Maths paper class 12 maths paper class 12

Time Allowed: 2 hours Maximum Marks: 40
General Instructions:
1. This question paper contains three sections – A, B and C. Each part is compulsory.
2. Section - A has 6 short answer type (SA1) questions of 2 marks each.
3. Section – B has 4 short answer type (SA2) questions of 3 marks each.
4. Section - C has 4 long answer-type questions (LA) of 4 marks each.
5. There is an internal choice in some of the questions.
6. Q 14 is a case-based problem having 2 sub-parts of 2 marks each.
Section A
Section B
1. Evaluate the integral: [2]
OR
Prove that:
∫ dx
1
( +2)( +5)
x
2
x
2
dx =
∫
0
π/2
cot x
√
( + )
tan x
√ cot x
√
π
4
2. Find the differential equation of the family of all straight lines. [2]
3. If and represent two adjacent sides of a parallelogram, then write vectors representing its
diagonals
[2]
a⃗ b
⃗
4. Find the vector equation f the plane passing through the point (1,1,1) and parallel to the plane
.
[2]
⋅ (2 − + 2 ) = 5
r ⃗ i
^
j
^
k
^
5. A speaks truth in 60% of the cases and B in 90% of the cases. In what percentage of cases are
they likely to agree in stating the same fact?
[2]
6. Determine P(E|F): A dice is thrown three times.E : 4 appears on the third toss, F : 6 and 5
appears respectively on first two tosses.
[2]
7. Evaluate: [3]
∫
π/2
0
dx
(1+ x)
cos
2
8. Solve the initial value problem: (y4 - 2x3 y) dx + (x4 - 2xy3) dy = 0, y(1) = 1 [3]
OR
Find the particular solution of the differential equation x(1 + y2)dx - y(1 + x2)dy = 0, given that y =
1, when x = 0.
9. If makes equal angles with the coordinate axes and has magnitude 3, find the angle
between and each of the three coordinate axes.
[3]
a
→
a
→
10. Find the shortest distance between the lines whose vector equations are [3]
= (1 − t) + (t − 2) + (3 − 2t)
r ⃗ i
^ j
^ k
^
TERM II
Sample Question Paper - 8
Mathematics (041)
Class- XII, Session: 2021-22

Section C
CASE-BASED/DATA-BASED
OR
using vectors, find the value of x such that the four points A(x, 5, -1), B(3, 2, 1), C(4, 5, 5) and D(4, 2,
– 2) are coplanar.
= (s + 1) + (2s − 1) − (2s + 1)
r ⃗ i
^ j
^ k
^
11. Prove that: .
[4]
log(tan x + cot x)dx
∫
0
π/2
= π(log 2)
12. Prove that the curves y2 = 4x and x2 = 4y divide the area of the square bounded by x = 0, x = 4,
y = 4 and y = 0 into three equal parts.
[4]
OR
Using integration, find the area of the region given below:
.
{(x, y) : 0 ⩽ y ⩽ + 1, 0 ⩽ y ⩽ x + 1, 0 ⩽ x ⩽ 2}
x
2
13. Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, -4,
-5) and B(2, -3, 1) intersects the plane 2x + y + z = 7.
[4]
14. In an office three employees Govind, Priyanka and Tahseen process incoming copies of a
certain form. Govind process 50% of the forms, Priyanka processes 20% and Tahseen the
remaining 30% of the forms. Govind has an error rate of 0.06, Priyanka has an error rate of
0.04 and Tahseen has an error rate of 0.03.
Based on the above information, answer the following questions.
i. The manager of the company wants to do a quality check. During inspection he selects a
form at random from the days output of processed forms. If the form selected at random
has an error, the probability that the form is NOT processed by Govind is
ii. Let A be the event of committing an error in processing the form and let E1, E2 and E3 be
the events that Govind, Priyanka and Tahseen processed the form. The value of
?
[4]
P ( ∣ A)
∑
i=1
3
Ei

