Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series :...
Transcript of Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series :...
![Page 1: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/1.jpg)
Studymate Solutions to CBSE Board Examination 2018-2019
Series : BVM/1
Candidates must write the Code on the title page of the answer-book.
Code No. 65/1/1
Roll No.
4 Please check that this question paper contains 21 printed pages.
4 Code number given on the right hand side of the question paper should be written on the title page of the answer-book by the candidate.
4 Please check that this question paper contains 29 questions.
4 Please write down the Serial Number of the question before attempting it.
4 15 minute time has been allotted to read this question paper. The question paper will be distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the question paper only and will not write any answer on the answer-book during this period.
MATHEMATICS[Time allowed : 3 hours] [Maximum marks : 100]
General Instructions:
(i) All questions are compulsory.
(ii) The question paper consists of 29 questions.
(iii) Marks for each questions are indicated against it.
(iv) Questions 1 to 4 in Section-A are Very Short Answer Type Questions carrying one mark each.
(v) Questions 5 to 12 in Section-B are Short Answer Type Questions carrying 2 marks each.
(vi) Questions 13 to 23 in Section-C are Long Answer I Type Questions carrying 4 marks each.
(vii) Questions 24 to 29 in Section-D are Long Answer II Type Questions carrying 6 marks each
(vii) Please write down the serial number of the Question before attempting it.
Disclaimer: All model answers in this Solution to Board paper are written by Studymate Subject Matter Experts. ThisisnotintendedtobetheofficialmodelsolutiontothequestionpaperprovidedbyCBSE. Thepurposeofthissolutionistoprovideaguidancetostudents.
![Page 2: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/2.jpg)
STUDYmate
2
Section – A
1. If A and B are square matrices of the same order 3, such that |A| = 2 and AB = 2I, write the value of |B|.
Ans. Given |A| = 2 and AB = 2I \ |AB| = 23|I| |A||B| = 23
\ |B| = 4
2. Iff(x)=x+1,find ddx
x(fof) .( )
Ans. Given f(x) = x + 1 then FOF(x) = f[f(x)] = f[x + 1] = x + 1 + 1 x + 2
\ ddx
FOF(x) = ddx
(x + 2)
= 1
3. Find the order and the degree of the differential equation x d ydx
dydx
22
2
2 4
1= +
.
Ans. Order = 2 Degree = 14. Ifalinemakesangles90°,135°,45°withthex,yandzaxesrespectively,finditsdirection
cosines.OR
Find the vector equation of the line which passes through the point (3, 4, 5) and is parallel to the vector 2 2 3i j k� � �+ − .
Ans. Directions cosines of line are l = cos(90º) = 0 m = cos(135º) = cos (90º + 45º) = – sin 45º
= −
12
n = cos (45º) =12
OR Given point (3, 4, 5)
and parallel vector b i j k��= + −2 2 3
^ ^ ^
\ Vector equation of line
r i j k i j k
= + + + + −( ) ( )^ ^ ^ ^ ^ ^
3 4 5 2 2 3λ
Section – B
5. Examinewhethertheoperation*definedonRbya*b=ab+1is(i)binaryornot.(ii)ifabinaryoperation, is it associative or not?
Ans. (a) ∀a,b∈R
![Page 3: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/3.jpg)
STUDYmate
3
ab∈R ⇒ab+1∈R ⇒a*b∈R ⇒Operationisbinary. (b) Leta,b,c∈R Now,(a*b)*c=(ab+1)*c=abc+c+1 ...(i) a*(b*c)=a*(bc+1)a(bc+1)+1=abc+a+1 ...(ii) By (i) and (ii), (a*b)*c≠a*(b*c) Not Associative.
6. Find a matrix A such that 2A – 3B + 5C = 0, where B and C=−
=
−
2 2 03 1 4
2 0 27 1 6
.
