MARCH 2018 PLUS ONE MATHS ANSWER KEY · 11.03.2018 · MARCH 2018 PLUS ONE MATHS ANSWER KEY...
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MARCH 2018 PLUS ONE MATHS ANSWER KEY P.A.PALANIAPPAN M.Sc., M.Phil.,B.Ed TYPE A TYPE B Q. No Option Answer Q. No Option Answer 1 b {1, -1} 1 d ab sinC 2 a 7 2 b {1, -1} 3 a P(AB) = P(A) + P(B) 3 d x + 2y =3 4 b 4 c 5 c 5 a -1 < x 1 6 a 6 b x 7 d 160, 640 7 a 8 a (-, 0) 8 a (x – 1) 2 + (y – 1) 2 = 1 9 a x – 2log(x + 1) + c 9 d f(x) = cosx in [-, ] 10 a 24 C 12 10 d tan -1 (e x ) + c 11 c 11 a P(AB) = P(A) + P(B) 12 d ab sinC 12 c 13 d x + 2y =3 13 a 24 C 12 14 a 20 14 a (-, 0) 15 d f(x) = cosx in [-, ] 15 d 160, 640 16 c 16 c 17 a (x – 1) 2 + (y – 1) 2 = 1 17 a x – 2log(x + 1) + c 18 b x 18 a 7 19 a -1 < x 1 19 a 20 20 d tan -1 (e x ) + c 20 b P.A.PALANIAPPAN M.Sc., M.Phil.,B.Ed PG ASST IN MATHS GOVT BOYS HR SEC SCHOOL, PATTUKKOTTAI CELL NO : 9443407917 www.Padasalai.Net www.TrbTnpsc.com http://www.trbtnpsc.com/2018/03/11th-public-exam-march-2018-question-papers-and-answer-keys-download.html www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasa www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasa www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasa www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasa www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasala www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasala www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasala www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasala
Transcript of MARCH 2018 PLUS ONE MATHS ANSWER KEY · 11.03.2018 · MARCH 2018 PLUS ONE MATHS ANSWER KEY...
MARCH 2018 PLUS ONE MATHS ANSWER KEY
P.A.PALANIAPPAN M.Sc., M.Phil.,B.Ed
TYPE A TYPE B Q. No Option Answer Q. No Option Answer
1 b {1, -1} 1 d
ab sinC
2 a 7 2 b {1, -1}
3 a P(AB) = P(A) + P(B) 3 d x + 2y =3
4 b
4 c
5 c
5 a -1 < x 1
6 a
6 b x
7 d 160, 640 7 a
8 a (-, 0) 8 a (x – 1)2 + (y – 1)2 = 1
9 a x – 2log(x + 1) + c 9 d f(x) = cosx in [- , ]
10 a 24C12 10 d tan-1 (ex) + c
11 c
11 a P(AB) = P(A) + P(B)
12 d
ab sinC 12 c
13 d x + 2y =3 13 a 24C12
14 a 20 14 a (-, 0)
15 d f(x) = cosx in [- , ] 15 d 160, 640
16 c
16 c
17 a (x – 1)2 + (y – 1)2 = 1 17 a x – 2log(x + 1) + c
18 b x 18 a 7
19 a -1 < x 1 19 a 20
20 d tan-1 (ex) + c 20 b
P.A.PALANIAPPAN M.Sc., M.Phil.,B.Ed
PG ASST IN MATHS
GOVT BOYS HR SEC SCHOOL,
PATTUKKOTTAI
CELL NO : 9443407917
www.Padasalai.Net www.TrbTnpsc.com
http://www.trbtnpsc.com/2018/03/11th-public-exam-march-2018-question-papers-and-answer-keys-download.html
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21 - A =
22
Let ABC be the triangle
= + +
= +
= =
23
1 5 5 1
Required no. of ways = 1 x 5 x 5 x 1
= 25 ways
24
an =(-1)n+1
n=7
a7 = (-1)7+1
=
25
y = 0 , PA = PB PA2 = PB2
(x1 – 7)2 + (0 + 6)2 = (x1 – 3)2 + (0 – 4)2 8x1 = 60 x1 = 15/2
Required point
26
(x, y) = (-1, -1) Equation of any line perpendicular to 3x – 5y + 11 = 0 is 5x + 3y + k = 0 k = 8, the required equation is 5x + 3y + 8 = 0
27
= x(-e-x)- = -xe-x + =-xe-x – e-x = -e-x(1 + x) + c
