APPLIED MATHEMATICS - hte-rajasthan-gov

1

Modal Solutions

APPLIED MATHEMATICS

(104)

DEPARTMENT OF SCIENCE & HUMANITIES

GOVT. R.C. KHAITAN POLYTECHNIC COLLEGE JAIPUR

Solved by: 1. B. M. Saini

2. Dr Lata Chanchlani Lecturer (Mathematics)

2

Govt. R. C. Khaitan Polytechnic College, Jaipur Duration 3 hr Question paper 2013 Max. Marks 70

Modal -Solutions

Subject- Applied Mathematics (104)

Q1(i). Find the centre and radius of the circle 2 2 4 6 12 0x y x y

Solution:- Comparing the given equation of circle with the standard equation of the circle 2 2 2 2 0x y gx fy c , we have

Centre , 2, 3g f

Radius= 2 2 4 9 12 25 5g f c

Q1(ii). If 15 8n nC C , then find the value of 21

nC .

Solution:- Using ,n nr n rC C we have 15 8

n nnC C 15 8 23n n

23 23

21 21 223 22Therefore 253.

2nC C C

Q1(iii). Find the modulus and argument of the complex number 1 i .

Solution:- Let cos sin 1r i i

cos 1, sin 1r r

2 1 31 1 2 2 and tan1 4 4

r r .

Modulus 2 , Argument 34

Q1(iv). Find the differential coefficient of xx w.r.t. x.

Solution: Let xy x

Taking Logarithm on both sides, we get

3

log loge ey x x

On differentiating both sides with respect to x , we get

1 1. . loge

dy x xy dx x

1 log 1 log Ans.x

e edy y x x xdx

Q1(v). Solve 2x y ydy e x edx

.

Solution:- 2y xdy e e xdx

2y xe dy e x dx

3

Ans.3

y x xe e C

Q2(i). Prove that tan 20 tan 40 tan 80 tan 60 .

Solution :- L.H.S.= sin 20 sin 40 sin80

cos 20 cos 40 cos80

= 2sin 20 sin 40 sin80

2cos 20 cos 40 cos80

= cos 20 cos 60 sin80

cos 20 cos 60 cos80

=

1cos 20 sin 80 sin 8021cos 20 cos80 cos802

= 2cos 20 sin 80 sin802cos 20 cos80 cos80

sin100 sin 60 sin80cos100 cos 60 cos80

4

sin100 sin 60 sin80cos100 cos 60 cos80

sin 180 80 sin 60 sin80

cos 180 80 cos60 cos80

sin 80 sin 60 sin 80 tan 60 R.H.S.cos80 cos 60 cos80

Q2(ii). If 2 3 , 1

2, 1x x a x

f xbx x

is differentiable at everywhere, then find the value of a

and b .

Solution: Since f x is everywhere differentiable , therefor it is continuous and differentiable at 1x . Since f x is continuous at 1x , we have

1 1

lim lim 1x x

f x f x f

2

1 1lim 3 lim 2 4x x

x x a bx a

2 4 2b a a b ….(1) Since f x is differentiable at 1x , we have

1 1

1 1lim lim

1 1x x

f x f f x fx x

2

1 1

3 4 2 4lim lim

1 1x x

x x a a bx ax x

1 1

2 2lim 4 lim

1x x

bx bx

x

Using (1)

15 lim 5 3

1x

bx b b ax

Hence 3, 5a b Ans.

Q3(i). Prove that:

1 1 11 1 11 1 1 1 .

1 1 1

ab abc

a b cc

5

Solution :- L.H.S.=

1 1 11 1 11 1 1

ab

c

Taking , and a b c common from 1 2 3, and C C C respectively, we get 1 1 11

1 1 11

1 1 1 1

a b c

abca b c

a b c

Taking 1 1 2 3 C +C C C we get

1 1 1 1 11

1 1 1 1 11 1

1 1 1 1 11 1

a b c b c

abca b c b c

a b c b c

1 11

1 1 1 1 11 1 1

1 11 1

b c

abca b c b c

b c

Taking 2 2 1 R R R and 3 3 1 R R R , we get 1 11

1 1 11 0 1 00 0 1

b cabc

a b c

On solving the determinant, we get

1 1 11 R.H.Sabca b c

6

Q3(ii). Find the inverse of Matrix 1 1 22 1 13 0 1

A

.

Solution :- Here 1 1 0 1 2 3 2 3 1 1 6 0A .

1 1 3 1 1 31 5 3 1 5 33 3 3 3 3 3

T

AdjA

1

1 1 31 1 5 36

3 3 3

AdjAAA

Ans.

Q4(i). Find the equation of a line passing through 4,5 and perpendicular to the line 2 3 5 0x y .

Solution :- Then the equation of parallel line to the given line will be 2 3 0.x y k

Since this line passes through the point 4,5 , it must satisfy the equation of the line, therefore we have

2 4 3 5 0 7k k

Substituting the valve of k, we get the required straight line as 2 3 7 0.x y Ans.

Q4(ii). Find the vertex, focus and length of Latus rectum of the parabola.

2 8 8y x y

Solution :- 2 28 8 8 16 8 16y y x y y x

24 8 2y x …(1)

On comparing the equation (1) with the general equation of the parabola 2 4Y AX ,

we have

2A , 2, 4X x Y y

7

Vertex: 0, 0

2 0, 4 02, 4

X Yx yx y

2, 4

Equation of axis: 0 4 0.Y y

Focus: , 0

2 2, 4 00, 4

X A Yx yx y

0, 4

Length of Latus rectum: 4 4 2 8A .