Ans. 2A – 3B + 5C = 0 \ 2A = 3B – 5C
2 3
2 2 03 1 4
52 0 27 1 6
A =−
−
−
2
6 6 09 3 12
10 0 1035 5 30
A =−
−
−
2
6 10 6 0 0 109 35 3 5 12 30
A =− − − −− − −
A =
−− − −
12
16 6 1026 2 18
A =
−− − −
8 3 513 1 9
7. Find: sec
tan
2
2 4
x
xdx
+∫
Ans. I x=
+∫
sec
tan x
2
2 4 Let tanx=t sec2 x dx = dt
I dt
tt t C x x C=
+= + + + = + + +∫ 2 2
2 2
24 4log log tan tan
8. Find: 1 24 2
− < <∫ sin ,x dx xπ π
OR
Find: sin− ( )∫ 1 2x dx
Ans. I x dx x= − < <∫ 1 24 2
sin , π π
= + −∫ sin cos sin cos2 2 2x x x x dx
= −( )∫ sin cosx x dx2 ...[Q sin x > cos x for
π π4 2< <x ]
= −( )∫ sin cosx x dx = –cos x – sin x + C
OR
I x dx= ( )−∫1 21.sin
![Page 4: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/4.jpg)
STUDYmate
4
= ( ) − ( ){ }− − ∫∫∫sin sin1 12 1 2 1x dx d
dxx dx dx
= ( ) − ×
−− ∫x x
xx dxsin 1
22 1 2
1 4
= ( ) −−
− ∫x x dtt
sin 1 2 14
...[Let1–4x2 = t; –8xdx = dt; 24
xdx dt=−
]
= ( ) + +−x x t Csin 1 2 142
= + − +−x x x Csin 1 22 121 4
9. Form the differential equation representing the family of curves y = e2x(a+bx),where‘a’and‘b’arearbitraryconstants.
Ans. Family of curve : y = e2x(a+bx)
ye
a bxx2 = +
Differentiating w.r.t. to x.
e y y e
eb
x x
x
2 2
2 2
2'−( )
=
y ye
bx
'−=
22
Again differentiating,
e y y y y e
e
x x
x
2 2
2 2
2 2 20
'' ' '−( ) − −( )( )
=
y y y y
e x
'' ' '− − −( )=
2 2 202
y'' – 2y' – 2y' + 4y = 0 y'' – 4y' + 4y = 0 Requireddifferentialequation.10. If the sum of two unit vectors is a unit vector, prove that the magnitude of their difference is
3 .OR
If a i j k b i j k and c i j k a b c�� � � � �� � � � � � � � �� �� �= + + = − + = − + +2 3 2 3 2, , find
Ans. Given: a b�� ��, are 2 vector.
Such that
a b a b�� �� �� ��= = + =1 1 1, , .
Q a b c�� �� �+ =
2 2
( ).( )a b a b�� �� �� ��+ + =1
a b a b�� �� �� ��2 2
2 1+ + =.
2 1a b�� ��. = − ...(1)
a b a b a b�� �� �� �� �� ��− = −( ) −( )2
.
= + −a b a b�� �� �� ��2 2
2 . = 1 + 1 + 1 {By equation (1)}
a b�� ��− =
23
\ a b�� ��− = 3
OR
![Page 5: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/5.jpg)
STUDYmate
5
a i j k��= + +2 3
^ ^ ^
b i j k��= − +^ ^ ^2
c i j k
= − + +3 2^ ^ ^
\ a b c�� �� � = −
−
2 3 11 2 13 1 2
= 2 (–4 –1) – 3(2 + 3) + 1 (1 – 6) = – 10 – 15 – 5 = – 3011. Adiemarked1,2,3inredand4,5,6ingreenistossed.LetAbetheevent“numberiseven”
andBbetheevent“numberismarkedred”.FindwhethertheeventsAandBareindependentor not.
Ans. Exp: A die is thrown \ S={1R,2R,3R,4G,5G,6G} A:NumberisEven \ A={2R,4G,6G}
\ P(A) = =36
12
B:Number marked red. B={1R,2R,3R}
P B( ) = =
36
12
\ P(A) × P(B) 12
12
14
× = ...(i)
Now (A∩B)={2R}
P(A∩B) =16 ...(ii)
By (i) and (ii) P(A∩B) ≠ P(A) P(B) \ A and B are not independent.12. Adieisthrown6times.If“gettinganoddnumber”isa“success”,whatistheprobabilityof
(i) 5 successes? (ii) atmost 5 successes?OR
TherandomvariableXhasaprobabilitydistributionP(X)ofthefollowingform,where ‘k’ issomenumber.
P X x
k if xk if xk if x
otherwise
=( ) =
===
,,,,
02 13 20
Determinethevalueof‘k’.Ans.No.ofrandomvariablei.e.,n=6. Success: getting odd no.
\ p = P(success) = =36
12
\ q = 1 – p =12
![Page 6: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/6.jpg)
STUDYmate
6
ByBinomialdistribution P(r success) = 6Cr p
r qn–r
(i) P(5 Success) = 6C5 12
12
5 1
= ×6 126
=664
=332
(ii) P(At most 5 success) = 1 – P(6 success)
= −
1 1
212
66
6 0
C
= −1 164
=6364.
