MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Summer 2015 Examination Subject & Code: Applied Maths (17301) Que. No.

Sub. Que.

Model Answer

Page No: 1/26

Model Answers

Important Instructions to the Examiners: 1) The Answers should be examined by key words and not as word-to-word as given in the model answer scheme. 2) The model answer and the answer written by candidate may vary but the examiner may try to assess the understanding level of the candidate. 3) The language errors such as grammatical, spelling errors should not be given more importance. (Not applicable for subject English and Communication Skills.) 4) While assessing figures, examiner may give credit for principal components indicated in the figure. The figures drawn by the candidate and those in the model answer may vary. The examiner may give credit for any equivalent figure drawn. 5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values may vary and there may be some difference in the candidate’s Answers and the model answer. 6) In case of some questions credit may be given by judgment on part of examiner of relevant answer based on candidate’s understanding. 7) For programming language papers, credit may be given to any other program based on equivalent concept.

Marks

Total Marks

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 1)

Sub. Que.

Page No: 2/26

Model Answers

Marks

Total Marks

Attempt any TEN of the following: a)

Ans.

At which point on the curve y = 3x − x 2 the slope of the tangent is -5. y = 3x − x2

dy = 3 − 2x dx But tangent is parallel to x-axis. ∴ 3 − 2 x = −5 ∴

½

∴x = 4

½

∴ y = 3 ( 4 ) − ( 4 ) = −4 2

½

∴ the point is ( 4, − 4 ) . b) Ans.

½

2

------------------------------------------------------------------------------------Divide 80 into two parts such that their product is maximum. Let x, y be the numbers. But x + y = 80 i. e., y = 80 - x

To maximize, p = xy = x ( 80 − x )

∴ p = 80 x − x 2 dp ∴ = 80 − 2 x dx d2 p ∴ 2 = −2 dx

½ ½

dp =0 dx or 80 = 2 x

For stationary values, ∴ 80 − 2 x = 0 ∴ x = 40

½

d2 p = −2 < 0 dx 2 ½ ∴ At x = 40, p has max imum value. ------------------------------------------------------------------------------------Evaluate: ∫ sin 3 x cos xdx At x = 40,

c)

3 ∫ sin x cos xdx

Ans.

= ∫ t 3 dt t4 +c 4 sin 4 x = +c 4 =

Put sin x = t ∴ cos xdx = dt

2

½ ½ ½ ½

2

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 1)

Sub. Que.

Page No: 3/26

Model Answers Evaluate

Ans.

∫ xe dx = x ∫ e dx − ∫  ∫ e dx  dx ( x ) ⋅ dx

x

x

x

x

d

= xe x − ∫ e x dx

½ ½

= xe x − e x + c

Ans.

Total Marks

∫ xe dx .

d)

e)

Marks

½+½

2

------------------------------------------------------------------------------------1 dx Evaluate ∫ ( x + 3)( x + 2) I =∫

1

( x + 3)( x + 2 ) 1

( x + 3)( x + 2 )

=

dx

A B + x+3 x+2

½

∴ A = −1 − − − − − − − − − − ( *)

B =1 1

( x + 3)( x + 2 )

=

½

−1 1 + x+3 x+2

1   −1 ∴I = ∫  + dx  x + 3 x + 2  = − log ( x + 3 ) + log ( x + 2 ) + c

½+½

Note (*): There are various methods to find the values of A and B to partially factorize the given expression including direct method. Students may apply any one of the methods. Take in count all such methods. OR

I =∫

1

( x + 3)( x + 2 )

dx

1 dx x + 5x + 6 1 =∫ dx 25 25 x2 + 5x + − + 6 4 4 1 =∫ dx 2 2 5 1   x +  −  2 2  =∫

2

½

½

2

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 1)

Sub. Que.

Page No: 4/26

Model Answers

Marks

5 1   x+ 2 − 2  1 ½ = log  +c  5 1 1  x+ +  2   2 2 2  x+2 = log  ½ +c  x+3 ------------------------------------------------------------------------------------f)

Evaluate

∫

log e 2

0

Total Marks

2

e 2 x dx log 2

Ans.

∫

log e 2

0

g) Ans.

 e2 x  e dx =    2 0

½

2x

½ e 2log 2 e0 = − 2 2 4 1 = − 2 2 3 1 = or 1.5 2 ------------------------------------------------------------------------------------Find the area between the line y = 2 x and x = 1 and x = 3 .

