Adobe Scan 19 Jan 2024
Adobe Scan 19 Jan 2024
Adobe Scan 19 Jan 2024
SAMPLE
QUESTION PAPER6
Mathematics
General nstructions Time:3 hrs Max. Marks:80
Each section is
question paper contains - five sections A, B, C, Dand E.
1.This choices in some questions..
compulsory. However, there are internal mark each.
A has18 MCQ's and 02 Assertion-Reason basd questions of 1
2. Section Answer (VSA)type questions of 2 marks each.
zSection has
B 5 Very Short
Section Chas 6 Short Answer (SA) type questions of 3marks each.
A marks each.
Long Answer (LA) type questions of 5
s Section D has 4
based/case/passage based/intergrated units of assessment
6 Section Ehas 3 source
sub-parts.
(4 marks each) with
ISection A 1mark
Choice Questions) Each question carries
(Multiple
4. is equal to
of the line
1 The cartesian equation Jo 1+ sin x
(c) 3n (d) r/2
(a) (b) 2n
Y=(2i+j)+ ali -j+4k) is disjoint from (-1,1
(a) I-2_y-_: (6)-2_y-2 5. If I be any interval
1 1 1 -1
4
4
the function f givenbyf(x)=x+-is
() -2y+1_2 then
x+2_y+1_2 -1 I
1 1
(a) strictly decreasing on
on /
is (b) strictly increasing
2. Ii P(A)=2 and P(B) =0, then B (c) decreasing on / true
(d) Only (a) and (c) are
equal to of the vector
(b) 0 6. Direction cosines
(a) 1
(c) not defined (d) 0.5 -2i +j -5k are -2 1 -5
2 1 5
independent events, (6) J0so V30
Aand Bare two (a) 30 30 30 -2 1 5
then P(An B) is equal to
(b) P(
-3 1-5
(c)-V30 30 V30 () T30 30 V30
(a) P(A) -P(A)P(B)
(d) P(A)-P(A)P(B)
(c) P(A)-P(A)P(B)
7. The least value of a, such the function f
given by f(x) =x+ax+lis strictly
16. Write the number
of
Mathema
increasing on (1, 2) is vect
perpendicular to both
theors oi
a=2i+j +2k and b=j+i vectorusnite
(a)-1 (b) - 2
(c) 0 (d) 1 (a) 1
(c) 3 (b) 2
8. If(i +3} +9k)x(3i - Ë +uk)=0, then à+ (d) 4
is equal to 17. Area of the
(a) 10 =4x, Y-axis andregion bounded by
(c) 0
(b) 18
(d) 1
the line y=3tishe
(a) 2 sq units 9
31 4 Sq units
9. If A=[2 -3 4], (c)sq units
B=|2, X-[1 2 3] and 3
9
(d) Sq units
2
18. A unit vector in the
Y=|3.then AB +XY is equal to a=2i +j +2k is direction ot vec:
1
(a) [28] (a)
(c) 28 (b) [24]
(d) 24 (b)
10. Abag contains 5 red
and 3 blue 1
drawn at randomballs.
If 3 balls are
replacement, then the without (a) i+j+k
getting
15
exactly15 one red ballprobability
is
of
11.
(a)
36 b)
46 () 5 56 Assertion-Reason
In the
folowing
Based Questions
(kxi).i +,-k is equal to (A) isfollowed by questions,
a
a statement of Assertir
(a)-1
(c) 0 (b) 1 the correct answer out statement of
of the Reason Choos
(R).
27
(d) -2 (a) Both A and R following
are true choices.
and R is the core
12. in x|dx is equal to explanation of A
(b) Both Aand R are true
(a) 1 explanation of A but R is not the
(b) 2 (c) A is
corTet
(c) 3 true
(d) 4 (a) Ais false but R is false
13. If y= Aex +Be-5x ,then d'y but R is true.
19.
is equal to Assertion (A) Scalar matrix
(a) 25y (b) Sy (c) -25y (d) 15y A=la,]= |0; ij
i=j where kis a scalar, is
14. an identity
then xy is equal matrix when k=1.
to Reasonmatrix.
scalar (R) Every
identity matrix is
(a) 1 not
(b) 2 20.
