1 Maths Soln - Vidyalankar Classes &...

1114/SY/Pre_Pap/2014/Comp/CP/MathsIII_Soln 1

S.Y. Diploma : Sem. III Applied Mathematics

Time: 3 Hrs.] Prelim Question Paper Solution [Marks : 100

Q.1 (a) log xe dx = x dx elog (x) = f(x)

= 2x

c2

Q.1 (b) 2 2

1 cos xdx

1 x sin x = 2

1 cos xdx

sinx sinx1 x

= 2

1cot x cosec x dx

1 x

= tan 1 x + cosec x + c

Q.1 (c) tanx 2e sec x dx

tan x = t sec2 x dx = dt

= te dt

= et + c = etan x + c

Q.1 (d) log x dx = log x1 dx

Integrating by part

= d

logx 1dx logx 1dx dxdx

= log x x 1

x dxx

= log x x x + c OR = x (log x 1) + c

Q.1 (e) 32 23

3

d y dyk

dxdx

Order = 3, Degree = 2

Q.1 (f) x dy y dx = 0 x dy = y dx

dyy

= dx

dxx

dyy

= dxx

log y = log x + c

Vidyalankar

dxdx

c

c

x1 dx

art

= logx 1dx logx 1dx dxdd

1dx1dxdxdx

1dx1dx1dxdd

logxlogxdd

logxlogxdddxdx

logxlogxdd

= log x x = log x x y11 x dxxx

= log x x = log x + c x + c OR = x (log x OR = x (log 1) + c + c

Q.1 (e) (e) 32 2233d yd y33 dydy

kkyydydyd yyd yy

dxdx3dx33d

Order = 3, Degree = Order = 3, D

Q.1 (f)Q.1 (f) x dy x dy y dx = 0 x dy = y d x Vdy

y

Vidyalankar : S.Y. Diploma Applied Mathematics

2

Q.1 (g) Equation of the curve y = 3x x2 (1) Differentiate wrt x

d

ydx

= 2d3x x

dx

dydx

= 3 2x

But, dydx

= slope of curve = 5

5 = 3 2x 2x = 3 + 5 2x = 8 x = 4 Put x = 4 in equation (1) y = 3 (4) (4)2 y = 12 16 y = 4 Point is (4, 4)

Q.1 (h) Two coins are tossed simultaneously, find the probability of getting atleast one head.

S = {HH, HT, TH, TT} n(S) = 4 A = {HH, HT, TH} n (A) = 3

n(A) 3

P(A) 0.75n(S) 4

Q.1 (i) 52

1n(S) C 52 A = card is a diamond 13

1n(A) C 13

n(A) 13 1P(A) 0.25

n(S) 52 4

Q.1 (j) S = {1, 2, 3, 4, 5, 6}

n(S) = 6 A = {2, 4, 6} n(A) = 3

P(A) = n(A)n(S)

= 36

= 12

= 0.5

Q.1 (k) xe dx = x

0

e1

= x

0

1

e

= 0

1 1

e e

= [0 1] = 1

Vidyalankar

sly, find the probability of getting atld the probability of n(S) = 4 n(S) = 4

n (A) = 3 n (A) = 3

0.75

a diamond ond

1 131

n(A) 13 1n(A) 130.250.25

n(A) 13 1n(A) 13n(S) 52 4(S) 52 4

) ) S = {1, 2, 3, 4, 5, 6} S = {1, 2, 3, 4, 5, 6} n(S) = 6 n(S) = 6 A = {2, 4, 6} = {2, 4, 6} n(A) = 3 A) = 3

P(A) = PVin(A)n(A)n(S)S)

= 36

= 1

Q.1 (k) xxe dxe x

Prelim Question Paper Solution

3

Q.1 (l) Let y2 = ax2

2ydydx

= 2 ax

y dydx

= 2

2

y

xx

dy

x ydx

= 0

Q.2 (a) y = sin (log x)

dydx

= cos(log x)

x

xdydx

= cos (log x)

x2

2

d y dydxdx

= sin(log x)

x

x2 2

2

d y dyx

dxdx = y

x2 2

2

d y dyx

dxdx + y = 0

Q.2 (b) 3x 2y 2 2ydye x e

dx

2y 3x 2dye (e x )

dx

2y

dy

e = (e3x + x2)

2ye dy = 3x 2(e x )dx

2y 3x 3e e x

c2 3 3

Q.2 (c) y = dy

x xydx

dy

x ydx

= x y

y1 dy 1

dx xy x (Put y = t)

