Class 12 Sample Paper 2022 Solution Maths Term 2

Solution of Practice Paper
Term -II (2021-22)
Class – XII
Mathematics (Code: 041)

VALUE POINTS
Q. No.

SECTION – A
1
Let I=∫ ( 1+1+xx )dx ∫( 1+1x )dx ∫( 1+xx )dx
2 = 2 + 2 -------(1)

( )
∫ 1+xx2 dx=tan−1 x [formula]

to evaluate ∫ ( )
x
1+ x
2
1
dx put 1+x 2=t on differentiating both sides we get 2xdx=dt or xdx= dt
2

( )
∫ 1+xx2 dx =∫( 12 dtt )= 12 log|t|= 12 log|1+ x 2
1
= log1+ x 2
2

1
Putting these values in equation( 1 )we have I= tan −1 x + log1+ x 2+c
2

OR

SOL-∫ ( 3 x − 6+6 −1
( x − 2)
2 ) (
dx =∫
3 ( x − 2 ) +5
( x − 2)
2
)
dx =3log|x-2||+5 ∫ ( x − 2 )−2 dx

( x −2 )−1 5
=3log|x-2|+5 +c=3log|x-2|- +c
−1 x−2
2 Given differential equation is
d dy 3
{
dx dx ( ) }=0

=>3 ( ) ( )
dy 3−1 d dy
dx dx dx
=0

( )
2
dy 2 d y
=>3 =0
dx dx 2

2
d y
clearly the highest order derivative occurring in the differential equation is 2 so its order is 2. Also it
dx
2
d y
is a polynomial equation in derivative and the highest power raised to is one so its degree is one .
dx 2
Hence the sum of the order and degree of the above given differential equation is 2+1=3

3 Sol-Let a^ and b^ be unit vectors such that a^ +b^ is also a unit vector
|a^ |=|b^ |=|a^ +b^ |----------------(1)
2 2 2
we know that |a^ + b^| =( a^ + b^ ) =( a^ )2 + ( b^ ) +2 a^ b^ -----------(2)
2 2 2
|a^ − b^| =( a^ − b^ ) =-( a^ )2+ ( b^ ) −2 a^ b^ -------(3)
adding equations (2) and (3) we have
2 2 2 2
|a^ +b^| +|a^ − b^| =2( a^ )2+2 ( b^ ) =|a^| +2|b^|
2

putting the values of |a^||b^||a^ + b^|(each=1)
2 2
1 +|a^ − b^| =2|a^| +2|b^| =2(1)+2(1)
2 2

|a^ − b^| =3 =>|a^ − b^|=√ 3
2

4 8 −4 3
, ,
√ 89 √ 89 √ 89
5
Sol-
Three balls are drawn one by one without replacement from a bag containing 5 whit and 4 green
balls .
Let X denote number of green balls (out of three green balls drawn)
=> X=0,1,2 and 3 only(and not upto 4 , the number of green balls in the bag
P(X=0)= Probability of getting green no green balls in the three draws i.e., all the three white
5 4 3 60 5
balls =P(WWW)= x x = =
9 8 7 504 42
P(X=1)= Probability of getting one green balls (and hence two white balls ) in three draws
4 5 4 5 4 4 5 4 4
=P(GWW)+P(WGW)+P(WWG)= x x + x x + x x
9 8 7 9 8 7 9 8 7
240 20
= =
504 42
similarly P(X=2)= Probability of getting two green balls (and hence one white ball ) in three
15
draws=
42
and similarly P(X=3)= Probability of getting all the three green balls =
4 3 2 2
P(GGG)= x x =
9 8 7 42
Hence probability distribution of X is
X 0 1 2 3

P(X) 5 20 15 15
42 42 42 42

6 Solution- There are 26 red cards and 26 black cards in a pack of 52 playing cards .
Let R and B denote the event of drawing red card and black card respectively
Now required probability=P(first card is red and second one is black)+P(first card is black and
second one is red)
=P(R)+ P ( )
B
R
+P(B)P
R
B ( )
26 26 26 26
= x + x
52 51 52 51
26

=
51

SECTION – B

( ) ( ) ( )
7 dx dx dx
Let I=∫ 3 =∫ =∫
2
1+ x +x + x
2
( 1+ x )+ x ( 1+ x ) ( 1+ x ) ( 1+ x 2 )

let ( dx
)=
A BX +C
+
( 1+x ) ( 1+ x ) 1+ X 1+ x 2
2 --------------(1)