∫

3

1

3

y ⋅ dx = ∫ 2 x ⋅ dx

2

½

1

3

 x2  = 2 ⋅   2 1

or

3

 x 2  1

= 32 − 1

½ ½

=8 ½ -------------------------------------------------------------------------------------

h)

Ans.

2

Find the order and degree of the following equation: d2y dy + 1+ =0 2 dx dx Order = 2

1

For degree, 2

 d2y  dy  2  = 1+ dx  dx  ∴ Degree = 2

1

2

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 1)

Sub. Que.

Page No: 5/26

Model Answers

i)

Form a differential equation if y 2 = 4ax .

Ans.

y 2 = 4ax

Marks

dy 1 = 4a dx dy ½ ∴ y2 = 2 y ⋅ x dx dy dy 2x ∴ y = 2x or =y ½ dx dx dy dy y or 2 x − y = 0 or − = 0. dx dx 2 x ------------------------------------------------------------------------------------From a pack of 52 cards one card is drawn at random. Find the probability of getting a king.

Total Marks

∴2y

j)

Ans.

n = n ( S ) = 52

2

½

m = n ( A) = 4

½

n ( A) ∴ p = p ( A) = n(S )

4 ½ 52 1 ½ = or 0.077 13 ------------------------------------------------------------------------------------An unbiased coin is tossed 5 times. Find the probability of getting three heads. =

k)

Ans.

1 = 0.5 2 Here n = 5 p=

∴q = 1− p =

1 = 0.5 2

2

½

∴ p = nC r p r q n − r 3

2

1 1 = C3     2 2 5 = or 0.3125 16 Note: Due to the use of advance non-programmable scientific calculators which is permissible in the board examination, writing directly the values of nCr or 5

Cr p r q n − r is permissible. No marks to be deducted for calculating directly the value. n

1

½

2

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 1)

Sub. Que.

Page No: 6/26

Model Answers

l)

A die is thrown, find the probability of getting an odd number.

Ans.

∴n = n(S ) = 6

Total Marks

½

∴ m = n ( A) = 3 ∴p=

Marks

½

n ( A) n(S )

3 ½ 6 1 = or 0.5 ½ 2 ------------------------------------------------------------------------------------=

2

Attempt any four of the following:

2) a)

Find the equation of tangent and normal to the curve y = x(2 – x) at (2, 0). y = x (2 − x)

Ans.

dy = 2 − 2x dx ∴ the slope of tangent at ( 2, 0 ) is ∴

1

½

m = 2 − 4 = −2 ∴ the equation of tangent is y − 0 = −2 ( x − 2 ) ∴ y = −2 x + 4

or

2x + y = 4

1 1 = m 2 ∴ the equation of normal is ∴ the slope of normal = −

½ ½ ½

b)

1 ½ ( x − 2) 2 ∴ 2 y = x − 2 or x − 2 y − 2 = 0 or − x + 2 y + 2 = 0 ½ ------------------------------------------------------------------------------------Find radius of curvature of the curve x = a cos3 θ , y = a sin 3 θ at

Ans.

θ=

y−0 =

π

4

x = a cos3 θ dx ∴ = −3a cos 2 θ sin θ dθ

½

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 2)

Sub. Que.

Page No: 7/26

Model Answers

Marks

y = a sin 3 θ dy ∴ = 3a sin 2 θ cos θ dθ dy dy / dθ 3a sin 2 θ cos θ ∴ = = = − tan θ dx dx / dθ −3a cos 2 θ sin θ d 2 y d  dy  dθ & =  × dx 2 dθ  dx  dx 1 = − sec2 θ × −3a cos 2 θ sin θ ∴ at θ =

and

Total Marks

½

½

½

π

, 4 π dy = − tan = −1 dx 4 2 π d y 1 = sec 2 × 2 dx 4 3a cos 2 π sin π 4 4 4 2 = 3a

½

½

3

  dy  2  2 3 2 2 1 +      1 + −1 ½   dx    ( )  = ∴ρ =  d2y 4 2 2 dx 3a 3a ½ = 2 ------------------------------------------------------------------------------------c)

Find the maximum and minimum value of y = x ³ −

Ans.

y = x³ −

15 x ² + 18 x . 2

15 x ² + 18 x 2

dy = 3x 2 − 15 x + 18 dx d2y ∴ 2 = 6 x − 15 dx

½

∴

For stationary values, ∴ 3x 2 − 15 x + 18 = 0 ∴ x = 2, 3

½

dy =0 dx ½+½

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 2)

Sub. Que.