Section A
1. Here we have,
Put x2 = t
Let
1 = A(t + 5) + B(t + 2)
Putting t = -5
1 = B(-5 + 2)
Putting t = -2
1 = A(-2 + 5) + B 0
OR
Let
.......... (i)
Using theorem of definite integral
, we have
........ (ii)
Adding eq.(i) and (ii)
I = ∫
dx
( +2)( +5)
x
2
x
2
∴ =
1
( +2)( +5)
x
2
x
2
1
(t+2)(t+5)
= +
1
(t+2)(t+5)
A
t+2
B
t+5
⇒ =
1
(t+2)(t+5)
A(t+5)+B(t+2)
(t+2)(t+5)
⇒
∴
⇒ B = −
1
3
∴ ×
⇒ A =
1
3
∴ I = ∫ − ∫
1
3
dx
+2
x
2
1
3
dx
+5
x
2
= ∫ − ∫
1
3
dx
+(
x
2
2
√ )
2
1
3
dx
+(
x
2
5
√ )
2
= ( ) − ( ) + C
1
3 2
√
tan
−1 x
2
√
1
3 5
√
tan
−1 x
5
√
y = dx
∫
0
π/2
cot x
√
( + )
tan x
√ cot x
√
y = dx
∫
π
2
0
cos x
sin x
√
+
sin x
cos x
cos x
sin x
√
√
y = dx
∫
π
2
0
cos x
sin x+cos x
f(x)dx = f(a + b − x)dx
∫
b
a
∫
b
a
y = dx
∫
π/2
0
cos( −x)
π
2
sin( −x)+cos( −x)
π
2
π
2
y = dx
∫
π/2
0
sin x
sin x+cos x
2y = dx
∫
π/2
0
cos x
sin x+cos x
+ dx
∫
π/2
0
sin x
sin x+cos x
2y = dx
∫
π/2
0
sin x+cos x
sin x+cos x
2y = 1dx
∫
π/2
0
2y = (x)
π
2
0
y =
π
4
Solution
MATHEMATICS 041
Class 12 - Mathematics

2. The general equation of the family of all straight lines is given by y = mx + c, where m and c are parameters.
Now,we have to solve y = mx + c = m
= 0
Therefore, the required differential equation is = 0
3. Given that represent the two adjacent sides of a parallelogram
In , using triangle law, we get
In using triangle law, we get
4. Position vector of the point (1,1,1) is
Any plane parallel to the plane is given by
since it passes through the point having position vector ,we have
2 - 1 + 2 + d = 0 d = 3
Therefore,the required equation of plane
.
5. Let E be the event that A speaks truth and F be the event that B speaks truth. Therefore, E and F are
independent events such that,
P(E) = and P(F) =
A and B will agree in stating the same fact in the following mutually exclusive ways:
i. A and B both speak truth
ii. A and B both tell a lie.
Therefore, required probability is given by,
P (A and B agree) =
=
= =
Therefore, A and B will agree in 58% cases.
6. Since a dice has six faces. Therefore
E = (1, 2, 3, 4, 5, 6) x (1, 2, 3, 4, 5, 6) x (4)
F = (6) x (5) x (1, 2, 3, 4, 5, 6)
= 1 x 1 x 6 = 6
⇒
dy
dx
⇒
y
d
2
dx
2
y
d
2
dx
2
,
a⃗ b
⃗
△ABC
+ =
AB
−
→
−
BC
−
→
−
AC
−
→
−
+ =
a⃗ b
⃗
AC
−
→
−
△ABD,
+ =
AD
−
→
−
DB
−
→
−
AB
−
→
−
+ =
b
⃗
DB
−
→
−
a⃗
− =
a⃗ b
⃗
DB
−
→
−
∴ Diagonals = +
AC
−
→
−
a⃗ b
⃗
= −
DB
−
→
−
a⃗ b
⃗
= + +
a⃗ i
^
j
^
k
^
⋅ (2 − + 2 ) − 5 = 0
r ⃗ i
^
j
^
k
^
⋅ (2 − + 2 ) + d = 0
r ⃗ i
^
j
^
k
^
+ + ∴ ( + + ) ⋅ (2 − + 2 ) + d = 0
i
^ j
^ k
^
i
^ j
^ k
^
i
^ j
^ k
^
⇒ ⇒
⋅ (2 − + 2 ) − 3 = 0
r ⃗ ı
^ ȷ
^ k
^
⇒ ⋅ (2i − + 2 ) = 3
r ⃗ ȷ
^ k
^
=
60
100
3
5
=
90
100
9
10
P ((E ∩ F ) ∪ ( ∩ ))
E
¯
F
¯
P (E ∩ F ) + P ( ∩ )
E
¯
F
¯
P (E)P (F ) + P ( )P ( )
E
¯
F
¯
× + × = =
3
5
9
10
2
5
1
10
29
50
58
100
n (S) = 6 × 6 × 6 = 216
⇒ n (F )