(c)-3 (d) -5 Assertlon (A) The relation Ron the
N×N
15. Let A and B be the
events associated with
the sample space S, then the
defined
a+d=b+ all c,
by (a, b)
for R(c, d)
set
value of (a, b), (c, d) eN
an equivalence relation.
P(A/B) lies in the interval xNiS
(a) (0, 1)
(c) (0, 1]
(b) [0, 1] Reason (R) Any
relationif itis an
equivalence relation,
(d) [o, 1)
symmetric and transitive. is
reflexive,
SectionB
c
sectioncomprises of very short answer type questions (VSA) of 2 marks each)
mis
If
AB=BA for any two square matrices, 23. Show that the points (a +5, a-4),
2.
then prOve by mathematical induction lie on a
that(AB)"=A"B". (u-2, a+3) and (a, a) do not
straight line for any value of a.
Or Or
4 21 then show that are
If(a, b), (a',b) and (a -a',b-b')
-1 1 collincar, then prove that ab'= a'b.
(A-21)(A-31)=0.
24. Find thegeneral solution of the
Findthe position vector of a point R which
differential equation dy e .
22. divides
the line joining the points dx
pi+2i-k) and Q(-i +Ë+k) in the ratio 25. The rcoordinate of a point
on the line
Q(5,1,-2)
2:1
() internally. (ii) externally. joining the points P(2, 2, ) and
is 4.Find its z-coordinate.
area of the
Tect 27. Using integration, find the which the
region bounded by thecurves y=|x+1|+1, 31. Find the value of a, for
x=-3, x=3 and y=0. function
WI+ ar -NI-ar if -1sx<0
ûand b is a
0. It the sum of two unit vectors is
magnitude f(r)= 2.r+1
unit vector, then show that the if 0<r<l
of their difference is 3.
continuous at r=0.
Or Iflä=10, 1~I=2 and a-b=12, then find Or
ota
the value of laxbl Ilx=a(21 -sint) and y = a(1- cost),
then
t 29. Find +x+1_ds. find dy when =.
d:
Vis
(u+1)°(* +2)
Section D
(This section comprises of long answer type questions (LA) of 5
32. Solve the following differential marks each
areZ=-50x+20
equation y, subject to
2x- y2-5,3x+y23,2x-3ysi)
(x +x'+x+)=2x' +x
dx
33. Evaluate p/4 sin x +cos x
and x>0, y20.
Or
constraints
Jo 9+16sin 2x-dx.
Find graphically, the
Or
Z=2x+5y, subject to maximum val.
Evaluate d.
below
constraints
Oacos x+bsin x 2x+4ys8, 3x +ys6,
34. Determine x+ ys4, x20, y 20.
value of thegraphically the minimum
objective function 35. If y=(log x)" +xog*, then find y
dr
SectionE
This section comprises of 3
36. If a real valued case-study/
function f(x) finitely
is
passage-based questions of 4 marks each
derivable at any point of its domain, it is
necessarily need
continuous at that
not be true. point. But
its converse
e.g. Every
are both polynomial,
continuous as constant
well as functions
differentiable and inverse trigonometric
functions are continuous and
differentiable in their domain etc.
Based on the above A relation R on
a set A is
information, answer
the following questions, equivalence relation on A
said to be an
iff it
(i) Write the interval in
which the
I.
Reflexive i.e. aRa or (a, a)e R,is
II. aE A
function f(x) =cos xis always Symmetric
or(a, b)e
i.e. aRb ’ bRa
continuous. III. Transitive R(b,a)e
i.e.
R,where a, be A.
if aRb and
(ii) Show that the or (a,b)eR bRC. then aRc
function and (b, c)e R
3x, forxs0
f(a) =,|0, for x>0 is continuous at ’(a,c)eR,
Based where a, b, ce A.
on the above
X=0. the following questions.