Q5(i). If 3 3 1x y tt

and 6 6 22

1x y tt

, then prove that 4 2 1.dyx ydx

Solution :- 6 6 22

1x y tt

23 3 3 3 2

2

12x y x y tt

23 3 2

2

1 12t x y tt t

23 3 2

2

1 12 2x y t tt t

3 3 1x y

……(1)

On differentiating with respect to x, we get

3 2 3 2.3 .3 0dyx y y x

dx

3 2 3 2. dyx y y x

dx

Multiplying by x both sides and using (1), we get

4 2. 1dyx y

dx

Hence proved.

8

Q5(ii). Prove that 5 4 35 5 10x x x has a maximum value for 1x and a minimum value for 3x .

This topic is not included in the revised syllabus of year 2017-18.

Q6(i). Solve the integral: sin1 sin

x dxx .

Solution :- Let sin 1 sin 11 sin 1 sin

x xI dx dxx x

11 sin1 sin

xdx dxx

If 1 2I I I

2 2

2 2 2 21 cos sin 2sin cosx x x xI dx

2 2cos sinx x dx

2 22 sin cosx x C

2

2 2

1 1cos sin1 sin x x

I dx dxx

2 2

1 11 12 .cos sin .2 2

x x

dx

4 2 2 4

1 1sin .cos sin .cos2 x x

dx

4 24 2

1 1 1 cossin2 2

xx

dx ec dx

4 2

1 .2log tan2

x C

Hence 2 2 4 22 sin cos 2 log tan .x x xI C

Q6(ii). Solve the integral: 2

4

11

xdx

x

.

9

Solution :- On dividing 2x ,both the sides the given integral becomes

2

22

11

1x dx

xx

2

2

11

1 2

x dxx

x

Let 2

1 11x t dx dtx x

, we have

22 2

dt

t

11 tan

2 2t C

1

11 tan2 2

xx C

211 1tan Ans.

2 2x C

x

Q7(i). Solve: 3dyx y xydx

.

Solution :- On dividing x , the given differential equation will become

31dy y ydx x

This is Barnouli’s differential equation of first order. On dividing the equation by 3y , we get

3 21 1dyy y

dx x

10

On taking 2y v , then 32 dy dvydx dx

, the equation becomes

1 1 21 22

dv dvv vdx x dx x

This is linear differential equation of first order of the form dy Py Qdx

.

Here 2 , 2P Q

x

Integrating factor =2

2log2

1e

dxPdx xxe e ex

. . . .y I F Q I F dx C

2 2

1 1. 2.v dx Cx x

2 2

1 2 Cx y x

Ans.

Q7(ii). Solve the following differential equation: 2 24 4 xD D y e .

Solution :- Auxiliary equation of the given differential equation will be

2

2

4 4 0

2 02, 2

m m

mm

21 2. . xC F c c x e

2

2

1. .4 4

xP I eD D

2

21

2xe

D

22

2xx e

11

2

2 21 2

. . . .

2

x x

y C F P Ixc c x e e

Ans.

Q8(i). Find the value: 4

0

log 1 tane x dx

Solution :- Let 4

0

log 1 taneI x dx

Using the property IV for definite integral, we get

4

0

log 1 tan4eI x dx

4

0

tan tan4log 1

1 tan tan4

e

xdx

x

4

0

1 tan 1 tanlog1 tane

x x dxx

4

0

2log1 tane dx

x

4

0

log 2 log 1 tane e x dx

4 4

0 0

log 2 log 1 tane edx x dx

4

0log 2e x I

2 log 24 eI

log 28 eI

Ans.

12

Q8(ii). Find the moments of the force ˆˆ3i k acting at a point ˆˆ ˆ2 3i j k about the point

ˆˆ ˆ2 .i j k

Solution :-Moment of force r F

Here ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ2 3 2 3 4r i j k i j k i j k

and ˆˆ3F i k

. Then 1 3 4 3 1 12 93 0 1

i j ki j k

3 12 9i j k Ans.

Solved by:

1. B. M. Saini

2. Dr Lata Chanchlani

Lecturer (Mathematics)

Govt. R. C. Khaitan Polytechnic College,

Jaipur.

13


Modal -Solutions


Q1(i). If 10 92r rP P , then find r.

Solution:- 10 92r rP P

10 ! 9 !2

10 ! 9 !r r

10. 9 ! 9 !2

10 . 9 ! 9 !r r r

10 210 r

10 2 10 5r r Ans.

Q1(ii). If 3 24 2

A

, find K so that 2 2 .A KA I

Solution:- 2 3 2 3 24 2 4 2

A A A

9 8 6 4 1 2

12 8 8 4 4 4

Using the value of 2A in 2 2A KA I , we have

1 2 3 2 1 0

24 4 4 2 0 1

K

1 2 3 2 2

14 4 4 2 2

K KK

K K

Ans.

14

Q1(iii). Find Projection of Vector ˆˆ ˆ2a i j k on the vector ˆˆ ˆ2 .b i j k

Solution:- Projection of Vector a on the vector .a bbb

2 2 1 5 Ans.

6 6

Q1(iv). Determine order and degree of the following differential equation:

322

2

2

1 dydx

Kd ydx

Solution:- On rationalizing the given differential equation, we have

3 22 2

21 dy d yKdx dx

Order of the highest order derivative =Order of diff. equation.=2.

Power of highest order derivative= Degree of diff. equation.=2

Q1(v). Differentiate 3 sin 5xx e x with respect to x.