OR
Sumofprobabilities=1k+2k+3k=1or6k=1⇒ k = 16
Theprobabiltiydistributionisasgivenbelow:
x 0 1 2
P(x)16
26
36
Section – C
13. ShowthattherelationRonRdefinedasR={(a,b):a≤b},isreflexive,andtransitivebutnotsymmetric.
OR Prove that the function f : N →N,definedbyf(x)=x2+x+1isone-onebutnotonto.Find
inverse of f : N → S, where S is range of f.Ans.R:R → R DefinebyR={(a,b):a≤b} For reflexive: Q ∀ x ∈ R x = x \ (x, x) ∈R \ Relationisreflexive. For symmetric: Q 1, 3 ∈ R 1 < 3 = (1, 3) ∈R But 3 < 1 ...[Not true] ⇒ (3, 1) ∉R ⇒ Relationisnotsymmetric. For transitive: Let(x,y)and(y,z)∈R
![Page 7: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/7.jpg)
STUDYmate
7
⇒ x ≤ y and y ≤ z ⇒ x ≤ z ⇒ (x, z) ∈R ⇒ Relationistransitive.
OR f : N → N Definedbyf(x)=x2 + x + 1 ...(i) For one-one: Letx1, x2 ∈ N Such that f(x1) = f(x2) \ x1
2 + x1 + 1 = x22 + x2 + 1
\ x12 + x2
2 + x1 – x2 = 0 \ (x1 – x2) (x1 + x2 + 1) = 0 \ x1 + x2 + 1 ≠ 0 ...[Q x1 and x2 ∈ N] \ x1 – x2 = 0 \ x1 = x2
\ Function is one-one. For onto: Lety∈ N such that f(x) = y \ x2 + x + 1 = y
\ x y+
− + =
12
14
12
\ x y+
= −
12
34
2
\ xy
+ =−1
24 32
\ xy
N y N=− −
∉ ∀ ∈4 3 1
2 \ Function is not onto. Now for f : N → S Functionwillbeone-oneandontoboth. \ f–1 : S → N
Defineby f yy− ( ) = − −1 4 3 12
14. Solve: tan tan− −+ =1 14 64
x x π
Ans. tan tan− −+ =1 14 64
x x π ... Q tan tan tan ,− − −+ =+−
<{ }1 1 1
11A B A B
ABAB
\ tan− +− ( ) ( )
=1 4 61 4 6 4
x xx x
π
\ 101 24 42
xx−
= tan π
\ 10x = 1 – 24x2
\ 24x2 + 10x – 1 = 0 \ 24x2 + 12x – 2x – 1 = 0
![Page 8: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/8.jpg)
STUDYmate
8
\ 12x(2x + 1) – 1(2x + 1) = 0 \ (12x – 1) (2x + 1) = 0
\ x = −112
12
,
Q 24x2 < 1 ⇒ x ∈ −
124
124
,
⇒ x =112
15. Using properties of determinants, prove that a a aa a a
2
32 2 1 1
2 1 2 13 3 1
1+ ++ + = −( )
Ans.LHS=a a aa a
2 2 2 1 12 1 2 13 3 1
+ ++ +
R1 →R1–R2,R2 →R2–R3
∆ =− −− −
a aa a
2 1 1 02 2 1 03 3 1
= −( )
+a
a1
1 1 02 1 03 3 1
2
R1 →R1–R2
= −( )
−a
a1
1 0 02 1 03 3 1
2
= (a – 1)2 [(a – 1) (1 – 0) – 0 + 0] = (a – 1)3
=R.H.S.
16. If log (x2 + y2) = 2 tan–1 yx
, show that
dydx
x yx y
=+−
Ans. log (x2 + y2) = 2 tan–1 yx
Differentiating w.r.t. to x.
ddx
x y ddx
yx
log tan2 2 12+( ) =
−
1 2 1
12 2
2 22
2x y
ddx
x yyx
ddx
yx+
+( ) = ×
+
\ 2 2 2 12 2
2
2 2 2
x yyx y
xx y
xy yx
++
=+
×− ×' '
x+yy'=xy’–y x + y = xy' – yy'
y x y
x y' = +
−
\ dydx
x yx y
=+−
OR
![Page 9: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/9.jpg)
STUDYmate
9
xy – yx = ab
Differentiate w.r.t to x.
