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The document outlines a marking scheme for a Class XII Mathematics exam, detailing solutions and correct answers for multiple-choice questions, very short answer questions, short answer questions, long answer questions, and case study questions. Each question is assigned a specific mark, and the correct answers are indicated for easy reference. The document serves as a comprehensive guide for evaluating student responses in the mathematics examination.
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CLASS-XIL
MATHEMATICS (041)
SECTION: A
(Solution of MCQs of 1 Mark each)
Q.No. Solution Marks
1. xty+z=9, x42 syez=
On solving above equations, we get x=2, y=4, 2=3
nxny+
Correct Answer is Option (c) 1 1
2 A(AdjA) =5T
A(AdjA)=|4II
« |Al=5
| Adj A|=|4
Correct Answer is Option (b) 25 2
3. | Let (AB’ -2B")’ =C
Consider C’ =(AB" 284")! = BA’ ~2AB" #C or -C
©. C isneither Symmetric matrix nor Skew symmetric matrix
Correct Answer is Option (c) Neither Symmetric matrix nor Skew
symmetric matrix 1
4 [Uf 1 x= Correct Answer is Option (b) 1/e 1
18.
y-V/=4e-D
Correct Answer is Option (d) 2, 1, - 6. 1
19. Correct Answer is Option (a)Both A and R true and R is the correct
explanation for A. 1
20. Correct Answer is Option (d)A is false but R is true. 1
Section -B
[This section comprises of solution of very short answer type questions
(VSA) of 2 marks each}
21 | cos *a+cos' B+cos'y=34
= costa = m,cos'B = 2 & cos
a=, -1 1
a(B +7) -B(y +a) +7 (a+)
= CI(-1-1) = (-1)(-1-1) + (-)(-1-1)
=2-242=2 7
oR
x xy]
ond) |
{ 1
sin cos?) |
23) J
42
4/2 22. _ | InisoscelesAABC, let AB = AC = a and BC = b (given),
4a __scm/sec.
di
1
+ Area is decreasing at the rate of 3b cm/sec. 1
OR
Let the side of a cube be x unit,
Volume of cube =V =?
S = 3x24 = & (constant)
ak
ane 1
Surface area = S = 6x*
Sai
a
a)
Hence, the surface area of the cube varies inversely as length of side 1
23. [ashe
a+b
12
12
12
12 24, xe2_ yell
Let P(3A—2,24-1,24 +3) be any point on a line “== 7 which
is at a distance of 5 units from the point Q(1,3,3)
According to the Question
=25
1" +(22-1-3) (2443-3 1
?_342=0
Required Point is -2,-1,3)or (4.3.7) 1
25.
1 Mark
For
Correct
as Figure
x- ~x
Bu =
Required Area= f° 2-dy => 12
“ ng z /
27 49 «
= FenO=F square units ape
Section —C
[This section comprises of solution short answer type questions (SA) of 3
marks each]
26. xey
y
x [0 0 [1 x [010
y | 20 10 [0 y | o [10
Ls
Marks
For
Correct
Figure
Comer points are A (0, 10), B(S, 5), C (15, 15) and D (0, 20) 42 Corner Points Value of Z
‘A, 10) Z=0+90=90
BG,5) Z= 15 +45 - 60 42
C5, 15) Z=45 + 135-180
D (0.20) Z=0+ 180-180
Minimum value of Z = 60 we
cr
n(S)=°C, ws 1s 1
Let X denote the larger of the two numbers obtained
“X=3,4,5,6,7 1/2
The Probability Distribution is
Xx POX)
3 wis
4 2s
5 B/S
6 4s
7 S/S 18
OR
Let P(A)=x and P(B)=y
‘According to the Question
1 1
P(AQB)== and P(A'OB')== 1
(ANB) == (NOB) = +
Paya) = 2 and. P()PB') =}
1 1
xy=4 and(—xy-y)=4 1
Zz add-9-y) = 5
‘on solving we get x=— or i 1
2B. cose
1=| ——— ar
Simao sin x)
cos.x
— pat sinx=¢
(sinx=1)(sinx—
cos x dxdt
1
1 nf
“J
“J
OR
sing
—
bin? ¢+2c0s6+3
sing
yi-cos* +206 +3
sing dg
{4+2cos¢—cos’ g
do
put cos
-sin gdg =at
29.
2yerdr=|
nea
(
dx_| 2xe’~y
dy z
dy
Compare with 2 + Py =
ih Ses Py=0
P=-3cotx , —_ Q=sin2x
1/2 The solution of given differential equation is
y(LF)=f Q(LP)dx
1 1
' =f sin2.
(ar) J sine
=f 2sin.xeos..—t
z ain’
dx
1
a2
30.
1s
Mark
For
Correct
Figure
12
= 9 1
= 5-9 +|(+6)-C-2)
zo r[e- IG |
= 4412=16squareunits 1
31. | x=asin?—beost y=acost +bsint
de wg
=< acost+bsint = y S = -asint +bcost =—x 1
dt dt
dy =x
au 2
ay 12
Section —D
[This section comprises of solution of long answer type questions (LA) of
S marks each]
3, 123
A=|2 3-3
32-4
|A]=1(-12+6)-2(-8-9)+3(4+9) 1
=-6+34439=6720
2
12
12 -3]fx] [4
2.3 2iyl=}a4
3-3 -4ji<} [Hs
AX=B
X=(4) B
=(A'yB
6 17 13-4
t}is 5 sila
as 9 -1}j-1s
244238195
| -s6+70+120
©) 604126415
12
12 33,
Let P(2.+3,2-+3,2)be any point on line
Let the line through origin and making an angle of Sith the given line be
along OP. Then direction ratios are proportional to 22+3—0,2+3-0,4-0
ie, 2A43,A43,2
Also, direction ratios of the given line are proportional to 2,1,1
cos 7 = 2A+9Q)+(A+ D+ OD)
3 faas3y +as3y + 2 +P
61+9
loa? +182 +18J6.