() If the information, answer
(ii) Show that the function f(x)
xE R, is continuous at x=2. =|x-2| (2,2), relation
(2, 3), (3,
R= {(1, 1),(1, 2),
1),(3, 2),(3,3)} (1, 3),
on the set
Or Show that the function f(x) =\cos 2x| R is A={1,2, defined
3}, then show
reflexive but neither
nor transitive.
that
is continuous at x="
4 (ii) If the symmetric
relation R=
(3, 1)} defined on {(1, 2), (2, 1), (1, 3),
37. In a classroom, a teacher teaches a topic
Relation on a set, which is defined below.
then show that Risthe set. A={1, 2, 3},
neither reflexive symmetric but
nor
OStagel
the
If
relation Ron the set Nof all
) naturalnumbers defined as
R={(z,y):y=x+5and x<4}, then
showthat.Risneither reflexive, nor
symmetric nor transitive.
Or
The.relation Rin the set Z of integers
givenby R={(a, b):2 divides (a-b)}, information, answer
showthat Ris an equivalence1relation. Based on the above
the following questions.
Ina.
school,teacher asks aquestion to (i)Find the total probability of
threestudents
8. Ravi, Mohit and Sonia. committing an error insolving the
probability of solving the question question.
The Mohit and Sonia are 30%, 25% i_checked by
Ravi, (ii) If the solution ofquestion then find
and.45%, respectively. The probability of teacher and has some erro,
question is not
makingerror by Ravi, Mohit and Sonia the probability that the
1.2%and 2%,. respectively. solved by Ravi.
are1%.
-2_Y-1-Z 1-sinxdx
1-1 4 Cos*
dr
Which is the required cartesian equation of line.
’ 2/ =sec-x - sec.xtan.x)
2 (c) It is given that 2/ =n[tanx -secxh
0-sec 0))
PA)=and1
P(B)=0 ’ 2/ = [(tan -secI)-(tan
2 ’ 2/=[(0 +1)-(0-1))
P(AnB) 2/ =2r
P(B) 1-1
Since, P(B) =0 5. (b) Given, f(x) = l()=1
is not defined. [on differentiating w.rt. x we get)
>1 then '(x) >0
(a) We have, P(AnB)= P(A)- P(B) If x<-1, then "(x) >0. If x interval / disjoint from
independent events] .f(z)isstrictly increasing on
|Aand Bare (-11):
P(A)-P(A): P(B)
PAnB)= P(A) [1- PIB)) =
6. (b) Let a=-2i+j-sk 10. (c) We have, 5 red and 3 blue bale
Mathernatics Ga;
.. Direction cosines of a are
Probability of getting exactly one
balls are drawn is red bal
-2
V-2+(1+(-5)? 5x3
56
15
56
1
11. (a) Given, (k xj)-i+j:k
N-2+ (1 +(-5
and -5 =(-i)i+j-k
i.e.
-2
/30/30 V30
-2+(1 +(-5°
1
y=3
and XY= [1 2 3|3|=P+6+ 12]= 20] ’X
Or
Ifgiven points are collinear, then
Aas =0, Ag2 =-SIn a, A33 = COs
-1
adjA = 0 -COS C. -Sin a
Mathematicsr.
8
ab' - a'b = 0
--ndr++2)d*
sin a then
0 sin a
|1 73
|A| =|0 cos a sin a
[expand by Ri] 2
|0 sina -cos a
=1.(- cosa -sina)
=(cosa + sina) =-1+0
’Aexists.
To find adj A,
(1)
28. Letc =â+6.
--0-9-|+-(-2)
>44 21 3
Hence, the requiredt=16
2 2 sq units
area is 16 sq units. (1)
and
...) (1) l=[Sint-2)
(i), we get sint
On squaring and adding Eqs. () and sint cos2 a-cost sin2at
laxbi'+(a b?=|a6'sin o+|aBoos'8 sint
29. We have,
+*+1
-dx l=cos 2a (dt - sin2a jcott ot
(r+1(* +2) =tcos2a - sin2alog |sint+C
*+*+1
iS aproper
rational Jcot wdr =log |sina]
The integrand
(*+ 1)'(* +2) sin2a-log |sin (x + )] +C
: l= (r+ a) cos2 a- (::t=+a] (1)
function.