ddxx d
dxyy x− = 0
Letxy = P and yx = Q
\ dPdx
dQdx
− = 0 ...(i)
Now, P = xy
taking log log P = y log x Differentiate w.r.t to x
1PdPdx
yx
x dydx
= + log
\ dPdx
x yx
x dydx
y= +
log ...(ii)
Q = yx
Taking log, log Q = x log y Differentiate w.r.t. to x
1QdQdx
xydydx
y= + log
dQdx
y xydydx
yx= +
log ...(iii)
Nowby(i),
x yx
x dydx
y xydydx
yy x+
− +
log log = 0
x x dy
dxxy dy
dxyy x xlog logy yxy− = −− −1 1
\ dydx
y y yxx x xy
x y
y x=−−
−
−
loglog
1
1
17. If y = (sin–1x)2, prove that (1 – x2) d ydx
x dydx
2
2 2 0− − = .
Ans. y = (sin–1 x)2
Differentiate w.r.t. to x
dydx
ddx
x= ( )−sin 1 2
dydx
x
x=
−
−2
1
1
2
sin
1 22 1− = −x dydx
xsin
Again differentiating with respect to x.
1 1 2
2 1
2
12
2
2 2 2− +
× −
−=
−x d ydx
x
x
dydx x
\ 1 2 022
2−( ) − − =x d ydx
x dydx
18. Find the equation of tangent to the curve y x= −3 2 which is parallel to the line 4x – 2y + 5 = 0. Also, write the equation of normal to the curve at the point of contact.
![Page 10: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/10.jpg)
STUDYmate
10
Ans. Given: y x= −3 2
dydx x
=−
32 3 2
= Slope of tngent
Q Tangent is parallel to the line 4x – 2y + 5 = 0
3
2 3 242x −
=−−
3 4 3 2= −x 9 = 16(3x – 2)
916
2 3+ = x
3 41
16x =
x =
4148
at x y= =4148
34
,
Equation of tangent
y x− = −
34
2 4148
4 34
2 48 4148
y x−=
−( )
24y – 18 = 48x – 41 48x – 24y – 23 = 0
Equation of normal, at 4148
34,
and slope =
−12
y x− =
−−
34
12
4148
4 34
12
48 4148
y x−=− −
96y – 72 = –48x + 41 48x + 96y – 113 = 0
19. Find: 3 53 182
xx x
fx++ −∫ .
Ans.Let I xx x
dx=+
+ −∫3 53 182
I
x
x xdx=
+
+ −∫32
2 103
3 182
=+ +
+ −∫32
2 3 13
3 182
x
x xdx
I xx x
dx dxx x
=+
+ −+ ×
+ −∫ ∫32
2 33 18
32
13 3 182 2
Let I I I= +32
121 2 ...(1)
I xx x
dx1 2
2 33 18
=+
+ −∫ Let x2 + 3x – 18 = t (2x + 3)dx = dt
![Page 11: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/11.jpg)
STUDYmate
11
I dt
t1 = ∫ = log |t| + C1 I2 = log|x2 + 3x – 18| + C1 ...(2)
Now I dtx x2 2 3 18
=+ −∫
=+ + − −
∫dt
x x2 2 32
94
94
18
=
+
−
∫dt
x 32
92
2 2 .
Ix
xC2 2
1
2 92
32
92
32
92
=+ −
+ ++log
I xx
C2 219
36
=−+
+log ...(3)
Using (1), (2) and (3), we get
I x x xx
C= + − +−+
+32
3 18 118
36
2log log where C = 9 + C2.
20. Prove that f x dx f a x dxa a
( ) ( ) ,0 0∫ ∫= − hence evaluate
x xl x
dxa sin
cos.
+∫ 20
Ans. To prove : f x dx f a x dxa a( ) ( )
0 0∫ ∫= −
R.H.S= f a x dxa( )−∫0
Let a – x = t x 0a
a0t dx = – dt
= −∫ f t dta( )
0
= ∫ f t dt
a( )
0 Q f x dx f x dx
a
b
b
a( ) ( )∫ ∫= −( )
= =∫ f x dx
a( ) L.H.S
0 Q f x dx f t dt
a a( ) ( )
0 0∫ ∫=( ) L.H.S.=R.H.S.