1 dy

dx2 y =

dtdx

1 dy

dxy =

2dtdx

Vidyalankar

2x )2

(e3x + x2) )

dydyy = 3x 22(e x )dx(e x )3x 22

d

2y 3x 33x 3e e xe2y 3x3x

c2 3 33 3

.2 (c)2 (c y = y = dydydyx xyx x

ydxdx

dydyx yy

yydx

= ddddx yVVVV1 dy y

dxy dxVVV1 d

2


4

dt t 1

2dx x x

dt 1 1

tdx 2x 2 x

I.F. = 1

dx2xe =

1log x

2e = 1 1

log2 xe =

1

x

1

tx

= 1 1

dx c2 x x

y

x =

1 1dx c

2 x

yx

= 1

logx c2

Q.2 (d) 2 2dy x y

dx xy

Put y = vx

dy dv

v xdx dx

2 2 2dv x v x

v xdx xvx

2 2

2

dv x (1 v )v x

dx vx

2 2dv 1 v v

xdx v

dx

vdvx

dx

vdvx

2v

log x c2

2 2y /x

log x c2

x = 1, y = 2 4

log(1) c2

c = 2

2 2y /x

log x 22

Vidyalankar

a2

2

v )2

vx

y2 22v v2

vv2

d

dxddvv

xxdydxdxx

vdvdv

Vidddy

2vlog x clog x c

22log xlog x

Vi

2 22 2y /x/x22

log x c2

log x

x = 1, y = 2 x = V44log(1

2log(

c = 2

y


5

Q.2 (e) M = ex + 2xy2 + y3 N = ay + 2x2y + 3xy2

My

= 4xy + 3y2 Nx

= 4x + 3y2

My

= Nx

Given D.E. is exact. solution is,

y = constant terms not

containing x

Mdx Ndy = c

x 2 3 y

y = constant

(e 2xy y )dx a dy = c

ex + x2y2 + xy3 + ya

loga = c

Q.2 (f) dy y ysin

dx x x

Let y

vx

y = vx

dy dv

v xdx dx

v + xdvdx

= v + sin v

xdvdx

= sin v

dv dx

sin v x

cosec v dv = dxx

dx

cosec v dvx

log | cosecv cotv | = logx + c

log y y

cosec cotx x

= logx + c

Q.3 (a) 40

2n(S) C 45 A = two of them valves are defective 4

2n(A) C 6

n(A) 6

P(A) 0.133n(S) 45

Vidyalankar

sin v

n v

ddy

dxsin v xv x

cosec v dv = ec v dv = dydxxxdydxdx

xcosec v dvosec

log | cosecv log | cosecv cotv |

log Vidycosec ccoseyx

c

Q.3 (a)) 402n(S) Cn(S) 402C2

A = two on(A


6

Q.3 (b) P = 3% = 0.03, n = 100

Mean (m) = np = 0.03 (100) = 3

m re m

P(r)r !

3 5e (3)

P(5) 0.10075!

Q.3 (c)Given, x 300, 25, x 350

Standard normal variate z is,

x x 350 300

z 225

Area = 0.5 (area between z = 0 to z = 2) = 0.5 0.4772 = 0.0228

Q.3 (d) 105n(S) C 252

A = {2 balls are white and 2 balls are black} 6 4

3 2n(A) C C 20(6) 120

n(A) 120

P(A) 0.4761n(S) 252

Q.3 (e) 52

2n(S) C 1326 (i) A = {both are spade} 13

2n(A) C 78

n(A) 78

P(A) 0.0588n(S) 1326

(ii) A = {one king and other queen} 4 4

1 1n(B) C C 16

n(A) 16

P(B) 0.0120n(S) 1326

Q.3 (f) The point (1, 1) is on the curve 2y3 = ax2 + b … (1) Putting x = 1, y = 1 in equation (1), we get, 2( 1)3 = a(1)2 + b

a + b = 2 Further differentiating (1) w.r.t. x, we get,

6y2 . dydx

= 2ax

dydx

= 2

2ax

6y =

2

ax

3y

Vidyalankar

s are black} lack} 12012

0.4761

26 are spade} ade}

132A) C 7878132C2

n(A) 78n(A) 78P(A) 0.0588P(A)

n(A) 78n(A) 78n(S) 1326n(S) 1326

) A = {one king and other queen} {one king and other q 4 44

1 11n(B) C C 16n(B) C C 164 441 11C CC1 11

n(A) 16A)

