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

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4 Please check that this question paper contains 29 questions.

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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.

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

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

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

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

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

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

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

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.

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

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

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

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

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.

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

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

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]

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

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

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

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