Multiply both sides by L.C.M ( 1+ x ) ( 1+x 2 )we get
1= A( 1+ x2 )+(BX+C) ( 1+ x )
OR 1=A+A x 2+Bx+B x 2+C+Cx
Equating coefficients of x 2,x and constant term ,we get
A+B=0
B+C=0
Putting values
A+C=1
1 1 −1
Solving A,B,C we get A= ,C= ,B=-A=
2 2 2
Putting values of A,B,C in (1) we get

( 1
( 1+x ) ( 1+ x )
2 )
1
= 2 + 2 (
−1 1
x+
2 )
1+ X 1+ X 2
1 1 1 x 1 1
= - 2+
2 1+ X 2 1+x 2 1+x 2
1 1 1
Hence I= log|( 1+ x )|- log( 1+ x2 )+ tan−1 x +c
2 4 2

8
The given differential equation is
𝑥𝑑 𝑦 −ydx= √ x 2 + y 2dx
dividing by dx
dy
x -y=
dx
=>x =y+√ x 2 + y 2=> - = √
dy dy y x 2 + y 2
dx dx x x
dy dv
Put y=vx=> =v+x
dx dx
Therefore
dv vx √ x 2 +v 2 x2 dv
v+x - = =>v+x -v=√ 1+v 2
dx x x dx

( dv
) ( ) | √ | √( )
2 2 2 2
=>∫
dx y y =log|Cx|=> y + x +y x
2 =∫ =>log |v+√ 1+v 2|=log|x|+log|c|=>log log + 1+ 2
√ 1+v x x x x x2
=Cx=>y+√ x 2 + y 2=Cx 2
OR

The given differential equation is
dY π
-3ycotx=sin2x given that y=2 when x=
dx 2
dY
Compairing with +Py=Q we have P=3cotx and Q=sin2x
dx

∫ Pdx
=-3∫ cotxdx =-3logsinx=log( sinx )−3
1
IF==∫ e =e log ( sinx ) =( sinx )−3= 3
pdx −3

sin x
The general solution is y(IF.)=∫ Q ( IF )dx+C
1 1
or y 3 =∫ sin 2 x 3 dx+C
sin x sin x
y ( 2 sinxcosx )
3 =∫ dx+C
sin x sin 3 x
cosx cosx
=2∫ 2 dx+C=2∫ dx+C=∫ 2 cosecx cotx dx=-2cosecx+C
sin x sinxcosx
or
y −2
3 = +C
sin x sinx
y=sin 3 x
multiplying every term by LCM=sin 3 x
π
y=-2sin 2 x +csin 3 x To find C putting y=2 when x= (given in (1)
2
2π 3 π
2=-2sin +csin
2 2
or 2=-2+c or c=4 puttung c=4 the required particular solution is
y=-2sin 2 x +4sin 3 x

9 ⃗a X⃗b =⃗c =>⃗c ⊥ ⃗aand⃗c ⊥ ⃗b
(By def of cross product) ----------(1)
similarly ⃗
b x ⃗c =⃗a=>⃗a ⊥ b⃗ and ⃗a ⊥ ⃗c -----------(2)
from (1) and (2) we have⃗ a ,⃗b ,⃗c are mutually at right angles.
Now ⃗a x ⃗b=⃗c (given)=>|⃗a x ⃗b| =|⃗c|
Using (2)|⃗ a||⃗b|sin 900=|⃗c|
|⃗a||⃗b|=|⃗c|-------(3)
b x ⃗c =⃗a=>|⃗b x ⃗c|=|⃗a|
similarly⃗
=>|⃗
b||⃗c|=|⃗a|----(4)
Dividing (3) by(4) ( to eliminate |⃗
b|) we have
|⃗a| |⃗c|
= =>|⃗a|2 =|⃗c|2therefore |⃗a|=|⃗c|------(5)
|⃗c| |⃗a|
dividing (3) by (5) |⃗
b|=1 putting it in (4) we have
|⃗c|=|⃗a|