P(B)P(B)n(A) 16A)n(S) 132n(S

Q.3 (f) ) The point (1, The point (1, 1) is on 2y 3 = ax a 2

+ b

Putting x = 1, y Putt 2( 2 1)3 = a

a + b = Further d

6


7

Slope of curve = Slope of tangent = 2

ax

3y

Slope of curve at (1, 1) = 2

a (1)

3 ( 1)

= a3

… (3)

Slope of x + y = 0 is 1

1 = 1 … (4)

Results (3) and (4) stand for the slope of the same thing.

a3

= 1

a = 3 Putting a = 3 in result (2), we get,

3 + b = 2 b = 2 + 3 = 1 b = 1

Thus, we have, a = 3, b = 1 Q.4 (a) Put log x = t

1x

dx = dt

= 2

dt

cos t

= 2sec t dt

= tan t + c = tan (log x) + c

Q.4 (b) 1x tan x dx = 1 1dtan x x dx tan x x dx dx

dx

= 2 2

12

x 1 xtan x dx

2 21 x

= 2 2

12

x 1 x 1 1tan x dx

2 2 x 1

= 2

12

x 1 1tan x dx dx

2 2 x 1

= 2

1 1x 1tan x [x tan x] c

2 2

Q.4 (c) Put t = x

tan2

dx = 2

2 dt

1 t cos x =

2

2

1 t

1 t

Vidyalankar

t

2ec t dt2

tan t + c c = tan (log x) + c g x)

11an x dxx dx11 = = 1tan xtan x11 dx dxx

ddd

= = 22

1 xx22

tan xtan x1 x22

= = 2

1 x2

tan xt 1

= ta

Put


8

= 2

2

2

2dt

1 t1 t

5 41 t

= 2 2

dt2

5 5t 4 4t =

2

dt2

t 9

= 11 t2 tan c

3 3 = 12 tan(x/2)

tan c3 3

Q.4 (d) Let I = /2

0

cos xdx

cos x sinx … (1)

= /2

0

cos x2 dx

cos x sin x2 2

…by property

I = /2

0

sinxdx

sinx cos x … (2)

(1) + (2)

2I = /2

0

sinx cos xdx

sinx cos x

2I = /2

/20

0

dx x

2I = 02

I = 4

Q.4 (e) y = 4x x2

0 = 4x x2

0 = x(4 x) x = 0 or x = 4

Area = 4

2

0

(4x x )dx = 43

2

0

x2x

3 = 2(4)2

340

3

= 323

or 10.66

Q.4 (f) x2 + y2 = 36

y2 = 36 x2

y = 236 x

Area = b

a

y dx

Vidyalankar

…by property …by prop

… (2) … (2)

y4

y = 4x = 4x x x22 0 = 4x = 4x x x2

0 = x(4 = x(4 x) x) x = 0 or x = 4 = 0 or x = 4

Area = Area = 4

2

0

(4x x )d(4x 2

= 323

Q.4 (f) xx22 + y + 2 = 36 y2 =


9

A = 6

2

0

36 x dx = 6

2 2

0

(6) x dx

= 62

2 2 1

0

x (6) x(6) x sin

2 2 6

= [0 + 18 sin 1 (1)] [0] = 18 ( /2) = 9

Area of circle = 4 (9 ) = 36

Q.5 (a) 3

x 1dx

x 4x

x 1dx

x(x 2)(x 2)

Consider, x 1

x(x 2)(x 2) =

A B Cx x 2 x 2

x + 1 = (x 2)(x 2)A x(x 2)B x(x 2)C Put x = 0 1 = ( 2) (2) A + 0 + 0

1 = 4A 1

A4

Put x = 2 3 = 0 + 2(4)B + 0

3 = 8B 3

B8

Put x = 2 1 = 0 + 0 + ( 2) ( 4)C

1 = 8C 1

C8

11x 1 8 84

x(x 2)(x 2) x x 2 x 2

x 1 1 1 3 1 1 1

dx dx dx dxx(x 2)(x 2) 4 4 8 x 2 8 x 2

1 3 1

log x log x 2 log x 2 c4 8 8

Q.5 (b) Let I = 0

x sin2 x dx … (1)

