Application of Derivatives -C.B.S

8/12/2019 Application of Derivatives -C.B.S

http://slidepdf.com/reader/full/application-of-derivatives-cbs 1/49

CISM CHENNAI INASTITUTE OF SCHOOL MATHEMATICSCCCIM

s.Murugaiyan ,Hindu colony II‐Main road ,Ullagaram ,chennai‐91: 044‐22423688

APPLICATIONS OF DIFERENTIATION

1. A ladder 5m long is leaning against a wall. The bottom of the ladder is

pulled along X-axis at the rate of 1.5 m /s. How fast is the height on the

wall decreasing when the foot of the ladder is 3m away from the wall?

O

A

B

x

y 5

Solution:

Let AB be a ladder 5m long.

Let OA = x m and OB = y m.

From the fig. 2 2 2OA OB AB+ =

2 2 25.......(1) x y+ = When the foot of the ladder is at 3 m away from the wall , x = 3m.

2 23 25 y+ =

2 16 y =

4 y =

Diff.Eqn.1 w.r.t ‘x’

2 2 0..........(2)dx dy x ydt dt + =

The bottom of the ladder is pulling away at the rate of 1.4 m / sec.

i.e.3

/ sec2

dxm

dt =

substitute x = 3, y = 4 and3

2

dx

dt = in (ii)

2(3)(3 / 2) 2(4) 0dy

dt + =

9

/ sec8

dy

mdt = −

2. A particle moves along the curve 3 26 12 75 5. y x x x= − + + Find the points

on the curve at which the y-coordinate is changing 5 times as fast as

the x-coordinate.





Solution:

Equation of the given curve,3 26 12 75 5. y x x x= − + +

Diff. w.r.t ‘t’,

26 3 24 75dy dx dx dx x xdt dt dt dt

= − +

Given 5dy dx

dt dt =

230 3 24 75dx dx dx dx

x xdt dt dt dt

⇒ = − +

230 3 24 75 x x⇒ = − + 2

8 15 0 x x⇒ − + =

( )( )3 5 0 x x− − =

3 (OR) 5 x x= =

When x = 3, y = 145/6 and when x = 5 and y = 65/2

Therefore Points on the curve are145 65

3, 5,6 2

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

3.

The surface area of a cube is increasing at the rate of 10 cm /sec. How fast the

volume increasing when the edge of the cube is 6 cm?

Solution:

Let x m be the length of an edge of a variable cube

i.e., Volume 3V x= surface area 26S x=

12dS dx xdt dt

⇒ = and 210 /dS cm sdt

=

10 12 dx

xdt

⇒ =

5/ sec

6

dxcm

dt x⇒ =

And 23dV dx

xdx dt

=

2 53

6

x

x

⎡ ⎤= ⎢ ⎥

⎣ ⎦

35/sec

3

xcm=

When 36 10 / secdV

x cm cmdx

= =





4. In an inverted conical funnel of height 10cm and base radius 5 cm, water is leaking at

the rate of 32 /cm s . How fast the water level changes when the depth of the water is

5/2 cm?

r

h

θ

Let r: radius of the water level

Let h : depth of the water level

Let θ be the semi bottom angle

5tan

10

r

hθ = =

2

hr ⇒ =

Volume of water at ‘t’ seconds

21

3V r hπ =

31

12hπ =

Diff. w.r.t ‘t’,

214

dV dhhdt dt

π =

Given5

2 and2

dV h

dt = − =

252

16

dh

dt

π ⎡ ⎤∴− = ⎢ ⎥⎣ ⎦

32/ sec

25

dhcm

dt π ⇒ = −

5.

Sand is pouring from a pipe at the rate of 312 /cm s . The falling snd form a

cone on the ground in such a way that the height of the cone is alwaysone-sixth of the radius of the base. How fast is the height of the sand cone

increasing when the height is 4 cm?





h

r

Solution:

Let h be the height of the sand cone.

Let r be the radius of the sand cone.

Given 6h = r

Volume of the sand cone

21

3V r hπ =

312V hπ = ( )6r h=∵

236dV dh

hdt dt

π ⇒ =

Given 312 /dV

cm sdt

= and h = 4 cm

( )12 36 16 dh

dt π ⇒ =

1

/sec48

dh

cmdt π =

6. A man 1.5 m tall is walking away from a lamppost 12 m high at the rate of

2 m / sec. how fast his shadow changes?

