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MATHEMATICS SAMPLE PAPER
SOLUTIONS
SECTIONA
1. Projection of a
on a.bbb
=
2 2
7 2 1 6 ( 4) 3 14 6 12 87492 6 3
+ + + = = =
+ +
2. Since a,b
and c
vectors are coplanar.
a b c 0 =
1 3 12 1 1 00 3
=
1( 3 ) 3(6 0) 1(2 0) 0 + + + + =
3 18 2 0 + + = 3 21 0 = 7 = 3. Let l, m, n, be direction cosine of given line.
ocos90 0; = =l o 1m cos602
= = and n = cos and n = cos
2 2 2m n 1+ + = l 2
210 cos 12
+ + =
23
cos4
= 3cos2
= ( is acute angle)
6pi =
4. 232 3 1 1
a2 2 2
= = =
5. Given family of curve is A Br
= +
Differentiating with respect to r, we get
2d Adr r
=
Again differentiating with respect to r, we get
2
2 3d v 2Adr r
= 2
2 2d v 2 A
.
dr r r =
2
2d v 2 dv
.
dr r dr
=
2
2d v dv
r 2dr dr
=
2
2d v dv
r 2 0dr dr
= + =
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6. Give differential equation is
2 xe y dx 1
dyx x
=
2 xe y dx. 1dyx
=
2 x
dx xdy e y
=
2 xdx e y
dx x
=
2 xdy e y
dx x x
= 2 xdy 1 e
.ydx x x
+ =
dx1
x 2 x21IF e dx e ex
= = =
SECTION-B
7.
2 0 1A 2 1 3
1 1 0=
2A A A=
2 0 1 2 0 12 1 3 2 1 31 1 0 1 1 0
=
4 0 1 0 0 1 2 0 0 5 1 24 2 3 0 1 3 2 3 0 9 2 5
2 2 0 0 1 0 1 3 0 0 1 2
+ + + + +
= + + + + + =
+ + +
Now, 2A 5A 4I +
5 1 2 2 0 1 1 0 09 2 5 5 2 1 3 4 0 1 00 1 2 1 1 0 0 0 1
= +
1 1 31 3 105 4 2
=
Now given 2A 5A 4I X 0 + + =
1 1 31 3 10 X 05 4 2
+ =
1 1 3X 1 3 10
5 4 2
=
1 1 3X 1 3 10
5 4 2
=
OR
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Given
1 2 3A 0 1 4
2 2 1
=
1 0 2A ' 2 1 2
3 4 1
=
A' 1( 1 8) 0 2( 8 3) 9 10 1 0= + = + = Hence, 1(A') will exist.
111 2
A 1 8 9;4 1
= = = 122 2
A ( 2 6) 83 1
= = =
132 1
A 8 3 5;3 4
= = + = 210 2
A (0 8) 84 1
= = + =
221 2
A 1 6 7;3 1
= = + = 231 0
A (4 0) 43 4
= = =
310 2
A 0 2 2;1 2
= = =
321 2
A (2 4) 22 2
= = =
331 0
A 10 12 1
= = =
Adj(A)
79 8 5 9 8 28 7 4 8 7 22 2 1 5 4 1
= =
1
9 8 2 9 8 21(A ') 8 7 2 8 7 21
5 4 1 5 4 1
=
8. Here f(x) = 2
a 1 0ax a 1ax ax a
Taking a common from 1C ,we get
f(x) = 2
a 1 0ax a 1ax ax a
applying 2 2 1,C C C + we get
f(x) =a2 2
1 0 0x a x 1x ax x a
+
+
Expanding along 1R , we get
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f(x) =a[1( 2 2a ax ax x ) 0 0}+ + + + f(x) =a( 2 2a 2ax x+ + )
f(x) =a(a+x)2
Now, f(2x)-f(x)=a(a+2x)2-a(a+x)2
=a{(a+2x)2-(a+x)2}=a(a+2x+a+x)(a+2x-a-x)
=ax(2a+3x)
9. Here
2 2
1 dxsin x sin 2x
1 1I dx I dxsin x 2sin x cos x sin x(1 2cos x)
sin x sin xI dx I dxsin x(1 2cos x) (1 cos x)(1 2cos x)
=+
= =+ +
= =+ +
Let cosx = z - sinx dx = dz
2