= 20

( x) sin ( x)dx by property

= 20

( x) sin (x) dx sin( ) = sin

B x(x 2)C2)Cx(xla11

A44

al33BB

8

+ 0 + ( 2) () ( 4)C 4

= 8C = 8C ya11CC8

dyyy11x 1 8x 1 44x(x 2)(x 2) x xx(x 2)(x 2) x

442)(x 2) x x(x 2) x

x 1 1x 1x(x 2)(x 2)x(x 2)(x 2)(x

x 1 1x 1dx

1x(x 2)(x 2)x(x 2)(x 22)(x 2)2)(x

Q.5 (b) Let I = Let


10

= 2 20

sin x x sin x dx

= 20

sin x dx I From (1)

2I = 20

sin x dx

2I = 0

1 cos 2xdx

2

= 0

cos 2x dx2 2

= sin2x

x2 2 2

= sin2( ) sin2(0)

( ) (0)2 2 2 2 2 2

2I = 2

(0)2 4

2I = 2

2

I = 2

4

Q.5 (c) The equation of the curve is

13x3 + 2x2y + y3 = 1 … (1) Differentiating w.r.t. x, we get,

13 (3x2) + 2 2 2dy dyx . y.(2x) 3y .

dx dx = 0

39x2 + 2x2 dydx

+ 4xy + 3y2 dydx

= 0

(2x2 + 3y2) dydx

= (39x2 + 4xy)

dydx

= 2

2 2

39x 4xy

2x 3y

dydx

at (1, 2) = 2

2 2

39 (1) 4(1)( 2)

2(1) 3( 2)

= 39 8

2 12=

3114

Slope of tangent at (1, 2) = 3114

Its equation in slope-point form is : y y1 = m(x x1)

Vidyalankar

urve is y + y3 = 1 = 1

r.t. x, we get, e ge

3x2) + 2 2 dy3y .2

d2 dydy

x .x .2 y2xx2 y.(2x)y.(2x)dx

x . y.(2x)x . y.(2x)dx

y.(2x)y.(2x)

39x39 2 + 2x + 2x2

dy

dydxdx

+ 4xy + 3y + 4x 2adydx

=

(2x(2x22 + 3y + 3y2) dydyydx

= (39x (3 2 +

Vid

dydydxdx

= d2 22x 3y2x2

Vidydydxx

at (1, a 2

Slop


11

Putting x1 = 1, y1 = 2, m = 3114

, we get,

y ( 2) = 3114

(x 1)

14y + 28 = 31x + 31 31x + 14x = 3

Now, Normal Tangent at (1, 2)

Slope of Normal at (1, 2) = 1

slope of tangent at (1, 2)

= 131

14

= 1431

= m , say.

Its equation in slope-point form is y y1= m (x x1)

Putting x1 = 1, y1 = 2, m = 1431

y ( 2) = 1431

(x 1)

31y + 62 = 14x 14 14x 31y 76 = 0

Q.5 (d) Let y = f (x) = x3 9x2 + 24x … (1)

Differentiating w.r.t. x. we get,

dydx

= f (x) = 3x2 18x + 24 … (2)

= 3 (x2 6x + 8) = 3 (x 2) (x 4) … (3)

Now, for Maxima or Minima. dydx

= 0

3 (x 2) (x 4) = 0 x = 2 or x = 4

Note that differentiation of dydx

is easy and therefore we will follow second

derivative test for Maxima or Minima. Further, differentiating relation (2) w.r.t. x, we get

2

2

d y

dx= 6x 18

(a) At x = 2

2

2x 2

d y

dx = 6 (2) 18 = 12 18 = 6 < 0

x = 2 gives maximum value of the function. From relation (1) ymax = f (2) = (2)3 9 (2)2 + 24 (2) = 8 36 + 48 = 20 The point of Maxima is : (2, 20)

Vidyalankar

24x … . we get, et,

= 3x2 18x + 24 8x + 24

= 3 (x (x22 6x + 8) 6x + 8) = 3 (x x 2) (x 2) (x 4) 4)

or Maxima or Minima. or Maxima or Minimaydydydx

= 0

3 (x 2) (x 2) (x 4) = 0 0 x = 2 or x = 4 = 2 or x = 4

Note that differentiation of Note that differentiation of d

derivative test for Maximarivative test for MFurther, differentiating Further, diffV2

2

d y2

dx= 6x 18

(a) At x = 2 (a) A

2d2


12

(b) At x = 4

2

2x 4

d y

dx= 6(4) 18 = 24 18 = 6 > 0

x = 4 gives minimum value of the function. From the relation (1) ymin = f(4) = (4)3 9 (4)2 + 24 (4) = 64 144 + 96 = 16 The point of Minima is : (4, 16)

Q.5 (e) Let ‘x’ be the breadth of a rectangular enclosure Length of enclosure = (100 2x)

(This is not divided by 2 because the fourth side is being a wall).