P (F) =
= (6, 5, 4)
= 1
And
Section B
7. Let I =
Dividing by cos2x in numerator and denominator, we get
I =
Consider I =
Put, tan x = t
sec2xdx = dt
I =
=
Let t = tan
= tan x
I =
=
=
Here, a = 1 and b =
Hence,
I =
8. The given differential equation is,
(y4 - 2x3 y) dx + (x4 - 2xy3) dy = 0
It is a homogeneous equation
Put y = vx and
So,
...(a)
1 - 2v2 = A(v3 + 1) + Bv(v2 - v + 1) + (Cv + D) (v2 + v)
1 - 2v2 = v3 (A + B + C) + v2 (-B + C + D) + v(B + D) + A
Comparing coefficients of like power of v,
A = 1 ...(i)
B + D = 0 ...(ii)
-B + C + D = 0 ...(iii)
A + B + C = -2 ...(iv)
=
n(F)
n(S)
6
216
∴ E ∩ F
n (E ∩ F )
∴ P (E ∩ F ) = =
n(E∩F)
n(S)
1
216
P (E|F ) = = =
P(E∩F)
P(F)
1/216
6/216
1
6
dx
∫
π
2
0
1
1+ x
cos
2
dx =
∫
π
2
0
x
sec
2
x+ x
sec
2
tan
2
dx
∫
π
2
0
x
sec
2
1+2 x
tan
2
dx
∫
π
2
0
x
sec
2
+ x
a
2
b
2
tan
2
⇒
dt
∫
π
2
0
1
+
a
2
b
2
t
2
dt
1
b
2
∫
π
2
0
1
+
a2
b
2
t
2
a
b
θ
dθ
1
b
2
∫
π
2
0
θ
a
b
sec
2
+ θ
a2
b
2
a2
b
2
tan
2
θ =
1
ab
( tan x)
1
ab
tan
−1 b
a
∣
∣
π
2
0
π
2ab
2
–
√
π
2 2
√
=
dy
dx
2 y−
x
3
y
4
−2x
x
4
y
3
x + v =
dv
dx
dy
dx
x + v =
dv
dx
2 vx−
x
3
v
4
x
3
−2
x
4
xv
3
x
3
x = − v
dv
dx
2v−v
4
1−2v
3
x =
dv
dx
+v
v
3
1−2v
3
dv =
1−2v
3
+v
v
3
dx
x
∫ dv = ∫ dx
1−2v
3
+v
v
3
1
x
= + +
1−2v
3
v(v+1)( −v+1)
v
2
A
v
B
v+1
Cv+D
−v+1
v
2