A

B

P

Q

12m1.5m

C

Solution:

Let AB be a lamppost 12 m high.

Let PQ be a person 1.5 m tall.

Let the length of the shadow be xm and his position after ‘t’ sec. is y m.

From the fig





12

1.5

7

AB x y

PQ x

x y

x

x y

+=

+=

=

AB x y

PQ x

+=

12

1.5

x y

x

+=

8 x x y= +

( )

2/3 2/3 2/3

1/3 1/3

...........1

2 2 , 2 2

2 2 03 3

x y a

a a

dy x ydx

− −

+ =

⇒ + =

7 x y=

Diff. w.r.t. ‘t’,

7 dx dy

dt dt = and given that 2 / sec

dym

dt = (rate of walking)

2/ sec

7

dxm

dt ⇒ =

7. Find the equation of Tangent and normal to the curve

( )2/3 2/3 2/3 at 2 2 , 2 2 x y a a a+ =

Solution:Equation of the curve : 2/3 2/3 2/3 ...........1 x y a+ =

Given point : ( )2 2 , 2 2a a

Diff. eqn.(1). w.r.t ‘x’

1/3 1/32 20

3 3

dy x y

dx

− −⇒ + =

dy y

dx x

⎛ ⎞= − ⎜ ⎟

⎝ ⎠

1

31

1

1 y

m x

⎛ ⎞⇒ = − = −⎜ ⎟⎝ ⎠

Equation of tangent

1

1

y y dy

x x dx

−=

− at ( )1 1

, x y





1/3

2 2 2 2

2 2 2 2

y a a

x a a

⎛ ⎞−= − ⎜ ⎟⎜ ⎟− ⎝ ⎠

2 21

2 2

y a

x a

−= −

−

4 2 0 x y a⇒ + − =

Equation of normal

1

1

y y dy

x x dx

−=

− at ( )1 1

, x y

1/3

1 1

1 1

y y x

x x y

⎛ ⎞−= ⎜ ⎟

− ⎝ ⎠

1/3

2 2 2 2

2 2 2 2

y a a

x a a

⎛ ⎞−= ⎜ ⎟⎜ ⎟

− ⎝ ⎠

2 21

2 2

y a

x a

−= −

−

0 x y⇒ − =

8. Find the equation of tangent to the curve 22 4 8 y x x= − + which is parallel

to the line 2 6 0 x y− + =

Solution:

Equation of the curve 22 4 8 y x x= − +

2 2 4dy xdx = −

Slope of the tangent at ( )1 1, dy

x ydx

= at ( )1 1, x y

1 2 x= −

Equation of the line : 2 6 0 x y− + = and a = 2 and b = -1

Given: Tangent is parallel to the line

dy

dx at ( )1 1

, x y a

b= −

1 2 2 x − =

1 4 x =

And 2

1 1 12 4 8 y x x= − +

16 16 8= − +

1 4 y =

∴ The point of contact : ( 4, 4 )





Equation of tangent

1

1

y y dy

x x dx

−=

− at ( 4, 4 )

42

4

y

x

−=

−

2 4 0 x y⇒ − − =

9. Find the equation of tangent to the curve 2 2 10 y x x= − + which is

perpendicular to the line 4 6 0 x y+ + =

Equation of the curve 2 2 10 y x x= − +

2 2dy

xdx

⇒ = −

Slope of the tangent =dy

dx at ( )1 1

, x y

m = 12 2 x −

Equation of the line : 4 6 0 x y+ + =

a = 1 and b = 4

Given : Tangent is perpendicular to the line

( )1 1, x y

dy b

dx a⇒ =

12 2 4 x⇒ − =

1 3 x⇒ =

2

1 1 12 10 y x x= − +

1 9 6 1 y⇒ = − +

1 4 y⇒ =

There fore point of contact : ( 3, 4 )

Equation of tangent :

( )

1

3,41

y y dy

x x dx

−=

−

44

3

y

x

−=

−

4 8 0 x y⇒ − − = 10. Find the equation of the normal to the curve y = x2 –6x+4 which is

parallel to the line x+4y+6=0Solution.Equation of the curve : y = x2 –6x+4

2 6dy

xdx

= −





⇒

Slope of the normal = dx

dy

⎛ ⎞−⎜ ⎟

⎝ ⎠ at ( x1 , y 1)

=1

1

2 6 x

−

−

Equation of the line : x+4y+6=0 and a= 1 and b = 4Given: Normal is parallel to the line.

