|
| Question 2 | | Find the inverse of the function
F(x) = x + 1, x ¹ 1
x - 1 | Ans.
Let us see whether the given function is invertable
Let x1 , x2 ¹ 1 x1 ¹ x2
Then x1 + 1 ¹ x2 + 1
And x1 - 1 ¹ x2 - 1
\ x1 + 1 ¹ x2 + 1
x1 - 1 ¹ x2 - 1
Þ f(x1) ¹ f(x2)
\ f is invertable function
Now y = x + 1 or xy - y = x + 1/X - 1
Or x(y - 1) = y + 1 or X = y +1/y - 1
we are given f = {(x,y) : y = x + 1/x - 1}
\ f-1 = {(y,x) : y ¹ 1 , x = y +1/y - 1 }
| | | Question 3 | | Find fof-1 and f -1of the function
F(x) = 1/x , x ¹ 0.
Also prove that fof-1 = f-1of | Ans.
f(x) = 1/x
\ f-1 = { ( 1/x = y,x)}
= { (y, 1/y) }
\ f-1 (x) = 1/x
\ fof -1 (x) = f(1/x) = 1 = x
1/x
Also f -1of (x) = f-1 (1/x) = 1 = x.
1/x
\ fof -1 (x) = f -1 of (x)
Hence f -1 of = f -1 of | | | Question 4 | | Evaluate lim 2x2 + 3x
x®¥ x2 +1 | Ans.
Put x = 1/y, so that when x®¥, y®0.
\ lim 2x2 + 3x = lim 2/y2 + 3/y
x®¥ x2 +1 y®0 1/y2 +1 = lim 2 + 3y = 2 + 0 = 2.
y®0 1 +y2 1 + 0 | | | Question 5 | | The profit function of a manufacturing firm is given by P(x) = -x3/3 + 729x - 2500. Find x so that the firm attains maximum profit, where x is the number of items manufactured. | Ans.
The profit function is
P(x) = -x3/3 + 729x - 2500
P1(x) = -x2 + 729; where p 1 is the d(p(x))/dx
\ P1 (x) = 0 give us -x2 + 729 = 0
or x = 27.
[ -27 is rejected as x is number of items]
Now P11 (x) = -2x
Which is -ve at x = 27
\ P(x) attain the maximum value at x = 27
\ Number of items (x) manufactured by firm = 27. | | | Question 6 | |
_______________________
If y = Öcosx + Ö cosx +Ö cosx+ ?..¥ then prove that (2y -1) dy/dx + sinx = 0.
| Ans.
we are given that
_______________________
y = Öcosx + Ö cosx +Ö cosx+ ?..¥
_______
= Öcosx + y
or y2 = cosx + y
Differentiating both sides w.r.t. x1 we get
2y dy = -sinx + dy
dx dx
or 2y dy - dy sinx = 0
dx dx \ (2y - 1) dy/dx + sinx = 0. | | | Question 7 | | Show that x/sinx is an increasing function in the intercal 0< x < p/2. | Ans.
Let f(x) = x/sinx
Then f1(x) = sinx - xcosx
sin2x
Now sin2x >0 for 0< x < p/2
[ In fact sin2 x> 0 < " x]
we know that
tanx > x for 0 < x < p/2
\ sinx > x for 0 < x < p/2
cos
\ sinx - x cosx > 0 for 0 < x < p/2
\ f1(x) = sinx - x cosx > 0 for 0 < x < p/2
sin2x
\ f (x) is increasing in the interval { 0, p/2} | | | Question 8 | | Find the deviative of cot( 2x +1) w.r.t. x using the first principles. | Ans.
Let y = cot [ 2 x +1] \ y +dy = cot {2(x + dx) + 1 }
or dy = cot {2 (x +dx) +1} - cot (2x +1) = cos (2x+1+ 2dx ) - cos(2x+1)
sin (2x+1+2dx) sin (2x+1)
= cos (2x+1+2ds) sin (2x+1) - sin(2x+1+2dx) cos (2x+1)
sin (2x+1+2fx) sin ( 2x+1 )
= sin (2x +1 - 2x -1 -2dx)
sin(2x +1 +2dx) sin(2x+1) \ dy = -sin (2dx)
sx sin(2x+1 + 2fx) sin(2x +1)dx \ lim dy = lim -sin(2dx) = lim-1
dx ®0 dx dx ®0 sin(2x+1+2dx) sin(2x+1)dx \ dy = -1 x 1 x 2
dx sin(2x+1) sin(2x +1) = -2
sin2 (2x +1) = -2 cosec2 (2x +1) | | | Question 9 | | ____
Use Lagrauge's mean value theorm to determine a point P on the curve y = Ö x - 2 defined in the interval [2,3] where the tangent is parallel to the chord joining the end points on the curve. Unit - 2 (Integral Calculus) | Ans. _____
Here f(x) = Ö x - 2
_____
Now Ö x - 2 is defined for all values of x ³ 2
\ f(x) is defined in the interval [2,3] again f1(x) = 1
2Ö x - 2 1 is defined for all values of x > 2.
