Application of Derivative Sloution

8/14/2019 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

S O L U T I O N S

LEVEL - 2 (Subjective)



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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

= =

Application of Derivative Sloution

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