28
t = sin-1x
dt =
dx
dx = dt
= et + c = + c
29
P(A) + P(B) + P(C) = 0.51 + 0.29 + 0.1
= 0.90 1 Not permissible
30
f(x) = sin x f(x) = - sin x f’(x) = cos x f’(x) = -cos x
f’(x) = cos x
31 2
+
=0
C1 C2 C1 C3
32
= -
=
5 = 5
= ( - )
33
= 6P1 + 6P2 + 6P3 + 6P4 + 6P5 + 6P6 = 6 + 30 + 120 + 360 +720 +720 = 1956
34
x2 – y2 + x – 3y – 2 = 0 compare with abc + 2fgh – af2 – bg2 – ch2 = 0
a = 1, b = -1, h =0, g = ½ , f = - , c = -2
= 2 – 9/4 + ¼ = 0 a + b = 1 + (-1)= 0
the angle is 900
35
pq = (tan + sin) (tan - sin)
= sintan
p2 – q2 = (tan + sin)2 – (tan - sin)2
= 4 sin tan
= 4
36
sec2(x + y) (1 +
)+ sec2(x – y) (1 –
) = 0
[sec2(x + y) + sec2(x – y)] + [sec2(x + y) – sec2(x – y)]
= 0
=
37
dx =
=
+
log
+ c
=
+ log + c
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38
=
dx P.A.PALANIAPPAN M.Sc., M.Phil.,B.Ed
=
=
[xlogx – x] + c
39
P(A) = P( ) =
P(B) = P( ) =
P(C) = P( ) =
P(exactly two hits) = P(AB ) + P(A C)+ P( BC)
=
+
+
=
=
40
At x = 0 , f(0) = 2
=
+ 1
=
+ 1
= 3 + 1 = 4
f(x) f(0)
f(x) is not continuous at x = 0
41 (a) =
Put a = 0
=
C2 C3
a is a factor of
b = 0, c = 0 = 0
a, b, c is a factor of Put a + b + c = 0 a + b = -c b + c = -a a + c = -b
=
= 0
(C1 C2 C3)
(a + b + c)2 is a factor of abc(a + b + c)2 is a factor of degree 5 Product of the diagonal elements 6 m = 6 – 5 = 1
k (a + b + c) is a other factor of
= k (a + b + c) abc (a + b + c)2 = k abc (a + b + c)3 ……….1 a = 1 b = 1 c = 1
= K (1 x 1 x 1) (1 + 1 + 1)3
4(16 – 1) – 1 (4 – 1) + 1(1 – 4) = K(3)3 60 – 3 – 3 = 27K 54 = 27K 2 = K Sub K = 2 in eqn (1)
= 2 abc(a + b + c)3
(b) Let = 4 + 5 +
= - -
= 3 + 9 + 4 P.A.PALANIAPPAN M.Sc., M.Phil.,B.Ed
= -4 + 4 + 4
= -
= - - - 4 - 5 -
= - 4 - 6 - 2
= -
= 3 + 9 + 4 - 4 - 5 -
= - + 4 + 3
= -
= - 4 + 4 + 4 - 4 - 5 -
= - 8 - + 3
Let = = - 4 - 6 - 2
= = - + 4 + 3
= = - 8 - + 3
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= m + n
- 4 - 6 - 2 = m(- + 4 + 3 ) + n(- 8 - + 3 )
-m + 4m + 3m - 8n - n + 3n Equating , -4 = -m – 8n ………………………(1) Equating , -6 = 4m – n ……………………… (2)
Equating , -2 = 3m + 3n ……………………… (3) Solve (2) and (3)
2 x 3 -18 = 12m – 3n
3 -2 = 3m + 3n -20 = 15m
-
= m
m = - sub in eqn(3) P.A.PALANIAPPAN M.Sc., M.Phil.,B.Ed
- 2 = 3(- ) + 3n
- 2 = -4 + 3n 2 = 3n
= n
m = - n = sub in eqn (3)
- 2 = 3 + 3
- 2 = -4 + 2 2 = 2
is a linear combination of and
given points are coplanar
42 (a)
P(n) = 72n + 16n – 1 N = 1 P(1) = 72 + 16 – 1 = 49 +15 = 64 P(n) is true Let us assume P(k) is true P(k) = 72k + 16K – 1 To prove: P(K + 1) is true P(K + 1) = 72(K+1) + 16 (K + 1) – 1 = 72K + 2 + 16K + 16 – 1
= 72k . 72 + 16K + 15 = 72k . 49 + 16K +15 + 49 – 49 = 49(72K – 1) + 16K + 64 = is divisible by 64 P(K+1) is true P(n) is true
(b) 117 = (1 + 10)7 =7C0(1)7 (10)0 + 7C1 (1)6 (10)1 + 7C2 (1)5 (10)2 + 7c3 (1)4 (10)3 + 7C4 (1)3 (10)4 + 7C5 (1)2 (10)5 + 7C6 (1)1 (10)6 + 7C7 (1)0 (10)7
= 1 + 70 +
102 +
103 +
104 +
105 +
7(10)6 + 107 = 1 + 70 + 2100 + 35000 + 350000 + 2100000 + 7000000 + 10000000 = 19487171
43 (a)
-
=
-
=
-
= x
- x
= x
- x
= x (1 + + …… ) – x
= x + - x -
+ …..
