DERIVATIVE OF A FUNCTION 1.5. DEFINITION OF A DERIVATIVE OTHER FORMS: OPERATOR:,,,
Application of Derivative Sloution
Transcript of Application of Derivative Sloution
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Application of DerivativesUNIT - 3
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S O L U T I O N S
1. Clearly = tandx
dySlope of normal = cot
Equation of normal at '' is )cossina(y )sin(cosax(cot += )
+ sincosasinasiny 2 ++= cossinacosacosx 2 asinycosx =+Clearly this is an equation of straight line which is at a constant distance a from origin.
2. 2dx
dye2
dx
dyey
)1,0(
x2x2 =
==
Tangent at (0, 1) is y - 1
= 2(x - 0) or 2x - y + 1 = 0.
It meets x-axis where y = 0 2x + 1 = 02
1x = .
Required point is
0,
2
1.
3. 3dx
dyx3
dx
dyxy
1x
23 ====
and 3dx
dy
1x
==
.
Hence tangents are parallel.
4. 2
2
2
3
a8x3
dxdy
a8xy == .
Slope of normal3
2
x3
a82
2
== (given)
a2xa4x22 == (Q x > 0 in first quadrant)
aa8
)a2(y
2
3
== .
Required point is (2a, a).
5. Let (x1, y1) be the required point.
)3x(2dx
dy=
Slope of tangent at (x1, y1) = 2(x1 - 3). Also, slope of chord joining (3, 0) and (4, 1) is
134
01=
2
7x1)3x(2 11 ==
413
27)3x(y
2
211 = ==
Required point is 41
,2
7.
LEVEL - 1 (Objective)
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6. xdx
dyx2
dx
dy2x3y2
2 ===
Slope of normal =x
1
Slope of normal at 111)1,1( ==
Equation of normal is y - 1 = 1(x - 1) x - y = 0.
7. Tangent is parallel to x-axis if 0dx
dy= .
0d
dxand0
d
dy0
d
dxd
dy
=
=
02cos302
2sin3
d
d==
.
422
=
= .
Also fordx
, e (sin cos ) 04 d
= = +
.
8. 2xdx
dy=
By given condition.x - 2 = -2(x - 2) x = 2.
9. 0dx
dy
b
y
b
n
a
x
a
n2
b
y
a
x1n1nnn
=
+
=
+
1nn
1nn
1n
1n
y.a
x.b
b
y
b
1
a
x
a
1
dx
dy
=
=
n n 1
n n 1
dy b a bAt (a,b),
dx a b a
= =
Tangent at (a, b) is y - b = )ax(a
b
2b
y
a
x1
a
x1
b
y=++=
which is independent of n.
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10. x26dx
dy=
3x0dx
dy== .
9918y == Point is (3, 9).
11. y = ax2 - 6x + b passes thorugh (0, 2)
2 = 0 - 0 + b b = 2
6ax2dx
dy=
6a36
2
3a2
dx
dy
23x
=
==
.
Since tangent is parallel to x-axis.
2a0dx
dy
2
3x
===
.
12. ,t2dt
dx=
Tangent is perpendicular to x-axis if 0t0dtdx == .
13. Let the point be ( , )
/ abe = ...... (1) (Q the point lies on the curve)
Also,a/x
ea
b
dx
dy =
/ a
( , )
dy be
dx a
= =
[By (1)]
aax
1y
=+
a1
y
a
x +=
+
Comparing with 1b
y
a
x=+ , we get
01a
1,b ==
+=
Point is (0, b).
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14. For the point (2, -1)
2 = t2 + 3t - 8 t2 + 3t - 10 = 0 t = 2, -5 and -1 = 2t2 - 2t - 5 2t2 - 2t - 4 = 0 t2 - t - 2 = 0 t = 2, -1 Common solution is t = 2
.2tfor76
3t22t4
dt
dxdt
dy
dxdy ==
+==
15. Coordinates of the point P are (at2, 2at).
Differentiating y2 = 4ax a4dx
dyy2 =
t1
at2
a2
dx
dy
y
a2
dx
dy
P
=== .