Area of the enclosure A = L × B = (100 2x) × x = 100x 2x2

A = 100x 2x2 … (1) Functional relation Differentiating w.r.t. x, we get,

dAdx

= 100 4x … (2)

Again differentiating w.r.t. x

2

2

d A

dx = 4 … (3)

Nor, for maximum area, dAdx

= 0

100 4x = 0 4x = 100 x = 25

Putting x = 25 in relation (3)

2

2x 25

d A

dx = 4 (which is independent of x) < 0

x = 25 certainly gives the maximum area of the rectangular enclosure. Form relation (1), Amax = 100(25) 2 (25)2 = 2500 1250 = 1250 sq.m.

Q.5 (f) The equation of the curve is y2 (2a x) = x3 … (1) Differentiating w.r.t. x, we get,

y2 ( 1) + (2a x) . 2y .dydx

= 3x2

2 (2a x) y dydx

= 3x2 + y2 … (2)

dydx

= 2 23x y

2(2a x) y

100 2x

Enclosur

100 2x

x x

Vidyalankar 6

x 1)

unctional relation al relatio

… (2)

… (3) … (3)

ea, adAdxx

= 0 = 0

= 25 in relation (3) ion

dd

2x 2525

AA

dx = 4 (which is independe4 (which is

x = 25 certainly gives the max = 25 certainly gives the Form relation (1), Form relation (1), A Amaxx = 100(25) = 100(25) 2 (25) 2 22

Q.5 (f) (f) The equation of the curvThe equation of y y22 (2a (2 x) = Differentiating w.r.tDifferentiati

y y2 ( 1) + (2

2 (2a

kakaka100 2x 2

sur

100 00 2x 2

x x


13

a,a

dydx

= 23(a) a

2(2a a) (a) =

2

2

4a

2a = 2

Further, differentiating (2) w.r.t. x, we get,

22

2

d y d y d y d y(2a x) y y ( 1) (2a x)

dx dx dxdx= 6x + 2y

dydx

Dividing throughout by 2, we get,

(2a x)22

2

d y dy dyy y 2a x

dx dxdx= 3x + y

dydx

Putting x = a, y = a, dydx

= 2 to find 2

2

d y

dx at (a, a), we get,

(2a a)2

22

d ya a (2) (2a a) (2)

dx= 3a + a (2)

a2

2

2

d y

dx 2a + 4a = 5a

a2

2

2

d y

dx = 3a

2

2

d y

dx =

2

3a

a=

3a

Then, the radius of curvature is :

=

3 22

2

2

dy1

dx

d y

dx

=

3 221 (2)

3a

= 3 2a(5)

3 =

a5 5

3 =

5 5 a3

units

Q.6 (a) T.T. = 2 2

2

(M.T.) ( 3x) 94 F.T 44x

OR

T.T. = 2 21 1 9

coeff.of x ( 3)2 2 4

= 2

1dx

9 9x 3x 2

4 4

= 2

1dx

3 1x

2 4

= 2

2

1dx

3 1x

2 2

Vidyalankar

et,

a (2)

vature is :

y2d y2

dx

= alyaa3 23 2221 (2)1 (2)

33aa

= = a 3 2a(5)

3

T. = dyy

2 22

2

(M.T.) ( 3x) 9M.T.) ( 3x) 92 22

4 F.T 44 F.T 424x OR

T.T. = .T. = 22

= =2x2


14

=

3 1x1 12 2log c

1 3 1 32. x2 2 2

= x 2

log cx 1

Q.6 (b) The two curves are y2 = 4x … (1)

x2 = 4y … (2)

Put y = 2x

4 in equation (1)

22x4

= 4x

x4 64x = 0 x (x3 64) = 0 x = 0, x = 4

When x = 0, y = 0 One point of intersection is (0, 0) When x = 4, y = 4 other point of intersection is (4, 4)

From y2 = 4x y1 = 2x1/2

From x2 = 4y y2 = 2x

4

The required area = 4

1 20y y dx

=1 242

0

x2x dx

4 =

4 43/2 3

0 0

x 1 x2

(3 / 2) 4 3

= 3/2 34 1[4 0]

3 12 =

32 163 3

= 163

sq. units

Q.6 (c) Let I = 3 2

0

x sin x cos xdx = 3 2

0

( x)sin ( x)cos ( x)dx

= 3 2

0

( x)sin x cos xdx = 3 2 3 2

0 0

sin x cos xdx x sin x cos xdx

= 3 2

0

sin x cos xdx I

I + I = 3 2

0

sin x cos xdx

Vidyalankar

int of intersection is (0, 0) ection is (0, 0) r point of intersection is (4, 4) of intersection is

4

= 4

1 210y y dx y y dx 1 21 2

== a1 22422

00

xx2x dx2x dx2 xx

4

= =y4 14 3/2 [43 123

3/2 13/2 [412

= y16163

sq. unit sq.