Solving eq. (i), (ii), (iii) and (iv) we get
A = 1, B = -1, C = -2, D = 1
Using eq. (a)
log v - log (v + 1) - log (v2 - v + 1) = log xc
log = log cx
= cx
Using the value of v, we get
At, x = 1, and y = 1, we have
OR
Given differential equation -
x(1 + y2) dx - y(1 + x2) dy = 0
x(1 + y2) dx = y(1 + x2)dy
Therefore,on separating the variables, we get,
On integrating both sides, we get
...(i)
Also, given that y = 1, when x = 0.
On substituting the values of x and y in Eq. (i),
we get,
On putting in Eq. (i), we get
which is the required particular solution of given differential equation.
9. Let and let A be the angle between and each of the coordinate axes.
Then, A is the angle between and each one of , and
cos A = a1 = 3 cos A [ = a1, = 1]
Similarly, a2 = 3 cos A and a3 = 3 cos A.
∫ dv − ∫ dv = ∫ dv = ∫ dx
1
v
1
v+1
2v−1
−v+1
v
2
1
x
( )
v
+1
v
3
( )
v
+1
v
3
= cx
y
+
x
3
y
3
c =
1
2
∴ =
y
+
x
3
y
3
x
2
⇒
dy = dx
y
(1+ )
y
2
x
(1+ )
x
2
∫ dy = ∫ dx
y
1+y
2
x
(1+ )
x
2
⇒ log 1 + = log 1 + + C
1
2
∣
∣ y
2
∣
∣
1
2
∣
∣ x
2
∣
∣
⎡
⎣
⎢
⎢
⎢
⎢
⎢
put 1 + = u ⇒ 2ydy = du
y
2
then ∫ dy = ∫ du = log |u|
y
1+y
2
1
2u
1
2
and put 1 + = v ⇒ 2xdx = dv
x
2
then ∫ dx = ∫ dv = log |v|
x
1+x
2
1
2
1
v
1
2
⎤
⎦
⎥
⎥
⎥
⎥
⎥
log |1 + (1 = log 1 + (0 | + C
1
2
)
2
∣
∣
1
2
∣
∣ )
2
⇒ log 2 = C
1
2
[∴ log 1 = 0]
C = log 2
1
2
log 1 + = log 1 + + log 2
1
2
∣
∣ y
2
∣
∣
1
2
∣
∣ x
2
∣
∣
1
2
⇒ log 1 + = log 1 + + log 2
∣
∣ y
2
∣
∣ ∣
∣ x
2
∣
∣
⇒ log 1 + − log 1 + = log 2
∣
∣ y
2
∣
∣ ∣
∣ x
2
∣
∣
⇒ log = log 2
∣
∣
1+y
2
1+x
2
∣
∣
[∵ log m − log n = log ]
m
n
⇒ = 2
1+y
2
1+x
2
⇒ 1 + = 2 + 2
y
2
x
2
⇒ − 2 − 1 = 0
y
2
x
2
= + +
a
→
a1
i
^ a2
j
^ a3
k
^ a
→
a
→
i
→
j
→
∴
→ ∧
= ⇒
a ⋅ i
| || |
a⃗ i
^
a1
3
∵ ⋅
a
→
i
^ | | = 3, | |
a⃗ i
^