Solution:-

3

33 3

sin 5

sin 5sin 5 sin 5

x

xx x

d x e xdx

dx de d xe x x x x edx dx dx

2 3 3sin 5 .3 sin 5 . .5cos5x x xe x x x x e x e x

2 3sin 5 sin 5 5 cos5xx e x x x x x Ans.

15

Q2(i). Find term independent of x in the expansion of 6

23 1 .2 3

xx

Solution :-Let the term independent of x be 1 thr term.

By using the formula of 1 thr term in the binomial expansion, we have

1n n r r

r rT C x a , we have

66 2

13 12 3

r r

r rT C xx

66 12 33 1

2 3

r rr

rC x

For the term independent of x, we must have 12 3 0 4r r . Hence the term independent of x will be the 5th term, which will be given by

6 4 46

5 43 1 6 5 3 3 1 5 Ans.2 3 2 4 3 3 3 3 12

T C

Q2(ii). Prove that

sin 5 2sin 3 sintan

cos5 cosx x x

xx x

.

Solution :- L.H.S.=

sin 5 2sin 3 sincos5 cosx x x

x x

2sin 3 cos 2 2sin 3

2sin 3 sin 2x x x

x x

2sin 3 cos 2 1 1 cos 22sin3 sin 2 sin 2

x x xx x x

22sin sin tan .2sin cos cos

x x xx x x

Q3(i). Solve the following system of linear equations by using Cramer’s Rule:

6;2 7;

3 12.

x y zx zx y z

16

Solution :- Here 1 1 11 0 2 1 2 1 1 6 1 13 1 1

2 5 1 4 0

6 1 17 0 2 6 2 1 7 24 1 7

12 1 1 12 17 7 12 0

x

1 6 11 7 2 1 7 24 6 1 6 1 12 213 12 1

17 30 9 4 0

y

1 1 61 0 7 1 7 1 12 21 6 13 1 12

7 9 6 8 0

z

According to Cramer’s Rule

12 4 83, 1, 2.4 4 4

x y zx y z

Ans.

Q3(ii). Prove that:

1 1 11 1 11 1 1 1 .

1 1 1

ab abc

a b cc

Solution :- L.H.S.=

1 1 11 1 11 1 1

ab

c

Taking , and a b c common from 1 2 3, and C C C respectively, we get

17

1 1 11

1 1 11

1 1 1 1

a b c

abca b c

a b c


1 1 1 1 11

1 1 1 1 11 1

1 1 1 1 11 1

a b c b c

abca b c b c

a b c b c

1 11

1 1 1 1 11 1 1

1 11 1

b c

abca b c b c

b c


1 1 11 0 1 00 0 1

b cabc

a b c



Q4(i). Find the equation of a straight line passing through the point 1,1 and perpendicular to the line 2 3 0x y .

Solution :- Let the slope of the given line be 1m , then

112

m .

18

We know that the condition of perpendicularity of two straight lines is 1 2 1.m m Therefore the slope of the straight line perpendicular to the given line will be

21

1 2mm

.

Then the equation of perpendicular line will be 2 .y x c


1 2 1 3c c .Substituting the valve of c, we get the required straight line as

2 3 0x y Ans.

Q4(ii). Find the vertex, focus and length of Latus rectum of the parabola.

2 4 4 0x x y

Solution :- 2 24 4 4 4 4 4x x y x x y

24 4 1x y …(1)

On comparing the equation (1) with the general equation of the parabola 2 4X AY ,

we have 1A , 2, 1X x Y y

Vertex: 0, 0

2 0, 1 02, 1

X Yx yx y

2,1

Focus: 0,

2 0, 1 12, 0

X Y Ax yx y

2,0

Length of Latus rectum: 4 4 1 4A .

Q5(i). Find

6

3 sin coslim

6x

x x

x

19

Solution :- Let 6

x h , when

6x then 0h .

6 6

3 sin cos3 sin cos 6 6lim lim

6 66x x

h hx x

hx

6

3 sin cos cos sin cos cos sin sin6 6 6 6lim

x

h h h h

h

6

1 3 3 13. cos 3. sin cos sin2 2 2 2lim

x

h h h h

h

6

2sinlim 2x

hh

Ans.

Q5(ii). If 2 1m

y x x

, then prove that:

2 22 11 0x y xy m y

Solution :- On differentiating y both sides with respect to x, we get

1

21 2

11 1 .22 1

my m x x x

x

21

21 2

111

m x xy m x xx

2

1 2 2

1

1 1

mx x myy m

x x

2 2 2

11x y m y

Again differentiating both sides with respect to x, we get

2 2 2

1 2 1 11 .2 2 2x y y xy m yy

20

2 2

2 11 0x y xy m y

Q6(i). Show that a cylinder of a given volume which is open at the top, has minimum total surface area, provided its height is equal to radius of its base.


Q6(ii). Evaluate: 5 4sin

dxx .

Solution :- 5 4sin

dxIx

2

22

2 tan5 41 tan

x

x

dx

22

22 2

1 tan

5 1 tan 8 tan

x

x x

dx

Let 22 2tan sec .

2x x

dxt dt , we have

22

5 1 8dtIt t

22

5 8 5dt

t t

2 2

25 4 3

5 5

dt

t

Let 45

t u dt du

2

2

25 3

5

dt

u

21

1

35

2 5. tan5 3

u C

12 5 5 4. tan5 3 3 5

t C

12 5 4tan3 3 3

t C

1 12

2 5 4tan tan3 3 3

x C

Ans.

Q7(i). Solve: logedyx y x xdx

.

Solution :- On dividing x , the given differential equation will become

1 logedy y xdx x


.