Q.6 (c) (c) Let I = I = 3 23

00

x sin x cos xx sin3 23

= 0

( x)

=


15

2I = 2 2

0

sin x cos x sinxdx = 2 2

0

(1 cos x)cos x sinxdx

Put cos x = t sin xdx = dt sin xdx = dt

2I = 1

2 2

1

(1 t )t ( dt)

2I = 1

2 4

0

(t t )dt

2I = 13 5

1

t t3 5

2I = 1 1 1 1

3 5 3 5

2I = 1 1 1 1

3 5 3 5

2I = 2 2

3 5

Q.6 (d) Let I = 5

4

5 xdx

x 4 5 x …(1)

I = 5

4

5 (9 x)dx

(9 x) 4 5 (9 x)

I = 5

4

x 4dx

5 x x 4 …(2)

Add (1) and (2)

I + I = 5

4

5 xdx

x 4 5 x +

5

4

x 4dx

5 x x 4

2I = 5

4

5 x x 4dx

x 4 5 x

2I = 5

4

dx

2I = 54x

2I = 5 4 = 1

I = 12

X 0 t 1 1

Vidyalankar

aaadx5 x

…(1) …(1) alalyayayayaa5 (9 x)9

dxdx(9 x) 4 5 (9 x)4 5 (9 x)

= = yydydydya

dya

dya5

44

x 44dx

5 x x 45 x x 4 …

d (1) and (2) and (2)

I + I = I + I = dydyidyidydydyidy55

4

5 x5dxdx

x 4 5 x4 5 x +

5

2I = 2I =iddidddididdid5544 5 x x 45 x

x 4 5

2I = 55

4

dx

2I = 2I =


16

Q.6 (e) dx

sinx sin2x =

dxsinx sinx . cos x

Put x

tan2

= t dx = 2

2dt

1 t

= 2 2

2 2 2

1 2dt.

2t 2t 1 t 1 t2 . .

1 t 1 t 1 t

= 2

2 3

1 t2 .dt

2t(1 t ) 4t 4t =

2

3

1 t2 .dt

2t 2t 4t

= 2

3

1 t. dt

3t t =

2

2

1 tdt

t(3 t )

= 2

2

1 tdt

t(3 t )

21 t

t 3 t 3 t =

A B Ct 3 t 3 t

1 + t2 = 3 t 3 t A t 3 t B t 3 t C

A = 13

, B = 23

, C = 2

3

21 t

dtt 3 t 3 t

= 1/ 3 2 / 3 2 / 3

dtt 3 t 3 t

= 1 2 2

log t log 3 t log 3 t c3 3 3

=1 x 2 x 2 x

logtan log 3 tan log 3 tan c3 2 3 2 3 2

Q.6 (f) 2 2

1dx

4 sin x 5cos x =

2

2 2

2 2

1

cos x dxsin x cos x

4 5cos x cos x

= 2

2

sec xdx

4 tan x 5

Put tan x = t sec2 x dx = dt

= 2

1dt

4t 5 =

22

1 1dt

4 t 5 /2

= 11 1 ttan c

4 5/2 5 /2

= 11 2 tanxtan c

2 5 5

Vidyalankar22

33

1 tt2 ..

33

1 ttddt

2t 2t 4t2t 4t33

= ka

22

2

1 t1 tdtdt

t(3 t )t( 2

C

3 t

3 t A t 3 t B tA t

=

alalaalaaaa1/ 3 2 / 3 2 / 33 2 / 3 2 / 3

dttt 3 t 3 t3 t 3

= = a1 2 21 2log t log 3 tg t log

2 223 33 3

g gg g

= ya1 x 2xlogtan logogtan

x 2x3 2 32

g gg

dy2 22 2

11dx

4 sin x 5cos xx 5cos x2 22 2 = yys

44

Put tan x = t ut tan x = t secec22 x dx = dt x

1 Maths Soln - Vidyalankar Classes &...

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