model question paper english 7.2.2019 final · 2019-02-11 · 1 higher secondary first year...
TRANSCRIPT
1
HIGHER SECONDARY FIRST YEAR
MATHEMATICS
MODEL QUESTION PAPER - SOLUTIONS
PART - I
(1) (2) 217
(2) (3) { }1R − −
(3) (2) 2
(4) (2) sec 1x <
(5) (1) 0
(6) (1) 45
(7) (3) 1
(8) (4) 4
(9) (2) ( )1, 2−
(10) (1) 3
(11) (4) 25
(12) (2) 60˚
(13) (2) ˆˆ ˆ2i j k− −
(14) (4) Does not exist
(15) (3) 0
(16) (3) 0
(17) (1) 6
(18) (2) log1
x
x +
(19) (4) 3 52
3log 2
x
c+
+
(20) (4) 2
3
PART - II
(21) y = cos x
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2
(22) Let log log logx y z
ky z z x x y
= = =− − −
log x = ( ) ( )k y zk y z x e
−− ⇒ = ... (1)
log y = ( ) ( )k z xk z x y e
−− ⇒ = ... (2)
log z = ( ) ( )k x yk x y z e
−− ⇒ = ... (3)
( ) ( ) ( )1 2 3× ×
xyz = ( ) ( ) ( )k y z k z x k x ye e e
− − −× ×
= ( )k y z z x x ye
− + − + −
= 0 1e = xyz = 1
(23) ( )tan 45 A° − = tan 45 tan
1 tan 45 tan
A
A
° −+ °
= 1 tan
1 tan
A
A
−+
(24) The required no. of ways = 6!
2!
= 6 5 4 3 2!
2!
× × × ×
= 360
(25) S = 4 7 10
15 25 125
+ + + +L
a = 1
1, 3,5
d r= =
s∞ = ( )21 1
a dr
r r+
− −
= 2
13
1 5 3 25 3551 4 5 16 1611 15 5
×+ = + × =
− −
(26) OAuuur
= ˆˆ ˆ2 3 5i j k+ −
OBuuur
= ˆˆ ˆ3 2i j k+ −
OCuuur
= ˆˆ ˆ6 5 7i j k− +
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3
ABuuur
= ˆˆ ˆ2 3OB OA i j k− = − +uuur uuur
BCuuur
= ˆˆ ˆ3 6 9OC OB i j k− = − +uuur uuur
= ( )ˆˆ ˆ3 2 3i j k− +
BCuuur
= 3ABuuur
Therefore, BCuuur
and ABuuur
are parallel vectors but B is the Common point
Hence, A,B,C are collinear
(27) ( )2 16
4
xf x
x
−=
+
( )f x is not defined at 4x = −
Therefore ( )f x is continuous for all { }4x R∈ − −
(28) y = 10 10log log logx x e
e=
dy
dx = 10
1loge
x
dy
dx = 10loge
x
(29) sin
1 cos
xdxx+∫
= logdu
u cu
−= − +∫
= log 1 cos x c− + +
(30) 4 2
1A
x
= −
( )( )2 3A I A I− − = 4 2 2 0 4 2 3 0
1 0 2 1 0 3x x
− × − − −
= 2 2 1 2
1 2 1 3x x
− − − −
Given ( )( )2 3A I A I− − = 0
2 2 1 2
1 2 1 3x x
− − − −
= 0 0
0 0
( )
( ) ( )( )2 2 4 2 3
1 2 2 2 3
x
x x x
− + −
− − − − + − − =
0 0
0 0
2 2x − = 0 x = 1
put
1 cos x u+ =
sin x dx du− =
sin xdx du= −
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4
PART - III
(31) {(1,1) (2,2) ( , )}R n n= L
{1,2,3, }S n= L
R is reflexive since ( , )a a R a S∈ ∀ ∈
R is symmetric since there is no pair ( , )a b R∈ such that ( , )b a R∈
R is transitive since ( , )a b and ( , ) ( , )b c R a c R∉ ⇒ ∉ .
Since R is reflexive, symmetric and transitive
the relation R is an equivalence relation.