1

1 1

2 6 4 x

−= −

−

12 6 4 x − =

1 5 x =

And 2

1 1 16 4 y x x= − +

⇒ y 1 = 25 –30 + 4 y 1 = - 1

∴ Point of contact: ( 5 , - 1 )Equation of Normal

1

1

y y dx

x x dy

⎛ ⎞−= −⎜ ⎟

− ⎝ ⎠ at ( 5 , - 1)

1 1

5 4

y

x

+= −

−

⇒ x + 4 y - 1 = 0

11. Find the equation of normal to the curve y = x2 – 8x + 2 which isperpendicular to the line 4x – y + 7 = 0.

Solution.

Equation of the curve : y = x2 – 8x + 2

2 8dy

xdx

= −

Slope of the Normal =dx

dy

⎛ ⎞−⎜ ⎟

⎝ ⎠at ( x1 , y 1 )

⇒ =1

1

2 8 x

−

−

Equation of the line : 4x – y + 7 = 0a = 4 and b = -1

Given: Normal is perpendicular to the line.

dx

dy− at (x1,y 1) =

b

a





1

1 1

2 8 4 x

− −=

−

⇒ 12 8 4 x − =

1 6 x =

And 21 1 18 2 y x x= − + ⇒ 1 36 48 2 y = − +

1 10 y = −

∴Point of contact : ( 6 , ‐ 10 )

Equation of Normal

1

1

y y dx

x x dy

⎛ ⎞−= −⎜ ⎟

− ⎝ ⎠ at ( 6 , ‐10 )

10 1

6 2(6) 8

y

x

+ −=

− −

⇒ 4 34 0 x y+ + =

12.Find the equations of Tangent and Normal to the curve2

2 2

2 1,

1 1

t t x y

t t

−= =

+ + at t = 2.

Solution.

Equation of curve ;2

2 2

2 1,

1 1

t t x y

t t

−= =

+ +

( ) ( )

( )

2

22

2 1 2 2

1

t t t dx

dt t

+ −

= +

=( )

2

22

2 2

1

t

t

−

+

and( )( ) ( )

( )

2 2

22

2 1 1 2

1

t t t t dy

dt t

− + − −=

+

( )2

2

4

1

t

t

−=

+

⇒ dy dy dt

dx dt dx

⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( )22 1

4

t

t

−=

−

( )21

2

t

t

− −=





When t = 2 , x1 = 4/5 and y 1 = - 3/5

Equation of Tangent

1

1

y y dy

x x dx

−=

− at t= 2

( )( )

1 43 / 5

4 / 5 2 2

y

x

− −+=

−

⇒ 15x – 20 y – 24 = 0Equation of normal

1

1

y y dx

x x dy

−= −

− at t = 2

3 / 5 4

4 / 5 3

y

x

+= −

−

20 x +15 y – 7 = 0

13. Find the equations of tangent and normal to the ellipse2 2

2 2 1 at

x y

a bθ + =

Solution.Parametric equations of ellipse

x= aCosθ and y = b Sinθ

⇒ sindx

ad

θ θ

= − and cosdy

bd

θ θ

=

cos

sin

dy b

dx a

θ

θ

−=

Equation of Tangent at ‘ θ’1

1

y y dy

x x dx

−=

− at ‘ θ’

sin cos

cos sin

y b b

x a a

θ θ

θ θ

− −=

−

cos sin1

x y

a b

θ θ + =

Equation of Normal at ‘ θ’

1

1

y y dx

x x dy

−=

− at ‘ θ’

sin sincos cos

y b a

x a b

θ θ

θ θ

− =−

2 2

cos sin

ax bya b

θ θ + = −

14. Prove that the curves y2 = x and xy = k cut at right angle if 8k2 =1





Solution.Equations of the curves : y 2 = x ……….( 1).and x y = k ……….(2 ).Now, xy = k

⇒ x2 y 2 = k2

x3 = k2 ( since y 2 = x) x3 = 1/ 8 ( since 8k2 – 1 )

⇒ x = 1 /2If ( x1 ,y 1) is the point of intersection then x1 = 1/2

From (1 ). 2 1dy

ydx

=

Slope of Tangent to the curve(1).