2Ö x - 2 \ f(x) is derivable in the interval ]2,3[.
Thus both the conditions of LMV therom are satisfied.
\ by LMV theorem there must exists at least one C in the interval ]2,3[ at which the tangent is parallel to the chord joining the end points on the curve, such that. F1 ( C ) = f(3) - f(2)
3 - 2
or 1 = Ö3 -2 - Ö2 - 2
2Ö c - 2 1 \ 1 = 1
2Ö c - 2 or 2Ö c - 2 = 1 \ 4(c - 2) = 1
or c - 2 = ¼
\ c = ¼ + 2 = 9/4 E ]2,3[
\ Now f (9/4) = Ö9/4 - 2 = Ö¼ = ½
Hence the tangent is parallel to the chord at the point (9/4, ½). | | | Question 10 | | Evaluate : ò14 |x - 2| dx | Ans.
½x - 2½ = x - 2 if x ³ 2.
And ½x - 2½ = -x - 2 if x < 2. \ ò14½x - 2½dx = ò12 - (x - 2) dx + ò24 (x - 2) dx
= ( -x2/2 + 2x)12 + (x2/2 + 2x)24 = [ (-4/2 + 4 ) - ( -1/2 + 2)] + [(16/2 -8) (4/2 -4)
= 2-3/2 + 0 +2 = 5/2. | | | Question 11 | | Evaluate
òxex/(x +1)2 dx
| Ans.
I = ò x e x dx
(x + 1)2 = òex [ 1/ (x +1)- 1/(x + 1)2 ] dx
= òex [f(x) + f1(x) ], where f(x) = 1/ (x+1)
= ex f(x) + c
= ex /(x+1) + C. | | | Question 12 | | Evaluate
______
ò Öx / Öa3 - x3 dx | Ans.
Let x 3/2 = t. then 3/2 Öx dx = dt
\ Öx dx = 2/3 dt \ I = òÖ x /Ö a3 - x3 dx = ò Ö2/3 dt/Öa3 - t2
= 2/3 ò dt/ Ö( a 3/2) 2 - t2
= 2/3 - sin-1 t/a3/2 + c {\ ò dx/Öa2-x2 = sin-1 x/a = c}
= 2/3 sin -1 (x/a)3/2 + c. | | | Question 13 | | Evaluate ò-11 ex dx as the limit of a sum. | Ans.
Let f (x) = ex . Let nh = 1- ( -1) = 2.
ò -11 exdx = lim h [ f (-1)+f(-1+h) +f(-1+2h)+ ------+f(-1+n-1h) ]
h ® o
= lim h[ e-1+e-1+h+e-1+2 h+-------+e-1+(n-1)h ]
h ® o
= lim h [e-1{ 1+eh+e2h+---------+e +(n+)h } ]
h ®o
= lim h [e-1{ 1-enh/1-eh } ]
h ® o
= lim h/1-eh . lim e-1 ( 1-enh)
h ® o h ®o
= (-1) [ e-1(1-e2)]
lim eh-1/h
h ® o
= e-1(e2 -1)}e2
1
= e-1
e | | | Question 14 | | Find the aver of the region bounded by the two parabolas y = x2 and x = y2 | Ans.
Sloving the equation y = x2 and x = y2 , the points of the intersection of the two curves are 0(0,) and P(1,1). The requried area is PROS in fig. (1.1) Y
\ A = ò y dx y = x2
\ Requried area A
= area OSPQ - area ORPQ
= ò01 Öx dx - ò01 x2 dx
=[2/3 x 3/2]01 - [x3/3]01
= 2/3 - 1/3 = 1/3 sq units. | | | Question 15 | | Prove that ò0p/4 log(1 + tanq) dq = p/8 log 2. | Ans.