Equation of tangent is
0atytx)atx(t
1at2y
22 =+= .
16. 145tany2
1
dx
dy1
dx
dyy2xy
02 ===== (given )
2
1y =
1x
4 =
Point is
2
1,
4
1.
17. The given line x + y = a.... (1) is tangent to the ellipse 116
y
9
x22
=+ ..... (2) if it intersects the ellipse at a
unique point.
Using (1) in (2), we get
144)xa(9x161)xa(16
1
9
x 2222
=+=+ .
25x2 - 18ax + 9a2 - 144 = 0. Its discriminant is zero. (18a)2 - 4 25 (9a2 - 144) = 0 18 18a2 - 4 25 9(a2 - 16) = 0 16a2 = 400
a = 5.
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18. 1dx
dyxcos
dx
dyxsiny
)0,0(
===
Slope of normal is -1 Equation of normal is y - 0 = -1(x - 0) x + y = 0.
19.t1
at2a2
dxdya2
dtdy,at2
dtdx ====
== 0tt
1point is (0, 0).
20. tsintcosa3dt
dx,tcostsina3
dt
dy 22 == .
tcos
tsin
tsintcosa3
tcostsina3
dx
dy2
2
==
Equation of tangent is )tcosax(tcos
tsintsinay
33 = .
x sin t + y cos t = a sin t cos t (cos2t + sin2 t).Divide by sin t cos t and get (b).
21. 2mx2dx
dyxy 1
2 ===
2
1
m2
1
dx
dy
x7y6 23
=== m1m2 = -1 at (1, 1)
2
= .
22. Solving y = 4 - x2 and y = x2, the point of intersection is 2,2 .
m1= slope of tangent = 22
m2= slope of tangent = 22
7
24
81
24tan =
= .
23. pyx22 = ........ (1)
qxy q or y
x= = .......... (2)
Differentiating (1),
y
x
dx
dy=
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Differentiating (2), 2x
q
dx
dy =
Product of slopes = 1qq
xy
q
x
q
y
x2
=
==
(by (2))
2isangle .
24. x3 - 3xy2 + 2 = 0 ..... (1)
3x2y - y3 - 2 = 0 ..... (2)
Differentiating (1) and (2), w.r.t. x, we get
xy2
yx
dx
dym
22
)1(From
1
=
=
22
)2(From
2yx
xy2dxdym ==
Since m1m
2= -1, curves cut at right angles.
25. 2 21 1 2 2A(bt , 2bt ), B(bt , 2bt )
b4dx
dyy2bx4y
2 ==
Slope of tangent at11A t
1
bt2
b2
dx
dyA ==
=
Slope of 21
2
2
12
btbt
bt2bt2AB
=
1t
1
btbt
bt2bt2
1
2
12
12 =
)tt(tt2t222
2112 =
1
1
2 tt2t = .
26.2
9y2
y2
182
dx
dy===
8
9xx18y
2 ==
Point is
2
9,
8
9.
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27. )1x(6)x(f =
Cx2
x6)x(f
2
+
=
Since 3)2(f = = C, (slope of tangent at (2, 1) = 3)
3x6x3)x(f2 +=
dx3x)x(f3 +=
It passes through (2, 1), therefore f(2) = 1 d = -1323
)1x(1x3x3x)x(f =+= .
28. =
=
cosad
dy,sina
d
dx
= tandydx
Equation of normal at is )]cos1(ax[cos
sinsinay +
=
0cosysin)ax( =
It passes through the point of intersection of lines x - a and y = 0. It passes through the point (a, 0).
29. Let c6bx6ax6)x(f2 ++=
dcx6bx3ax2)x(f 3 +++=Here, f(0) = d = f(1)
0)x(f = , will have at least one root in (0, 1)
6ax2 + 6bx + 6c = 0 will have at least one root in (0, 1) ax2 + bx + c = 0 will have at least one root in (0, 1).