dz dzI I(1 z )(1 2z) (1 z)(1 z)(1 2z)
= = + + +
Here , integrand is proper rational function. Therefore by the form of partial function ,
we can write 1 A B C
(1 z)(1 z)(1 2z) 1 z 1 z 1 2z= + ++ + + + (i)
1 A(1 z)(1 2z) B(1 z)(1 2z) C(1 z)(1 z)(1 z)(1 z)(1 2z) (1 z)(1 z)(1 2z)
+ + + + + + =
+ + +
1= A(1-z)( 1+2z) +B(1+z)(1+2z)+C(1+z)( 1 - z ) (ii)
Putting the value of z=-1 in (ii) we get
11 2A 0 0 A2
= + + =
Again, putting the value of z=1 in (ii),we get
1=0+B.2.(1+2)+0 1=6B B=16
Similarly, putting the value of z=12
in(ii) ,we get
1 31 0 0 C2 2
= + +
1=
34
C c=43
Putting the value of A,B,C in (i) we get
1 1 1 4
(1 z)(1 z)(1 2z) 2(1 z) 6(1 z) 3(1 2Z)= + ++ + + +
1 1 4I dz2(1 z) 6(1 z) 3(1 2Z)
= + + + +
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1 1 4I dz2(1 z) 6(1 z) 3(1 2Z)
=
+ +
1 1 4I log 1 z log 1 z log 1 2z C2 6 3 2
= + + + +
putting the value of z , we get
1 1 2log 1 cos x log 1 cos x log 1 2cos x C2 6 3
= + + + +
OR
Let
2 2
2 2
2)
2 2 2
2
2 2
2 1 1 2
1 2 2
x 3x 1 x 1 2 3xdx dx1 x 1 x
(1 x dx xdxdx 2 31 x 1 x 1 x
dx xdx1 x dx 2 31 x 1 x
1 1x 1 x sin x 2sin x 3 1 x C
2 23 1
sin x x 1 x 3 1 x C2 2
+ + = =
= +
= +
= + + +
= + +
10. Here 2(cos ax sinbx) dxpi
pi
=
2 2bx
2 2
cos ax sin 2cosaxsinbx)dx
cos axdx sin axdx 0
pi
pi
pi pi
pi pi
= +
= +
[First two integranda are even function while third is odd function]
2 2
0 0
0 0
0 0 0 0
I 2 2cos axdx 2sin bxdx
(1 cos2ax)dx (1 cos2bx)dx
dx cos2axdx dx cos2bxdx
pi pi
pi pi
pi pi pi pi
= +
= + +
= + +
[ ]00 0
sin2ax sin2bxI 2 x2a 2b
sin2a sin2b22a 2b
pi pipi
= +
pi pi = pi+
11. Let E,F and A three events such that
E = selection of Bag A and F=selection of bag B
A= getting one red and one black ball of two
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Here ,p(E)=P(getting 1 or 2 in a throw of die)=2 16 3
=
1 2p(F) 13 3
= =
Also, P(A/E)=P (getting one red and one black if bag A is selected)=6 4
1 110
2
C C 2445C
=
and P(A/F)=P(getting one red and one black if bag Black if bag B is selected)=3
1 110
2
C 7C 2145C
=
Now, by theorem of total probability,
p(A)=P(E).P(A/E)+P(F).P(A/F)
1 24 2 21 8 14 22p(A)3 45 3 45 45 45
+ = + = =
OR
Let number of head be random variable X in four tosses of a coin .X may have values
0,1,2,3 or 4 obviously repeated tosses of a coin are Bernoulli trials and thus X has
binomial distribution with n=4 and p= probability of getting head in one toss=12
q=probability of getting tail (not head) in one toss= 1 - 12
=12