1

1

1

2m

y=

From (2). 0dy

y x

dx

+ =

Slope of Tangent to the curve (2).

12

1

ym

x= −

Now1 2

1

1

2m m

x= −

Sub.1

1

2 x =

⇒ m 1 m2 = -1 This proves that the curves cut at right angle .

15. Prove that the curves y2= 4x and xy = k cut at right angle if k2 = 32 .

Solution.

Equations of curves : y 2 = 4x … ( 1)

xy = k … ( 2)From ( 2) , x2 y 2 = k2

⇒ 4 x3 = 32 (since k2 = 32 )

⇒ x1 = 2If ( x1 ,y 1 ) is the point of intersection of the curves. Then x1 = 2.

From ( 1) 2 4dy

dx=

2dy

dx y=

Slope of the curve (1 ),1

1

2m

y=





From (2 ) , 0dy

y xdx

+ = dy y

dx x⇒ = −

Slope of the curve (2 ) 12

1

ym

x= −

Now, 1 2

1

2m m

x= −

⇒ 1 2 1m m = − ( Since x1 = 2 )

This proves that the two curves cut at right angle.

16.Find the equation of Tangent to hyperbola2 2

2 2 1

x y

a b− = at ( aSecθ , bTanθ ).

Solution.

Equation of hyperbola :2 2

2 2 1

x y

a b− =

2 2

2 2

0

x y dy

a b dx− =

dy bx

dx ay=

Slope at the point ( )sec , tana bθ θ

sec

tan

bm

a

θ

θ =

Equation of Tangent

tan sec

sec tan

y b b

x a a

θ θ

θ θ

−=

−

⇒ sec tan

1 x y

a b

θ θ

− = .

17. Find the points on the curve y = x3 at which the slope of the Tangentis equal to the y coordinate of the point.Solution.

Equation of the curve : y = x3

⇒ 23

dy x

dx=

Slope of the Tangent , m = 3x12

and1m y=

( ∵ slope of the tangent = y-coordinate )

⇒ y 1 = 3x12

and y 1 = x13 ( Since y = x3 )

⇒ x13 =3x1

2

⇒ x1 = 3and y 1 = 27

∴ Point of contact: ( 3 , 27 ).





INCEASING AND DECREASING FUNCTION

18. Find the interval in which the function f(x) = x3 +3x2 - 24x +15 isincreasing or decreasing.Solution.

Equation of the curve : f(x) = x3 +3x2 - 24x +152( ) 3 6 24 f x x x′ = + −

( ) 3( 4)( 2) f x x x′ = + −

If ( ) 0 f x′ > then 3 (x+4) (x – 2) > 0

⇒ (x+4) (x – 2) > 0

⇒ x ∈ ]-∝ , -4 [ ∪ ] 2 , ∝ [

∴ f(x) is increasing in the intervals ]-∝ , -4 [ ∪ ] 2 , ∝ [ .If ( ) 0 f x′ < then 3 (x+4) (x – 2) < 0

⇒ (x+4) (x – 2) < 0

⇒ x ∈ ] –4 , 2 [

∴f(x) is decreasing in the interval ] –4 , 2 [ .

19.Find the intervals in which the function f(x) = 10 +45x +3x2 – x3 isincreasing or decreasing.

Solution.

Equation of the curve : f(x) = 10 +45x +3x2 – x3

( ) f x′ = 45 + 6x –3x2

( ) f x′ =- 3 ( x-5) ( x +3)

If ( ) 0 f x′ > then - 3 ( x-5) ( x +3) > 0⇒ ( x-5) ( x +3) < 0

⇒ x ∈ ] –3 , 5[

∴ f(x) is increasing in the interval ] –3 , 5[If ( ) 0 f x′ < then - 3 ( x-5) ( x +3) < 0

⇒ ( x-5) ( x +3) > 0

⇒ x ∈ ] -∝ , -3[ ∪ ]5 , ∝[

∴f(x) is decreasing in the intervals ] -∝ , -3[ ∪ ]5 , ∝[ .