Let I = ò 0 p/4 log(1 + tanq) dQ
= ò0 p/4 log [ 1 + tan(p/4 -Q) dQ
[\ ò0a f(x) dx = ò0a f ( a - x) dx] = ò0 p/4 log [1 - tanQ/1 + tanQ] dQ
= ò0 p/4 log [ 2/1 + tanQ ] dQ
= ò0 p/4 logdQ - ò0 0 p/4 log [1 - tanQ] dQ
\ I = log 2 [0] 0 p/4 - I
or 2 I = log 2 [ 0 p/4]
or I = p/8 log 2. | | | Question 16 | | Evaluate:
òx2 + 1/x4 + 1 dx | Ans.
ò x2 + 1/x4 + 1 =dx
= ò(1+1/x2)/(x2 + 1/x2) dx
= ò 1 + 1/x2 dx
x2 + 1/ x2 - 2 +2
Let x - 1/x = z, so that ( 1+1/x2) dx = dz
\ òx2 + 1/x4 + 1 dx = ò dz/ z2 + 2 = ò dz / z2+(Ö2)2 = (1/Ö2)( tan-1 z/Ö2 )+ c [\ ò 1/x2 + a2 dx = 1/a tan -1 x/a]
= 1/Ö2 tan-1 x - 1/x/Ö2 + c
= 1/Ö2 tan-1 x2 - 1/Ö2x + c | | | Question 17 | | Evaluate :
òx4/x4 - 1 dx | Ans.
Let x4 º x4
x4 - 1 (x-1)(x+1)(x2+1) = 1 + A + B + (x + D) ?..(1)
x-1 x+1 x+1 Multiplying by x4 - 1 we get
º (x4 - 1) + A (x+1) (x2+1) +B (x - 1) (x2 +1)
+ (x + D) (x + 1) ( x - 1 ) ?. (2) Putting x = 1, -1 in (II) , we get
1= A (2) (2) And 1 = B(-2)(2)
or A = ¼ and B = -1/4
Equating ¥ efficients of x3 in (2) , we get
A + B +C = 0
But a = ¼ and b = -1/4
\ c = 0
Puuting x = 0 in (2) , we get
0 = -1 + A -B - D
or D = -1 +A -B
= -1 + ¼ + ¼ = -1/2
\from (1) x4 = 1 + 1 - 1 - 1 .
x4 -1 4(x-1) 4(x+1) 2(x2+1) x4/x4 -1 dx = 1 + 1/4(x-1) - 1/4/x+1 - 1/2(x2+1).dx
= x + ¼ log (x -1 ) -1/4 log (x + 1) -1/2 tan-1 x + c
= x + ¼ log x - 1 -1/2 tan -1 x + c
x+1 | | | Question 18 | | Find the area of the region bounded by the curve c:y = tanx, tangent drawn to C at x =p/4 and the x - axis. | Ans.
Requried area = shaded area on the figure
= Area (OPR) - area (DPQR)
= òo p/4 tanx dx - DPQR ??..(1)
To find DPQR, we proceed as follows :
Now Point P is ( p/4 , 1) Y
![]()
\ y = tan x
\ dx/dy = sec2x
\Slope of tangent at P = sec2p/4 = 2
i.e., tanQ = 2
\ QR = PR cot Q
= 1 x ½ = ½
\ Area of DPQR = ½ x ½ x 1 = ¼ sq.unit. q R
\ from (1)
Required area = (log sec x )p/4 - ¼
= log 02 - ¼
=[1/2 log2 - ¼ ] sr.units. | | | Question 19 | | Evaluate : òop/4 (sinx + cosx)/(9+16 sin2x) dx | Ans.
Let I = òo p/4 sinx + cosx dx
9+16 sin2x = òo p/4 cosx + sinx dx
9+16 { 1-(sinx - cosx)2} Let sinx - cosx = t , then ( cosx +sinx) dx = dt
When x = 0, t = -1 and when x = p/4, t = 0
\ I = ò-1o dt
9 + 16 ( 1 - t2) = ò-1o dt
25 - 16t2 = ò-1o dt
52 - (4t)(4E)2 = 1/2x5 x ¼ log [ 5 + 4t/5 - 4t]o-1
= 1/40 [log1 - log(1/9)]
= 1/40 (log9) = 1/40 log 32
= 1/20 log3. | | | Question 20 | | _______
Evaluate ò(3x - 2) Öx2 +x+ 1 dx. | Ans.