(32) L.H.S
= cot(180 ).sin(90 )cos( )
sin(270 ).tan( ).cosec(360 )
θ θ θθ θ θ+ − −
+ − +
= cot cos cos
( cos ) ( tan ) cosec
θ θ θθ θ θ
× ×− × − ×
= 2cos cos sin
cot cossinsin
cos
θ θ θθ θ
θθθ
×× =
= R.H.S
Hence proved.
(33) 5 questions to be attempted out of 9 questions.
2 questions are compulsory.
∴ No. of ways = 37C
= 7 6 5
3 2 1
× ×× ×
= 35 .
(34) 2
3
1, , 10a x b n
x= = =
1rT + = n r r
rnC a b− ×
= 2 10
3
110 ( )
r
r
rC xx
− ×
= 20 2 310 r r
rC x x− −×
1rT + = 20 510 r
rC x−
20 5r− = 15
5r− = 5−
1r =
1 1T +∴ = 20 5
110C x −
= 15
110C x
Here the coefficient of 15x is 10 .
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5
(35) From the figure,
1m = tan 60 3° =
2m = tan120 tan(180 60 )° = − °
= 3−
∴ Equation of lines are
y = 1m x c+ ,
3 7y x= + , 3 7y x= − +
(36) L.H.S. 2 2 2
1 1 1
x y z
x y z
=
2 2 1c c c→ −
3 3 1c c c→ −
=2 2 2 2 2
1 0 0
x y x z x
x y x z x
− −
− −
=2
1 0 0
( )( ) 1 1y x z x x
x y x z x
− −
+ +
Taking ( )y x− and ( )z x− from 2c and 3c
= ( )( )[ ]y x z x z x y x− − + − −
= ( )( )( )x y y z z x− − −
=R.H.S.
(37) Given : | | 3 | | 4 | | 5a b c= = =rr r
| |a b c+ +rr r
= 0
2| |a b c+ +rr r
= 2 2 2| | | | | | 2( )a b c a b b c c a+ + + ⋅ + ⋅ + ⋅r r rr r r r r r
0 = 9 16 25 2( )a b b c c a+ + + ⋅ + ⋅ + ⋅r rr r r r
50− = 2( )a b b c c a⋅ + ⋅ + ⋅r rr r r r
25 .a b b c c a− = ⋅ + ⋅ + ⋅r rr r r r
(38) logx x dx∫
logu x= dv∫ = x dx∫
du =1dxx
v = 2
2
x
60°
30°
120°
O
y
x
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6
udv∫ = uv vdu− ∫
logx x dx∫ = 2 21log | |
2 2
x xx dx
x− ×∫
= 2 2
log | |2 4
x xx c− +
(39) ( )P A = 3
8
1( )
8P B =
( )P A = 1 ( )P A−
= 3 5
18 8
− =
( )P A B∪ = ( ) ( )P A P B+ ( A∴ and B are mutually exclusive ( ) 0P A B∩ = )
= 3 1 4 1
8 8 8 2+ = =
( )P A B∩ = ( ) ( )P B P A B− ∩
= 1 1
08 8
− = .
(40) 0
2 2 2 2lim
2 2x
x x
x x→
+ − + +×
+ +
=( )0
2 2 1 1lim
2 2 2 22 2x
x
x x→
+ −= =
++ + .
PART - IV
(41)(a) | |x =if 0
if 0
x x
x x
− ≤
>
( )f x = if 0
if x > 0
x x x
x x
− + ≤
+
( )f x =0 if 0
2 if 0
x
x x
≤
>
( )g x = if 0
if 0
x x x
x x x
− − ≤
− >
( )g x =2 if 0
0 if 0
x x
x
− ≤
>
When 0x ≤
( )( )f g xo = [ ( )] ( 2 )f g x f x= −
when 0 ( )x f g x> o = [ ( ) (0) 0f g x f= =
When 0x >
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7
( )( )g f xo = [ ( )] [2 ]g f x g x= =0
when 0x ≤
( )( )g f xo = [ ( )] (0) 0g f x g= =
(b) 3
3
x y
x y
+ ≥
+ =
2 5
2 5
x y
x y
− ≤
− =
2 3
2 3
x y
x y
− + ≤
− + =
x 0 3 x 0 2.5 x 0 -3
y 3 0 y -5 0 y 1.5 0
(0,3), (3,0)A B are the 2 points on the line 3x y+ = and (0,0) is not in the solution region
of 3x y+ ≥ .