Now, by using partial fraction,
B .) C 30. Let tan = 0,
+x+1 A /3
Let (1)
(*+1(*+2) (*+) (r+ 1) x+2) ’tan 0,=- tan 6
1
tan 0 =
+1
’+x+1= A(x++1)(*+2)+ B(x+2)+C(r »tan0, =
(:: tan(- 0)= - tan 0)
*+x+1=A+3* +2)+B(*+2)
+Clu+2x+1)
*+*+1=(A+C+(3A+B+2C)x
+2A+2B+C)
106 OStage l Proflclency Level
tan
1 T
6
(1) = lim
h-0V1-ah + /
RHL = lim
2a
t ah
Mathermatle
1+ å
f(x)= limt
h--’0
(0 +
1 2h+1 0+1 h-0
Again, let cot0, col, = Ti1 = lim
h-0 h-1 0--1
2(0)+ 1
Now f(0) =
cot , = cot 3 =-1
f(x)is ContinuoUs at x= 0.
. LHL = RHL =f(O)
-1 1
Cot
v3 3
From Egs. (i). (i) arnd (i), we get
a=-1
and lat tan an- Or
Conside, x = a(21 -sint)
tan - sin=, (sin(- 6) =- sin Ondifferentiating w.rt. t.we get
=a2-cost)
tan(-1) =O3
Consicder, y = a(1-cost)
tan , =-1 On differentiating w.rt. 1,we get
tan 8, =- tan dy = asint
V1+ ax -/1-ax
2r*+x
if -1sx<0 It is a variable
31. Given, f(x) =
2x+1
separable type differential equat
x-1
if 0<x<1 dy = 21+
LHL
r=0
= lim0 f(*)= h-0
lim f(0-h) =h-»0
lim f(-h) On
integrating both++t
sides, we get
/1+a(-h)--a-) 2.+x
= lim
h0 -h ++t 1d
V1-ah -1+ ah v1-ah + 1+ ah 21*+
= lim
h-0 -h
X
V1-ah + V1+ ah r+ 1) +1(u + 1)
1-ah-(1+ ah) (r+ 1)(*+
= h0-h(/1-ah
lim + 1+ ah) ..)(1)
Using partial fraction
-2ah
Let 2r+ A
=h-0-h1-ah
lim +1+ ah) (x+ 1)(r + 1) 1+ 1 Bx +C ...0
x²+1
Levef 107
SamplePaper 06
OSlago Profliclency
Alx+ 1) +(Bx + C)(r +1)
(x+ 1)(r+ 1)
2x+x =A(r+
-1)+(Br + C)(r +1)
2x'+x=A(r?2+1)+ B(*+x)+C(*+ 1)
2x+x=(A+ B)x+ A+ (B+C)* +C (1) -4tren!
r a -1ir
Lower limit Whern g
comparngthe coefficients of r2,x and constant
Upper limit When s
on sides, we get
frormboth
(ems A+B=2,B+C=1 Sin -
f=c05
A=-C
A+C=0
Onsohingabove
and equations, we get
1 3 1
8=
and C= (1)
16/1-')
substiutingthe values of ABand Cin Eq. (0).
On
1
21²+r E2 + 2 2
+ 1 241
(r+ 1)( +1) 5+ 4!
get
bothsides,we 5
2xx16
n integrating dx
2:+x -dx
2+1
+ )e+1)
..(i) (1)
1
40 5--9
1
[trom Eq. (0] 40
1 lcg9cç t
40
(log9-iog) ,
dt 2
1
Pt+1=t 2dr =dt ’0dr = 2 log(3
[where, C=C+ Cl
(1)
which is the required solution.
3 Let -4sin x+Cos x
Jo 9+ 16 sin 2x
Sinx + COS * (n
dx
Jo 9+ 16(1 + sin 2x -1)
16 in denominator (0).we get
ladding and subtracting On adhng Eqs 0) aKt
(+ )
Sin x + COS * (1)
2= cos & si
0 9+ 16[1 - (1- sin 2r)]
sin x + COS x
1-(cosx+ sin x a'cosx + b snx
-2sin x COSx)| We know that
" cos²0+sin0=1 [)dr =2 [r)dr, itt2a - x) =f()
cos 0|
|andsin 20 = 2 sin0
108 OStage Proficiency Level
Jo a+ b 2
dt ’
0<12,
So, the half plane is towardswhich
is true.
the oriain
dt Also, x 20 and y0, so
dt
Ist quadrant.
the region lies
On drawingthe graph of eachlinear
1
the following graph. Intirst
has no intersection point. requation,WE
quadrant, these en
tan+c| 2x-y=-5
ab
-[tan o - tan 0] 8
- 5x+2/=-30
-0 tan =tantan tA0.5)
abl2
and tan0= tan(tan 0) =0 3B (0,3) -3y=12
(6.0)
2
../=
2ab
1
c1,0)
(2) [o 6-5 -4 -3/2 -14 1\2 3 466 7 8
34. Given objective function is -2
Minimise Z=-50x + 20y
Subject to constraints, 2x - y2-5 .)
