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Class – XII
MATHEMATICS (041)
SQP Marking Scheme (2019-20)
TIME: 3 Hrs
...
25
1
5
(c) (2,3)
1
6
(b)
1
7
(c)
1
8
(b) sin
+c
(a) 0
(b) ⃗ = − ̂ + ̂ +
1
9
10
11
12
13
14
1
3×p
−
1
+ ( ̂+
̂)
= (−2) = 2
2
1
1
1
1
=2
1
1
−3
2
OR
15
decreasing at rate of 72 units/sec
...
e
...
}
y = ae + be ………………(1)
= 2ae − be
…………
...
(3)
putting values on LHS
1
= − − 2y
=(4ae + be ) − (2ae − be ) − 2(ae + be )
=4ae + be − 2ae + be − 2ae − 2be
=0
1
2
23
x = 2y ………(1)
⟹ 2x = 2
(given
dx
dx
⟹ 2x = 2
dt
dt
⟹x=1
from (1)
y=
so point is
24
=
)
1
1
1,
= a⃗ − b⃗
...
b⃗ × c⃗ − b⃗ × a⃗ − c⃗ × c⃗ + c⃗ × a⃗
= a⃗ − b⃗
...
b⃗ × c⃗ + a⃗ × b⃗ + c⃗ × a⃗
= a⃗
...
a⃗ × b⃗ +a⃗
...
b⃗ × c⃗ − b⃗
...
(c⃗ × a⃗)
= a⃗
...
(c⃗ × a⃗)
= a⃗
...
(c⃗ × a⃗)
=0
(STP remains same if vectors a⃗, b⃗ , c⃗ are changed in cyclic order)
1
1
OR
⃗+ ⃗+ ⃗
...
⃗ = 0
...
⃗ = 0
⟹ 3 + 5 + 7 + 2 ⃗
...
⃗ = −(9 + 25 + 49)
83
⟹ ⃗
...
⃗
⃗ ⃗
3ı + 4ȷ + 5k
...
e ∈ R − {2} which is same as given set = R − {2}
(co-domain=range)
Also = f(x)
3( + 1 )
f(x) = f
( − 2)
2
=
(
)
(
(
)
)
(
)
+3
since f(x) =
−3
(
Let
= sin
⟹ 2 cos
,
1
2
2x + 3
x−3
2(3y + 3) + 3(y − 2) 9
=
=
3y + 3 − 3y + 6
9
Thus for every y ∈ B,there exists x ∈ A such that f(x) =
Thus function is onto
...
(
)
( )=
Inverse is given by =
28
1
1
)
1 − x + 1 − y = a(x − y)
= sin
1 − sin + 1 − sin = a(sin − sin )
cos + cos = a(sin − sin )
+
−
+
−
cos
= 2a cos
sin
2
2
2
2
−
−
⟹ cos
= a sin
2
2
1
2
1
4
−
2
⟹ cot
=a
⟹
⟹ − = 2 cot
⟹ sin
− sin
= 2 cot
differentiating w
...
t
...
(1)
y = a(sin 2θ − 2θ cos 2θ)
dy
⇒
= a(2cos 2θ + 4 θ sin 2θ − 2cos 2θ)
dθ
⇒ = a(4 θ sin 2θ)………………(2)
using (1)and (2)
a(4 θ sin 2θ)
⇒
=
a(4θcos 2θ)
sin 2θ
⇒
=
= tan 2θ
cos 2θ
Differentiating again with respect to x, we get
d y
dθ
⇒
=2
2θ
...
dx
a(4θcos 2θ)
d y
π
1
=2
...
r
...
x
⇒
=
1
+
put in (1)
5
⇒
+
=
⇒
+
=
+√
+
x
( + √1 +
⇒
=
+ 1+
⇒
=
1+
1
)
−
⇒
=
√1 +
integrating both sides
⇒
=
√1 +
⇒ log + 1 +
= log + log
⇒ log
= log
1
+ 1+
⇒
⇒
+ 1+
+ 1+
1
2
=
⇒
30
Consider I=∫ |
+
−2 |=
−( − 2 )
( −2 )
−2 |
+∫ |
−2 |
I=∫ −(
−2 )
+∫ (
−2 )
+
I=− − +
=
1
I=∫ |
−
+
−2 |
|
I=−
=
ℎ
ℎ
1≤ <2
2≤ ≤3
1
1
−
+
1
I= = 2
31
Let X denotes the smaller of the two numbers obtained
So X can take values 1,2,3,4,5,6
P(X=1 is smaller number)
P(X=1)=
=
1
2
=
(Total cases when two numbers can be selected from first 7 numbers
are 7 )
P(X=2)=
=
P(X=3)=
=
P(X=4)=
=
P(X=5)=
=
P(X=6)=
=
1
=
2
2
3
4
5
6
6
6
21
6
21
Mean =∑
+
=
4
21
12
21
5
21
10
21
+
+
+
3
21
12
21
+
=
2
21
10
21
1
21
6
21
1
2
1
=
OR
Let
= event of selecting a two headed coin
= event of selecting a biased coin, which shows 75% times Head
=event of selecting a unbiased coin
...
1
∴ ( )= ( )= ( )=
3
(coin
...
e
...
e
...
To check draw 150 + 200 < 1350 i
...
(1)
x + y = 1…………………
...
−0 −
2 2
−
35
Let
1
1−
1
2
square units
1
1
2
1
2
be the radius and ℎbe the height of half cylinder
Volume =
ℎ = (constant)……………
...
(3)
maxima/minima = 0
−1 4
⇒ (2 ) +
+2
⇒ V=
1
⇒ (2
)=
⇒
=
1
=0
1
4 +2
2+
1
………………
...
=
= ℎ ……………………
...
r
...
h
⇒ℎ=
1
⇒
⇒
ℎ
= 12
ℎ
= 18
so Z=
⇒
= 12ℎ − 12ℎ
− 27
3
2
− 12
3
2
=−
is maximum when ℎ =
1
is maximum when ℎ =
when ℎ =
,
= 2ℎ − ℎ = 2
...
⃗lie on required plane
...
ı̂ − 2k = 5
∴normal to given plane ⃗ = ı̂ − 2k lie on required plane
...
Where ⃗ = ( − 2)ı̂ + (y − 1)ȷ̂ + (z − 2)k
⃗ = =2ı̂ − 3ȷ̂ − k
⇒Scaler triple product
⃗
⃗
1
⃗ =0
1
12
−2
−1
−2
2
−3
−1 = 0
1
0
−2
⇒ ( − 2)(6 − 0) − ( − 1)(−4 + 1) + ( − 2)(0 + 3) = 0
⇒ 6 − 12 + 3 − 3 + 3 − 6 = 0
⇒ 2 + + = 7…………………
...
(2)
⇒General point on the line is (2 + 3, −3 + 4,5 + 1)
As line (2) crosses plane (1) so point Q should satisfy equation(1)
∴ 2(2 + 3) + (−3 + 4) + (5 + 1) = 7
4 +6−3 +4+5 +1 = 7
6 = −4
2
=−
3
(− + 3,2 + 4, − + 1)=
, 6, −
1
13