since, we know that P(X=r)= n r n rrC q , r 0,1, 2..........n
=
therefore,
P(X=0)= 0 4 0 4
40
1 1 1 1C 1 12 2 2 16
= =
( )
( )
( )
( )
1 4 1 1 44
1
2 4 2 2 24
2
2 4 3 3 14
3
44
1 1 1 1 4C 42 2 2 2 161
1P X 1 4
3P X 2 8
1P X 3 4
P X 4
1 1 1 6C 62 2 2 2 161 1 1 1 4C 42 2 2 2 161
C2
= =
= =
= =
= = =
= = =
= = =
= =
4 4 4 4 01 1 1
112
6
2 2 1
= =
Now required probability distribution of X is
x 0 1 2 3 4
4P(x) 116
14
38
14
1
16
Required mean = i ix p =
=
1 1 3 1 10 1 2 3 416 4 8 4 16
1 3 3 1 8 24 4 4 4 4
+ + + +
= + + + = =
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variance = 2
2 2 2x i i i i i ix p x p X p
= =
= 2 2 2 2 21 1 3 1 10 1 2 3 4 2
16 4 8 4 16
+ + + +
1 3 9 1 44 4 4
= + + +
1 3 9 34 4 41 6 9 12 4 1
4 4
= + +
+ + = = =
12. Here
Now
( )( ) r i r j xy {(xi yj zk) i}.{(xi yj zk) j} xy ( yk zj).(xk zi) xy (0i zj yk) ( zi 0 j xk) xy
0 0 xy xy 0
+ = + + + + +
= + + = + + + +
= + + =
13. Let P ( , , ) be the point of intersection of the given line (i) and plane (ii)
x 2 y 1 z 23 4 12 +
= = ..(i)
and x y + z = 5 .(ii)
since ,point P ( , , ) lies on line (i)( therefore it satisfy(i)
2 1 23 4 12
3 2; 4 1; 12 2
+ = = =
= + = = +
Also point P ( , , ) lie on plane (ii) 5 + = (iii) putting the value of , , in (iii) we get
3 2 4 1 12 2 511 5 5 0
2; 1; 2
+ + + + = + = = = = =
hence the coordinate of the point of intersect ion p is (-2,-1,2)
therefore ,required distance= d = 2 2 2(2 1) ( 1 5) (2 10)+ + + + + 9 16 144 169 13units+ + = = 14. Here sin[ 1cot (x 1) + ]= 1cos(tan x) let 1cot (x 1) + = cot x 1 = + 2 2 2cos ec 1 cot 1 (x 1) x 2x 2 = + = + + = + +
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12 2
1 1sin sin
x 2x 2 x 2x 2
= =
+ + + +
1 12
1cot (x 1) sin
x 2x 2
+ =
+ +
1
2 2
1
2 2
1 1
2
tan x tan x
sec 1 tan 1 x
1 1cos cos
1 x 1 x1
tan cos1 x
again
= =
= + = +
= =
+ +
=
+
now equation (i) becomes
1 12 2
1 1sin sin cos cos
x 2x 2 1 x
= + + +
2 2
2 2
2 2
1 1x 2x 2 1 x
x 2x 2 1 xx 2x 2 1 x 2x 2 1
1x
2
= + + = ++ + +
+ + = + + =
=
Or
Here
21 2 1 2
21 2 1 2
2 21 2 1 2 1
2 21 2 1
21 2 1
5(tan x) (cot x)8
5(tan x) ( tan x)8
5(tan x) (tan x) tan x4 8
52(tan x) tan x 04 832(tan x) tan x 88
pi+ =
pi + pi =
pi pi + + pi =
pi pi pi + =
pi pi =
let 1tan x y =
22 2 2
2 2
1 1
32y y 0 16y 8 y 3 08
16y 12 y 4 y 3 0 4y(4y 3 ) (4y 3 ) 03(4y 3 )(4y ) 0 y or y
4 43
tan x [ does not belongs to domain of tan x i.e, , ]4 4 2 4
x tan 14
pi pi = pi pi =
pi + pi pi = pi + pi pi =
pi pi pi + pi = = =
pi pi pi pi =
pi = =
15.