(0,-4) y 3xty=3
3r + y3 ...i) Thus, we get the common shaded
2x-3y s 12 which gives the feasible region ABQ
...i) region and it is unboun
The corner points of
C(1, 0) and D(6 0) feasible region are A(0, 5) B(0 3
and *>0,y 0 ...()
Table for line 2x-y=-5ís The value of Z at corner
points are given below
-5/2 Corner points
5 Z=-50x +20y
(1) A(0, 5)
Z=-50(0) +20(5) =100
So, the line passes through the points BIO, 3) Z=- 50(0) + 20(3) =60
C(1, 0) Z =- 50(1) +
(0, 5)
D(6, 0) 20(0) =- 50
Onputting (0, 0) in the inequality 2x - y 2-5, we get Z=-50(6) +20(0) =- 300
0-02-5 Here, feasible region is
and unbOunded so the minimum
0-5 which is true. maximum
Now, we drawa value may or may not exist.
So, the half plane is towards
the oriain. dotted line of
Table for line 3r +y=
3is Here,
-
we see that50x+20y <-300 inequation
or -5x+2y <-30
- 5* + half plane
region. 30has a point determined
feasible2y<-
1
in by
3
Hence, no minimum value common with io
SamplePaper O6
OStage Proficiency Level 109
the
have.
following LPP The intersecion point of lines () and (i) is E
We
Maximise, Z=2x + 5y
Subjectio COnstraints Thus, the corner points are
8 6)
+2ys4 O0, 0), A(2, 0), B|C(0. 2).
2x+ 4y S8Orx
1)
3r+ ys6
Ihe values of Z at corner points are as TollÛMs
I+yS4
Corner points Z = 2x + 5y
20.y20
(1) Z-0+0=0
and O(0, 0)
consideringtheinequations as equations, we Z -2x2+ 5x0 = 4
NoW. A(2, 0)
get 6 46 -92
x+2y = 4 .) z=2x2+55
3r +y=6 ...0)
*+y=4 C(0, 2) Z=2x0+ 5x2 = 10 (maximum)
..()
and (1)
+2y=4is Hence,the maximum value of Z is 10.
forlinex
TAble
0 35. We have, y=(log x +°9*
Letu = (log and v=09*
0 2
(0, 2)>
6t(0, 6)
3
-(0,4)
=k+x 2l0gx
210g v=9]..()
(i), we get
X From Egs. ().() and
(2)
|log*
(1)
9-b+b-cis divisible by 2.
() Required probability = P(A)
a-c)isdivisible by 2 ’ (a, c)e R 3
transitive, - SPE,) P(A/E,)
So,R is reflexive, symmetric and
transltive. P(A/Eg)
Thus,Ris
an equivalence relation, (1) = P(E;)P(A/E,) + P(Ez) +PIEg) PA/E3)
Hence, Ris
1.2 45 2
30 1 25
100 100
eventthat question has Some error
B Let 100 100 100 100
The (2)
A= event that question is solved by Ravi =0.003+ 0.003 0 009 =0.015
E,=The
eventthat question is solved by Mohit (i) Required probability = 1- P(E,/A)
Ez=The
The event
that question is solved by Sonia
P(E) P(A/E)
=1
Then, we have
25 45
P(E;)P(A/E,) + P(E)P(A/Ez2)
30
P(E2)= 100 P(Es)= 100 + P(E3) P(A/E3)
PIE)= 100
30
1.2 100 100
PIA/E;)= 100 P(A/E2) = 100
=1
30 1 25 1.2 45 2
o100 100100o
100 10010o
2
0.003 =1-0.2= 0.8
and P(A/E3)=100 =1 (2)
0.015