Here 1 , logeP Q xx


logedxPdx xxe e e x

. . . .y I F Q I F dx C

. log .ey x x x dx C

2

21 1log . .2 2exyx x x dx C

x

2 2

log2 4ex xyx x C

Ans.



22

2 3 2 01 2 0

1, 2

m mm m

m

2

1 2. . x xC F c e c e

2

1. .3 2

xP I eD D

1

1 2xe

D D

1

1 1 2xe

D

xxe

21 2

. . . . x x x

y C F P Ic e c e xe

Ans.

Q8(i). For vectors a and b

prove that :

2 22 2 . ;a b a b a b where ;a a b b

.

Solution :- 22ˆ. sin . ;a b a b n

22 2 ˆ ˆ. sin . . ;a b n n

22 2 1 cos .1;a b

2 22 2 2 cos ;a b a b

222 2 2 cos ;a b a b

Q8(ii). Use Simpson’s 38

rule to evaluate 1

20 1

dxx .


23

Solved by:

1. B. M. Saini



Govt. R. C. Khaitan Polytechnic College, Jaipur.


Modal -Solutions


Q1(i). Find the modulus and argument of the complex number 1; i .

Solution:- Let cos sin 1r i i

cos 1, sin 1r r

2 11 1 2 2 and tan1 4

r r .


Q1(ii). If , then prove that 2 1 cos .dyy xdx

Solution: Here siny x y

On squaring both sides, we get

2 siny x y

On differentiating both sides with respect to x , we get

2 . cosdy dyy x

dx dx

cos Ans.2 1

dy xdx y

24

Q1(iii). Solve 2x y ydy e x edx

.


2y xe dy e x dx

3

Ans.3

y x xe e C

Q1(iv). Find the centre and radius of the circle 2 2 12x y x


Centre , 6,0g f

Radius= 2 2 36 6g f c

Q1(v). Out of 10 flowers, in how many ways 4 flowers can be selected for the worshipping of the God.

Solution:- No. of ways flowers can be selected for worshipping= 104C 10 9 8 7 210.

4 3 2 1

Q2(i). If tan cos cot sin , then prove that : 1cos4 2 2

.

Solution :- tan cos cot sin

sin cos cos sincos cos sin sin

cos cos cos sin sin cos sin sin 0 cos cos sin 0

cos sin 2 12

n

2 1cos sin

2n

2 11 1cos . sin .

2 2 2 2n

25

2 1

cos .cos sin .sin4 4 2 2

n

2 1 1 1cos

4 2 2 2 2 2 2 2n n

Hence proved.

Q2(ii). If 2 , 1, 1

x xf x

ax b x

is differentiable at , then find the value of a and b .

Solution: Since f x is everywhere differentiable at 1x , therefore it is continuous and differentiable at 1x .

Since f x is continuous at 1x , we have

1 1

lim lim 1x x

f x f x f

2

1 1lim lim 1x x

x ax b

1 a b ….(1) Since f x is differentiable at 1x , we have

1 1

1 1lim lim

1 1x x

f x f f x fx x

2

1 1

1 1lim lim1 1x x

x ax bx x

1 1

1lim 1 lim1x x

ax bxx

0 0

1 1lim1 1 limh h

a h bh

h

2 Using (1)a

1b

Hence 2, 1a b Ans.

Q3(i). Prove that:

1 1 11 1 11 1 1 1 .

1 1 1

ab abc

a b cc

26

Solution :- L.H.S.=

1 1 11 1 11 1 1

ab

c

Taking , and a b c common from 1 2 3, and C C C respectively, we get 1 1 11

1 1 11

1 1 1 1

a b c

abca b c

a b c


1 1 1 1 11

1 1 1 1 11 1

1 1 1 1 11 1

a b c b c

abca b c b c

a b c b c

1 11

1 1 1 1 11 1 1

1 11 1

b c

abca b c b c

b c


1 1 11 0 1 00 0 1

b cabc

a b c



Q3(ii). Find the equation of a line passing through 4,5 and parallel to the line 2 3 5 0x y .

27

Solution :- Then the equation of parallel line to the given line will be 2 3 0.x y k


2 4 3 5 0 7k k

Substituting the valve of k, we get the required straight line as 2 3 7 0.x y Ans.

Q3(ii). Prove that the line 0lx my n touches the parabola 2 4 ,y ax if 2ln am .

Solution :- From the equation of the line we have, lx nym

.

Substituting this value of y in equation of the parabola 2 4 ,y ax we have

2

4lx n axm

2 24lx n axm

2 2 2 22 4l x lxn n axm

2 2 2 22 4 0l x x ln am n

Since the line touches the parabola, therefore two points of contact must coincide. Hence two roots of the obtained quadratic equation will be same. For which we must have

2 4 0B AC

22 2 22 4 4 0ln am l n

2 2 2 2 4 2 24 16 16 4 0l n lm na a m l n

2 2 4lm na a m

2ln am Hence proved.

Q4(i). Differentiate log sinexxy x x w.r.t. x.

Solution :- Let loge xu x and sin xv x , then dy du dvdx dx dx

.

28

Taking Logarithm on both the sides in u and v , we get

log log loge e eu x x , log log sine ev x x

On differentiating , we get

1 1 1. log .2 e

du xu dx x

and 1 1. log sinsine

dv xv dx x

log 1

log2

e x

edu x xdx

and 1sin log sinsin

xe

dv x xdx x

log 1 1log sin log sin

s

2 in

e xx

e edydx

x x x xx

Ans.

Q4(ii). A window is in the shape of a rectangular with a semi-circle covering the top. If the perimeter of the window be a fixed length P, find its maximum area.


