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KENDRIYA VIDYALAYA SANGATHAN
VARANASI REGION
MARKING SCHEME (SECOND PRE-BOARD)
SUB:- MATHS
1) 8,12 1
2)2 π3 1
3) 5×2 14) -15 1
SECTION-B5) f ( x )=x2−3 x+2
f (f ( x ))=(x¿¿2−3 x+2)2−3(x¿¿2−3 x+2)+2¿¿ 1Final answer after correct calculationx4−6 x3+10 x2−3 x 1
6) Let f ( x )=sin−1 √1−x2∧g ( x )=tan−1 √1+x
df (x)dg(x )
=
df (x)dx
dg (x)dx
½
df (x )dx
=−2 ½
dg(x )dx
= 12+x
½
correct answer=−2(2+x ) ½
7) Let x=25∧h=0.05 ½ f ( x )=√x
f ( x+h )=f ( x )+h df (x)dx
½
for evaluating f ( x )=5∧df (x )dx
= 110
½
correct answer=5.005 ½
8) I=∫0
π2
sinxsinx+cosx
dx
I=∫0
π2
cosxsinx+cosx
dx using properties of definite integral ½
adding above
2 I=∫0
π2
sinx+cosxsinx+cosx
dx ½
2 I=π2 ½
I=π4 ½
9) Co-ordinates of the centre (-3,a) ½ Radius of the circle=aEquation of the circle(x+3)2+(y-a)2=a2
After solving it
a=(x+3)2+ y2
2 y 1
After differentiating once required solution is as follows2 y ( x+3 )+[ y2−( x+3 )2] dy
dx=0 ½
10) Area of the parallelogram ¿ 12|a × b| ½
To evaluate (a × b)=−2 i+3 j+7 k 1area of the parallelogram=√62
2 ½
11) zmin=10 x+4 y ½ Where x be the amount of chemical from the supplier S y be the amount of chemical from the supplier T4 x+ y≥ 80 2 x+ y ≥ 60 , x , y≥ 0 1 12
marks¿be given for chil d' s own view
12) tan−1( cosx1+sinx
)
¿ tan−1cos2 x
2−sin2 x
2
(cos( x2 )+sin( x
2 ))2 ½
¿ tan−1cos❑ x
2−sin❑ x
2
(cos( x2 )+sin( x
2 ))❑ ½
¿ tan−1 tan( π4− x
2 ) 1 ½
¿( π4− x
2 )
SECTION-C13) giventhat f ( x )= x
1+|x|
f ( x )= x1+x
if x≥ 0 and f ( x )= x1−x
if x<0 ½
Case-1 if x1 and x2 are positive and prove that function is 1-1 1Case-2 if x1 and x2 are negative and prove that function is 1-1 1For onto:- ∀ y∈ (−1,1 )i . e . rangeof the function
∃ x∈R∧x= y1− y
if y ≥0∧x= y1+ y
if y<0
such that f ( x )= y therefore function is onto 1 ½
ORReflexive:- we know that 4∨0⇒ 4∨(a−a )∀ a∈ A
⇒ (a ,a)∈ R therefore relationis reflexive 1Symmetric :- let (a , b)∈R ⇒ 4∨(a−b )∀ a ,b∈ A ⇒∃m∈Z suchthat 4m= (a−b )∀ a ,b∈ A ⇒ 4(−m)= (b−a )∀ a , b∈ A ⇒ 4∨(b−a )∀ a ,b∈ A ⇒ (b ,a)∈ R 1 ½
similarly for transitivity 1 ½
14) Let 6 x= y
sin−1 y+sin−1√3 y=−π2
⟹ sin−1 √3 y=¿− π2−sin−1 y ¿ ½
⟹√3 y=sin (−π2
−sin−1 y )
⟹√3 y=−cos ¿ ) 1⟹√3 y=−cos ¿ ) ½
⟹ y=± 12 1
but y=1/2 is not acceptable ½
therefore y=−12 and accordingly x=-1/12 ½
OR
L .H . S .=cos ¿
¿cos (tan−1 2/748 /49
) 1
¿cos (tan−1 724
)
¿ 2425 1
R . H . S=sin ¿
¿sin(2 tan−1 34) 1
¿ sin( tan−1 247
) ½
¿ 2425 ½
15) L .H .S=|1+a2−b2 2ab −2 b2ab 1−a2+b2 2a2b −2 a 1−a2−b2|
Applying c1⇢c1−bc3∧c2⇢c2+ac3 we get
¿| 1+a2+b2 0 −2 b0 1+a2+b2 2 a
b (1+a2+b2) −a(1+a2+b2) 1−a2−b2| 1 ½
taking out ( 1+a2+b2 ) common¿c1∧c2 ½
applying R3 → R3−b R1+a R2
¿(1+a2+b2)2|1 0 −2b0 1 2 a0 0 1−a2−b2| 1
after expanding∧findingout the correct result 1
16)LHL= lim
x →−π6
√3 sinx+cosx
x+ π6
¿ limx→− π
6
2( √3 sinx2
+ cosx2
)
x+ π6
1
¿2. limx →−
π6
sin ( π6
+x)
x+ π6
1 ½
¿2 1
Therefore k=2 ½
17) Give that x=a sinpt∧ y=b cospt dxdt
=ap cospt ½
dydt
=−bp sinpt ½
dydx
=−bp sinptap cospt ½
y ' '= ddt (−b sinpt
acospt ) . dtdx ½
y ' '=−b /a2cos3 pt ……………(i) ½ Now consider(a¿¿2−x2) . y=a2 bcos3 pt ¿ …………..(ii) 1multiplying (i )∧(ii ) we get(a¿¿2−x2). y . y = {-b} ^ {2¿ ½ Therefore(a¿¿2−x2) . y . y + {b} ^ {2} =¿ proved .
