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Second Year Higher Secondary Model substituting , then b=5−2 a , b=1
Examination, February 2023 5. f (x)=x 2−4 x (3)
Mathematics (Science) f ' (x)=2 x−4
Answer Key f ' (x)=0⇒ 2 x=4 x = 2
Answer any 6 questions . Each carries 3 scores. So R divides into two intervals (-∞, 2] and [2, ∞)
a11 a12 a13
1. A=
[ a21 a22 a23 ] (3) f ' (−1)=2(−1)=−2<0
f(x) is decreasing on (-∞, 2]
a11 =2−2=0 a21=4−2=2
f ' (3)=2(3)−4=2>0
a12=2−4=−2 a22=4−4=0 f(x) is increasing on [2, ∞)
a13=2−6=−4 a23=4−6=−2 a
6. A unit vector in the direction of ⃗
A= 0 −2 −4
a
[
2 0 −2 ] is given by a^ =
⃗
a|
|⃗
(3)
2. R={(1,3) ,(2,6) ,(3,9),(4,12)} (3) |⃗a|=√ 12+12 +22= √1+1+4= √6
R is not reflexive since (a , a) ∉ R ^i + ^j+2 k^ 1
a^ = = ^i+ 1 ^j+ 2 k^
for every a ∈ A √6 √ 6 √ 6 √6
R is not symmetric since (1,3) ∈ R 7. Angle between two lines is given by,
b⃗1 . b⃗2
but (3,1) ∉ R .
R is not transitive since (1,3),( 3,9) ∈ R
cos θ =
| |
|b⃗1||b⃗2|
(3)
but (1,9) ∉ R . b⃗1= ^i +2 ^j+2 k^ , b⃗2=3 ^i +2 ^j+6 k^
1 2=2 4 |b⃗1|= √12 +22+ 22=√ 9=3
3. 2 A=2
4 2 8 4[ ][ ] (3)
|b⃗2|= √32 +22 +62 =√ 49=7
|2 A|=8−32=−24
b⃗1 . b⃗2=1.3+2.2+2.6=19
|A|=2−8=−6
19 19
4| A|=4 (−6)=−24 |3.7
cos θ = |= 21
|2 A|=4 |A|
19
θ =cos ( ) −1
4. f(x) is continuous at a, then LHL=RHL=f(a) (3) 21
f(x) is continuous at x= 2 P ( A∩B) P( A). P( B)
8. i) P( A/ B)= = =P( A)
LHL=lim f ( x)=lim (5)=5 P( B) P( B)
– –
x→2 x→2
Therefore P( A/ B)=P( A)=0.3 (1)
RHL=lim f ( x)=lim (ax+ b)=2 a+b
ii) P( A∩B ' )=P ( A ). P(B' ) (2)
✛ ✛
x→2 x→2
Therefore 2 a+b=5 since A and B are independent events.
f(x) is continuous at x= 10 P( B' )=1−P(B)=1−0.6=0.4
LHL= lim f (x)= lim (21)=21 Now ,
x→10
–
x→10
– P( A∩B ')=0.3 x 0.4=0.12
RHL= lim f (x )= lim (ax +b)=10 a+b Answer any 6 questions. Each carries 4 scores.
✛ ✛
x→10 x→10
9. i) y (1)
Therefore 10 a+b=21
ii) If f ( x1 )=f ( x 2) for all x1, x2∈ R (3)
subtracting them , we get 8 a=16 , a=2
then 3−4 x 1=3−4 x 2
Second Year Higher Secondary Model substituting , then b=5−2 a , b=1
Examination, February 2023 5. f (x)=x 2−4 x (3)
Mathematics (Science) f ' (x)=2 x−4
Answer Key f ' (x)=0⇒ 2 x=4 x = 2
Answer any 6 questions . Each carries 3 scores. So R divides into two intervals (-∞, 2] and [2, ∞)
a11 a12 a13
1. A=
[ a21 a22 a23 ] (3) f ' (−1)=2(−1)=−2<0
f(x) is decreasing on (-∞, 2]
a11 =2−2=0 a21=4−2=2
f ' (3)=2(3)−4=2>0
a12=2−4=−2 a22=4−4=0 f(x) is increasing on [2, ∞)
a13=2−6=−4 a23=4−6=−2 a
6. A unit vector in the direction of ⃗
A= 0 −2 −4
a
[
2 0 −2 ] is given by a^ =
⃗
a|
|⃗
(3)
2. R={(1,3) ,(2,6) ,(3,9),(4,12)} (3) |⃗a|=√ 12+12 +22= √1+1+4= √6
R is not reflexive since (a , a) ∉ R ^i + ^j+2 k^ 1
a^ = = ^i+ 1 ^j+ 2 k^
for every a ∈ A √6 √ 6 √ 6 √6
R is not symmetric since (1,3) ∈ R 7. Angle between two lines is given by,
b⃗1 . b⃗2
but (3,1) ∉ R .
R is not transitive since (1,3),( 3,9) ∈ R
cos θ =
| |
|b⃗1||b⃗2|
(3)
but (1,9) ∉ R . b⃗1= ^i +2 ^j+2 k^ , b⃗2=3 ^i +2 ^j+6 k^
1 2=2 4 |b⃗1|= √12 +22+ 22=√ 9=3
3. 2 A=2
4 2 8 4[ ][ ] (3)
|b⃗2|= √32 +22 +62 =√ 49=7
|2 A|=8−32=−24
b⃗1 . b⃗2=1.3+2.2+2.6=19
|A|=2−8=−6
19 19
4| A|=4 (−6)=−24 |3.7
cos θ = |= 21
|2 A|=4 |A|
19
θ =cos ( ) −1
4. f(x) is continuous at a, then LHL=RHL=f(a) (3) 21
f(x) is continuous at x= 2 P ( A∩B) P( A). P( B)
8. i) P( A/ B)= = =P( A)
LHL=lim f ( x)=lim (5)=5 P( B) P( B)
– –
x→2 x→2
Therefore P( A/ B)=P( A)=0.3 (1)
RHL=lim f ( x)=lim (ax+ b)=2 a+b
ii) P( A∩B ' )=P ( A ). P(B' ) (2)
✛ ✛
x→2 x→2
Therefore 2 a+b=5 since A and B are independent events.
f(x) is continuous at x= 10 P( B' )=1−P(B)=1−0.6=0.4
LHL= lim f (x)= lim (21)=21 Now ,
x→10
–
x→10
– P( A∩B ')=0.3 x 0.4=0.12
RHL= lim f (x )= lim (ax +b)=10 a+b Answer any 6 questions. Each carries 4 scores.
✛ ✛
x→10 x→10
9. i) y (1)
Therefore 10 a+b=21
ii) If f ( x1 )=f ( x 2) for all x1, x2∈ R (3)
subtracting them , we get 8 a=16 , a=2
then 3−4 x 1=3−4 x 2
