maths - key answer€¦ · 1. 5*7 = lcm of 5, 7 = 35 2. -1range of the function y=sec x is 0, 2 3....
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II PUC / BOARD EXAMINATION / MATHEMATICS‐KEY ANSWER 1
( ) . ( )
3 1 3
5 5 25
P A P B
x
P(A B)
II PUC MATHEMATICS - KEY ANSWER
------------------------------------------------------------------------------------------
PART-A 1. 5*7 = lcm of 5, 7 = 35
2. Range of the function y=sec-1x is 0,2
3. 1x5, 5x1
4. x2 –36 = 36–36
x2 –36=0
6x36x2
5. y=tan√x sec √x √
6. 2x e dx + ex + C
7. A vector whose magnitude is the same as that of a given vector, but
direction is opposite to that of it, is called negative vector of the given
vector.
8. l = cos900=0, m=cos1350 = √, n cos45
√
9. Any feasible solution which optimizes (maximum or minimum) the
objective function is called its optimal solution.
10.
PART-B
11. f(x) = cosx, g(x) = 3x2
(i) gof(x) = g[f(x)] = g[cosx] = 3cos2x
(ii) fog (x) = f[g(x)], f[3x2] = cos3x2
12. cot-1(–x)=θ
–x = cot θ
X =–cot θ
X = cot( θ
cot-1x = θ cot -1(–x)
cot-1(–x) = cot -1x
II PUC / BOARD EXAMINATION / MATHEMATICS‐KEY ANSWER 2
1 1
2 2
3 3
1 2 3 11 1
1 3 2 12 2
1 1 8 1
12 ( 2 8 ) 3 (3 1) 1( 2 4 2 )
21 3 0
2 0 1 2 2 2 ] 1 52 2
1 5 .
x y
x y
x y
A rea o f th e tr ia n g le is sq u n its
13 2sin sin sin sin ( )
5 5
2sin
5
2,
5 2 2
1
1sin
13.
14.
15. Sin2x + cos2y = 1
2sinx . cosx – (2siny . cosy)y| = 0
sin2ysin2xy
2siny.cosy2sinxcosxy ||
16. y = xx
logy = logxx
logy = xlogx
1 1. log .1
(1 log ) (1 log)x
dyx x
y dx x
dyy x x
dx
17. f(x) = x2 – 4x = 6
f|(x) = 2x – 4
for f(x) strictly decreasing f| (x)<0
42 4 0
2x x
2 ( , 2)x x
II PUC / BOARD EXAMINATION / MATHEMATICS‐KEY ANSWER 3
18. I = cot . xlog sinx dx Let log sinx t
I = log sinx cotx. dx . cosxdx dt
I = tdt C cotx dx = dt
= 2 + C
19. I = x sec x dx
I = xtanx – tanx 1 dx
I = xtanx – xlog |secx|+C
20. Order = 2 Degree doesn’t exist
21. ˆ ˆˆ ˆ ˆ ˆ
a i 3j 7k, b=7i- j+8k+
2 2 2
.Projection of
(1) (7) (3) ( 1) (7) (8)
7 ( 1) 8
7 3 56 60
49 1 64 114
ona b
b
a b
22. Let ˆ ˆˆ ˆ ˆ ˆ3 , 2 7a i j k b i j k
ˆˆ ˆ
ˆˆ ˆ1 1 3 ( 1 21) (1 6) ( 7 2)
2 7 1
i j k
a xb i j k
ˆˆ ˆ20 5 5i j k 2 2 220 5 ( 5) 400 25 25A a xb
450
15 2 .A sq units
II PUC / BOARD EXAMINATION / MATHEMATICS‐KEY ANSWER 4
23. Equation of the plane is
123z4y6x
14z
3y
2xtherefore
4c3,b2,ahere1cz
by
ax
24. 1Σpi i = 1 0.1 + K + 2K + 2K + K = 1 6K = 1–0.1
0.90.15
6K
K = 0.15
PART-C
25. 1 2( , )T T R Triangle T1 is similar to triangle T2.
(i) Triangle T1 is similar to triangle 1 1 2( , )T T T R ∴ R is reflexive.