Q5(i). Prove that:

3

3

3

11 .1

a ab b a b b c c a a b cc c

Solution :- L.H.S.=

3

3

3

111

a ab bc c

Taking 2 2 1 3 3 1R R -R , R R -R , we get 3

3 3

3 3

111

a ab a b ac a c a

Taking b a common from 2R and c a from 3R , we get

3

2 2

2 2

10 10 1

a ab a c a b a ab

c a ac

Taking 3 3 2 R R -R , we get

29

3

2 2

10 10 0

a ab a c a b a ab

c a a b c

Taking c a a b c common from 3R , we get

3

2 2

10 10 0 1

a ab a c a c b a b c b a ab


R.H.S.b a c a c b a b c

Q5(i). Solve the following equations by using Inverse matrix method:

6;2;

2 1.

x y zx y z

x y z

Solution :- The given system of equations can represented in matrix form as

AX B

where 1 1 1 61 1 1 , , 22 1 1 1

xA X y B

z

Then 1(1 1) 1( 1 2) 1(1 2) 0 3 3 6 0A .

0 3 3 0 2 22 3 1 3 3 02 0 2 3 1 2

T

AdjA

1

0 2 21 3 3 06

3 1 2

AdjAAA

1

0 2 2 6 61 1 3 3 0 2 126 6

3 1 2 1 18

X A B

30

6 11 12 26

18 3X

123

xyz

. Hence 1, 2, 3x y z

Q6(i). Solve the integral: 3tan secx x xdx .

Solution :- Let 1 2tan tan and secx t x t x xdx dt

1tan .I t tdt

2 21

2

1tan . .2 1 2t tt dt

t

2 21

2

1 1 1tan2 2 1t tt dt

t

21

2

1 1tan 12 2 1t t dt

t

2

1 11tan tan2 2t t t t C

2 1 1tan tan

2 2 2x x x x C

Q6(ii). Solve the integral: 22 1dx

x x .

Solution :- 22 1dx

x x

2

112

2 2

dxxx

2

12 1 1 1

4 2 16

dx

x

31

2

12 1 9

4 16

dx

x

Let 14

x t dx dt

2

2

12 3

4

dx

t

3434

1 1. log32 2.4

t Ct

314 4

314 4

1 log3

x Cx

1 2 1log3 2 1

x Cx

Q7(i). Solve the differential equation: tan secdy y x xdx

.

Solution :- This is linear differential equation of first order of the form dy Py Qdx

.

Here tan , secP x Q x

Integrating factor = tan log sec secePdx xdx xe e e x

. . . .y I F Q I F dx C

.sec sec .secy x x xdx C

sin cosy x C x

Ans.



32

2 7 12 04 3 0

3, 4

m mm m

m

3 4

1 2. . x xC F c e c e

2

2

1. .7 12

xP I eD D

2 21 1

4 14 12 30x xe e

3 4 21 2

. . . .1 30

x x x

y C F P I

c e c e e

Ans.

Q8(i). Find the value:

2

0

sinsin cos

xx x

.

Solution:- Let

2

0

sinsin cos

xIx x

…..(1)

Using IV property of definite integral, we have

2 22

0 02 2

sin coscos sinsin cos

x xIx xx x

…..(2)

On adding the equations (1) and (2), we get

2

0

cos cos2cos sin cos sin

x xI dxx x x x

2

2

00 2

dx x

Hence Ans.4

I

33

Q8(ii). Find a vector of magnitude 9, which is perpendicular to both the vectors ˆˆ ˆ4 3i j k

and ˆˆ ˆ2 2 .i j k

Solution :-Let ˆˆ ˆ4 3a i j k and ˆˆ ˆ2 2b i j k

ˆˆ ˆ

ˆˆ ˆ4 1 3 2 3 8 6 4 22 1 2

i j ka b i j k

ˆˆ ˆ2 2a b i j k

1 4 4 9 3a b

Unit vector of perpendicular to both the vectors will be

ˆˆ ˆ2 2

ˆ3

i j ka bna b

Hence vector of magnitude 9, which is perpendicular to both the vectors will be

ˆˆ ˆ9 2 2 ˆˆ ˆˆ9 9 3 6 6

3

i j ka bn i j ka b

Ans.

Solved by:

1. B. M. Saini




Jaipur.

34

Govt. R. C. Khaitan Polytechnic College, Jaipur

Duration 3 hr Question paper 2016 Max. Marks 70

Modal -Solutions


Q1(i). Write the complex number 12i in polar form.

Solution:- Let 1cos sin2ir i

1 1cos , sin2 2

r r

2 1 1 1 1 and tan 12 2 4

r r .

Polar form 4 41 cos sin

2i i

Q1(ii). Prove that the matrix 1 111 12

i iA

i i

is a unitary matrix.

Solution:- 1 111 12

i iA

i i

* 1 11

1 12T i i

A Ai i

35

*

2 2 2 2

2 2 2 2

1 1 1 11 11 1 1 12 2

4 01 1 1 11 10 44 41 1 1 1

1 00 1

i i i iAA

i i i i

i i i ii i i i

I

*AA I . Hence A is a unitary matrix.

Q1(iii). If tan ,f then prove that 2

22

1

ff

f

Solution:- tan 2 tan 2f f

2 2

22 tan1 tan 1

f

f

Q1(iv). Evaluate : x x

x x

e e dxe e

.

Solution:- Let x x x xe e t e e dx dt

log log x xdt t c e e ct

.