Page No: 8/26

Model Answers

d2y = 6 ( 2 ) − 15 = −3 < 0 dx 2 ∴ At x = 2, y has max imum value and it is At x = 2,

Marks

Total Marks

½

15 ½ ( 2 ) ² + 18 ( 2 ) = 14 2 d2y ½ At x = 3, = 6 ( 3) − 15 = 3 > 0 dx 2 ∴ At x = 3, y has min imum value and it is 15 y = ( 3) ³ − ( 3) ² + 18 ( 3) = 13.5 ½ 2 ------------------------------------------------------------------------------------y = ( 2) ³ −

d)

Evaluate

e x ( x + 1) ∫ cos ²( xe x )dx

Ans. e ( x + 1) x

∫ cos ²( xe )dx x

=∫

4

Put xe x = t

∴ ( xe x + e x ⋅1) dx = dt

1

∴ e x ( x + 1) dx = dt

dt cos 2 t

1 ½

= ∫ sec2 tdt = tan t + c

1

= tan ( xe ) + c x

e)

Ans.

------------------------------------------------------------------------------------- ½ sec 2 x Evaluate ∫ dx 3 tan 2 x − 2 tan x − 5

sec 2 x ∫ 3 tan 2 x − 2 tan x − 5 dx

Put tan x = t ∴ sec2 xdx = dt

½

1 dt 3 t − 2 t− 5 2 5  ∴3 t2 − 2 t− 5 = 3 t2 − t−  3 3  2 1 1 5  = 3 t2 − t+ − −  3 9 9 3  =∫

2

 1  2  4  2  = 3  t −  −     3   3  

1

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 2)

Sub. Que.

Page No: 9/26

Model Answers

Marks

1

∴I = ∫

dt  1  2  4  2  3  t −  −     3   3   1 1 = ∫ dt 2 3  1   4 2  t−  −    3  3  1 4  t− 3 − 3  1 1 = ⋅ ⋅ log  +c 1 4 3 4   t − + 2   3 3 3  5  t− 3  1 = ⋅ log  +c 8  t+1    1  3 t− 5  = ⋅ log  +c 8  3 t + 3 

½

1

½

1  3 tan x − 5  = ⋅ log  +c ½ 8  3 tan x + 3  ------------------------------------------------------------------------------------f)

Evaluate

Total Marks

4

x sin −1 x ∫ 1 − x² dx

Ans. x sin −1 x ∫ 1 − x² dx

Put sin −1 x = t 1 ∴ dx = dt 1 − x² Also x = sin t

= ∫ sin t ⋅ t ⋅ dt = t ∫ sin tdt − ∫

1

½

( ∫ sin tdt ) dtd (t ) dt

= t ( − cos t ) − ∫ ( − cos t ) ⋅1 ⋅ dt

½ 1

= −t cos t + ∫ cos t ⋅ dt = −t cos t + sin t + c

= − sin −1 x ⋅ cos ( sin −1 x ) + x + c

-------------------------------------------------------------------------------------

½ ½

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 3)

Sub. Que.

Page No: 10/26

Model Answers

Marks

Total Marks

Attempt any four of the following: a)

π /2

∫

Evaluate

0

Ans.

dx 9 − 4 x²

The given problem cannot be solved within the given limits because for the integrating the given function within the prescribed limits the function must be well defined on the given 1 π interval. For example at x = , is a non-real number 2 9 − 4 x² 4 and hence the function is not defined on the interval 0, π  . 4  -------------------------------------------------------------------------------------

4

π /3

b)

Ans.

Evaluate

sin x dx sin x + cos x /6

∫ π

Re place x → π / 2 − x ∴ sin x → cos x & cos x → sin x

π /3

sin x I= ∫ dx sin x + cos x π /6 ∴I =

π /3

cos x dx cos x + sin x /6

∫ π

π /3

sin x + cos x dx sin x + cos x /6

∫ π

∴ 2I =

1

π /3

∫ 1⋅ dx

∴ 2I =

½

π /6

∴ 2 I = [ x ]π / 6 π /3

π

= ∴I =

1

3

π

−

½

π ½

6

½

12 OR

I=

π /3

sin x dx sin x + cos x /6

∫ π

( )

) (

sin π − x 2 = ∫ dx π − x + cos π − x π / 6 sin 2 2 π /3

(

∴I =

½

)

π /3

cos x dx cos x + sin x /6

∫ π

½

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 3)

Sub. Que.