Let, I x xxdx=
+∫sincos1 20
π ...(i)
I x xxdx=
−+∫
( )sincos
ππ
1 20 ...(ii)
Q f x dx f a x dxa a( ) ( )
0 0∫ ∫= −( ) Adding (i) and (ii), we get,
2I = ππ sin
cosxxdx
1 20 +∫
2I = 2 1 20
2π
π sincos
/ xxdx
+∫ Q f x dx f x dx f x f a xa a( ) ( ) ( ) ( )
0
2
02 2∫ ∫= = −( )if
Let cosx = t sin x dx = – dt
x 01
p/20t
I dt
t= −
+∫π 1 21
0
= −
−π tan 1
1
0t
![Page 12: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/12.jpg)
STUDYmate
12
= − −
ππ04
I = π2
4
21. Solve the differential equation: xdy – ydx = x y dx2 2+ , given that y = 0 when x = 1.OR
Solve the differential equation: (1 + x2) dydx
xy x+ − =2 4 02 , subject to the initial condition y(0) = 0.
Ans. Given : xdy – ydx = x y dx2 2+
xdy y x y dx= + +( )2 2
dydx
y x yx
=+ +2 2
...(1) Lety = vx ...(2)
dydx
v xdvdx
= + ...(3)
Using (1), (2), & (3), we get,
v xdv
dxvx x x v
x+ =
+ +2 2 2
v xdv
dxv v+ = + +1 2
dvv
dxx1 2+
= ∫∫
L.H.S.=dvv1 2+
∫ , Letv = tan q
= ∫secsec
2 θθdv
L.H.S.= ∫sec θ θd
= +log sec tanθ θ
= + +log v v1 2
\ log log| | log| |v v x c+ + = +1 2
So, v v cx+ + =1 2
yx
x yx
cx++
=2 2
y x y cx+ + =2 2 2
...(4) Given: y = 0 and x = 1 0 + 1 = c c = 1
So, solution of differential equation,
y x y x+ + =2 2 2
OR
![Page 13: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/13.jpg)
STUDYmate
13
( )1 2 4 02 2+ + − =x dy
dxxy x
dydx
xx
y xx
++
=
+2
1412
2
2
dydx
Py Q+ =
P = 2
1412
2
2
xx
Q xx+
=+
&
I.F. = ∫e Pdx
= ∫ +exxdx2
1 2
= +e xlog| |1 2
I.F. = 1 + x2
So, solution of differential equation,
y x x
xx dx( )
( )( )1 4
112
2
22+ =
++∫
y x x dx( )1 42 2+ = ∫
y x x c( ) .1 4
32 3+ = +
at x = 0, y = 0.
0 4
30= +( ) c
c = 0
∴ + =y x x( ) ( )1 4
32 3
y x
x=
+4
31
3
2( )
22. If i j k i j i j k^ ^ ^ ^ ^ ^ ^ ^
, ,+ + + + −2 5 3 2 3 and i j k^ ^ ^- -6 respectively are the position vectors of points A,
B,CandD,thenfindtheanglebetweenthestraightlinesABandCD.Findwhether AB� ��� and
CD� ���
are collinear or not.Ans. Given:
A i j k B i j C i j k and D i j k� � � � � � � � � � �+ +( ) +( ) + −( ) − −( ), ,2 5 3 2 3 6
AB i j i j k i j k� ��� � � � � � � � �= +( ) − + +( ) = + −2 5 4
CD i j k i j k i j k� ��� � � � � � � � � �= − −( ) − + −( ) = − − +6 3 2 3 2 8 2
AnglebetweenAB and CD� ��� � ���
cos .
θ = =− − −
×=−
= −AB CDAB CD
� ��� � ���� ��� � ���
2 32 23 2 6 2
3636
1
q = p
Q AnglebetweenAB and CD� ��� � ���
is p.
⇒ AB and CD� ��� � ���
are opposite and collinear vector.
23. Find the value of l, so that the lines 13
7 14 32
−=
−=
−x y zλ
and 7 73
51
65
−=
−=
−x y zλ
are at
rightangles.Also,findwhetherthelinesareintersectingornot.Ans. Given lines are
![Page 14: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/14.jpg)
STUDYmate
14
13
7 14 32
−=
−=
−x y zλ
⇒ x y z−−
=−
=−1
32
7
32λ ...(i)
and 7 73
51
65
−=
−=
−x y zλ
⇒ x y z−−
=−
=−−
137
51
65λ
...(ii)
If lines (i) and (ii) are at right angle then,
−( ) −
+ ( ) + −( ) =3 3
7 71 2 5 0λ λ
⇒ 97 7
10 0λ λ+ − =
⇒ 10l – 70 = 0 ⇒ l = 7 Now from (i),
x y z−−
=−
=−
=13
21
32
δ ...(Say)
Then general point P (–3d + 1, d + 2, 2d + 3). From (ii),
x y z−−
=−
=−−
=13
51
65
µ ...(Say)
Then general point in the line is Q (–3m + 1, m + 5, –5m + 6) If lines intersect then, –3d + 1 = –3m + 1 ⇒ –3d + 3m = 0 ⇒ d = m ...(iii) d + 2 = m + 5 ⇒ d – m = 3 ...(iv) and 2d + 3 = –5m + 6 ⇒ 2d + 5m = 3 ...(v) Solving (iv) and (v) 2(3 + m) + 5m = 3 6 + 7m = 3
µ δ=−
=37
187
,
Herem and d doesnot satisfy (iii). \ Linesarenotintersecting
Section – D
24. If A =
1 1 11 0 23 1 1
, findA–1.Hence,solvethesystemofequations
x + y + z = 6, x + 3z = 7, 3x + y + z = 12. OR
Find the inverse of the following matrix using elementary operations.