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2 21
2 2
2 2 2 21
2 2 2 2
1 x 1 xy tan1 x 1 x
1 x 1 x 1 x 1 xtan
1 x 1 x 1 x 1 x
+ + = +
+ + + + = + + +
=
4 41 1
2 2 2
41
2
2 2 1 x 2 2 1 xtan tan
1 x 1 x 2x
1 1 xtan
x
+ + = + +
+ =
let x2 = sin 1 2sin (x ) = putting the value of x2,we get
21
2
1 1
1 1
1 2
1 1 sintan
sin
2cos1 cos 2tan tansin 2sin cos
2 2
tan cot tan tan2 2 2
sin x2 2 2
+ =
+
= =
pi = =
pi pi= =
20 x 1
sin 0 sin sin2
0 02 2 2
02 4
2 2 2 2 4
2 2 2 4
, ,
2 2 4 2 2 2
pi <
pi pi pi > >
pi pi pi pi pi
differentiating both sides with respect to x, we get
4
dy 2xdx 2 1 x
=
4
dy xdx 1 x
=
16. Given x= a cos +bsin
dx
a sin b cosd
Also, y a sin b cos
= +
=
dya cos bsin
ddy
dy a cos bsinddxdx a sin b cosd
= +
+ = = +
dy xdx y
= 2
2 2
dyy x.d y dxdx y
=
2 2
2 22 2
d y dy d y dyy y x y x y 0dx dxdx dx
= + + =
17. Let A be the area and a be the side of an equilateral triangle.
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23A a
4=
Differentiating with respect to t we get
dA 3 da2adt 4 dt
=
dA 3 2a 2dt 4
= [Given dadt
= 2cm/sec.]
dA 3adt
= a 20cm
da 20 3sq cm / sdt
=
=
18. Let 2I (x 3) 3 4x x dx= +
Let x + 3 = 2dA (3 4x x ) B
dx +
x + 3 = A (-4 2x) +B x + 3 =- 4A =-2Ax + B x + 3= (-4A + B) =- 2Ax [By comparing coefficients]
-2A = 1 A = 12
=
Again, -4A + B = 3
14 B 32
+ = 2 +B = 3 B = 1
Here, 1
x 3 ( 2 4) 12
+ = +
21I ( 2x 4) 1 3 4x x dx2
= +
2 21I ( 2x 4) 3 4x x dx 3 4x x dx2
= +
1 21I I I2
= + (i), where..
Now 21I ( 2x 4) 3 4x x dx Let 3-4x x2 = z (-2x 4) dx = dz
23
1 12I zdz (z) C3
= + 3
2 22 1
2I (3 4x x ) C3
+=
Again 22I 3 4x x dx
22I (x 4x 3)dx= + { }22I x 2) 7 dx= + 2 22I ( 7) (x 2) dx= +
2 12 21 7 x 2I (x 2) 3 4x x sin C2 2 7
+= + + +
Putting the value of 1I and 2I in (i), we get
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3
2 2 1122
2
C1 2 7 x 2I (3 4x x ) (x 2) 3 4x x sin C2 3 C 2 7
+= + + +
3
2 2 122
1 1 7 x 2I (3 4x x ) (x 2) 3 4x x sin C3 2 2 7
+= + + + +
2 2 11 1 7 x 23 4x x (3 4x x ) (x 2) sin C3 2 2 7
+ = + + + +
2 1x 7 x 23 4x x (2x 11) sin C,6 2 7
+= + + + where 12
CC C2
=
19. The number of handmade fans, mats and plates sold by three school A, B and can be
represented by 3 3 matrix as
A 40 50 20X B 25 40 30
C 35 50 40
=
And their selling price can be represented by 3 1 matrix as
25 Handmade fansY 100 Mats
50 Plates
=
Now, the total funds collected by each school is given by the matrix multiplication as
A 40 50 20 25XY B 25 40 30 100
C 35 50 40 50
=
40 25 50 100 20 50 7000XY 25 25 40 100 30 50 XY 6125
35 25 50 100 40 50 7875
+ +
= + + = + +
Hence, total funds collected by school A = Rs.7000
Total funds collected by school B = Rs.6125
Total funds collected by school C = Rs.7875
Total funds collected for the purpose = Rs.(7000+6125+7875)
= Rs. 21000
Value: Students are motivated for social service.