Page No: 11/26

Model Answers ∴ 2I = ∴ 2I =

Marks

Total Marks

π /3

sin x + cos x dx sin x + cos x /6

∫ π

1

π /3

∫ 1⋅ dx π

½

/6

∴ 2 I = [ x ]π / 6 π /3

π

=

3

−

π

π

∴I =

½ ½

6

½

c)

12 ------------------------------------------------------------------------------------Find the area bounded by two curves y 2 = x and x 2 = y

Ans.

Given y 2 = x and x 2 = y

4

∴( x2 ) = x 2

∴ x = 0, x = 1 A=∫

b

a

½+½

( y2 − y1 ) dx 1

1

= ∫  x − x 2  dx 0 2 x3  =  x 3/ 2 −  3 0 3

1

2 13  =  ⋅13/ 2 −  − 0 3 3 1 = or 0.333 3

½

1

OR

∴ x = 0, x = 1 A=∫

b

a

½

4

½+½

( y2 − y1 ) dx

1

= ∫  x 2 − x  dx 0

1

1

 x3 2  =  − x3/ 2  3 3 0 13 2  =  − ⋅13/ 2  − 0 3 3  1 =− or −0.333 3 1 ∴ the area = or 0.333 3

1

½

½

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 3)

Sub. Que.

Model Answers

(

Solve xy 2 dy − x3 + y 3 dx = 0 given y = 0 when x = 1.

Ans.

xy 2 dy − ( x3 + y 3 ) dx = 0

∴

dy dv =v+x dx dx

3 dv x + ( vx ) 1 + v 3 ∴v + x = = 2 dx xv 2 x 2 v 3 dv 1 + v ∴x = 2 −v dx v dv 1 ∴x = 2 dx v 1 ∴ ∫ v 2 dv = ∫ dx x 3 v ∴ = log x + c 3 y3 ∴ 3 = log x + c 3x At x = 1 & y = 0, c = 0 3

Ans.

Total Marks

dy x 3 + y 3 = dx xy 2

Put y = vx

e)

Marks

)

d)

∴

Page No: 12/26

1 ½

½

½

½ ½

y3 ½ ∴ 3 = log x 3x ------------------------------------------------------------------------------------2 dy Solve the differential equation ( x + y ) = a2 dx Put x + y = v dy dv = or dx dx  dv  ∴ v 2  − 1 = a 2  dx  ∴1 +

dv a2 −1 = 2 dx v 2 dv a a2 + v2 ∴ = 2 +1 = dx v v2  v2  ∴  2 2  dv = dx a +v 

dy dv = −1 dx dx

½ ½

∴

½

½

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 3)

Sub. Que.

Page No: 13/26

Model Answers

 v2  ∴ ∫  2 2  dv = ∫ dx  a +v   a2  ∴ ∫ 1 − 2 2  dv = ∫ dx  a +v  1 v ∴ v − a 2 ⋅ tan −1   = x + c a a  x+ y ∴ x + y − a tan −1   = x+c  a 

Marks

Total Marks

½ 1

or

 x+ y y − a tan −1  =c  a 

½

4

------------------------------------------------------------------------------------f)

Ans.

Solve x

dy − y = x2 dx

dy − y = x2 dx dy 1 ∴ − ⋅y=x dx x 1 ∴ P = − and Q = x x pdx ∴ IF = e ∫ x

=e

1

∫ − x dx

= e − log x 1 = x

1

∴ y ⋅ IF = ∫ Q ⋅ IF ⋅ dx + c ∴y⋅

1 1 = ∫ x ⋅ ⋅ dx + c x x

y = 1.dx + c x ∫ y ∴ = x+c x ∴

-------------------------------------------------------------------------------------

1 1 1

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 4)

Sub. Que.

Page No: 14/26

Model Answers

Marks

Total Marks

Attempt any Four of the following: 5

∫

9− x dx 9− x + 3 x+3 3

a)

Evaluate

Ans.