Let I = ò(3x - 2) Öx2 +x+ 1 dx.
Here d/dx (x2 +x+ 1) = 2x +1
\ we express 3x - 2 in terms of 2x +1
Now 3x - 2 = 3/2 (2x+1) -7/2
\ I = ò{ 3/2 (2x + 1) - 7/2 } Öx2 +x+ 1 dx.
= 3/2 ò (2x +1) Öx2 +x+ 1 dx - 7/2 òÖx2 +x+ 1 dx.
= 3/2 I1 - 7/2 I2 + c
Now I1 = ò(2x + 2) Öx2 +x+ 1 dx.
Let x2 +x+ 1 = t , then (2x +1) dx = dt.
\ I1 = ò Öt dt = 2/3 t 3/2 = 2/3 (x2 +x+ 1)3/2
and I2 = ò Ö x2 +x+ 1 dx = ò Ö ( x + ½)2 + (Ö3/2)2 dx.
= (x + ½) Öx2 +x+ 1 + ¾ log [(x + ½) Ö x2 +x+ 1] +c
2 2 Ö3/2
= (2x + 1) Ö x2 +x+ 1 + 3 log [(2x + 1) Ö x2 +x+ 1] +c
4 8 Ö3
\ from (1)
I = 3/2 . 2/3 (x2 +x+ 1)3/2 - 7/2[(2x+1) Ö x2 +x+ 1 +
4
3/8 log [{2x +1 +2 Öx2 +x+ 1}] +c
= (x2 +x+ 1)3/2 - 7/8 (2x +1) Öx2 +x+ 1 - 21/16 log [ 2x + 1 + 2 Ö x2 +x+ 1] +c
Ö3 | | | Question 21 | |
Prove that òop/2 Öcotx/(ÖCotx + Ötanx) dx = p/4
| Ans.
Let I = òop/2 Öcotx / (ÖCotx + Ötanx) dx
= òop/2 Öcotx (p/2 -x) _____ dx
ÖCotx (p/2 -x) + Ötanx (p/2 -x) [\òoa f(x) dx = òoa f(a - x)dx] = òop/2 Ötanx dx / Ötanx + Öcotx
\ I + I = òop/2 Öcotx/ Ötanx + Öcotx dx + òop/2 Ötanx/Öcotx + Ötanx dx
= òop/2 Öcotx + Ötanx/Öcotx + Ötanx dx
= òop/2 1.dx
= [x]op/2
= p/2
\2I = p/2 or I = p/4. | | | Question 22 | | _____
Solve the differential equation (Ö a + x ) dy/dx + x = 0. | | Ans.
____
The given equation is( Öa + x) dy/dx + x = 0 Þ dy = -x /Öa + x dx
Þ òdy = - ò x dx / Öa + x
Þ òdy = - ò a + x - a / Öa + x dx
____
Þ y = - ò Öa + x dx + òa (a+ x) -1/2 dx Þ y = - (a + x)3/2 + a . (a + x) ½ + C
3/2 1/2 Þ y = -2/3 (a +x) 3/2 + 2a Öa + x + C.
which is the required solution. | | | Question 23 | | Find the differential equaion which corresponds to the equaion Y = ke sin-1x | Ans.
We are given Y = k . e sin-1x???.(1)
Defferentiating w.r.t. x we get
dy/dx = k.esin-1x 1/Ö1- x2 ???(2)
Putting k = y from (1) in (2) , we get
Sin-1 x
dy/dx = y . esin-1x
esin-1x Ö1-x2 or dy/dx = y/Ö1-x2
Which is the required differential equation. | | | Question 24 | | Solve the differential equation.
dy/dx + y cot x = sin2x. | Ans.
The given equation is
dy/dx + y cot x = sin2x. ???..(1)
Here P = cotx and Q = sin2x on coparing
(1) with dy/dx + Py = Q
\ I.F. = e òcot x dx
=e log sin x = sinx
\ Solution is
y.sinx = òsin2x sinxdx +c.
or y.sinx = 2òsin2x cosxdx +c.
or y.sinx = 2sin3x +c.
3 | | | Question 25 | | Solve the differential equation
y2 dx +(x2 +xy + y2) dy/dx = 0. | Ans.