20.Find the intervals in which the function f(x) = log x3 –( x - 3) is increasing or

decreasing.Solution.

Equation of the curve : f(x) = logx3 – x + 33

( ) 1 f x x

′ = −

3'( )

x f x

x

−=





If ( ) 0 f x′ > then3

0 x

x

−> ⇒ ( )3 0 x x− <

⇒ x ∈ ] 0 , 3[ , Since x >0

∴ f(x) is increasing in the interval ] 0 , 3[ .

If ( ) 0 f x′ < then

3

0

x

x

−

< ⇒ ( )3 0 x x− >

⇒ ( ) ( ),1 3, x ∈ −∞ ∪ ∞ and 0 x >

∴ f(x) is decreasing in the interval ] 3 , ∝ [ .

21.Find the intervals in which the function f(x) = a 2x - 2x log a isincreasing or decreasing.

Solution.

Equation of the curve : f(x) = a 2x - 2x log a

⇒ ( ) f x′ = (2 log a) a 2x – 2log a( ) f x′ = 2log a ( a 2x – 1 )

If ( ) f x′ > 0 then 2log a ( a 2x – 1 ) > 0

⇒ ( a 2x – 1 ) > 0

⇒ ( a x – 1) (a x+1) > 0

⇒ ( a x – 1) > 0

⇒ x∈ ]0 , ∝ [

∴ f(x) is increasing in the interval ]0 , ∝ [If ( ) f x′ < 0 then 2log a ( a 2x – 1 ) < 0

⇒ ( a x – 1) (a x+1) < 0

⇒ ( a x – 1) <0⇒ x∈ ] -∝ ,0 [

∴ f(x) is decreasing in the interval ] -∝ ,0 [

22. Find the intervals in which the function log y = Sin x + x is increasing

or decreasing.

Solution.

Equation of the curve : log y = Sin x + x

1cos 1

dy x

y dx

⎛ ⎞= +⎜ ⎟

⎝ ⎠

( )cos 1dy

y xdx

⎛ ⎞= +⎜ ⎟

⎝ ⎠

If 0dy

dx> then y(Cos x + 1)> 0 ⇒ (Cos x + 1) > 0 , sin 0

x x y e x R

+= > ∀ ∈∵





⇒ (Cos x + 1) > 0

⇒ (Cos x + 1) > 0 for all x : since -1 ≤ Cos x ≤ 1

∴ The functions is always an increasing function x R∀ ∈

23.Show that ( ) 2

log 1

2

x y x

x

= + −+

is an increasing function of x for all

values of x > -1

Solution.

Equation of the curve : ( ) 2

log 12

x y x

x= + −

+

⇒ ( )

2

1 4

1 2

dy

dx x x= −

+ +

( ) ( )

( ) ( )

2

2

2 4 1

1 2

x xdy

dx x x

+ − +=

+ +

( )( )

2

21 2

x

x x=

+ +

If f’(x) > 0 then 0dy

dx>

⇒ dy

dx ( ) ( )

2

21 2

x

x x=

+ + > 0

⇒ 11 x+

> 0 , Since x2 > 0 and ( 2+x) 2 > 0 for all x

⇒ 1+x > 0

⇒ x > - 1

∴ The function f(x) is increasing for all x > -1

24. Find the least value of ‘a’ Such that the function f(x) = x2 +ax +1 isstrictly increasing on (1,2) .

Solution.

Equation of the given curve ; f(x) = x2 +ax +1

⇒ ( ) f x′ = 2x + a

If f(x) is strictly increasing function on ( 1 , 2),then ( ) f x′ > 0 , for all x in ( 1 ,2)

2x + a > 0 ∀ x ∈ ( 1 ,2)

⇒ 2 0a+ > and 4 0a+ > 2, 4a a> − > −

The least value of a is -2.





∴f(x) is a strictly increasing function for the least value of a = -2

25.Find the greatest value of ‘a’ Such that the function f(x) = 2x2 + ax + 5is strictly decreasing function on ( 2 ,3 ).