Now, 2 = 9
9
9cos2 A + 9 cos2 A + 9 cos2 A= 9
27 cos2 A = 9
cos2 A =
cos A =
10.
units.
OR
A(x, 5, –1), B(3, 2, 1), C(4, 5, 5), D(4, 2, –2)
i.e., (x - 3) (-9) -3(-7) -2(-3) = 0
x = 6
Section C
11.
Let the given integral be, y =
Let, I = .... (i)
Use King theorem of definite integral
| | = 3, | |
a⃗ a
→
⇒ + +
a
2
1
a
2
2
a
2
3
⇒
⇒
⇒
1
3
⇒
1
3
√
⇒ A = ( )
cos
−1 1
3
√
= − 2 + 3 + t (− + − 2 )
r ⃗ i
^ j
^ k
^
i
^ j
^ k
^
= − − + s ( + 2 − 2 )
r ⃗ i
^ j
^ k
^
i
^ j
^ k
^
= − 2 + 3
a⃗1 i
^
j
^
k
^
= − + − 2
b
⃗
1 i
^
j
^
k
^
= − −
a⃗2
i
^ j
^ k
^
= + 2 − 2
b
⃗
2
i
^ j
^ k
^
− = − 4
a⃗2
a⃗1
j
^ k
^
× =
b
⃗
1 b
⃗
2
∣
∣
∣
∣
∣
i
^
−1
1
j
^
1
2
k
^
−2
−2
∣
∣
∣
∣
∣
= 2 − 4 − 3
i
^ j
^ k
^
× =
∣
∣
b
⃗
1 b
⃗
2
∣
∣
+ +
(2)
2
(−4)
2
(−3)
2
− −
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
√
= 29
−
−
√
( × ). ( − ) = (2 − 4 − 3 ). ( − 4 ) = −4 + 12 = 8
b
⃗
1 b
⃗
2 a⃗2 a⃗1 i
^
j
^
k
^
j
^
k
^
d = =
∣
∣
∣
( − ).( × )
a⃗
2 a⃗
1 b
⃗
1 b
⃗
2
×
∣
∣
b
⃗
1 b
⃗
2
∣
∣
∣
∣
∣
8
29
√
= (x − 3) + 3 − 2
BA
−
→
−
i
^ j
^ k
^
= + 3 + 4
BC
−
→
−
li
^
j
^
k
^
= 1 + 0 − 3
BD
−
→
−
i
^
j
^
k
^
⎫
⎭
⎬
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
= 0
∣
∣
∣
∣
x − 3
1
1
3
3
0
−2
4
−3
∣
∣
∣
∣
y = log( + )dx
∫
π
2
0
sin x
cos x
cos x
sin x
y = log( + )dx
∫
π
2
0
sin x
cos x
cos x
sin x
− ( log sin xdx + log cos xdx)
∫
π
2
0
∫
π
2
0
log sin xdx
∫
π
2
0
f(x)dx
∫
b
a
= f(a + b − x)dx
∫
b
a
I = log sin( − x)dx
∫
π
2
0
π
2
I = log cos xdx
∫
π
2
0

Adding eq.(1) and eq.(2)
Let, 2x = t
2 dx = dt
At x = 0, t = 0
At x = , t =
Similarly,
y =
Hence proved.
12. The given curves are y2 = 4x and x2 = 4y
Let OABC be the square whose sides are represented by following equations
Equation of OA is y = 0
Equation of AB is x = 4
Equation of BC is y = 4
Equation of CO is x = 0
On solving equations y2 = 4x and x2 = 4y, we get A(0, 0) and B(4, 4) as their points of intersection.
The Area bounded by these curves
sq units
Hence, area bounded by curves y2 = 4x and x = 4y is sq units ......(i)
2I = log sin xdx
∫
π
2
0
+ log cos xdx
∫
π
2
0
2I = log dx
∫
π
2
0
2 sin x cos x
2
2I = log sin 2x − log 2dx
∫
π
2
0
⇒
π
2
π
2I = log sin tdt − log 2
1
2
∫
π
0
π
2
2I = log sin xdx − log 2
2
2
∫
π
2
0
π
2
2I = I − log 2
π
2
I = log sin xdx = − log 2
∫
π
2
0
π
2
log cos xdx = − log 2
∫
π
2
0
π
2
− ( log sin xdx + log cos xdx)
∫
π
2
0
∫
π
2
0
y = log 2 + log 2
π
2
π
2
y = π log 2
= [ − ] dx
∫
4
0
y( parabola =4x)
y
2 y( parabola =4y)
x
2
= (2 − ) dx
∫
4
0
x
−
−
√
x
2
4
= [2 ⋅ − ]
2
3
x
3/2 x
3
12
4
0
= [ − ]
4
3
x
3/2 x
3
12
4
0
= ⋅ (4 −
4
3
)
3/2 64
12
= ⋅ −
4
3
( )
2
2
3/2
64
12
= ⋅ (2 −
4
3
)
3 64
12
= −
32
3
16
3
=
16
3
16
3

