
4
(C) (D) buesa ls dksbZ ugha
9
7 9 4
If P ( A ) = , P (B ) = and P (A ∩ B ) = , then P(A/B) is :
13 13 13
2 5
(A) (B)
9 9
4
(C) (D) None of these
9
(xiii) rk'k dh 52 iÙkksa dh xM~Mh ls ,d iÙkk fudkyk x;k vkSj fQj fcuk cnys nwljk iÙkk
fudkyk x;k gSA fudkys x;s nksuksa iÙks csxe gksus dh çkf;drk gS % 1
1 1
(A) (B)
17 221
1
(C) (D) buesa ls dksbZ ugha
13
A card is drawn from a pack of 52 cards and then a second card is
drawn without replacement. The probability that both cards drawn
are queens is :
1 1
(A) (B)
17 221
1
(C) (D) None of these
13

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

(8) 3631/(Set : A)
(xiv) ;fn A vkSj B nks LorU= ?kVuk,¡ bl çdkj gSa fd P(A ∪ B) = 0.60 vkSj
P(A) = 0.2, rks P(B) dk eku gS % 1
(A) 0.5 (B) 0.6
(C) 0.7 (D) buesa ls dksbZ ugha
If A and B are two independent events such that P(A ∪ B) = 0.60
and P(A) = 0.2, then P(B) is :

(A) 0.5 (B) 0.6

(C) 0.7 (D) None of these
→ →
(xv) a = iˆ + 2 ˆj − 3kˆ vkSj b = 3iˆ − ˆj + 2kˆ lfn'kksa ds chp dk dks.k gS % 1

 5   9 
(A) cos −1   (B) cos −1  
 14   14 
−5
(C) cos −1   (D) buesa ls dksbZ ugha
 14 
→ →
The angle between the vector a = iˆ + 2 ˆj − 3kˆ and b = 3iˆ − ˆj + 2kˆ is :

 5   9 
(A) cos −1   (B) cos −1  
 14   14 
−5
(C) cos −1   (D) None of these
 14 

(xvi) ;fn nks js[kkvksa ds fnd~-dksT;kvksa ds vuqikr 4, 3, 2 vkSj 1, −2, 1 gS]a rks js[kkvksa ds
chp dk dks.k gS % 1

(A) 90° (B) 60°

(C) 45° (D) buesa ls dksbZ ugha
If direction cosines of two lines are proportional to 4, 3, 2 and 1,
−2, 1, then the angle between the lines is :
(A) 90° (B) 60°
3631/(Set : A)

(9) 3631/(Set : A)
(C) 45° (D) None of these

[k.M – c
SECTION – B

1 , x >0

2. n'kkZb, fd f (x ) =  0 , x = 0 ,dSdh ugha gSA 2
− 1 , x < 0

1 , x >0

Show that f (x ) =  0 , x = 0 is not one-one.
− 1 , x < 0


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

;fn A = 
3 1
4. , rc f (A) Kkr dhft,] tgk¡ f (x ) = x 2 − 5x + 7 . 2
− 1 2
 3 1
If A =  , then find f(A), where f (x ) = x 2 − 5x + 7 .
− 1 2
5. f=Hkqt dk {ks=Qy Kkr dhft, ftlds 'kh"kZ (0, 0), (−2, 3) vkSj (10, 7) gSaA 2
Find the area of the triangle whose vertices are (0, 0), (−2, 3) and (10,
7).

6. (tan x )cot x dk x ds lkis{k vodyt Kkr dhft,A 2

Find the derivative of (tan x )cot x w. r. t. x.

dy
7. Kkr dhft,] tcfd x = e 2t . cos t , y = e 2t . sin t . 2
dx

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

( 10 ) 3631/(Set : A)
dy
Find , when x = e 2t . cos t , y = e 2t . sin t .
dx

8. eku Kkr dhft, % 2

−1
∫ tan x dx

Evaluate :
−1
∫ tan x dx .