Solution.

Equation of the curve : f(x) = 2x2 + ax + 5

⇒ f’(x) = 4x + aIf f(x) is strictly decreasing function on ( 2 , 3) , then f’(x) < 0

⇒ 4x+a < 0 , ∀ x ∈ ( 2 ,3 )

⇒ 8+a < 4x+a < 12+ a

⇒ If f’(x) < 0 , 8 0,12 0a a+ < + < and 8, 12a a< − < − then the greatest value of

a = -12.

26.Find the intervals in which the function4sin 2 cos

( )

2 cos

x x x x f x

x

− −=

+

is

increasing (or) decreasing

Solution.

Equation of the curve :4sin 2 cos

( )2 cos

x x x x f x

x

− −=

+

⇒ ( ) ( )( )

( )2

4cos 2 cos sin 4sin 2 cos sin( )

2 cos

x x x x x x x x x f x

x

− − + − − − −′ =

+

( )

2

2

4 cos 4 4 cos

2 cos

x x

x

+ − −=

+

( )

2

2

4 cos 4 4 cos

2 cos

x x

x

+ − −=

+

If f’(x) > 0 , then( )

( )2

cos 4 cos0

2 cos

x x

x

−>

+

⇒ Cos x ( 4 – Cos x ) > 0

⇒ Cos x > 0 , ∵ ( 4 – Cos x ) is always positive for all ‘x’ .

⇒ - π/2 < x < π/2

∴f(x) is increasing for x∈ ] 3π/2 , 2π [ ∪ ] 0 , π/2 [ .

If ( ) f x′ < 0 , then Cos x < 0 , As above

⇒ π/2 < x < 3π/2





∴ f(x) is decreasing in the interval ] π/2 , 3π/2 [ .

27. Find the intervals in which the function f(x) = Sin x + Cos x is

increasing or decreasing in [ 0 , 2 ]

Solution.

Equation of the function : f(x) = Sin x + Cos x

⇒ ( ) f x′ = Cos x – Sin x

⇒ ( ) f x′ = √2 Cos( x+π/4 )

If ( ) f x′ > 0 , then √2 Cos( x+π/4 ) > 0 ⇒ Cos( x+π/4 ) > 0

⇒ - π/2 < x+π/4 < π/2

⇒ - 3π/4 < x < π/4

∴ f(x) is increasing in the interval] 5π/4, 2π] ∪ [0 , π/4 [

If ( ) f x′ < 0 , Then √2 Cos( x+π/4 ) < 0 ⇒ Cos( x+π/4 ) < 0

⇒ π/2 < x+π/4 < 3π/2⇒ π/4 < x < 5π/4

∴f(x) is decreasing in the interval ] π/4 , 5π/4 [.

MAXIMUN AND MINIMUM

28.Find the absolute maximum and absolute minimum values of the

function f(x) = Cos2x+ Sin x , x∈ [ 0 , ].

Solution.

Equation of the curve : f(x) = Cos2x+ Sin x

⇒ ( ) f x′ = -2 Sin x Cos x + cos x

⇒ ( ) f x′ = Cos x( 1 – 2Sin x)

Let ( ) f x′ = 0 ⇒ Cos x( 1 – 2Sin x) = 0

⇒ Cos x = 0 (or) Sinx = 1/2

If Cos x = 0 , then x = π/2 ∈ [ 0 ,π ]. And

If sin x = 1/2 , then x = π/6 ∈ [ 0 ,π ].

If we include π/2 , π/6 with the extreme points 0 and π , thenf(0) = Cos 20 + sin 0 = 1

f(π/6) = Cos

2

π/6 + Sin

π/6 = 5/4f(π/2) = Cos2π/2 + Sinπ/2 = 1

f(π) = Cos2π + Sin π = 1

∴ Absolute Maximum =5

max 1, ,14

⎧ ⎫= ⎨ ⎬

⎩ ⎭= 5/4





and Absolute minimum =5

min. 1, ,14

⎧ ⎫= ⎨ ⎬

⎩ ⎭= 1 .

29. Find both maximum value and ,minimum value of f(x) = 3x4 – 8x3 +12x2 – 48x + 1 in the interval [1 , 4 ].

Solution.