OR
Consider y=log [ x+√x2+a2 ]
dydx
= 1[ x+√ x2+a2 ] [1+ x
√ x2+a2 ] 1
dydx
= 1[ √x2+a2 ] 1
Again y ' '=−12
(x2+a2)−3 /2× 2 x 1
( x2+a2 ) y ' '=−x dydx 1
( x2+a2 ) y ' '+x dydx
=0
18) Let ( x+1 )=A . dd x
(1−x−x2 )+B ½
x+1=A (−1−2 x )+B
After solving the equation A=−12
∧B=12 ½
x+1=−12
(−1−2 x )+12
∫(x+1)√1−x−x2dx=∫ [−12
(−1−2 x )+ 12 ]√1−x−x2dx ½
¿∫−12
(−1−2 x ) √1−x− x2 dx+∫ 12 √1−x−x2dx
¿∫−1
2(−1−2 x ) √1−x−x2 dx⏟
+∫ 12 √1−x−x2dx⏟
Consider ∫−12
(−1−2x ) √1−x−x2dx⏟
Put t=1-x-x2 and dt = (1-2x)dx.
∫−12
(−1−2 x ) √1−x−x2dx=−12 ∫ √t dt=−1
2( t3 /2
3/2)
¿−13(1−x−x2)
32……………………………………………(i) 1
Again ∫ 12 √1−x−x2 dx=∫ 1
2√¿¿¿
¿18
(2 x+1 )√1−x−x2+ 516
sin−1 2 x+1√5
……………………..(ii) 1
adding (i )∧(ii ) we get required answer ½
OR
∫ x2+1(x2+4)(x¿¿2+25)dx
¿
put y=x2
x2+1
(x2+4)(x¿¿2+25)= y+1( y+4 ) ( y+25 )
¿ ¿A
y+4+ B
y+25
1
After solving A=−17
∧B=87 1
Now y+1
( y+4 ) ( y+25 ) ¿ −1/7
y+4+ 8/7
y+25
x2+1
(x2+4)(x¿¿2+25)= −1/7
(x¿¿2+4)+ 8/7(x¿¿2+25)¿
¿¿
½
∫ x2+1(x2+4)(x¿¿2+25)dx
¿ =∫ −1/7(x¿¿2+4)dx
+∫ 8 /7(x¿¿2+25)dx
¿¿
¿− 114
tan−1 x2+ 8
35tan−1 x
5+C 1 ½
19) I=∫0
π /2 xsinx . cosxsin4 x+cos4 x
dx
I=∫0
π /2 ( π2−x)sinx . cosx
sin4 x+cos4 xdx ½
Adding above
2 I=π2 ∫
0
π /2 sinx . cosxsin4 x+cos4 x
dx ½
¿ π2 ∫
0
π /2 sinx . cosxcos4 x
sin 4 xcos4 x
+ cos4 xcos4 x
dx ½
¿ π2 ∫
0
π /2 tanx . sec2 xtan 4 x+1
dx ½
put tan2 x=t
tanx . sec2 xdx=12
dt and limit changes ¿0¿∞ 1
2 I=π4∫0
∞ 11+t2 dt
¿ π4
¿ ½
¿ π2
8
I= π2
16 ½
20) The given differential equation is( x+ y ) dy+ ( x− y ) dx=0
dydx
=−x− yx+ y
For proving D.E. is homogeneous 1
Put y=vx
dydx
=v+x dvdx ½
Substituting above in the given D.E.,
We get
v+x dvdx
= v−1v+1
x dvdx
=−1+v2
1+v
1+v1+v2 dv=−dx
x ½
Integration both the sides of the equation we get
∫ 1+v1+v2 dv=∫−dx
x
∫ 11+v2 dv+¿∫ v
1+v2 dv=¿∫−dxx
¿¿
tan−1 v+12
log (1+v2 )=−log ( x )+c 1 ½
Now change the answer in x and y we get required answer
log ( x2+ y2 )+2 tan−1 yx=c ½
21) Give that
[ a⃗ , b⃗ , c⃗ , ]=0 ½
consider
[ a⃗+ b⃗ , b⃗+c⃗ , c⃗+ a⃗ , ]=( a⃗+b⃗ ) .¿ ½
¿ (a⃗+ b⃗ ) .¿ ½
= a⃗ . (b⃗ × c⃗ )+a⃗ . (b⃗× a⃗ )+a⃗ . ( c⃗× a⃗ )+ b⃗ . ( b⃗× c⃗ )+b⃗ . (b⃗× a⃗ )+b⃗ . ( c⃗ ×a⃗ ) 1
¿ [ a⃗ , b⃗ , c⃗ , ]+[ b⃗ , c⃗ , a⃗ , ] 1
¿ [ a⃗ , b⃗ , c⃗ , ]+[ a⃗ , b⃗ , c⃗ , ] ¿2 [ a⃗ , b⃗ , c⃗ , ]=0 ½
22) Let the direction ratios of the required line is a,b and c
Equation of the line passing through the given point
x−1a
= y−2b
= z+4c 1