30. As f(1) = -2 and ]6,1[x2)x(f
Applying Lagranges mean value theorem,
2)c(f5
)1(f)6(f=
)1(f10)6(f +
210)6(f
8)6(f .
31. Here, 03
12fand)3(f)1(f =
+=
a + b + 11 - 6 = 27a + 9b + 33 - 6 and
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0113
12b2
3
12a3
2
=+
++
+
26a + 8b = -22 and 113
b2b4
3
4
34a3 =++
+
1+
13a + 4b = -11 and 113
24b)3413(a =
+++
13a + 4b = -11 and 0b3
2a34 =+
13a + 4b = -11 and 12a + 2b = 0 13a + 4b = -11 and 6a + b = 0Solving we get, a = 1, b = -6.
32. Here f(1) = f(2) and 03
4f =
c2b48cb1 ++=++ and 0c3
4b2
3
43
2
=+
+
Solve and get (b).
33. Applying Rolles theorem to F(x) = f(x) - 2g(x),
F(0) = 0, F(1) = f(1) - 2g(1)
0 = 6 - 2g(1) g(1) = 3.34. Since f(x) = x3 - 6x2 + ax + b defined on [1, 3] satisfies Rolles theorem
f(1) = f(3) 1 - 6 + a + b = 27 - 54 + 3a + b a = 11. It is independent of b
Rb .
35.
0
0
)0(f)x(f
)x(f)x(flim
2
0x
2
x 0
2xf (x ) f (x)lim
f (x)
=
(By LHospitals Rule)
1)x(f
)x(fx2lim1
2
0x=
+=
( f (0) 0 Q as f is strictly increasing)
36. f(x) = x2 - 6x + 8, 4x2 f(2) = 0 = f(4)
By Rolles theorem, )4,2(c such that 0)c(f =3c06c2 == .
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37. 3 2 2f (x) x 3x f (x) 3x 3 3(x 1)= = =
),1(]1,(xif.e.i1xif0)x(f2
and 2f (x) 0 if x 1 i.e. if x ( 1, 1)
38. )10x3(xx10x3dx
dy
1x5xy
223
+=+=+=
=
3
10x)0x(3
dx
dy
( )0x3
10x3
=
0)0x(
3
10xif0
dx
dy
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43. We must have 0b
f,0)0(f =
=
010secb0sina2 =++
and 016
secb6
sina2 =+
+
3
2a,1b ==
44. )xlog1(x
1
x
1.
x
1xlog
x
1
dx
dy
x
xlogy
22=+==
.ex1xlog0dx
dy===
For x < e log x < 1 and for x > e log x > 1
Atdx
dy,ex = changes sign from +ve to -ve.
y is maximum at x = e and its value is1
ee
elog = .
45. If x = t, then2
ty
2
= are the parametric equations of the parabola
2
t,t
2
and if its distance from (0, 5) is d, then
22
225
2
ttdz
+==
052
tt2t2
dt
dz 2=
+=
2t(t 8) 0 t 0, 2 2
2 = = , 22tfor0)8t3(
2
1
dt
zd 22
2
=>=
z and hence d is min. when 22t = y = 4.
Points are 4,22 .46. Let two parts be x and 64 - x
2233)x64(3x3
dx
dS)x64(xS =+= and
2
2
d S6x 6(64 x) 0
dx= + > for each x.
S is min. when 0)x64(3x30dx
dS 22 ==
(x + 64 - x)(x - 64 + x) = 0 x = 32 Parts are 32, 32.
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47. xy x log y x log x= =
)xlog1(ydx
dyxlog1
dx
dy
y
1+=+=
0dx
dy= 1ex0xlog1 ==+
48. 10x5x5xy345 +=
234x15x20x5
dx
dy += and )3x6x2(x10x30x60x20dx
yd 2232
2
+=+=
3,1,0x0)3x)(1x(x50dx
dy 2 ===
2 2 2
2 2 2
x 0 x 1 x 3
d y d y d y0, 0, 0
dx dx dx= = =
= < >
Max. at x = 1, Min. at x = 3.