Equation of the curve : f(x) = 3x4 – 8x3 +12x2 – 48x + 1

⇒ ( ) f x′ = 12x3 –24x2 +24x - 48

( ) f x′ = 12 (x2+2) (x – 2)

Let ( ) f x′ = 0 ⇒ 12 (x2+2) (x – 2) = 0

⇒ x –2 =0 , ∵12(x2+2) > 0 for all ‘x’ .

⇒ x = 2If we include the extreme points x = 1 , x = 4 of [ 1 , 4 ] with the turning point x = 2 ,then

X= { 1 , 2 , 4 }f(1) = 3 – 8 = 12 – 48 + 1 = -40.f(2) = 48 – 64 + 48 – 96 + 1 = -63f(4) = 768 – 512 +192 –192 + 1 =257

∴ Maximum value = 257and Minimum value = - 63.

30. Find the values of local maximum and local minimum of the function

[ ]3 2( ) 2 3 120 24 x -6,5 f x x x x= + − + ∀ ∈

Solution.

Equation of the curve : 3 2( ) 2 3 120 24 f x x x x= + − + 2( ) 6 6 120 f x x x′ = + −

⇒ If ( ) f x′ = 0 , then 6x2 +6x – 120 = 0

⇒ 6(x+5) (x – 4) = 0

⇒ x= 4 , x= -5 are the points of local maximum or points of local minimum.Now, ( ) f x′′ = 12x + 6

( ) f x′′ = 6( 2x+1)

When x =4 , (4) f ′′ > 0 ⇒ x=4 is a point of local minimum and the

corresponding local minimum value f( 4) = 128+48 –480 + 24= -280

When x =-5 , ( ) f x′′ < 0 ⇒ x =- 5 is a point of local maximum and the

corresponding local maximum value f( -5) = -250 + 75 +600 + 24= 449.





31. Find the local maximum and local minimum of the function f(x) = 2x3 +3x2 –36x +10 ,using first derivative test.

Solution.

Equation of the curve : f(x) = 2x3 +3x2 –36x +10⇒ ( ) f x′ = 6x2+ 6x - 36

( ) f x′ = 6 (x+3) (x –2)

If ( ) f x′ > 0 , then 6 (x+3) (x –2) > 0 ⇒ x < -3 and x > 2 and

If ( ) f x′ > 0 , Then 6 (x+3) (x –2) < 0⇒ -3 < x < 2

The sign of ( ) f x′ changes from + ive into – ive across x = -3

∴ x =-3 is the point of local maximum and f(-3) is the local maximum value.i.e f(-3 ) = 91

The sign of ( ) f x′ changes from – ive into + ive across x = 2

∴ x = 2 is the point of local minimum and the local minimum value ,

f(2) = -36.

ROLLE’S THEOREM AND LAGRANGE’S MEAN VALUE THEOREM

32. Verify Rolle’s theorem for the function f(x) = x2 –7x +10 in [ 3 , 4].

Solution.

The function f(x) = x2 –7x +10 is defined in [ 3 , 4].( i) f(x) is a polynomial function and it is continuous n [3 , 4](ii) f(x) is differentiable in ] 3 , 4 [

i.e ( ) f x′ = 2x – 7

(iii) f(3) = -2 and f( 4) = -2 ⇒ f(3) = f( 4)

There must exists a least one point c ∈ ]3 , 4[ such that ( ) f c′ = 0

Let ( ) f x′ = 0 ⇒ 2x –7 =0

⇒ x = 7/2 ∈ ]3 , 4[

∴ f(x) verifies Rolle’s theorem in [3,4] for c = 7/2 .

33. Verify Rolle’s theorem for the function ( ) 4 4sin cos in 0,2

f x x x π ⎡ ⎤

= + ⎢ ⎥⎣ ⎦

Solution.