(0, 5), (2.5,0)C D− are the 2 points on the line 2 5x y− = and (0,0) is in the solution region
of 2 5x y− ≤ .
(0,1.5), ( 3,0)E F − are the 2 points on the line 2 3x y− + = and (0,0) is in the solution
region of 2 3x y− + ≤ .
∴∆ GHI is the required solution set of the given linear inequalities.
(42) Given A B C π+ + =
(a) L.H.S. cos cos cosA B C= + +
= 2cos cos cos2 2
A B A BC
+ − ⋅ +
2 2 2
A B Cπ+ = − Q
= 2cos cos cos2 2 2 2
C A BC
π − ⋅ − +
= 22sin cos 1 2sin2 2 2 2
C A B C ⋅ − + −
= 1 2sin cos sin2 2 2 2
C A B C + − −
= 1 2sin cos cos2 2 2 2 2
C A B Cπ + − − −
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8
= 1 2sin cos cos2 2 2 2 2
C A B A B + − − +
= 1 2sin 2sin sin2 2 2
C B A + ⋅ ⋅
= 1 4sin sin sin2 2 2
A B C + ⋅ ⋅
= R.H.S
(b) L.H.S. cos cos cosa A b B c C= + +
= 2 sin cos 2 sin cos 2 sin cosR A A R B B R C C+ +
sin
a
AQ = 2
sin sin
b cR
B C= = .
= [sin 2 sin 2 sin 2 ]R A B C+ +
= 2 2 2 2
2sin cos sin 22 2
A B A BR C
+ − ⋅ +
= [ ]2sin( ) cos( ) sin 2R A B A B C+ ⋅ − +
= [ ]2sin( )cos( ) sin 2R c A B Cπ − − +
[2sin .cos( ) 2sin cos ]R C A B C C= − +
= 2 sin [cos( ) cos ]R C A B C− +
= 2 sin [cos( ) cos[ ( )]R C A B A Bπ− + − +
= 2 sin [cos( ) cos( )]R C A B A B× − − +
= 2 sin 2sin sinR C A B×
= 2 sin 2sin sinR A B C
= 2 sin sina B C
L.H.S = . .R H S
Hence proved.
(43) (a) Let 2( ) 2 4 6 2P n n n n= + + + + = +L n N∀ ∈
Step (1)
Put n = 1
(1)P = 22 1 1 1× = +
2 = 1
(1)P∴ is true.
Step (2)
Put n = k
Let ( )P k = 22 4 2k k k+ + + = +L be true.
Step (3)
Put n = 1k +
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9
( 1)P k + = 22 4 2( 1) ( 1) ( 1) ( 1)( 2)k k k k k+ + + + = + + + = + +L
L.H.S. = 2 4 2 2( 1)k k+ + + + +L
= ( 1) 2( 1)k k k+ + +
= ( 1)( 2)k k+ +
( 1)P k∴ + is true.
Hence by principle of mathematical induction, ( )P n is true for all n N∈ .
(b) Let the initial amount of bacteria, 30a = .
Since the amount of bacteria is doubled in every one hour, the sequence is
1 230,30 2 ,30 2× × L
It is in G.P.
End of 2nd hour = 30 2 1202× =
End of 4th hour = 430 2 480× =
∴ End of nth hour = 30 2n× .
(44) (a) L.H.S.
log log log
log 2 log 2 log 2
log3 log3 log3
x y z
x y z
x y z
2 2 1 3 3 1,R R R R R R→ → → −
log log log
log 2 log 2 log 2
log3 log3 log3
x y z
=
log log log
log 2 log3 1 1 1
1 1 1
x y z
× 2 3R R∴ ≅
= 0
L.H.S = R.H.S
Hence proved.
(b) ar = ˆˆ ˆ2 3i j k− +
br = ˆˆ ˆ2 3 4i j k− + −
cr = ˆˆ 2j k− +
ar = lb mc+
r r
ˆˆ ˆ2 3i j k− + = ˆ ˆˆ ˆ ˆ( 2 3 4 ) ( 2 )l i j k m j k− + − + − +
1 = 2l− 1
2l⇒ = −
2− = 3l m− ... (2)
3 = 4 2l m− + ... (3)
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10
Solving (1) and (2)
1
32
m − −
= 2= −
3
22
− + = m
1
2m =
Put the values of l and m in equation (3)
1 1
4 22 2
− × − + × = 3
2 1+ = 3
Thus one vector is a linear combination of other 2 vectors.