49. A point on hyperbola 1b
y
a
x2
2
2
2
= is )tanb,seca( .
Here 23b,62a == .
p = perpendicular distance from 3x + 2y + 1 = 0
13
1tanb2seca3 ++=
p is max. or min. if p2 is max. or min.
Let )1tanb2seca3(13
1z ++=
0secb2tanseca30d
dz 2 =+=
3
1
623
232
a3
b2sin =
==
3
2cos = . Putting these values of sin and cos , the points are (6, -3), (-6, 3).
Also p is min. at (-6, 3).
50. f(x) = x1/x, x > 0.
Since x = e is a point of maxima, f(e) > f(x) for all x > 0.
)(f)e(f > > /1e/1e
( ) ( ) e/1ee/1e > ee > .
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1. Integratingdy
dx= 3x2 - 4x + A ;
dy
dx x
= 1= 0 A = 1
Hencedy
dx= 3x2 - 4x + 1 ;
Integrating again,
y = x3 - 2x2 + x + B ; ]yx = 1 B = 5 . Thus y = x
3 - 2x2 + x + 5
Alsody
dx= 0 given x =
1
3and z = 1 f(1/3) =
139
27; f(1) = 5
also f(0) = 5 ; f(2) = 7 .Hence Global Maximum Value = 7 ; Global Minimum Value = 5.
2. = 1 +f x
x
( )3 , Limitx 0 ln
13
1
+
f x
x
x( )
/
= 2
f(x) have co-efficient of x3, x2, x or constant term zero in order that the limit may exist .
= ln eLimitx x
f x
x 01
3.
( )
= Limitx 0
f x
x
( )4 = 2 =
Limitx 0
a x b x cx
x
6 5 4
4
+ += 2 c = 2
Hence f(x) = ax6 + bx5 + cx4
f (x) = x3 (6ax2 + 5bx + 8) f (1) = 0 and f (2) = 0 gives 6a + 5b + 8 = 0 and 24a + 10b + 8 = 0
a =2
3; b = -
12
5 f(x) =
2
3x6 -
12
5x5 + 2x4
3. f (x) = 10 x - 5Ax -6 and f (x) = 10 + 30Ax -6 > 0 i.e. f (x) = 0 gives a minima
x7 =A
2 x =
A
2
1 7
/
Since A > 0
we get only one minima and no maxima
Hence smallest value of f(x) will be at x =A
2
1 7
/
i.e. f(x)]min
= 5A
AA
2 2
2 7 5 7
+
/ /
= 24 or 5A A
22
2
2 7 2 7
+
/ /
= 24 A = 224
7
7 2
/
4. f(x) = sin3x + sin2x
xcosxsin2.xcosxsin3)x(f2 += )2xsin3(xcosxsin +=
For an extremum 0)x(f = if it exists. sin x = 0 or cos x = 0 or 3 sin x + 2 = 0
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0xcos2
x2
i.e.
2
3,
2
3but if 0= , x = 0 and hence only one solution
0
23
,00,2
3
For this value of there are two distinct solutions,Since f(x) is continuous, these solutions give one maximum and one minimum because for a continuous
function, between two maxima there must lie one minimum and vice versa.
5. Here 2xbxxlog8
1)x(f += is defined and continuous for all x > 0.
Then x2b
x8
1)x(f += or
x8
1bx8x16)x(f
2 +=
for extrema let 0)x(f = 16x2 - 8bx + 1 = 0
so,4
1bbxor
162
)1b(64b8x
22 =
=
Obviously the roots are real if b2 - 1 0 b > 1 [as b > 0]
Hence, when b > 1, then using number line rule for )x(f as shown in figure we know )x(f changes sign
from +ve to -ve at 4
1bb
x
2
=
4
1bbxat)x(f
2
max
= and )x(f changes sign from -ve to +ve at
4
1bbx
2 +=
4
1bbxat)x(f
2
min
+= also if b = 1
x
)1x4(
x
1x8x16)x(f
22 =
+= no change in sign.