Q1(v). Find the perpendicular distance from the point 5,5 to the straight line 3 4 10 0.x y

Solution:- distance from the point 5,5 to the straight line 3 4 10 0x y

2 2

3 5 4 5 103 ( 4)

15 20 10 5 1 Ans.525

Q2(i). Prove that:

2

0

sin4sin cos

xx x

.

36

Solution:- Let

2

0

sinsin cos

xIx x

…..(1)

Using IV property of definite integral, we have

2 22

0 02 2

sin coscos sinsin cos

x xIx xx x

…..(2)

On adding the equations (1) and (2), we get

2

0

cos cos2cos sin cos sin

x xI dxx x x x

2

2

00 2

dx x

Hence Ans.4

I

Q2(ii). Prove that 1cos 20 cos 40 cos 60 cos8016

.

Solution :- L.H.S.= 1cos60 2cos 20 cos 40 cos802

.= 1 1 Cos 20 40 Cos 20 40 cos802 2

.= 1 1cos 20 cos804 2

= 1 1cos 20 cos80 cos804 2

= 1 1 1cos 80 20 cos 80 20 cos804 2 2

= 1 cos100 cos 60 cos808

= 1 1cos 180 80 cos808 2

= 1 1cos80 cos808 2

= 1 R.H.S.16

37

Q3(i). Solve the following equations by using Inverse matrix method:

2 3 9;6;2.

x y zx y zx y z

Solution :- The given system of equations can represented in matrix form as

AX B

where 2 1 3 91 1 1 , , 61 1 1 2

xA X y B

z

Then 2 1 1 1 1 1 3 1 1 4 0 6 2 0A .

2 0 2 2 2 42 1 1 0 1 14 1 3 2 1 3

T

AdjA

1

2 2 41 0 1 12

2 1 3

AdjAAA

1

2 2 4 9 18 12 81 1 0 1 1 6 0 6 22 2

2 1 3 2 18 6 6

X A B

2 11 4 22

6 3X

6 1

1 12 26

18 3X

123

xyz

. Hence 1, 2, 3x y z

Q3(ii). Prove that:

3

3

3

11 .1

a ab b a b b c c a a b cc c

Solution :- L.H.S.=

3

3

3

111

a ab bc c

Taking 2 2 1 3 3 1R R -R , R R -R , we get

38

3

3 3

3 3

111

a ab a b ac a c a

Taking b a common from 2R and c a from 3R , we get

3

2 2

2 2

10 10 1

a ab a c a b a ab

c a ac

Taking 3 3 2 R R -R , we get

3

2 2

10 10 0

a ab a c a b a ab

c a a b c

Taking c a a b c common from 3R , we get

3

2 2

10 10 0 1

a ab a c a c b a b c b a ab


R.H.S.b a c a c b a b c

Q4(i). Find the equation of the circle passing through the points (1,2); (3,-4) and (5,6).

Solution :- General equation of a circle is given by

2 2 2 2 0x y gx fy c

Since the circle passes through the points (1,2); (3,-4) and (5,6), therefore the points will satisfy the equation of circle, we have

2 4 5g f c ….(1)

6 8 25g f c ….(2)

10 12 61g f c ….(3)

On solving the equations (1), (2) and (3) for f, g and c, we get

1 13, and 102 2

f g c

39

On substituting the values of f, g and c in the equation of the circle, we get the required equation of the circle as

2 2 13 10 0x y x y

Q4(ii). Find the centre, focus, length of axes and length of Latus rectum of the hyperbola.

2 22 2 8 1 0x y x y

Solution :- 2 22 2 4 1x x y y

2 22 1 2 4 4 1 ( 7)x x y y

2 21 2 2 6x y

2 21 21

6 3x y

…(1)

On comparing the equation (1) with the general equation of the hyperbola 2 2

2 2 1Y Xb a

,

we have

22 2

2

66, 3 1 1 33

aa b eb

1, 2X x Y y

Centre: 0, 0

1 0, 2 01, 2

X Yx yx y

1, 2

Focus:

0,

1 0, 2 3 31, 2 3

X Y be

x yx y

1,5 and 1, 1

Length of axes: Transverse axis = 2b= 2 3

Conjugate axis = 2a= 2 6

Length of Latus rectum: 22 2 6 4 3

3ab

.

40

Q5(i). Prove that the curves 2 2 1ax by and 1 2 1 2 1a x b y , intersects orthogonally if

1 1

1 1 1 1a b a b

.

Solution :- Given equations of curves are 2 2 1ax by …(1)

and 1 2 1 2 1a x b y …(2)

On differentiating equations (1) and (2) w.r.t. x, we have

12 2 0 (Say)dy dy axax by mdx dx by

11 1

212 2 0 (Say)dy dy a xa x b y mdx dx b y

The two curves intersects orthogonally if

…(3)

From equations (1) and (2), we have

2 2 1 2 1 2

2 1 2 1

ax by a x b y

x a a y b b

12

2 1

b bxy a a

…(4)

On eliminating 2

2

xy

from equations (3) and (4), we have

1 1 1 1

1 1 11

b b bb b b a aaa bb aaa a

1 1 1 1

1 1 1 1 1 1 1 1b b a a a b a b

Hence proved.

Q5(ii).Solve the following integral 21

xxe dxx .

1 2 1

1 2 1 2 11 1ax a x x bbm mby b y y aa

41

Solution :-

2 2

1 11 1

xx x exe dx dxx x

2 2

1 11 1

xx e dxx x

2

1 11 1

x xe dx e dxx x

On using the integration by parts for the first integral, we have

2 2

1 1 11 1 1

x x xe e dx e dxx x x

2 2

1 1 11 1 1

x x xe e dx e dxx x x

11

xe Cx

Ans.