since the line is perpendicular to the lines
x−83
= y+19−16
=
z−107
∧x−15
3= y−29
8= z−5
−5
Therefore
3 a−16 b+7c=0∧3 a+8b−5 c=0 1
After solving above two equations we get
a80−56
= b21+15
= c24+48
=k
½
a24
= b36
= c72
=k
a2=b
3= c
6=k ½
Substituting the values∈therequired equation
x−12
= y−23
= z+46 ½
Change the following equation in vector form ½
23) Let E1¿ first drawn ball is white
E2¿ first drawn ball is not white
E=second ball drawnis white 1
P ¿E1)¿6
10∧P ( E 2 )= 4
10 ½
P( EE 1 )=5
9∧P( E
E 2 )=69 1
The required probability
P( E 1E )=
P(E 1)× P( EE 1
)
P ( E1 )× P( EE 1 )+P(E 2)× P( E
E 2)
½
after substituting the above values in above we get
P( E 1E )=5
9 ½
SECTION-D
24) [−4 4 4−7 1 35 −3 −1][1 −1 1
1 −2 −22 1 3 ]=[8 0 0
0 8 00 0 8] ¿8 Ι 1 ½
inverse of [1 −1 11 −2 −22 1 3 ]=1
8 [−4 4 4−7 1 35 −3 −1] 1
To express the system in matrix form AX=B
Where A=[1 −1 11 −2 −22 1 3 ]∧B=[491 ] 1
X=[ xyz ]=A−1 . B=
18 [−4 4 4
−7 1 35 −3 −1] [491 ] 1 ½
¿ [ 3−21 ] therefore x=3, y = -2 and z =1 1
25) Diagram ½
Let 2x be the length and y be the width of the window
Then radius of the opening = x
Since the perimeter of the window is 10 m
2x+y+y+2 πx
2=10
y=10−x(π+2)2
……………………(i) 1
hence A=2 xy+12
π x2
A=2 x (10−x ( π+2 )2 )+ 1
2π x2 1
differentiating above we get
r
hO
r
y
Bx
D
A
C
dAdx
=10−2 x ( π+2 )+πx
¿10−πx−4 x 1
put dAdx
=0
x= 10π+4 1
again differentiating d2 Ad x2 =−π−4
Which is negative 1
Therefore function attain its maxima for the above value of x
Substituting the value of x in (i) we get
y= 10π+4 ½
OR
Diagram
½
Let the radius of the cone be x and height be y
h= y+r ………….(i)
and x2=r2− y2 ………(ii) ½
volume of the cone V=13
π r2 h
V=13
π (r2− y2 )( y+r ) using (i) and (ii) 1
Differentiating above we get
dVdy
=π3
[ (r2− y2 )+ ( y+r ) . (−2 y )]
=π3
[ (r+ y ) (r− y−2 y )]
¿ π3
[ (r+ y ) (r−3 y )] 1 ½
dVdy
=0 ⟹ (r+ y ) (r−3 y )=0
Therefore r=− y∧r=3 y ½
Therefore h=4 r3
d2Vd x2 =
π3
(r+ y ) (−3 )+(r−3 y )
d2Vd x2 (at y= r
3 )=−4 π3
. r3
.3+(r−r )=−4 πr3
1
Which is negative
Therefore volume of the cone is maximum when altitude =4r/3
And volume of the cone ¿π3
¿
¿ π3
¿
¿ 827
( 43
π r3)
¿ 827
(volume of the sphere) 1
26) Let X denote the number of red balls drawn
X=0,1,2,3,4 , ½
Probability of red ball=2/3
probility of no red ball=13 ½
P ( X=0 )=4C 0¿
P ( X=1 )=4 C1 ¿
P ( X=2 )=4 C2 ¿
P ( X=3 )=4C3 ¿
P ( X=4 )=4 C4 (23)
4
=1681
2