Neither maximum nor minimum if b = 1.
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Thus,
=
>+
=
>
=
1bwhenimumminnorimummaxneither)x(f
1band4
1bbxwhen)x(f
1band4
1bbxwhen)x(f
)x(f2
min
2
max
6. Since f(x) the cubic curve vanishes at x = -2 therefore we can choose
y = f(x) = (x + 2) (ax2 + bx + c)
0)cbxax(1)bax2()2x(dx
dy 2 =+++++= at x = -1 and x = 1/3
-a + c = 0 and 15a + 24b + 9c = 0 or 24a + 24b = 0 or a + b = 0 c = a, b = -a,
y = f(x) = a(x + 2)(x2 - x + 1)
Again
=1
1
)given(3
14dx)x(f
=++1
1
2
3
14dx)1xx()2x(a =
=++1
1
23
3
14dx)2xxx(a =+=
1
0
2
3
14dx)2x(a2
3
14a
3
14or
3
14x2
3
xa2
1
0
3
==
+
a = 1, b = -1, c = 1
Hence the cubic equation is (x + 2)(x2 - x + 1) = 0
7. f(1) = 2(1) - 3 = -1
The function f(x) = 2x - 3 is increasing on [1, 3] f(1) = -1 is the smallest value of f(x) at x = 1.
Also,)2b3b(
)1bbb(x)x(f
2
233
+++
+= is decreasing in [0, 1) for fixed values of b.
Its smallest value occurs when
2b3b
1bbb1
2b3b
1bbbxlim)x(flim
3
23
2
233
1x1x
++
++=
++
++=
This value must be greater than or equal to -1.
0)1b()2b(
)1b()1b(.e.i0
2b3b
1bbb2
2
23
+++
++
+
i.e. 01b0)1b()2b(
)1b( 2 +++
Q
By the method of intervals, it follows that b must assume the values in (-2, -1) or 1b , in order that
f(x) = -1, the smallest value. Possible values of b are ),1[)1,2(
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8. If x = a, then from the equation of the curve, y2(a + a) = a2(3a - a)
i.e. y2 = a2 i.e., y = + a
Hence, the points on the curve where x = a are P(a, a), Q (a, -a)Differentiating the equation of the curve w.r.t. x, we get
2 2dy2y (a x) y 6ax 3x
dx
+ + =
2 2
dy 6ax 3x y
dx 2y(a x)
=
+
Hence, the slope of the tangetnt at P(a, a) =(a,a)
dy 1
dx 2
=
Slope of normal at P(a, -a) =(a,a)
dy2
dx
=
Equation of tangent at P, (y-a) = (x-a) i.e. x - 2y + a = 0Equation of the normal at P, y - a = -2(x-a) i.e. 2x + y - 3a = 0
Again, Slope of tangent at Q(a, -a) =( a , a)
dy 1
dx 2+
=
Slope of normal at Q(a, -a) =( a , a )
dx2
dy
=
Hence, equation of tangent at Q, y + a = -(x - a) i.e., x + 2y + 3a = 0Equation of normal at Q, y + a = 2(x - a) i.e., 2x - y - 3a = 0.
9. We havedx
dy= cosx - 2 sin 2x = cos x - 4 sin x cosx
0dx
dy= when cos x = 0 or sin x =
4
1
The given function in periodic with period 2 . Now
cos x = 0 x =2
and
2
3and
sin x =4
1 4
1sinand
4
1sinx 11 =
sin-1
4
1 lying between 0 and
2
Again 2
2
dx
yd= - sin x - 4cos2x
03dx
yd
2x
2
2
>=
=
05dx
yd
2
3x
2
2
>==
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x =2
1 2
2
1 d y 15sin sin x 4(1 2sin x) 0
4 dx 4
= =