9− x I =∫3 dx 9− x + 3 x+3 1

3

1

5

3

5

∴I = ∫

Re place x → 6 − x ∴9 − x → x + 3

1

& x+3→9− x

x+3 dx x+3 + 3 9− x 3

3

1

1

9− x + 3 x+3 dx 3 9− x + 3 x+3

5 3

∴ 2I = ∫ 1 5

∴ 2 I = ∫ 1⋅ dx 1

∴ 2 I = [ x ]1

½

∴ 2I = 5 − 1 ∴I = 2

1

5

½

4

OR 5

I =∫

9− x dx 9− x + 3 x+3 3

3

1 5

∴I = ∫ 1

3 3

5

∴I = ∫

9 − (6 − x)

9 − (6 − x ) + 3 (6 − x) + 3

dx

½

x+3 dx x+3+ 3 9− x

½

9− x + 3 x+3 dx 3 9− x + 3 x+3

1

3

3

1

5 3

∴ 2I = ∫ 1 5

∴ 2 I = ∫ 1 ⋅ dx 1

∴ 2 I = [ x ]1

½

∴ 2I = 5 − 1 ∴I = 2

1

5

-------------------------------------------------------------------------------------

½

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 4)

Sub. Que. b) Ans.

Page No: 15/26

Model Answers

Marks

Total Marks

dx

∫ 4 cos ² x + 9sin ² x

Evaluate

dx

∫ 4 cos ² x + 9sin ² x dx / cos ² x 4cos ² x + 9sin ² x cos ² x 2 sec xdx =∫ 4 + 9 tan ² x =∫

Put tan x = t ∴ sec2 xdx = dt

dt 4+9t² dt =∫ 4  9 + t ²  9  dt 1 = ∫ 9  2 2   +t² 3 1 1  t  = ⋅ tan −1  +c 9 2/3  2/3 1  3 tan x  = tan −1  +c 6  2  ------------------------------------------------------------------------------------=∫

c)

x2 y2 Using integration find the area of the ellipse 2 + 2 = 1 a b

Ans.

x2 y2 + =1 a2 b2  x2  ∴ y 2 = b 2 1 − 2   a  2 b ∴ y 2 = 2 ( a2 − x2 ) a b 2 ∴y = a − x2 a Now y = 0 gives a 2 − x 2 = 0 i. e., x = a, − a a

½

½

1

1

1

1

∴ A = 4 ∫ ydx 0

= 4∫

a

0

b 2 a − x 2 dx a

1

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 4)

Sub. Que.

Page No: 16/26

Model Answers

Marks

Total Marks

a

b x 2 a2  x  2 = 4⋅  a − x + sin −1    a 2 2  a  0 =

1

 4b  a 2 0 + sin −1 (1)  − 0  a  2 

½

4b  a 2 π  ⋅ a  2 2  = π ab ½ ------------------------------------------------------------------------------------=

d)

Ans.

(

)

(

4

)

Solve 2 x 2 + 6 xy − y 2 dx + 3x 2 − 2 xy + y 2 dy = 0

M = 2 x 2 + 6 xy − y 2 ∂M ∴ = 6x − 2 y ∂y

1

N = 3x 2 − 2 xy + y 2 ∂N = 6x − 2 y ∂x ∴the equation is exact.

½

∴

∫

Mdx +

y cons tan t

∫ ( 2x

2

∫

½

Ndy = c

terms free from x

+ 6 xy − y 2 ) dx + ∫ y 2 dy = c

1

x3 x2 y3 + 6 y ⋅ − y2 x + =c 3 2 3 2 3 y3 2 2 or x + 3x y − y x + =c 3 3

∴ 2.

1

OR

( 2x

2

+ 6 xy − y 2 ) dx + ( 3x 2 − 2 xy + y 2 ) dy = 0

dy 2 x 2 + 6 xy − y 2 =− 2 dx 3x − 2 xy + y 2 dy dv Put y = vx ∴ = v + x dx dx 2 2 dv 2 x + 6vx − v 2 x 2 ∴v + x = − 2 dx 3x − 2vx 2 + v 2 x 2 ∴

1

½

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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_____________________________________________________________________________________________________

Subject & Code: Applied Maths (17301) Que. No. 4)

Sub. Que.

Page No: 17/26

Model Answers dv 2 + 6v − v 2 =− dx 3 − 2v + v 2 dv 2 + 6v − v 2 ∴x = − −v dx 3 − 2v + v 2 2 2 dv − ( 2 + 6v − v ) − v ( 3 − 2v + v ) ∴x = dx 3 − 2v + v 2 dv −2 − 9v + 3v 2 − v 3 ∴x = dx 3 − 2v + v 2  3 − 2v + v 2  dx dv = ∴ 2 3 x  −2 − 9v + 3v − v 

Marks

Total Marks

∴v + x

 3 − 2v + v 2  dx ∴∫  dv = ∫ 2 3 x  −2 − 9v + 3v − v  ∴−

½

½

1  −9 + 6v − 3v 2  dx dv = ∫  2 3 ∫ 3  −2 − 9v + 3v − v  x

1 1 ∴− log ( −2 − 9v + 3v 2 − v3 ) = log x + c 3 2 3  1 y  y  y  ∴− log  −2 − 9 ⋅ + 3   −    = log x + c ½ 3 x  x   x    -------------------------------------------------------------------------------------

(

)

e)

Solve 1 + x 2 dy − x 2 ydx = 0

Ans.