The given equation can be written as
y2 dx +(x2 +xy + y2) dy/dx = 0. ?..(1)
It is a homogenous differential equation of degree 2 rewriting (1) y2 + dy = 0 ...........(2)
x2 +xy + y2 dx let y = zx so that dy/dx = z + x dz/dx.
\(ii) reduces to z2x2 + z + x dz/dx = 0
x2 +x2z + z2y2
or Z2 + z + x dz/dx = 0.
1+Z+Z2
or z2 +z+z2+z3 + x dz/dx = 0
1+Z+Z2 or z(1+ 2z +z2) + x dz/dx = 0
1+Z+Z2
or dx/x + (1+ z +z2) dz = 0
z(1+2Z+Z2)
or dx/x + 1+ 2z +z2-z dz = 0
z(1+2Z+Z2) or dx/x + 1 - 1 dz = 0
z 1+2Z+Z2 or dx/x + 1 - 1 dz = 0
z (1+Z )2 Integrating, ò dx/x + ò ½ dz - ò1/(1+x)2 dz = C.
or logx + logz = 1/1+z = C.
or logx + log y/x + 1/1+y/x = C.
or log y + x/x+y = C.
Which is the required solution. | | | Question 26 | | Prove that
® ® ® ® ® ®
(a x b)2 = a2 b2 - (a . b)2 | Ans.
® ® ® ® ® ®
(a x b)2 = (a x b) . (a x b)
® ®
= (|a||b| sinq n)2
= a2 b2 sin2q
= a2 b2 (1-cos2q)
= a2 b2 - a2 b2 cos2q
® ® ® ®
= a2 b2(|a||b| cosq )2
® ® ® ®
= a2 b2 - (a . b)2 | | | Question 27 | | Find the vector moment of three forces î + 2 î - 3k, 2î + 3 î + 4k, and î - î - k, acting on a particle at a point P (0,1,2) about the point A (1,-2,0). | Ans.
Let F be the resutant of the given three forces,then
®
F = î + 2 j - 3k + 2î + 3 j + 4k - î - j - k,
= 2î + 4 j - 2k
®
and AP = -î + 3 j + 2k
\ Vector moment of the forces about A
® ®
= AP x F = (-î + 3 î + 2k) x (2î + 4 î + 2k)
= -2î + 6 j - 10k
® ®
\ Moment = |AP x F| = Ö140 | | | Question 28 | | Using section formula, prove that the points (-2,3,5), (1,2,3) and (7,0,-1) are collinear. | Ans.
Let A (-2,3,5), B(1,2,3) and C (7,0,-1) be the given points. Let the point c (7,0,-1) divided the joint of AB in the ratio K:1.
\ K-2 , 2k+3 , 3k+5 Become the wordinates of C.
k+1 k+1 k+1 \ K-2 = 7, 2k+3 = 0, 3k+5 = -1
k+1 k+1 k+1 \ K-2 = 7 Þ 6K + 9 = 0 Þ 2K + 3 = 0.
k+1 \ K = -3/2
This value of k = -3/2 satisties the other two equations. \ C lies on the line segment joining A and B and divides it in the ratio -3/2 : 1 or 3:2 externally. | | | Question 29 | | Find the acute angle between the places.
r .(3î - 2 j + 6k) + 2 = 0 and r . (4î - 20 j + 5k) = 3. | Ans.
®
Let n1 be normal to the plane
®
r .(3î - 2 j + 6k) + 2 =
®
and n2 be normal to the palne r . (4î - 20 j + 5k) = 3.
® ®
\ cos q = n1 n2
½n1½½ n2½
= (3î - 2 j + 6k). (4î + 20 j - 5k)
Ö9+ 4+36 x Ö16+ 400 +25
= 12 - 40 - 30 = -58
7 x 21 147 This is the cosiue of the obfuse angle between the planes. | | | Question 30 | | Find the reflection of the point P(-1,-1,3) in the plane 2x + 3y - 4z - 10 = 0. | Ans.
Equaion of the line through the point P(-1,-1,3) and perpendicular to the plane
2x + 3y - 4z - 10 = 0. Is
x + 1 = y + 1 = Z - 3 = r (say)
2 3 -4 Any point on this line is P1( 2x-1, 3x-1, -4x+3). If this point is the reflection of P in the plane , then mid-point of PP1 lies in the plane.