(ii) 1 2( , )T T R Triangle T1 is similar to Triangle T2
Triangle T2 is similar to Triangle T1 1 2( , )T T R ∴ R is symmetric and T2
(iii) 1 2( , )T T R Triangle T1 is similar to Triangle T2
2 3( , )T T R Triangle T2 is similar to triangle T3
Triangle T1 is similar to triangle T3 1 3( , )T T R ∴R is transitive
Therefore R is an equivalence relation
26. 1 1 1 1 1 1
4 11 1 4 1 313 72 tan tan tan tan tan tan
4 12 7 3 7 1713 7
27. F(x) F(y)=
cos cos sin sin cos sin sin cos 0
sin cos cos sin sin sin cos cos 0
0 0 1
x y x y x y x y
x y x y x y x y
y)F(x1000y)cos(xy)sin(x0y)sin(xy)cos(x
II PUC / BOARD EXAMINATION / MATHEMATICS‐KEY ANSWER 5
2
4 3
32
2 4
4
4
4
dxx at at
dtdy
y at atdt
dy atat
dx at
28.
29. f(x) = x2–4x–3 and [1, b] =[1, 4]
Being a polynomial function f(x) is continuous in [1, 4] and hence differentiable in (1, 4) f(b)=f(4)=-3, f(a)=f(1)=-6 and f|(x)=2x-4
| ( ) ( ) 5( ) 2 4 1 (1,4)
2
f b f af c c c
b a
Hence Mean Value Theorem is verified.
30. f(x)=√ where x=36 and ∆ 0.6 f(x + ∆ = f(x)+ f|(x) ∆
1
21 0.1
36.6 6 (0.6 6 6 0.05 6.0512 2
x x x xx
31. Problem must be dx1)(x
e3)(x3
x
dx1)(x
21)(x1xedx
1)(xe3)(x
33x
3
x
2 3 2
1 2
( 1) ( 1) ( 1)
xx e
e dx cx x x
32.
2π
0
2π
0
2
2sin2xx
21dxcos2x)(1
21dxcos | 2
0
1 1
(sin )2 2 2 4
II PUC / BOARD EXAMINATION / MATHEMATICS‐KEY ANSWER 7
Required equation is (3 2 4) ( 2) 0
.(2,2,1) (1) (6 2 2 4) (2 2 1 2) 2
2 3 0
2
32
(1) , (3 2 4) 9 2) 03
7 5 4 8 0
x y z x y z
Sub in we get
becomes x y z x y z
x y z
37.
38. E1 : Event that six occurs E2 : Event that six does not occur A=Event that man reports that six occurs in the throwing of the die.
83
245
243
243
41
65
43
61
43
61
)P(A/E)P(E)P(A/E)P(E)P(A/E)P(EA)/P(E
43
EAP,
65BP,
61)P(E
2211
111
121
PART-D
39. Given function is f(x)=4x+3, x ∈ R
x4
334x3)(4xf
(f(x))f(x)f).(fL.H.Sx(x)f).(fproveTo:Step(2)
43y(y)f
34xyf(x)yx(y)fLet
(y)ffindTo:Step(1)
1
11
1
1
1
1
II PUC / BOARD EXAMINATION / MATHEMATICS‐KEY ANSWER 8
43y(y)fwithinvertibleisf
If.fandIf.f:Step(4)
y343-y4
43-yf(y))f(f(y))f. (fL.H.S
y(y))f. (fproveTo:Step(3)
1
11-
11
1
40.
1 2 3 3 1 2 4 1 2
5 0 2 4 2 5 0 3 2
1 1 1 2 0 3 1 2 3
4 1 1 4 1 2 0 0 3
9 2 7 0 3 2 9 1 5
3 1 4 1 2 3 2 1 1
3 1 2 4 1 2 1 2 0
4 2 5 0 3 2 4 1 3
2 0 3 1 2 3 1 2 0
B
A B C
B C
A
1 2 3 1 2 0 0 0 3
( ) 5 0 2 4 1 3 9 1 5
1 1 1 1 2 0 2 1 1
( ) ( )
A B C
A B C A B C
413
729
114
CB)(A
II PUC / BOARD EXAMINATION / MATHEMATICS‐KEY ANSWER 9
41.
42.
43.
1z;2y;1x
121
zyx
34
5
71151135935
401X
71151135935
Adj.(A);40213
121332
A
(A)adj.A1AwhereD,.AX
34
5D,
zyx
X;213
121332
AwhereD,AX
11
2y)x(12xy)x(1)x(1
1.2(2x)yy)x(1
x2tany)x(1x1
1x)2(tany
x)(tany
12
222
2122
11
2
21
1
21
sec/cm48
1dtdh
2)4(36
12dtdh
dtdh.2h36
dtdv
3h12V
h2r31V,conetheofVolume
cm4handh6rr.61h;sec/3cm12
dtdv
II PUC / BOARD EXAMINATION / MATHEMATICS‐KEY ANSWER 10
44.