SECTION-C
20 Reflexivity: By commutative law under addition and multiplication
B + a = a +b a,b N Ab=ba a,b N Ab(b+a)=ba(a+b) a,b N (a,b)R(a,b) Hence, R is reflexive
Symmetry: Let(a,b)R(c,d)
(a,b)R(c,d) ad(b + c) = bc (a + d) bc(a + d) = ab(b + c)
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cb(d + a) da(c + b) [By commutative law under addition and multiplication]
(c + d) R(a, b) Hence, R is symmetric.
Transitivity: Let(a, b) R (c, d) and (c, d) R (e, f)
Now, (a, b) R (c, d) and (c, d) R (e, f)
ad( b +c) = bc(a + d) and cf(d + e) de( c + f)
b c a dbc ad+ +
= and d e c fde cf+ +
=
1 1 1 1c b d a
+ = + and 1 1 1 1e d f c
+ = +
Adding both, we get
1 1 1 1 1 1 1 1c v e d d a f c
+ + + = + + +
1 1 1 1b e a f
+ = + e b f abe af+ +
=
af(b+e) = be(a +f) (a, b) R (e, f) [c,d 0] Hence, r is transitive.
In this way, r is reflexive symmetric and transitive
Therefore, r is an equivalence relation.
21. Given circle is 2 2x y 4+ =
dy2x 2y 0dx
+ = [By differentiating]
dy xdx y
=
Now, slope of tangent at (1, 3 )
dy 1(1, 3) .dx 3
= =
Slope of normal at (1, 3) 3= Therefore, equation of tangent is
y 3 1x 1 3
=
x 3y 4+ = Again, equation of normal is
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y 3 3x 1
=
y 3x 0 =
To draw the graph of the triangle formed by the lines x-axis, (i) and (ii), we find the
intersecting of these three lines which give vertices of required triangle. Let O, A, B be
the intersecting of these lines.
Obviously, the coordinate of O, A, B are (0, 0), ( )1, 3 and (4, 0) respectively. Required area = area of triangle OAB
= area of region OAC + area of region CAB
1 40 1y dx y dx= + [Where in 1st integrand y 3x= and in 2nd 4 xy
3
= ]
1 42 21 10 0
0 1
4 x x 1 (4 x)3xdx dx 32 23 3
= + =
3 1 90
2 23
=
3 9 12 2 3
2 2 3 2 3= + = = sq units.
Or
3 3 3
2 3x 2 2 3x 2
1 1 1
(e x 1)dx e e dx (x 1)dx + + = + +
2 1 2e .I I= + . (i), 3 3
3x 22 2
1 1
I e dx;I (x 1)dx= = +
We have, b
h 0a
f (x)dx lim h{(a b) f (a 2h) ... f (a nh)}
= + + + + + +
For 1I 3xf (x) e ,= a = 1, b = 3
b a 2h h nh 2n n
= = =
Now, 3
3x
h 01
e dx lim h{f (1 h) f (1 2h) ...f (1 nh)}
= + + + + +
{ }3(1 h) 3(1 2h) 3(1 nh)h 0lim h e e ... e + + +
= + + +
{ }3 3h 3 6(1 2h) 3nhh 0lim h e .e e .e ... e +
= + + +
{ }3h 3h 2 3h nh 0lim h e (e ) ... (e )
= + + +
3h 3h n
33hh 0
e ( 1 (e )e .lim h
1 e
=
3h 3nh
33hh 0
e (1 e )e .lim h
1 e
=
[Applying formula for sum of GP]
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3h 6
3 3 6 03h3hh 0
3h 0
e ((1 e ) 1e .lim e (1 e ).e .