Midpoint of PP1 is (r-1, 3r-2, -2r +3)
2
This point must lie in the plane 2x +3y - 4z -10 = 0.
\2(r-1) + 3(3r-2) -4 (-2r +3) -10 = 0
2
or 4r -4 + 9r -6 _ 16r -24 -20 =0
or 29r -54 =0
or r = 54/29
\ Point P1 is (97/29, 133/29 ,-129/29) | | | Question 31 | | Find the equation of the plane through the points (4,5,1), (3,9,4) and (-4,4,4) | Ans.
Equation of any plane through the point (4,5,1) is ,
a(x-4) +b(y-5) +c (z-1) =0 ??(1)
Let P,Q,R be the points(4,5,1), (3,9,4) and (-4,4,4) respectively.
®
\ PQ = -î + 4 j + 3k
®
and QR = - 7î - 5 j
® ®
\ PQ x QR = | | ? | ? | ? |  | | î | j | k | | -1 | 4 | 3 | | -7 | -5 | 0 | = 15î - 21 j + 33k ?..(2)
From (2) the direction ratio of the normal to the plane (1) are 15, -21 and 33.
\putting a = 15, b = -21 , c = 33 in (1), we get
15(x - 4) -21 (y-5) + 33 (z - 1) =0
or 15x - 60 021y + 105+33z - 33 = 0
or 15x -21y + 33z + 12 = 0
Hence 5x-7y + 11z +4 =0 is the equation of the requried plane. | | | Question 32 | | ® ® ® ® ® ®
If a . b x c ¹ 0 and a = b x c
® ® ®
a. b x c
® ® ®
b = c x a
® ® ®
a. b x c
and ® ® ®
c = a x b
® ® ®
a. b x c
® ® ® ® ® ®
then show that a . a1 + b . b1 + c . c1 = 3 | | Ans.
® ® ® ® ® ® ® ®
a . a1 = a . b x c = a . b x c = 1
® ® ® ® ® ®
a . b x c a . b x c
| ® | ® | ® | ® | ® | ® | ® | ® | ® | ® | ® | | | | b . | b = | b . c x a | = b . c x a | = a . b x c | | | | | | | | | ® ® ® ® ® ® ® ® ®
a . b x c a . b x c a . b x c
® ® ® ® ® ® ® ® ® ® ®
c . c = c . a x b = c . a x b = a . b x c
® ® ® ® ® ® ® ® ®
a . b x c a . b x c a . b x c
® ® ® ® ® ®
\ a . a1 + b . b1 + c . c1 = 1+1+1 = 3. | | | Question 33 | | ® ®
Prove that the two vectors a and b are equal if and only of their components along the x & y- axes are equal. | Ans.
® ®
Let a = aî1 + a2j and b = b1î + b2j be two vector where a1, a2 and b1 , b2 are the components
® ®
of a and b along x & y-axis respectively.
Necessary Condition :-
® ®
a = b Þ a1î + a2j = b1î + b2j
\ (a1- b1) î = (b2 - a2)j
It shows either (a1 - b1) î and (b1 - b2) î are parallel or each is a zero vector .
But they are not parallel.
\ (a1- b1) î = (b2 - a2) j = 0
\ a1- b1 = 0 and b2 - a2
Sufficient condition :-
In this case a1= b1 and a2 = b2. We have to show
® ®
that a = b. a1= b1 and a2 = b2.
\a1- b1 = 0 and b2 - a2 = 0
\ (a1- b1) î = 0 = (b2 - a2)j
or a1î + a2j = b1î + b2j
® ®
\ a = b | | | Question 34 | | Consider a group of 10 families each having 5 children of which 2 are boys and 3 are grils.One child is selected from each family, each family child of the family being equally likely to be selected. Calculate the prbability that amoung th 10 children so selected there are exactly 5 girls. | Ans.
We may consider this as the case of 10 independent experiments in each of which the probebility of success ( selected child is a girl) is 3/5. The number of success is thus a biominal random variable with distribution B (10,3/5).
Hence the required probability is
C (10,5) (3/5)5 (2/5)5 = 0.2 (approx.) | | | Question 35 | | In a bolt factory , A,B andC manufacture respectively 25%, 35% and 40% of the totalk bolts. Of their outputs 5,4 and 2 per centr are respectively from the product.
1. If the bolt drawn is found to be defective, what is the probality that it is manufactured by the machine B?
2. What is probability that the bolt drawn is defective. | Ans.
1.Let the events H1,H2 and H3 be the following:
H1: the bolts is manufactures by machine A.