Probability distribution of X is given by ½
X 0 1 2 3 4
P(x) 1/81 8/81 24/81 32/81 16/81
mean=∑ X . P( X)
¿ 83 1
variance=∑ X2 P ( X )−mean2
¿ 64881
−¿ 1 ½
27)
Product A(x) B(y) Available time (hours)
I 3 hr 2 hr 12
II 3 hr 1 hr 9
profit 7/- per item 4/- per item
½
Let x items of product A and y items of Product B be manufactured
Z=7 x+4 y ½
3 x+2 y ≤12 ½
3 x+ y≤ 9 ½
x , y ≥ 0
Correct graph 1 ½
Correct shading 1
Corner point Z=7x+4y
1 (0,6) =24
2 (2,3) =26 maximum
3 (3,0) =21
4 (0,0) 0
Therefore x=2 and y=3 is the correct answer for maximum profit 1 ½
28)
Let P' (x , y , z) be the image of the point P(1,0,0)
Let M be the foot of perpendicular PM on the line AB
Given equation of the line x1= y
2= z
3=∝ 1
x=∝ , y=2∝ , z=3∝ 1
Co-ordinates of Point M (∝ ,2∝, 3∝¿
DR’s of PM are ∝−1,2∝ ,3∝ 1
and DR’s of AB is 1,2,3 and PM is perpendicular to AB therefore.
a 1a 2+b1 b2+c1 c 2=0
1(∝−1¿+2 (2∝ )+3 (3∝ )=0
∝= 114 1
Thus the co−ordinates of the point M ( 114
, 17
, 314
) ½
Now since M is the mid point of PP ' therefore
( x+12
, y2
, z2 )=( 1
14, 17
, 314
)
x=−67
, y=27∧z=3
7
therefore required image¿6/7,2/7,3/7) 1 ½
OR
P (1,0,0)
P’ (x,y,z)
MA B
O
A
BM
Equation of the plane containing the point (1,-1,2) is
a ( x−1 )+b ( y+1 )+c ( z−2 )=0 …………………(i) 1
Where a,b and c are the direction ratios normal to the plane
Since plane (i) is perpendicular to the given planes therefore
2a+3b−2c=0 ……………………..(ii) ½
a+2b−3 c=0 ……………………….(iii) ½
From equation (ii) and (iii) we get
a−9+4
= b−2+6
= c4−3
a−5
=b4= c
1=∝(say)
a=−5∝ , b=4∝∧c=∝ 1 ½
Substituting above values of a,b and c in equation (i)
We get −5∝ ( x−1 )+4∝ ( y+1 )+∝ ( z−2 )=0
5 x−4 y−z=7 1
Which is the required equation of the plane
Now let d be the distance between the plane and the point (-2,5,5)
d=|5 × (−2 )+ (−4 )× 5−5−7
√52+(−4 )2+(−1)2 | 1
d=√42 ½
29) Diagram 1
Areaof the shaded region=2 ( areaOAMO )+2(area MABM )
givenequatio ns are 4 x2+4 y2=9 ………….(i)
And y2=4 x ……………….(ii)
Solving equation (i) and(ii) we get the co-ordinates of the point of intersection
x=12∧x=−9
2 1 ½
Similarly for x=12
the value of y=±√2
¿ for x=−92
the value of y doesnot exist
Area of OAMO +areaof MABM=∫0
1/2
2√x dx+∫1/2
3/2
√¿¿¿¿ 1
¿¿ 1
¿ 43
¿ ½
¿ 43
. 12√2
+ 9 π16
−√24
−98
sin−1 13
¿ 23√2
+ 9 π16
−√24
−98
sin−1 13
½
Required area = 2(Area of OAMO +areaof MABM ¿
¿2( 23√2
+ 9 π16
−√24
−98
sin−1 13)
¿ √26
+ 9π8
− 94
sin−1 13
½