9. eku Kkr dhft, % 2

dx
∫ 9x 2 − 1
Evaluate :
dx
∫ 9x 2 − 1 .
10. vody lehdj.k (3xy + y 2 ) dx + (x 2 + xy ) dy = 0 dks gy dhft,A 2
Solve the differential equation :
(3xy + y 2 ) dx + (x 2 + xy ) 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 x;h gSA çkf;drk Kkr dhft, fd nksuksa yky gksAa 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 red.

[k.M – l
SECTION – C
12. fl) dhft, % 4

3631/(Set : A)

( 11 ) 3631/(Set : A)
 1 + x 2 − 1 1
tan −1   = tan −1 x ; x ≠ 0
 x  2
 
Prove that :
 1 + x 2 − 1 1
tan −1   = tan −1 x ; x ≠ 0
 x  2
 

13. n'kkZb, fd x = 2 ij Qyu f(x) = |x − 2|, x ∈ R larr gS ijUrq vodyuh; ugha gSA 4
Show that the function f(x) = |x − 2|, x ∈ R is continuous but not
differentiable at x = 2.
14. t fcUnq ij oØ x = a sin 3 t, y = b cos 3 t dh Li'kZjs[kk dk lehdj.k Kkr dhft,A 4
Find the equation of tangent at the point t to the curve x = a sin 3 t, y =
b cos 3 t.
15. ,d flDds dh pkj mNkyksa esa iVksa (tails) dh la[;k dk çkf;drk caVu Kkr dhft,A 4
Find the probability distribution of the number of tails in four tosses of
a coin.

16. f=Hkqt dk {ks=Qy Kkr dhft, ftlds 'kh"kZ (1, 2, 3), (2, 5, −1), (−1, 1, 2) gSaA
4
Find the area of triangle whose vertices are (1, 2, 3), (2, 5, −1),
(−1, 1, 2).

[k.M – n
SECTION – D

17. fuEu lehdj.kksa dks vkO;wg fof/k }kjk gy dhft, % 6

x + 2y − 3z = − 4,
2x + 3y + 2z = 2,
3x − 3y − 4z = 11.

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

( 12 ) 3631/(Set : A)
Solve the following equations by matrix method :
x + 2y − 3z = − 4,
2x + 3y + 2z = 2,
3x − 3y − 4z = 11.

18. js[kk y = x + 2 vkSj oØ y = 1 x 2 + 2 ls f?kjs {ks= dk {ks=Qy Kkr dhft,A 6
3

Find the area enclosed between the straight line y = x + 2 and the curve
1
y = x2 + 2.
3
vFkok
OR
eku Kkr dhft, %
π
x
∫ 1 + sin2 x dx
0

Evaluate :

π
x
∫ 1 + sin2 x dx
0

19. fcUnq (3, −1, 11) ls js[kk x = y − 2 = z − 3 ij yEc dk lehdj.k Kkr dhft,A yEc
2 3 4
dk ikn Hkh Kkr dhft,A 6

Find the equation of the perpendicular from the point (3, −1, 11) to the
x y −2 z −3
line = = . Also find the foot of perpendicular.
2 3 4

3631/(Set : A)

( 13 ) 3631/(Set : A)
vFkok
OR

fcUnqvksa (2, 1, 0), (3, −2, −2) vkSj (3, 1, 7) ls xqtjus okys ry (plane) dk lehdj.k
Kkr dhft,A
Find the equation of the plane passing through the points (2, 1, 0), (3,
−2, −2) and (3, 1, 7).

20. fuEu L.P.P. dks xzkQh; fof/k }kjk gy dhft, % 6
U;wure % Z = 18x + 10y
O;ojks/kksa ds vUrxZr %
4x + y ≥ 20,
2x + 3y ≥ 30,
x, y ≥ 0.
Solve graphically the following L. P. P. :
Minimize : Z = 18x + 10y
subject to constraints :
4x + y ≥ 20,
2x + 3y ≥ 30,
x, y ≥ 0.

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