3(24+3)
6a? 43243
=> VP 43043 =(2443)
‘Squaring both sides, we get
2 43A43=(2443)*
= PsBAt
> Aad
Therefore, the coordinates of point P(L,2,
Hence Equations of required lines are
x
Dor P(-1,1,-2)
‘The lines are
x42
1
Let P(.-2,22-+3,42—Ibe any point on line (1) and Q(2u+1,31+2,4u+3)
be any point on line (2). Also, the given point is (11,1).
For some definite values of and jt, the required line passes through
A,Pand Q
The direction ratios of AP are 4—3,22+2,4%-2
The direction ratios of AQ are 2,3,1+1,4u+2
=3_ 242 _ 42-2
“On Sus aps
223242 _M-1
Qu Suet QusT
Qk , 2A+2=3uk-+k , 2
=k (let)
2h
+13
kan42 2413
=h4232=9
Also k=242=11 a
Hence The direction ratios of AP are 6,20,34 i.e. 310,17 aw
‘Therefore, Equation of required line is
3 0 17 we
3 Net r= [73 ae ceseeee (1)
I Seex+tanx
Using fF (x)de=f f(a-a)de
[aa :
hb Seo(m—x)-+tan(n—x)
-x)(-tans)
} =See.x—tanx
2(n=x)tan x
= 1 =f ey 1
b seox stan
2(x+n-s)tanx
1)+(2)a2r= pees
(+2) 20» ane 1
x x
1-2 dx
> 2 Tesins
_mpe_sina(I~sins)
© 240 (1¥sinx)(I-sinx)
==)" ldx
2*\cos*x cos? x.
Bf, (tan xsecx—tan® x)dx
= FI [seextan.x—see? x +t}a
sec x—tan + aJf
(-1-0+z)-(1-0+0)]
1= Ffn-2]
OR
“10g (1+cos x)dx ay
[;log[1+cos(n—x) dx => 1 =f log(1-cosx)de (2)
Adding (1) and (2)
21 = [log{ (1+ cos x)(1-cos x) ]de
=I 4 fft9e(1-cos* x)dx
SIs + {"logsin® xd
" logsin xdx
Since log[sin(n—x) ]=logsinx
1=2{ ogsin.xde 3)
s}ar
= 152 flogeosxae (4)
Adding (3) and (4)
w= f; “log sin xcos xdx
logsin 2vdr—log2 ("Ld
log 2fx]f
x
—Fhog2 5
Files (5)
logsin xd
(Changing 1 to x)
[*logsin xa
From (5) => I= log2
=n log2
35)
‘One — one: Let x,,x, © R, such that
FR) = fF)
9x3 +65,
29x 46x, —
5)=0
es ){9(x, + x,)+6}=0
3 9(x8 x2) +6(,
a
or 9x, +9x,+6=0 which is not possible
+f isone-one
Onto: Let y=9x? +6x—5
= 9x" +6x-(G+y)=0
5 _ 6 POH AOS+Y) _-62 V36JT454y
209) 18
p= AY 6) _ -1t fy F6
3
63)
Now,xeR,=>x20and so x= is rejected
ool dor6
3
Now x20=.
= fy+621= y+621
=> yrs
“R, ={y:y €[-5,%)} =codomain of f.
2 f is onto.
Hence fis one one and onto function.
12
12
Section —E
[This section comprises solution of 3 case- study/passage-based questions
of 4 marks each with two sub parts. Solution of the first two case study
questions have three sub parts (i),(i),(ii) of marks 1,1,2 respectively.
Solution of the third case study question has two sub parts of 2 marks
each.)
36.
VQ = 0 = 3t? + 3t — 100
( No, the above function cannot be used to estimate number of vehicles in
the year 2020 because for 2020 we have t= 0 and
V(0)=0-0+0-100 = ~100
Which is not possible (ii) V(20) = (20)? — 3(20)? + 3(20) — 100
‘Therefore, the estimated number of vehicles in the year 2040 are 6760. 1
Gi)
VW) = 3¢?-6t +3 1
3(t? —2t+1)
= 3(t-1)?=0.
Hence V(t)is always increasing function. 1
37. | Let jis the event that a student is regular
E, is the event that a student is irregular
‘Ais the event that a student attains grade A
PE) = , rey =
100 2100
Pase)=*2 , pase =
100 100
(Required Probability = P(A/ E, a 1
(ii) Required Probability = P(A)
= P(E, )P(A/E,) + P(E: )P(A/ Es)
= 20, 80 70 10 _ at 1
100°100 ~ 100°100 100
(iii) Required Probability = P(B,/ A)
PU )PALE) 1
P(E, )P(A/E,) + P(E, )P(A/
a
100100 ps 1
30 80,70 1031
OR
(iti) Required Probability = P(,/ A)
PCE, )P(AL 1
1
38. | @ Let length, breadth and height of the tank are x,xand y respectively
‘According to the Question
500
u2
u2 For maxima or minima, =
4000
=2
} co 0)
Surface Area is minimum when x=10m
Minimum Surface Area =100 + 22 ~ 300m?
+
(i) If x=10m then y=Sm
and Volume of the tank = x" y = (10)'(5)=500m*
New Volume = (2x) y =4x"y = 4(10)°(S)
Increase in Volume of the tank:
©. % Increase in Volume of the tank=300%
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