|
|
| 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
| | |
Moderation Team
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