e 11 e lim 33h
= =
3 6e (1 e )
3
=
For 2I F(x) = x2 + 1, a = 1, b = 3 b ah
n
= 2hn
= b
nh=2
Now, { }3
2
h 01
(x 1)dx lim h f (1 h) f (1 2h) .... f (1 nh)
+ = + + + + + +
( ){ } ( ){ } ( ){ }2 2 2h 0limh 1 h 1 1 1 2h 1 ... 1 nh 1 = + + + + + + + + + ( ) ( ) ( )2 2 2 2 2
h 0limh n 1 h 2h 1 4h 4h 1 9h 6h ... (1 n h 2nh)
= + + + + + + + + + + + + +
( ) ( )2 2 2 2 2h 0limh n n h 1 2 3 ...n 2h 1 2 3... n
= + + + + + + + + +
2
h 0
h n(n 1)(2 1) 2h n(n 1)lim h 2n6 2
+ + += + +
h 0
h nh(nh h)(2nh h)lim 2nh nh(nh h)6
+ + = + + + +
h 0
2(2 h)(2 2 h) 2(2 0)(4 0)lim 4 nh(nh h) 4 2(2 0)6 6
+ + + + = + + + = + + +
16 8 324 4 86 3 3
= + + = + =
Putting the value of 1I and 2I in (i), we ge,
2 3 6 1 6 1 7e .e (1 e ) 32 e (1 e ) 32 32 (e e )I
3 3 3 3 3
+ = + = + =
22. The given differential equation can be written as
1
2 2
dx x tan ydy 1 y 1 y
+ =+ +
Now (i) is linear differential equation of the form 1 1dx P x Qdy
+ = ,
where , 1
1 12 2
1 tan yp and Q1 y 1 y
= =
+ +
Therefore, I.F = 12
1 dytan y1 ye e
+=
Thus , the solution of the given differential equation is
1 1
1tan y tan y
2
tan yxe e dy C
1 y
= +
+
Let 1
1tan y
2
tan yI e dy1 y
=
+
substituting 1tan y t = so that 21 dy dt, we get
1 y
= +
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t t t t t tI t e dt t e 1.e dt t e e e (t 1)= = =
1tan y 1I e (tan y 1) = substituting the valiue of I in the equation (ii) , we get
1 1tan y tan y 1
x.e e (tan y 1) C = + or 11 tan yx (tan y 1) Ce = + which is the gernal solution of the given differential equation
OR
Given differential equation is
2 2dy xydx x y
=
+ ..(i)
Let y= x dy d
xdx dx
= +
Now (i) becomes
2 2
2 2 2 2 2
d x d xx x
dx dxx x x (1 v )
+ = + =+ +
2 2d d
x xdx dx1 x 1
+ = =
+ +
3 3
2 2
d dx x
dx dx1 1
= =+ +
2
3dx 1 dx
+ =
Integrating both sides, we get
2
3 31 dx d d dxd
x xv
+ = + =
21 log log x C
2 + = +
putting the value of yx
= ,we get
2
2
x ylog log x Cx2y
+ + = 2
2
x log y log x log x C2y
+ + =
2
2
x log y C2y
+ =
put y=1 and x=0 in(ii) 0+log 1 =C C = 0
Therefore required particular solution is 2
2
x log y 02y
+ =
23. Let the given lines
x 1 y 1 z 1
2 3 4 +
= = (i)
and x 3 y k z
1 3 1
= = (ii)intersect at P ( , , ) P lie in (i)
1 1 1
2 3 2 +
= = = (say)
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2 1, 3 1, 4 1 = + = = +
again, P lie on (ii)also
3 k
1 2
= =
2 1 3 3 1 k 4 1
1 2 1 + +
= =
2 2 3 1 k 4 1
1 2 1 +
= =
I II III
from I and II
3 1 k2 2 4 4 3 1 k
2
= =
k=3 -1-4 +4 k=- +3
k =32
+3 =92
Now, we know that equation of plane containing lines.