= -
= (app)
(b) 2 tan - cot = -1
2 tan -
= - 1
2 tan2 + tan - 1 = 0
(2 tan - 1) (tan + 1) = 0
2tan - 1 = 0
tan = ½
when tan = - 1 = -tan
tan = tan(-
)
= n + (-
) P.A.PALANIAPPAN M.Sc., M.Phil.,B.Ed
= n -
; n Z
When tan = ½ = tan()
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= n +
= n + tan-1(½)
= n -
(or)
= n + tan-1(½) ; n Z
44 (a)
tan
=
cot
tan
=
cot
tan
=
cot
Proof :
tan
=
cot
RHS
cot =
cot
=
cot
=
cot
=
cot
= cot tan
cot
= tan tan
cot
= tan
RHS = LHS
(b) g(x) = x2 + 2x + 1, g[f(x)] = 4x2 – 12x + 9 let f(x)= u g[f(x)] = u2 + 2u + 1 = 4x2 – 12x + 9 u2 + 2u = 4x2 – 12x + 8 u(u + 2) = 4(x2 – 3x + 2) = 4(x – 1) (x – 2) = (2x – 2) (2x – 4) = (2x – 4) (2x – 4 + 2)
= u (u + 2) P.A.PALANIAPPAN M.Sc., M.Phil.,B.Ed
u = 2x – 4
f(x) = 2x - 4
45 (a)
x2 + y2 – 2x + 6y + 6 = 0 x2 + y2 – 5x + 6y + 15 = 0 2g = -2, 2f = 6, c = 6 2g = -5, 2f = 6, c = 15
g = -1 , f = 3 g = - , f = 3
The centre A is, Centre is =
(-g, -f) = (1, -3) r2 =
r1 = =
= =
r1 = 2 r2 = ½
Distance =
=
=
=
= r1 – r2
The circles touch each other
(b)
= 2
=
= 2
=
= 2
46 (a)
x -2 -1 0 1 2
f(x) -4 -2 0 2 3
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LHS
=
=
= 2
RHS
=
= 2
LHS = RHS
It is differentiable at x = 1
(b) dx
u = - x
= - x -1
du = - x -1 dx
. dx =
= - = -eu + c = + c
47 (a)
f(x) = x2 [a, b] = [1, 3]
x =
=
f(a + rx) = f(1 + )
=
=
=
=
=
=
2
=
2
= 2(1 + 2(1 + 0) +
(1+0)(2+0))
= 2
=
sq. units
(b) P(A1) = ½ P =
=
P(A2) = ½ P =
=
P(B) = P(A1) . P + P(A2) . P
= ½ x
+ ½ x
= ½
= ½
P.A.PALANIAPPAN M.Sc., M.Phil.,B.Ed
PG ASST IN MATHS
GOVT BOYS HR SEC SCHOOL,
PATTUKKOTTAI
CELL NO : 9443407917
www.Padasalai.Net www.TrbTnpsc.com
http://www.trbtnpsc.com/2018/03/11th-public-exam-march-2018-question-papers-and-answer-keys-download.html
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