Q6(i). Find the value of the integral

2

4

1

1

xdx

x

.

Solution :-

2 2

42

2

11111

x xI dx dxx x

x

Let 2

1 11x t dx dtx x

, we have

2

2 2

dtIt

By using the formula 2 2

1 1 log2

x a Cx a a x a

, we have

42

1 21 log 12 2 2

xxI C

xx

2

2

1 2 1log2 2 2 1

x xI Cx x

Ans.

Q6(ii). Solve the following differential equation: 2cos tandyx y xdx

.

Solution :- On dividing 2cos x , the given differential equation will become

2 2sec tan secdy y x xdx


.

Here 2 2sec , tan secP x Q x x

Integrating factor =2sec tanPdx dx xe e e

. . . .y I F Q I F dx C

tan 2 tantan secx xye x xe dx C

Let 2tan , secx t xdx dt

tan t t tant t tan 1x t xye e dt C e e C e x C

tan tantan 1x xy e x Ce Ans.

Q7(i). Solve the following differential equation: 2 3 2 sin 2xD D y e x .


2 3 2 01 2 0 1, 2

m mm m m

2

1 2. . x xC F c e c e

43

2

1. . sin 23 2

xP I e xD D

2 2

1 1 sin 23 2 3 2

xe xD D D D

1 1 sin 2

1 ( 1) 2 4 3 2xe x

D D

1 1 sin 2

1 3 2xe x

D D

2

3 2 sin 29 4

x Dxe xD

3 2 sin 2

9( 4) 4x Dxe x

1 3 2cos 2 2sin 240

xxe x x

1 3cos 2 sin 220

xxe x x

21 2

. . . .1 3cos 2 sin 220

x x x

y C F P I

c e c e xe x x

Ans.

Q7(ii). Find the 10th term in Binomial expansion of 121 .x

x

Solution :- By using the formula of 1 thr term in the binomial expansion, we have

1n n r r


9

12 12 910 9 1 9

1T T C xx

12 39 9

1C xx

6

12 11 10 13 2 1 x

6

220x

Ans.

44

Q8(i). If vectors ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ2 3 ; 2 3 and 3 5i j k i j k i Pj k are coplanar, then find the value of

P.

Solution :- By using the condition of co planarity of three vectors A

, B

and C

, we have

. 0A B C

2 1 11 2 3 03 5P

2 10 3 1 5 9 1 6 0P P

20 6 14 6 0P P

7 28 0 4P P Ans.

Q8(ii). Find the moments of the force ˆˆ3i k about the point ˆˆ ˆ2i j k which passes through

the point ˆˆ ˆ2 3i j k .


Here ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ2 3 2 3 4r i j k i j k i j k

and ˆˆ3F i k

. Then 1 3 4 3 1 12 93 0 1

i j ki j k

3 11 9i j k Ans.

Solved by:

1. B. M. Saini




Jaipur.

45


Modal -Solutions


Q1(i). Find the centre and radius of the circle 2 2 4 6 12 0x y x y


Centre , 2, 3g f

Radius= 2 2 4 9 12 25 5g f c

Q1(ii). Solve 2x y ydy e x edx

.


2y xe dy e x dx

3

Ans.3

y x xe e C

Q1(iii). Find the modulus and argument of the complex number 1 3i .

Solution:- Let cos sin 1 3r i i

cos 1, sin 3r r

2 31 3 4 2 and tan1 3

r r .


Q1(iv). In how many ways 8 persons seated around a round table?

Solution:- Circular permutation = 1 ! 8 1 ! 7! 5040 Ans.n

46

Q1(v). 2 2sin cosdxx x

Solution:- 2 2

2 2 2 2

sin cossin cos sin cos

x xdx dxx x x x

2 2

2 2 2 2

sin cossin cos sin cos

x x dxx x x x

2 2

1 1cos sin

dxx x

2 2sec cosxdx ec xdx

tan cotx x C

Q2(i). Prove that the middle term in the expansion of 21 nx is 1.3.5... 2 1.2

!n nnx

n

.

Solution :- Total number of terms in the expansion of 21 nx = 2 1n

Middle term 2 1 1 12

thn n

By using the formula of 1 thr term in the binomial expansion, we have

1n n r r


1 thn in the expansion of 21 nx 22 2 !1

! !n nn n n

n

nC x x

n n

1.2.3.4.5.6.....2

! !nn x

n n

1.3.5.... 2 1 . 2.4.6....2

! !nn n

xn n

1.3.5.... 2 1 .2 1.2.3....

! !

nnn n

xn n

1.3.5.... 2 1 .2 !

! !

nnn n

xn n

47

1.3.5.... 2 1 .2

Ans.!

nnn

xn

Q2(ii). Prove that 3sin10 sin 50 sin 60 sin 7016

.

Solution :- L.H.S.= 1sin 60 2sin10 sin 50 sin 702

.= 3 1 Cos 10 50 Cos 10 50 sin 702 2

.= 3 1Cos 40 sin 704 2

= 1 1Cos 40 sin 70 sin 704 2

= 3 1 1sin 70 40 sin 70 40 sin 704 2 2

= 3 1sin110 sin 708 2

= 3 1sin 180 70 sin 708 2

= 3 1sin 70 sin 708 2

= 3 R.H.S.16

Q3(i). If the function f x is continuous at 1x , then find the value of k

, 0 12 , 1 2

x xf x

k x x

Solution :- Since f x is continuous at 1x , we have

1 1

lim lim 1x x

f x f x f

01 1

lim lim lim 1 1hx x

f x x h

1 1 0

lim lim 2 lim 2 1 2x x h

f x k x k h k

1 2f k 1 2 2

3k k

k Ans.