![Page 15: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/15.jpg)
STUDYmate
15
A =−
−−
1 2 21 3 00 2 1
Ans. A =
1 1 11 0 23 1 1
|A| = 1(–2) – 1 (1 – 6) + 1(1) |A| = –2 + 5 + 1 |A| = 4
Cofactor matrix, C =−
−− −
2 5 10 2 22 1 1
adj A( ) =−
− −−
2 0 25 2 11 2 1
A adj A− = ( )1 1
|A|
A− =−
− −−
1 14
2 0 25 2 11 2 1
Let,B =
1 1 11 0 23 1 1
, which is equal to A.
\ B–1 = A–1
B− =−
− −−
1 14
2 0 25 2 11 2 1
X
xyzand C=
=
6712
BX=C X=B–1C
xyz
=
−− −
−
14
2 0 25 2 11 2 1
6712
xyz
=
− +− −+ −
14
12 2430 14 126 14 12
xyz
=
312
\ x = 3, y = 1, z = 2.OR
![Page 16: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/16.jpg)
STUDYmate
16
A =−
−−
1 2 21 3 00 2 1
A = IA
1 2 21 3 00 2 1
1 0 00 1 00 0 1
−−
−
=
A
R2 →R2+R1
1 2 20 5 20 2 1
1 0 01 1 00 0 1
−−
−
=
A
R1 →R1+2R3andR2 →R2+2R3
1 2 00 1 00 2 1
1 0 21 1 20 0 1
−
−
=
A
R1 →R1+2R2andR3 →R3+2R2
1 0 00 1 00 0 1
3 2 61 1 22 2 5
=
A
I = BA
B A= =
−1
3 2 61 1 22 2 5
25. Atankwithrectangularbaseandrectangularsides,openatthetopistobeconstructedsothatits depth is 2 m and volume is 8 m3.Ifbuildingoftankcosts`70persquaremetreforthebaseand `45 per square metre for the sides, what is the cost of least expensive tank?
Ans.LetVbethevolumeoftankandbasedimensionsareaandbmeter. Given:
2 m
a m
b m
V = 8 m3
V=2×a×b V=2abm3
2ab=8 ab=4
ba
=4 ...(i)
Areaofthebase=abm2
Costforbuildingbase=`70ab Areaofthesides=2(2b)+2(2a) Areaofthesides=4(a+b)m2
Costforbuildingsides=45×4(a+b)=`180(a+b) Toalcostfortankbuilding=70ab+180(a+b)
C aa
aa
= × × + +
70 4 180 4 ...[Using (i)]
C a
a= + +
280 180 4
dCda a
= + −
0 180 1 4
2
![Page 17: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/17.jpg)
STUDYmate
17
dCda a
= −
180 1 4
2
For minima or maxima,
dCda
= 0
180 1 4 02−
=a
a=2andb=2 ...[Using(i)]
d Cda a
2
2 3180 8=
d Cda aa
2
22
3
180 8 0=
=×
>
d Cda
2
2 0>
\ Costforbuildingminimumata=2mandb=2m.
Minimum cost = ` 280 180 2 42
+ +
= ` 1,000
26. Usingintegration,findtheareaoftriangleABC,whoseverticesareA(2,5),B(4,7)andC(6,2).OR
Findtheareaoftheregionlyingabovex-axisandincludedbetweenthecirclex2 + y2 = 8x and insideoftheparabolay2 = 4x.