1 1 1
1 1 1
x x y y z za b c
= =
and 2 2 2
2 2 2
x x y y z za b c
= = is
1 1 1
1 1 1
2 2 2
x x y y z za b c 0a b c
=
Therefore , required equation is
x 1 y 1 z 12 3 4 01 2 1
+
=
(x-1)(3-8)-(y+1)(2-4)+(z-1)(4-3)=0
-5(x-1)+2(y+1)+(z-1)=0
-5x+2y+z+6=0 5x-2y-z-6=0
24. we know that if A and B are two independent events then
P(A B) = P(A).P(B)
Also, since A and B are two independent events A ,B and A, B are also independent events .
P( A B) = P( A ).P(B) P(A B ) = P( A ).P( B ) Now, let P(A) = x and P(B) = y
P( A ) = 1 x and P( B ) = 1 y
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Given P( A B) = 215
and P(A B ) = 16
P( A ).P(B) = 215
and P(A).P( B )= 16
(1-x).y =2
15 and x.(1 y )=
16
y xy =2
15 ..(i) and x xy =
16
(ii)
From(i) y. (1-x) =2
15
2y15(1 x)=
Putting the value of y in (ii) ,we get
x - x2
15(1 x) = 16
215x 15x 2x 1
15 15x 6
=
6(-15x2+13x)=15-15x -90x2+78x=15-15x
- 90x2 + 93x 15 = 0 30x2 31 x + 5=0
30x2 25x 6 x + 5=0 5x( 6 x 5 ) 1 (6 x 5 )=0
(6x-5)(5x-1)=0 x = 56
or x= 15
Now, x = 56
y = 2 12 4
5 15 515 16
= =
and x =15
2 1
1 61y
1
55
=
=
Hence P(A)= 56
and P(B)= 45
or P(A) =15
P(B) = 16
25. Given f(x) = sinx cosx f(x)=cosx+sinx
for critical points
f(x)=0 cosx+sinx=0
sinx = -cosx tanx=-1
tanx = tan34pi
x = 3
n4pi
pi + ,nZ
x = 34pi
,74pi
[other value does not belong to (0, 2pi )]
Now f(x) = - sinx + cosx
( ) 3x
4 f '' x pi
=
= - sin34pi
+cos 34pi
=1 1 2 2 02 2 2
= =
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1 1 2 22 2 2
= =
Again ( ) 7x
4 f '' x pi
=
= - sin74pi
+ cos 74pi
=1 1 2 2 02
2 2
+ = = >
i.e., f(x) is minimum at x=74pi
Local minimum value of f(x) =f(74pi
)=sin74pi
- cos74pi
=1 1 2 22 2 2
= =
Therefore, local maximum and local minimum values are 2 and - 2 respectively. 26. Given constraints are
2x+4y 8 ..(i) 3x+y 4 (ii) x+y 4 ..(iii) x 0, y 0 ..(iv) from graph of 2x+4y 8
we draw the graph of 2x+ 4y = 8 as
X 0 4
y 2 0
2 0 4 0 8 + (0,0) origin satisfy the constraints.
Hence , fesible region lie origin side of line 2x + 4y = 8
For graph 3x+y 4 we draw the graph of line 3x+ y =6
X 0 2
y 6 0
3 0 0 6 +
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origin (0,0) satisfy 3x+ y 6 hence , fesible region lie origin side of line 3x + y =6
for graph of x+ y 4 we draw the graph of line x+ y=4
X 0 4
y 4 0
0 0 4+ origin (0,0) satisfy x + y 4 hence feasible region lie origin side of line x+4= 4
also x 0, y 0 says feasible region is in ist quadrant. therefore, OABC is required feasible region.
Having corner point O(0,0) , (0,2) , B (8 6
,
5 5) , C(2,0)
here feasible region is bounded.
NOW the value of objective function Z = 2x + 5y is obtained as
Corner point Z=2x+5y
O(0,0) 0
(0,2) 2 0 5 2 =10
B (8 6
,
5 5)
8 62 55 5
=9.2
C(2,0)
2 2 5 0 =4
Hence, maximum value of Z is 10 at x =0, y =2