10 Solution-Here ⃗ ^ ^j− 2 k^ and ⃗
b1=i+2 ^ ^j− 4 k^ =2(i+2
b 2=2 i+4 ^ ^j− 2 k^ )
=2⃗ b1Therefore , the lines are parallel now using the formula for distance between parallel lines =
|( ⃗a2 −⃗a1) × ⃗b|
|⃗b|
Shortest distance is 2√5
OR
Sol-
Given point is ⃗ ^ ^j− 4 k^ we know that vector along the line r⃗ =
a1=(1 ,2 ,-4 ) =i+2

i+2 ^ ^j+6 k^ )is ⃗
^ ^j − 4 k^ =λ ( 2 i+3 b1= ( 2 i^ +3 ^j+6 ^k )and a vector along the line ⃗ ^ ^j − k^ )
b 2=( i+
we know vector equation of the plane passing through point ⃗a and parallel to the two given lines is (r⃗
-⃗
a1).⃗n =0
where ⃗n =⃗b1x⃗
b 2------------------(1)
now ⃗n =⃗b1x⃗
b2

|
i^ ^j k^
=2 3 6
1 1 −1 |
=i^ (-3-6)- ^j(-2-6)+ ^k (2-3)=− 9 i+8 ^ ^j − k^
from equation (1)r⃗ . ⃗n −⃗ a1 . n⃗ =0 => r⃗ . ⃗n =⃗a1 . n⃗ -------(2)
⃗
a
putting values of 1and⃗n in (2) equatoin of required plane is
^ ^j − 4 k^ ).( − 9 i^ +8 ^j− k^ )=-9+16+4
^ ^j − k^ )=( i+2
i.e.,r⃗ .(− 9 i+8
^ ^j − k^ )=11
or r⃗ .(− 9 i+8 (Vector form of the plane )
^ ^ ^ ^ ^
or( x i+ y j+z k ).( − 9 i+8 j− k )=11^
or 9x+8y-z=11
or 9x+8y-z-11=0 (Cartesian form of the plane)

SECTION – C
2
11
Let I=∫ |x − x|dx
3

−1

Again let f(x) = x 3 − x =x( x 2 −1 )= x ( x − 1 )( x+1 )
Now break the limit at x=0,1 (because on putting f(x)=0 we get x=0,1,-1)
It is clear that x 3 − x ⩾ 0 on[-1,0]

3
x − x ⩽ 0 on[0,1]
x 3 − x ⩾ 0 on[1,2]
Hence the interval of the integral can be subdivided as
2 0 1 2

∫|x 3 − x|dx =∫ ( x 3 − x ) dx -∫ ( x3 − x ) dx +∫ ( x3 − x ) dx
−1 −1 0 1
0 1 2
11
==∫ ( x − x ) dx +∫ ( x − x ) dx +∫ ( x − x ) dx =
3 3 3

−1 0 1
4

12 Given equation of the circle is x 2 + y 2 =16 and √ 3 y =x, represents a line through origin.
1
The line y= x intersect the circle .
√3
x2
Therefore x 2 + =16
3
2 2
3 x +x
=16=>4 x 2 =48
3
=>x 2 =12=> x=± 2 √ 3
2 √3
When x=2 √ 3 then y= =2
√3
1
Required area shaded in first quadrant =(Area under the line y= x from x=0 to 2 √ 3) +
√3
( Area under the circle from x=2 √ 3to x=4
2 √3 4
1
=∫ x dx + ∫ √ 16 − x dx
2

0 √3

2 √3
4π
after solving it we get required area = sq units
3

OR

Required area (shaded)
= Area under curve y = x 3 with respect to ‘x’ axis from x = -2 to x = 1

13 x y z
We know that d.r.’s of the line = = are is denominators 2 ,3 ,-6
2 3 −6
therefore d.r.’s of any line parallel to it are also 2 ,3 ,-6 therefore equation of the line through P
x y z
(1 ,-2 ,3 ) and parallel to the given line = = are
2 3 −6
x y z x −1 y+2 z −3
= = = = (= λ)--------(1)
2 3 −6 2 3 −6
let this line meet the given plane x-y+z=5 in the point Q say ) from equation (1) x-1=2 λ y+2=3
λ,z-3=-6 λ=>x=2 λ+1 ,y=3 λ-2
z=-6 λ+3
therefore point Q is Q( 2 λ+1 , 3 λ-2 , -6 λ+3 ) for some real λ
But this point Q lies on the plane x-y+z=5 therefore
(2 λ+1)-(3 λ-2)(-6 λ+3 )=
1 x y z
2 λ+1-3 λ+2-6 λ+3 =5=>-7 λ=-1=> λ= = =
7 2 3 −6

1 9 − 11 15
putting λ= in (3) coordinates of point Q are ( , ) P (I -2, 3)
7 7 7 7
Required distance is PQ

√( ) ( ) ( )
2 2 2
9 − 11 15
−1 + +2 + −3 =1

7 7 7 Q

√( ) ( ) ( )
2 2 3 2 −6 2
7
+
7
+
7

14
√ 4 9 36
+ +
49 49 49
(a)0.9 2.
=
√49
49
=1

(b)0.083