(1 + x ) dy − x 2

2

ydx = 0

1 x2 ∴ dy − dx = 0 y 1 + x2 ∴∫

1 x2 dy − ∫ dx = c y 1 + x2

∴∫

1 1   dy − ∫ 1 − dx = c 2  y  1+ x 

∴ log y − ( x − tan −1 x ) = c or

4

log y − x + tan −1 x = c

-------------------------------------------------------------------------------------

1

1

1 1 4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

(ISO/IEC - 27001 - 2005 Certified)

_____________________________________________________________________________________________________

Subject & Code: Applied Maths (17301) Que. No. 4)

Sub. Que. f)

Model Answers

Marks

d2y dy +x +y=0 2 dx dx

y = sin ( log x )

1

1 dy = cos ( log x ) ⋅ dx x dy ∴ x = cos ( log x ) dx d 2 y dy 1 ∴x 2 + = − sin ( log x ) ⋅ dx dx x 2 d y dy ∴ x2 2 + x = − y dx dx 2 d y dy ∴ x2 2 + x + y = 0 dx dx ∴

1

1 1

------------------------------------------------------------------------------------5)

Attempt any Four of the following: a)

The probability that A can shoot at a target is shoot at the same target is

5 and B can 7

3 . (A and B shoot independently.) 5

Find the probability that i) The target is not shot at all. ii) The target is shot by at least one of them. Ans.

Total Marks

Show that y = sin ( log x ) is solution of differential equation

x2

Ans.

Page No: 18/26

5 5 2 ∴ P ( A ') = 1 − = 7 7 7 3 3 2 ∴ P ( B ') = 1 − = P ( B) = 5 5 5 i ) P ( target is not shot ) = P ( A '& B ') P ( A) =

½ ½

= P ( A ') ⋅ P ( B ' ) 2 2 = ⋅ 7 5 4 = or 35

1 0.114

½

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

(ISO/IEC - 27001 - 2005 Certified)

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Subject & Code: Applied Maths (17301) Que. No. 5)

Sub. Que.

Page No: 19/26

Model Answers

Marks

Total Marks

ii ) P ( at least1shoot ) = 1 − p(target is not shot) = 1−

4 or 35

1 − 0.114

1

31 ½ or 0.886 35 ------------------------------------------------------------------------------------Note for Numerical Problems: For practical purpose, generally the values of fractional numbers are truncated up to 3 decimal points by the method of rounded-off. Thus the solution is taken up to 3 decimal points only. If answer is truncated more than 3 decimal points, the final answer may vary for last decimal points. Thus 31/35 is actually 0.885714285 but can be taken as 0.886. Due to the use of advance calculators, such as modern scientific nonprogrammable calculators, the step 31/35 may not written by the students and then directly the answer 0.012 is written. In this case, no marks to be deducted. ------------------------------------------------------------------------------------If 30% of the bulbs produced are defective, find the probability that out of 4 bulbs selected, a) one is defective, b) at the most two are defective. =

b)

Ans.

4

30 = 0.3 100 ∴ q = 0.7 Here n = 4 p=

i ) p = nC r p r q n − r = 4C1 ( 0.3) ( 0.7 ) 1

3

1 1

= 0.412 ii ) p = p ( 0 ) + p (1) + p ( 2 ) = 4C0 ( 0.3) ( 0.7 ) + 4C1 ( 0.3) ( 0.7 ) + 4C2 ( 0.3) ( 0.7 ) 0

4

1

3

= 0.24 + 0.412 + 0.265 = 0.916

2

2

1 1

OR p = 1 −  p ( 3) + p ( 4 ) 

4

OR

3 1 4 0 = 1 −  4C3 ( 0.3) ( 0.7 ) + 4C4 ( 0.3) ( 0.7 )    = 1 − [ 0.076 + 0.008]

1

= 0.916

1

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

(ISO/IEC - 27001 - 2005 Certified)

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Subject & Code: Applied Maths (17301) Que. No. 5)

Sub. Que.