48

Q3(ii). Solve the following system of linear equations by using Cramer’s Rule:

2 3 6;2 4 7;3 2 9 14.

x y zx y zx y z

Solution :- Here 1 2 32 4 1 1 36 2 2 18 3 3 4 123 2 9

34 30 24 20 0

6 2 37 4 1 6 36 2 2 63 14 3 14 56

14 2 9 204 98 126 20 0

x

1 6 32 7 1 1 63 14 6 18 3 3 28 213 14 9

49 90 21 20 0

y

1 2 62 4 7 1 56 14 2 28 21 6 4 123 2 14

42 14 48 20 0

z

According to Cramer’s Rule

20 20 201, 1, 1.20 20 20

x y zx y z

Q4(i). Prove that:

2

2 2 2 2

2

4 .a ab ac

ab b bc a b cac bc c

49

Solution :- L.H.S.=

2

2

2

a ab acab b bcac bc c

Taking , and a b c common from 1 2 3, and C C C respectively, we get a a a

abc b b bc c c

Taking , and a b c common from 1 2 3, and RR R respectively, we get

2 2 2

1 1 11 1 11 1 1

a b c

Taking 2 2 1 3 3 1 R R +R and R R +R , we get

2 2 2

1 1 10 0 20 2 0

a b c


2 2 2 2 2 20 2=4 Ans.

2 0a b c a b c

Q4(ii). Find the equation of a straight line passing through the point 3,4 and perpendicular to the line 2 3 7 0x y .

Solution :- Let the slope of the given line be 1m , then

123

m .

We know that the condition of perpendicularity of two straight lines is 1 2 1.m m Therefore the slope of the straight line perpendicular to the given line will be

21

1 32

mm

.

Then the equation of perpendicular line will be 3 .2

y x c

50


3 14 32 2

c c

Substituting the valve of c, we get the required straight line as 3 2 1 0x y Ans.

Q5(i). If y x yx e then prove that

2log

1 loge

e

xdydx x

Solution :- y x yx e ….(1)

Taking Logarithm on both sides inequation (1), we get

logy x x y

….(2)

log 1 logy x y x y x x

1 logxy

x

….(3)

On differentiating equation (2) both sides with respect to x , we get

1. log 1dy dyy xx dx dx

log 1dy dy yxdx dx x

1 log dy x yxdx x

1 logdy x ydx x x

….(4)

Using the value of y from equation (3) into equation (4), we get

2

1 log log1 log 1 log

xxxdy x

dx x x x

Ans.

51

Q5(ii). Show that the semi-vertical angle of a cone of maximum volume and given slant height is

1tan 2 .


Q6(i). Solve the integral 1

2

sin1

x x dxx

.

Solution :- 1

2

sin1

x xI dxx

Let 1sin sin cosx t x t dx tdt , we have

sin . .cos . sincos

t t t dtI t tdtt

On integrating by parts, we get

cos cosI t t tdt C

cos sinI t t t C

2 11 sinI x x x C Ans.

Q6(ii). Solve the integral 21

xxe dxx .

Solution :-

2 2

1 11 1

xx x exeI dx dxx x

2 2

11 1

x xx e e dxx x

21 1

x xe edx dxx x

On integrating by parts the first integral, we get

2 21 1 1

x x xe e eI dx dxx x x

52

1

xeIx

Ans.

Q7(i). Solve the following differential equation: log 2loge edyx x y xdx

.

Solution :- On dividing logx x , the given differential equation will become

1 2loge

dy ydx x x x


.

Here 1 2,loge

P Qx x x


log log (log ) loge e edxPdx x x x

ee e e x

. . . .y I F Q I F dx C

2log .log .e ey x x dx Cx

Let 1log ,e x t dx dtx

22log 2 t 2.

2ety x dt C C t C

2log loge ey x x C

Ans.

Q7(ii). Solve the following differential equation: 3

23 3 2 xd y dy y e

dx dx .


3

2

3 2 0

1 2 0

1 1 2 0 1,1, 2

m m

m m m

m m m m

53

21 2 3. . x xC F c c x e c e

2

3

1. .3 2

xP I eD D

2

21

1 2xe

D D

2

21

2 1 2xe

D

21

9 2xe

D

2

9xx e

2 21 2 3

. . . .

9

x x x

y C F P Ixc c x e c e e

Ans.

Q8(i). Find the value: 4

0

log 1 tane x dx

Solution :- Let 4

0

log 1 taneI x dx

Using the property IV for definite integral, we get

4

0

log 1 tan4eI x dx

4

0

tan tan4log 1

1 tan tan4

e

xdx

x

4

0

1 tan 1 tanlog1 tane

x x dxx

54

4

0

2log1 tane dx

x

4

0

log 2 log 1 tane e x dx

4 4

0 0

log 2 log 1 tane edx x dx

4

0log 2e x I

2 log 24 eI

log 28 eI

Ans.

Q8(ii). A force ˆˆ ˆ2 4 2i j k is passing through point 0,1, 2 . Find the moment of force w.r.t.

the point 1, 2,0 .


Here ˆ ˆˆ ˆ ˆ ˆ0 1 1 2 2 0 3 2r i j k i j k

and ˆˆ ˆ2 4 2F i j k

.

Then 1 3 2 6 8 2 4 4 62 4 2

i j ki j k

2 6 10i j k Ans.

Solved by:

1. B. M. Saini



Govt. R. C. Khaitan Polytechnic College, Jaipur.

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