Area bounded by curve x2 = 4y and the lines x = 0, x = 4 and X-axis
= sq units ........(ii)
The area bounded by curve y2 = 4x, the lies y = 0, y = 4 and Y-axis
= sq units .......(iii)
From Equations. (i), (ii) and (iii), area bounded by the parabolas y2 = 4x and x2 = 4y divides the area of square
into three equal parts.
OR
.....(1)
y = x + 1.......(2)
Solving (1) and(2),we get, x = 1 and y = 2.
Area
13. The direction ratios of line joining are
The equation of line passing through (3, -4, -5) and having Direction ratios (-1, 1, 6) is given by
Suppose
The general point on the line is given by
Line intersect the plane . So,
General point on the line satisfy the equation of plane.
The point of intersection of line and plane is
.
Distance between (3, 4, 4) and (1, -2, 7)
CASE-BASED/DATA-BASED
= dx
∫
4
0
y( parabola =4y)
x
2
= dx
∫
4
0
x
2
4
= [ ]
x
3
12
4
0
=
64
12
16
3
= dy
∫
4
0
x( parabola =4x)
y
2
= dy
∫
4
0
y
2
4
= [ ]
y
3
12
4
0
=
64
12
16
3
y = + 1
x
2
= ( + 1)dx + (x + 1)dx
∫
1
0
x
2
∫
2
1
= [ + x + [ + x
x
3
3
]
1
0
x
2
2
]
2
1
= [( + 1) − 0] + [(2 + 2) − ( + 1)]
1
3
1
2
=
23
6
A(3, −4, −5) and B(2, −3, 1)
[(2 − 3), (−3 + 4), (1 + 5)] = (−1, 1, 6)
= =
x−3
−1
y+4
1
z+5
6
[∵ − − ]
x−x1
a
y−y
b
z−z1
c
= = = λ( say )
x−3
−1
y+4
1
z+5
6
⇒ x = −λ + 3, y = λ − 4 and z = 6λ − 5
(3 − λ, λ − 4, 6λ − 5)
2x + y + z = 7
(3 − λ, λ − 4, 6λ − 5)
∴ 2(3 − λ) + λ − 4 + 6λ − 5 = 7
⇒ 6 − 2λ + λ − 4 + 6λ − 5 = 7 ⇒ 5λ = 10
λ = 2
(3 − 2, 2 − 4, 6 × 2 − 5) = (1, −2, 7)
= (3 − 1 + (4 + 2 + (4 − 7
)
2
)
2
)
2
− −
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
√
= = = 7units
4 + 36 + 9
− −
−
−
−
−
−
−
√ 49
−
−
√

14. Let A be the event of commiting an error and E1, E2 and E3 be the events that Govind, Priyanka and Tahseen
processed the form.
i. Using Bayes' theorem, we have
Required probability
ii.
= 1 [ Sum of posterior probabilities is 1]
P ( ∣ A) =
E1
P( )⋅P(A∣ )
E1 E1
P( )⋅P(A∣ )+P( )⋅P(A∣ )+P( )⋅P(A∣ )
E1 E1 E2 E2 E3 E3
= =
0.5×0.06
0.5×0.06+0.2×0.04+0.3×0.03
30
47
∴ = P ( ∣ A)
Ē1
= 1 − P ( ∣ A) = 1 − =
E1
30
47
17
47
P ( ∣ A) = P ( ∣ A) + P ( ∣ A) + P ( ∣ A)
∑
i=1
3
Ei
E1
E2
E3
∵

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