∴ The given vectors are coplanar.
(45) (a) The function ( )h x is defined at all points of the real line ( , )R = −∞ ∞ , for any 0 0x ≠ .
0
lim ( )x x
h x→
=0
1lim sinx x
xx→
= 0 0
0
1( )
sinx h x
x=
For 0x = 1
0, ( ) sinh x xx
=
x− ≤ 1
sinx xx
≤
( )g x = ,x− 1 1
( ) sin , ( )f x h x xx x
= =
0
lim ( )xg x
→ = 0 ,
0lim ( ) 0xh x
→= and
0
1lim sin 0x
xx→
=
∴ By Sandwich theorem
0
limx→
1
sinxx
= 0 (0)h=
( )h x∴ is continuous in the entire interval.
(b) sin y = sin( )x a y+
x = sin
sin( )
y
a y+
Differentiating
1 = 2
sin( ) cos sin cos( )
sin ( )
dy dya y y y a y
dx dx
a y
+ ⋅ ⋅ − × + ⋅
+
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11
2sin ( )a y+ = [sin( ) cos cos( ) sin ]dy
a y y a y ydx
+ ⋅ − + ⋅
= [sin( )]dy
a y ydx
+ −
2sin ( )
sin
a y dy
a dx
+=
(46) (a) I = 6
2 1
xdx
x⋅
+∫ .
= 2 1
62
tt dt
t
−×∫
= 3 2( 1)t dt−∫
= 3
33
tt
−
= 3 3
33
t tc
−+
= 2( 3)t t c− +
= 2 1(2 1 3)x x c+ + − +
= 2 1(2 2)x x c+ − +
= 2( 1) 2 1x x c− + +
(b) Let 1A and 2A be the events of job done by engineer I an II .
Let B be the event that the error occurs in the work.
1( )P A = 60
100 1
3( / ) 0.03
100P B A = =
2( )P A = 40
100 2
4( / ) 0.04
100P B A = =
Applying Baye’s theorem,
1( / )P A B = 1 1
1 1 2 2
( ) ( / )
( ) ( / ) ( ) ( / )
P A P B A
P A P B A P A P B A
⋅⋅ + ⋅
=
60 3 18
100 100 100060 3 40 4 18 16
100 100 100 100 1000 1000
×=
× + × +
= 18 1000 9
1000 34 17× =
Put 2 1x t+ =
2 1x + = 2t
2dx = 2tdt
Now x = 2 1
2
t −
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12
2( / )P A B =
40 4
100 10034
1000
×
= 16 1000 8
1000 34 17× =
1( / )P A B >
2( / )P A B
∴ The serious error would have been done by Engineer I.
(47) (a) Let ,a v and s be the acceleration, velocity and distance respectively.
v = ds
dt
a = 28 /dv
m sdt
= −
dv∫ = 8 dt−∫
18v t c= − +
When the brakes are applied, 0t = and 24 /v m s=
24 = 18 0 c− × +
1c = 24
v∴ = 8 24t− +
ds
dt = 8 24t− +
ds∴ = ( 8 24)t dt− +∫
s = 2
2
824
2
tt c− + +
s = 2
24 24t t c− + +
when t = 0 , 0s =
0 = 2c
s∴ = 24 24t t− +
when the bike stops, v = 0
8 24t− + = 0
3t =
when t = 23, 4 3 24 3s = − × + ×
= 36 72− +
s = 36 metres
The bike stops at a distance 4 metres to the barriers.
(b) The equation of pair of straight lines is 2 22 3 6 19 20 0x xy y x y− − − + − = .
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13
Consider
2 22 3x xy y− − = (2 3 )( )x y x y− +
2 22 3 6 19 20x xy y x y− − − + − = (2 3 )( )x y l x y m− + + +
Equating the coefficient of ;x y , constant term
6− = 2l m+ ... (1)
19 = 3l m− ... (2)
20− = lm ... (3)
Solving (1) and (2),
2l m+ = 6−
3l m− = 19
5m = 25−
5m = −
( 10)l∴ + − = 6−
l = 4
∴ Separate equations are
2 3 4x y− + = 0
5x y+ − = 0
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