Ans.FirstwefindtheequationsofthesidesoftriangleABCbyusing y yy yx x
x x− =−−
−12 1
2 11( )
The equation of AB is y x− =−−
−5 7 54 2
2( )
⇒ x – y + 3 = 0 ...(i)
The equation of BC is y x− =−−
−7 2 76 4
4( )
⇒ 5x – 2y + 34 = 0 ...(ii)
The equation of side AC is y x− =−−
−5 2 56 2
2( )
⇒ 3x + 4y – 26 = 0 ...(iii) Clearly, Area of ∆ABC = Area ADB + Area BDC AreaADB:TofindareaADB,wesliceitintoverticalstrips.Weobservethateachverticalstrip
has its lower end on side AC and the upper end on AB. So, the approximating rectangle has Length=(y2 – y1),Width=∆x and Area = (y2 – y1) ∆x. Since the approximating rectangle can
move from x = 2 to x = 4.
\ Area ADB ( )y y dx2 12
4
−∫
⇒ Area ADB = ( )x x dx+ −−
∫ 3 26 3
42
4
[Q P(x, y1) and Q(x, y2) lie on (iii) and (i) respec. Q 3x + 4y1 – 26 = 0 and y2 = x + 3]
Similarly, we have
⇒ Area BDC = −∫ ( )y y dx4 34
6
⇒ Area BDC =−
−
−
∫
34 52
26 344
6 x x dx [QR(x,y3) and S(x, y4) lie on (iii) and (ii) respec. Q 3x + 4y3 – 26 = 0 and 5x + 2y4 – 34 = 0]
![Page 18: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/18.jpg)
STUDYmate
18
\ Area of ∆ABC = + −−
+−
−
−
∫ ( )x x dx x x3 26 3
434 52
26 342
4
∫
4
6
dx
⇒ Area of ∆ABC = −
+ −
1472
14 42 72
2
2
4 2
4
6x x x x
= 7 square units. OR
The given equation of the circle x2 + y2=8xcanbeexpressedas(x–4)2 + y2 = 16. Thus, the centreofthecircleis(4,0)andradiusis4.Itsintersectionwiththeparabolay2 = 4x gives
x2 + 4x = 8x or x2 – 4x = 0 or x (x – 4) = 0 or x = 0, x = 4 Thus, the points of intersection of these two curves are O(0, 0)
andP(4,4)abovethex-axis. From the Fig 8.16, the required area of the region OPQCO included
betweenthesetwocurvesabovex-axisis = (area of the region OCPO) + (area of the region PCQP)
= +∫ ∫ydx ydx
0
4
4
8
= + = − −∫ ∫2 4 40
4 2 2
4
8x dx x dx( )
= ×
+ −∫2 2
34
32
0
4
2 2
0
4
x t dt, where, x – 4 = t
= + − + ×
−323 2
4 12
44
2 2 2 1
0
4t t tsin
= + × + ×
= + + ×
= + = +−32
342
0 12
4 1 323
0 82
323
4 438 32 1sin (π
π ππ) square units
27. Find the vector and Cartesian equations of the plane passing through the points (2, 2, –1), (3,4,2)and(7,0,6).Alsofindthevectorequationofaplanepassingthrough(4,3,1)andparalleltotheplaneobtainedabove.
OR
Find the vector equation of the plane that contains the lines r i j i j k
= + + + −( ) ( )^ ^ ^ ^ ^
λ 2 and the point (–1,3,–4).Also,findthelengthoftheperpendiculardrawnfromthepoint(2,1,4) to the plane,
thusobtained.Ans. Given A(2, 2, –1), B(3, 4, 2) and C(7, 0, 6) equation of plane:
x x y y z zx x y y z zx x y y z z
− − −− − −− − −
=1 1 1
2 1 2 1 2 1
3 1 3 1 3 1
0
x y z− − +− − +− − +
=2 2 1
3 2 4 2 2 17 2 0 2 6 1
0
x y z− − +
−=
2 2 11 2 35 2 7
0
ExpendingalongR1, (x – 2) (20) – (y – 2) (–8) + (z + 1) (–12) = 0 5(x – 2) + 2(y – 2) – 3(z + 1) = 0
![Page 19: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/19.jpg)
STUDYmate
19
5x – 10 + 2y – 4 – 3z – 3 = 0 5x + 2y – 3z – 17 = 0 Cartesian form
r i j k
.( )^ ^ ^
5 2 3 17 0+ − − = vector form
The equation of the plan parallel to the plane r i j k
.( )^ ^ ^
5 2 3 17 0+ − − = is r i j k d
.( ) .^ ^ ^
5 2 3 0+ − − =
Since it passes through (4, 3, 1). Therefore, ( ).( )^ ^ ^ ^ ^ ^
4 3 5 2 3i j k i j k d+ + + − = 20 + 6 – 3 = d d = 23.
