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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
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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
Prelim Question Paper Solution
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
Vidyalankar : S.Y. Diploma Applied Mathematics
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
Prelim Question Paper Solution
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
Vidyalankar : S.Y. Diploma Applied Mathematics
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 =
Prelim Question Paper Solution
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
Vidyalankar : S.Y. Diploma Applied Mathematics
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
Prelim Question Paper Solution
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
Vidyalankar : S.Y. Diploma Applied Mathematics
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
Prelim Question Paper Solution
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
Vidyalankar : S.Y. Diploma Applied Mathematics
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)
=
Prelim Question Paper Solution
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 =
Vidyalankar : S.Y. Diploma Applied Mathematics
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