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