\ the required equation of plane is r i j k� � � �. 5 2 3 23+ −( ) =
OR
Given: line r i j i j k
= + + + −( ) ( )^ ^ ^ ^ ^
λ 2
b i j k1 2���
= + −^ ^ ^
AvectorjoiningpointA(1,1,0)andB(–1,3,–4)
AB b i j k� ��� � ��
= = − + −2 2 2 4^ ^ ^
\ normalvectortotheplaneisgivenby n b b
�� ��� � ��= ×1 2
= −− −
i j k^ ^ ^
1 2 12 2 4
= − + − − − + +i j k^ ^ ^( ) ( ) ( )8 2 4 2 2 4
n i j k��= − + +6 6 6
^ ^ ^.
Equation of plane passing through (–1, 3, –4)
( ).r a n� �� ��− = 0
( ( )).( )^ ^ ^ ^ ^ ^
r i j k i j k
− − + − − + + =3 4 6 6 6 0
r i j k
.( ) ( )^ ^ ^
− + + − + − =1 3 4 0
r i j k
.( )^ ^ ^
− + + = 0 – x + y + z = 0 or x – y – z = 0.
PQ =
− −
+ − + −
| |
( ) ( )
2 1 4
1 1 12 2 2
PQ = =33
3 units.
28. AmanufacturerhasthreemachineoperatorsA,BandC.ThefirstoperatorAproduces50%ofdefectiveitems,whereastheothertwooperatorsBandCproduces5%and7%detectiveitemsrespectively.Aisonthejobfor50%ofthetime,Bonthejob30%ofthetimeandConthejobfor20%ofthetime.Alltheitemsareputintoonestockpileandthenoneitemischosenatrandomfromthisandisfoundtobedefective.WhatistheprobabilitythatitwasproducedbyA?
Ans.Let E1 = Machine operator A E2 = Machine operator B E3 = Machine operator C and F = Item is defective
![Page 20: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/20.jpg)
STUDYmate
20
\ P E P F
P E P F
P E
1 1
2 2
3
50100
510
1100
30100
310
5100
( ) = = =
( ) = = =
(
( /E )
( /E )
)) = = =20100
210
71003P F( /E )
Then, PP
P E(E /F)
(E ).P(F/E )(E ).P(F/E ) P(E ).P(F/ ) P(E ).P(F1
1 1
1 1 2 2 3
=+ + //E )3
=+ +
=+ +
=
510
1100
510
1100
310
5100
210
7100
55 15 14
534
.
. . .
29. A manufacturer has employed 5 skilled men and 10 semi-skilled men and makes two models A andBofanarticle.ThemakingofoneitemofmodelArequires2hoursworkbyaskilledmanand2hoursworkbyasemi-skilledman.OneitemofmodelBrequires1hourbyaskilledmanand3hoursbyasemi-skilledman.Nomanisexpectedtoworkmorethan8hoursperday.Themanufacturer’sprofitonanitemofmodelAis`15 and on an item of model B is `10.Howmanyofitemsofeachmodelshouldbemadeperdayinordertomaximizedailyprofit?FormulatetheaboveLPPandsolveitgraphicallyandfindthemaximumprofit.
Ans.LetthenumberofitemsofmodelAandBbexandyrespectively. ThenmaximizeprofitZ=15x+10y ...(i) Subjecttocontraints 2x + y ≤ 40 ...(ii) 2x + 3y ≤ 80 ...(iii) and x ≥ 0, y ≥ 0 Let2x+y=40 and 2x+3y=80
x 0 20 x 10 40y 40 0 y 20 0
![Page 21: Code No. 65/1/1 · 2019-05-31 · Studymate Solutions to CBSE Board Examination 2018-2019 Series : BVM/1 Candidates must write the Code on the title page of the answer-book. Code](https://reader030.fdocuments.us/reader030/viewer/2022040400/5e6dc6ec2cff3c03aa2811e3/html5/thumbnails/21.jpg)
STUDYmate
21
Corner Point Optimum Value (15x + 10y)(20, 0) ` 300(10, 20) `350 (Maximum)
0 803
,
`
8003
No.ofitemofmodelAandBshouldbemanufactured10and20respectivelyandmaximumprofit=`350.
Bycornerpointtheoremoptimumvaluewillbemaximumatcornerpoint(10,20).
vvvvv