C1)(xtandx11)(x
1dx22xx
1
11)(x112xx22xx
Caxtan.
a1dx
ax1
θseca
1)θ(tanaaxCθ.a1
θdθsecadxθdθseca.θseca
1
θtanaxPutdxax
1I
1222
2222
122
22
2222
2222
22
45. 2interceptyand3interceptxmakes1
2y
3xlineThe
2b3,114y
9xisEllipse
22
II PUC / BOARD EXAMINATION / MATHEMATICS‐KEY ANSWER 11
32 2
0
32 22 2 1
0
2 22 2 1
2 2Re , 3 (3 )
3 3
2 33 sin 3
3 2 2 3 2
2 3 3 3 33 3 sin 3(3) 0
3 2 2 3 2
2 9 9 39 3 .
3 2 2 2 2
quired Area A x x dx
x x xx x
sq units
46.
Cx161
4xx)(logyx
Cdx4x.
x1
4x.xlogyx
Cdxxlogx)y(x
CdxQ(IF)y(IF)issolutiongeneralThe
xeeeIF
xlogxQ;x2P
xlogxyx2
dxdy
44
2
442
32
22logxdxx2
Pdx
47.
II PUC / BOARD EXAMINATION / MATHEMATICS‐KEY ANSWER 12
Let a be the position vector of the given point A with respect to the origin 0 of the rectangular co-ordinate system. Let l be the line which passes through the point A and is parallel to a given vector b. Let r be the position vector of an orbitrary point P on the line.
Then is parallel to the vector b
i. e. AP λb, where λ ∈ R
OP – OA λb
r a λb is the vector form
The cartesian form is as follows:
Let A A(x1, y1, z1) and the direction ratios of the line be a, b, c
Consider P P(x, y, z) Then,
czz
byy
axxisformcartesianThe
λczz;λ
byy;λ
axx
c)b,(a,λ)z,y,xz)y,(x,bλar
ckbjaibandkzjyixazk,yjxir
111
111
111
111
48.
50
0050
050
4949|150
150
xxnx
n
100991
1001
10099C1
0XP11)P(X11)P(X(b)
10099
21
1001
1009950
1001
10099C1)P(X(a)
2,......501,0,x,pqCx)P(X10099p1q,
1001p50,n
II PUC / BOARD EXAMINATION / MATHEMATICS‐KEY ANSWER 13
PART-E
49. Pf: First we prove that
1
1
4545
a
a
a
a
a
0
a
0
a
0
a
0
a
0
a
a
a
0
a
0
0
a
a
0
a
a
0
a
a
0
a
a
a
0
function)oddanisxcosxsin(0dxxcosxsinNow
0dxf(x)
f(x)x)f(thenfunction,oddisf(x)If:(2)Case
dxf(x)2dxf(x)dxf(x)dxf(x)
f(x)x)f(thenfunction,evenanisf(x)If:(1)Case
dxf(x)dxx)f(dxf(x)
dxf(x)dtt)f(
dxf(x)dt)(t)f(
0t0xatax
dtdx
txputintegral1sttheIndxf(x)dxf(x)dxf(x)
oddisf(x)if0,
evenisf(x)if,dxf(x)2dxf(x)
II PUC / BOARD EXAMINATION / MATHEMATICS‐KEY ANSWER 14
49. b.
2x)(44)(5xx400
0x402x2x1
4)(5x
4x2x12x4x12x2x1
4)(5x
4x2x45x2x4x45x2x2x45x
4x2x2x2x4x2x2x2x4x
L.H.S.
50.
x+y=50 3x+y=90
X 50 0
Y 0 50
X 30 0
Y 0 90
II PUC / BOARD EXAMINATION / MATHEMATICS‐KEY ANSWER 15
The feasible region is as shaded in the graph, it is a bounded set, hence the optimal solution exists at the corner points.
Corner points Z = 4x + y
(x, y) = (0, 0) Z = 0
(x, y) = (30, 0) Z = 120
(x, y) = (20, 30) Z = 110
(x, y) = (0, 50) Z = 50
∴ The maximum value of the function is Zmax=120 and it occurs at the point (x, y) = (30, 0)
50. Since f(x) is continuous at x = , we have,
2k
11kπ
)(xx(cosx)Lt
)(xx
1)(kxLt
)(xx
f(x)lim
)(xx
f(x)lim
************** End ***************
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