Page No: 20/26

Model Answers

Marks

Total Marks

Note: Due to the use of advance non-programmable scientific calculators which is permissible in the board examination, writing directly the values of nCr or Cr p r q n − r is permissible. No marks to be deducted for calculating directly the value. ------------------------------------------------------------------------------------In a certain examination 500 students appeared, mean score is 68 and S. D. is 8. Assuming data is normally distributed, find the number of students scoring, a) less than 50 b) more than 60. n

c)

(Given that area between Z = 0 to Z = 2.25 is 0.4878 and area between Z = 0 & Z = 1 is 0.3413.)

Ans.

Given x = 68

N = 500

50 − 68 = −2.25 8 σ ∴ p ( x ≤ 50 ) = p ( z ≤ −2.25) i) z =

x−x

σ =8

=

½

= p ( 2.25 ≤ z )

= 0.5 − p ( 0 ≤ z ≤ 2.25 )

½

= 0.5 − 0.4878 = 0.0122

½

∴ no. of students = N ⋅ p = 500 × 0.0122 = 6.1 i.e., 6 60 − 68 = −1 8 σ ∴ p ( 60 ≤ x ) = p ( −1 ≤ z ) ii ) z =

x−x

=

= p ( −1 ≤ z ≤ 0 ) + p ( 0 ≤ z ) = 0.3413 + 0.5 = 0.8413

½

½

½

½ ∴ no. of students = N ⋅ p = 500 × 0.8413 ½ = 420.65 i.e., 421 -------------------------------------------------------------------------------------

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 5)

Sub. Que. d)

Page No: 21/26

Model Answers π

Evaluate

Marks

Total Marks

dx

∫ 5 + 4cos x 0

Ans.

x =t 2 2dt 1− t2 ∴ dx = x = and cos 1+ t2 1+ t2 Put tan

x t

0 0

½

π

½

∞

π

∞

dx ∴∫ = 5 + 4 cos x ∫0 0

1 2dt ⋅ 2  1− t  1+ t2 5 + 4 2   1+ t 

∞

1 dt + 9 t 0

= 2∫

½

½

2

∞

1 dt t + 32 0

= 2∫

2

∞

1  t  = 2 ×  tan −1    3  3  0

2  tan −1 ∞ − tan −1 0  3 2 π  =   32

=

=

1

½

π

½ 3 -------------------------------------------------------------------------------------

e)

Ans.

Evaluate

∫x

2

4

x dx + 3x − 4

x dx + 3x − 4 x =∫ dx ( x − 1)( x + 4 )

∫x

2

 1/ 5 4 / 5  = ∫ + dx (Please refer next note)  x − 1 x + 4  1 4 = log ( x − 1) + log ( x + 4 ) + c 5 5

2

1+1

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 5)

Sub. Que.

Page No: 22/26

Model Answers

Marks

Total Marks

Note: To find the partial fractions, traditional partial fraction method is generally used. But apart from this direct method of partial fraction is also allowed here.

-------------------------------------------------------------------------------------

dy + y = z log x dx

f)

Solve x log x

Ans.

(Considering z as constant.) dy + y = z log x dx dy 1 z ∴ + y= dx x log x x 1 z ∴P = and Q = x log x x x log x

∴ IF = e ∫

1

pdx

=e

∫ x log x dx

=e

log ( log x )

= log x

1

∴ y ⋅ IF = ∫ Q ⋅ IF ⋅ dx + c z ∴ y ⋅ log x = ∫ ⋅ log x ⋅ dx + c x 1 ∴ ⋅ dx = dt Put log x = t x ∴ y ⋅ log x = z ∫ t⋅ dt + c ∴ y ⋅ log x = z ⋅ ∴ y ⋅ log x = z ⋅

1 ½ ½

2

t +c 2

( log x )

½ 2

½ +c 2 -------------------------------------------------------------------------------------

Attempt any Four of the following:

6) a)

1 2 2 If P ( A ) = , P ( B ') = and P ( A ∪ B ) = , find P ( A '∩ B ' ) and 2 3 3 P A . B

( )

Ans.

P ( A) =

1 2

2 1 P (B) = 1− = 3 3

½

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 6)

Sub. Que.