H2 :the bolts is manufactured by machine B
H3: the bolt is manufactured by machine C.
Clearly H1 , H2 and H3 are mutually exclusive and exhaustive.
Let the event E be :
E: the bolt is defective.
The event E occurs with H1 or with H2 or with H3.
Now P(H1) = Probability that the bolt drawn is manufactured by machine A
= 25% = 0.25
Similarly, P(H2) = 0.35
And P(H3) - 0.40
Again P(E/H1) = Probabily that the bolt drawn is defective given the condition that it is manufactured by machine A
= 5% = 0.05
Similarly
P(E/H2) = 0.04 and P(E/H3) = 0.02
We are requested to find P(H2/E), i.e. given the condition that bolt drawn is defective , what is the probality that it was manufactured by machine B.
We have the Bayes formula
P(H2/E) = P(H2) P(E/H2)
P(H1) . P(E/H1) + P(H2) . P(E/H2) +P(H3) . P(E/H3)
Substituting in the above expression.
P(H2/E) = 0.32. x 0.04
0.25 x 0.05+0.35 x 0.04 + 0.40 x 0.02 = 0.0140 = 28
0.0345 69 (2)Probability that the bolt drawn is defective = P(E)
= P(H1).P(E/H1)+P(H2).P(E/H2)+P(H3)+ P(H3).P(E/H3)
=0.25 x 0.05 + 0.354 x 0.04 + 0.40 x 0.02
= 0.0345 | | | Question 36 | | Find the probability distribution of the number of success in two tosses if a die where success is defined as a number greater that 4'. Sketch its graph. | Ans.
Let x be the number of successes .
Then x can take values 0,1 or2
Now P(x = 0) = P (no success)
= 4/6 x 4/6 = 4/9 P(x = 1) = P( 1 auccess and 1 faliure)
= P(success and failure) + P ( failure and success)
= 2/6 x 4/6 + 4/6 x 2/6 = 4/9
P(x = 2) = P (both successes )
= 2/6 x 2/6 = 1/9
\ Required probability Distribution is x | 0 1 2__
P(x)| 4/9 4/9 1/9 Its graph is shown below  | | | Question 37 | | There are 2% defective nuts in a large bulk of nuts .Write the probability distribution for getting defective nuts in a randum sample of 10 nuts. | Ans.
Let p be the probability of defective nuts.
\ p = 2/100 = 1/50
and q = 1 - 1/50 = 49/50 Here n= 10
Then the probability distribution is given by (q+p)n
\(q+p)n = (49/50 + 1/50)10
= (49/50)10 + 10C1 (49/50)9 (1/50) + 10C2 (49/50)8 (1/50)2+???.+(1/50)10. | | | Question 38 | | An unbiased die is thrown again and again until three sixes are obtained. Find the probability of obtaining third six in the 6th throw of the die. | Ans.
Required Probability = P( throwing 2 sixes in first throws and a six in the 6th throw)
= P (throwing six twice in 5 throw) x P ( throwing a six in one throw)
= 5c2 p2q3 x 1/6 ( p = 1/6 \q = 5/6)
= 5 x 4 (1/6)2 (5/6)3 x 1/6
1 x 2
= 10 x 1/36 x 125/216 x 1/6 = 625/23328 | | | Question 39 | | 3 bad arlicles are mixed with 7 good ones. Find the probablity distribution of number of bad articles if 3 are drawn at random without replacement from this lot. Sketch its graps. | Ans.
Let X be the number of bad articles. Then x can take values 0,1,2,or 3
Now P (x=o) = p (3 good articles)
=7C3 = 7
10C3 24 P(x=1) = P(1 bad and 2good articles )
= 3C1 . 7C2 = 21
10C3 40 P(x = 2) = P(2 bad and 2good articles )
= 3C2 . 7C1 = 7
10C3 40
P(x = 3) = P(3 bad articles )
= 3C3 = 1
10C3 120 \Required P.D.is X | 0 1 2 3
P(x) | 7/24 21/40 7/40 7/120 Its graph is shown below  | Hope you will find them useful please comment I will post more if you like it |
|
About the Author:
|
Back to Community Shelf
No Nickels Awarded![[Post New]](/templates/default/images/icon_minipost_new.gif){kind=icon}
342