Page No: 23/26

Model Answers

Marks

Total Marks

i ) P ( A '∩ B ') = P ( A ∪ B ) ' = 1− P ( A ∪ B) 2 3

= 1−

½

1 or 0.333 3 ii ) P ( A ∩ B ) = P ( A ) + P ( B ) − P ( A ∪ B )

½

=

1 1 2 + − 2 3 3 1 = or 6 P ( A ∩ B) ∴P A = B P (B) =

½ ½

0.167

( )

1 1 = 6 1 3 1 0.5 = or ½ 2 ------------------------------------------------------------------------------------b)

Ans.

If the probability that an electric motor is defective is 0.01, what is the probability that the sample of 300 electric motors will contain exactly 5 defective motors? ( e−3 = 0.0498 ) p = 0.01

n = 300

∴ m = np = 0.01× 300 = 3 −m

⋅m r! −3 e ⋅ 35 ∴ p ( 5) = 5! = 0.101 p (r ) =

e

1

r

1½ 1½ 4

------------------------------------------------------------------------------------c)

4

Fit a Poisson distribution for the following observation: x f

20 8

30 12

40 30

50 10

60 6

70 4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 6)

Sub. Que.

Page No: 24/26

Model Answers

Marks

Total Marks

Ans. x 20 30 40 50 60 70

∴ mean m = ∴p=

f 8 12 30 10 6 4 70

xy 160 360 1200 500 360 280 2860

2860 = 40.857 70

1

2

e− m m r r!

e−40.857 ( 40.857 ) 1 = r! ------------------------------------------------------------------------------------r

d)

A metal wire 36 m long is bent to form a rectangle. Find its dimensions when its area is maximum.

Ans.

Let x and y be the sides of rectangle.

∴ 2 x + 2 y = 36 or ∴ y = 18 − x

x + y = 18 ½

But area A = xy = x (18 − x ) = 18 x − x 2 dA ∴ = 18 − 2 x dx d2A ∴ 2 = −2 dx For stationary values,

1 ½ ½

dA =0 dx

∴18 − 2 x = 0 ∴x = 9 d2A = −2 < 0 At x = 9, dx 2 ∴ At x = 9, A has max imum value and the other side is y = 18 − x = 9

4

½ ½

½

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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Subject & Code: Applied Maths (17301) Que. No. 6)

Sub. Que.

Page No: 25/26

Model Answers

e)

1 1 Find the equation of tangent to the curve x = , y = 1 − , when t t t=2.

Ans.

1 1 x = , y = 1− t t ∴ y = 1− x dy ∴ = −1 dx ∴ at t = 2, x = 0.5 and y = 0.5 and slope m = −1

Marks

Total Marks

1

½+½

∴ the equation is, y − b = m( x − a)

1

∴ y − 0.5 = −1( x − 0.5) ∴ y − 0.5 = − x + 0.5 ∴ x + y −1 = 0

1

4

OR 1 1 x = , y = 1− t t dx 1 dy 1 ∴ = − 2 and = dt t dt t 2 dy 12 dy ∴ = dt = t dx dx −12 dt t dy ∴ = −1 dx ∴ at t = 2, x = 0.5 and y = 0.5

½

½ ½+½

and slope m = −1 ∴ the equation is, y − b = m( x − a)

∴ y − 0.5 = −1( x − 0.5 )

1

∴ y − 0.5 = − x + 0.5 ∴ x + y −1 = 0

-------------------------------------------------------------------------------------

1

4

MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION (Autonomous)

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_____________________________________________________________________________________________________

Subject & Code: Applied Maths (17301) Que. No. 6)

Sub. Que.

Page No: 26/26

Model Answers

f)

Find the area between the parabola y = 4 − x 2 and the x-axis.

Ans.

y = 4 − x 2 and

Marks

Total Marks

x − axis i. e., y = 0

∴4 − x = 0 2

∴ x = −2, 2

½+½

b

∴ A = ∫ ydx a

=∫

2

−2

( 4 − x ) dx

½

2

2

 x3  = 4 x −  3  −2 

1

3 −2 )   3 23   ( 3 =  2 −  −  ( −2 ) −  3   3  

=

32 3

or

10.667

-------------------------------------------------------------------------------------

-------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Important Note In the solution of the question paper, wherever possible all the possible alternative methods of solution are given for the sake of convenience. Still student may follow a method other than the given herein. In such case, FIRST SEE whether the method falls within the scope of the curriculum, and THEN ONLY give appropriate marks in accordance with the scheme of marking. ------------------------------------------------------------------------------------------------------------------------------------------------------------------------

½

1

4