( ) 4 4sin cos f x x x= + is defined in [ 0 ,π/2 ]

( i) f(x) is sum of two continuous functions and therefore f(x) is also a

continuous function in[ 0 ,π/2 ]

(ii ) f(x) is differentiable in ] 0 ,π/2[.

i.e ( ) 3 34sin cos 4cos sin f x x x x x′ = −





2 24 sin cos sin cos x x x x⎡ ⎤= −⎣ ⎦

sin4 x= −

(iii) ( ) 4 40 sin 0 cos 0 1 f = + = and 4 4sin cos 12 2 2

f π π π ⎛ ⎞

= + =⎜ ⎟⎝ ⎠

( )02

f f π ⎛ ⎞= ⎜ ⎟⎝ ⎠

There must exists at least one point c ∈ ] 0 ,π/2[.

Let ( ) 0 sin 4 0 f x x= ⇒ − =

sin 4 0 x =

4 x π =

⇒ 0,4 2

x π π ⎤ ⎡

= ∈ ⎥ ⎢⎦ ⎣

∴ f(x) verifies Rolle’s theorem in [ 0 ,π/2 ] for the point c= π/4∈ ] 0 ,π/2[.

34. Examine the validity of Rolle’s theorem for the function

] [1 1 33cos 0,

8 x π

− += ∈ ( ) sin sin 2 f x x x= − on [0 , ].

Solution.

( ) sin sin2 f x x x= − is defined on [0 ,π].

( i ) f(x) is difference of two continuous functions and therefore f(x) is also a

continuous function on [0 ,π].

( ii ) f(x) is differentiable in ] 0 ,π [.

i.e., ( ) cos 2cos2 f x x x= −

24cos cos 2 x x⎡ ⎤= − − −⎣ ⎦

( iii ) ( )0 sin 0 sin 0 0 f = − = and ( ) sin sin 2 0 f π π π = − =

( )0 ( ) f f π =

There must exists at least point c∈ ] 0 ,π [ such that f ’(c) = 0

Let ( ) 20 4cos cos 2 0 f x x x= ⇒ − − =

1 33cos

8 x

+=

⇒ ] [1 1 33cos 0,8

x π − +

= ∈

∴ f(x) is valid for the truth of Rolle’s theorem on[0 ,π].

35. Discuss the applicability of Rolle’s theorem for the function

( ) ( )2log 4 log 5 f x x= − − on[ -3,3]

Solution.





( ) ( )2log 4 log 5 f x x= − −

1 33

2c

− += ∈ is defined on [ -3,3].

( i) f(x ) is a difference of two continuous functions and therefore f(x) is also acontinuous function on [ -3 ,3 ].

( ii ) f( x) is differentiable in ] –3 , 3 [.

i.e ( ) 2

2

4

x f x

x=

−

( iii ) f( -3) = 0 and f( 3 ) = 0 ⇒ f( -3 ) = f(3)

There must exists at least one point c ∈ ] –3 , 3 [ such that f’( c) = 0 .

Let ( ) 2

20 0

4

x f x

x= ⇒ =

−

2x = 0

⇒ x = 0 ∈ ] –3 , 3 [∴ The function is applicable for Rolle’s theorem on [ -3 ,3 ] .

36. Find the values of ‘a’ and ‘b’ if ( ) 2 3 f x ax bx= + + satisfies Rolle’s

theorem for c = -3/4 and f( -2) = f( ½) = 5 .

Solution.

By Rolle’s theorem , f’(c ) = 0

( ) 2 3 f x ax bx= + +

( ) 2 f x ax b′ = +

and3

4 f

−⎛ ⎞′⎜ ⎟⎝ ⎠

=0

⇒ 2a ( -3/4) +b = 0

⇒ 3a = 2b …… .( i )Given : f( -2) = 5

⇒ 4a –2 b +3 = 54a –3a = 2 From ( i)

∴ a = 2 and b = 3 .

37. Discuss the applicability of Rolle’s theorem for the function

f(x) =( 1 –Cos 2x ) a x on [0 , ].

Solution.

( ) ( )1 cos2 x f x x a= − is defined on [0 ,π ].





( i ) f(x) is product of two continuous functions and therefore f(x) is also

continuous in [ 0 ,π ].

( ii ) f(x) is differentiable in ] 0 ,π[.

i.e., ( ) ( )2sin 2 1 cos 2 log x x f x xa x a a′ = + −

2

4sin cos 2sin log

x x

x xa x a a= + ⋅

2 sin (2cos sin log ) x

a x x x a= +

( iii ) f( 0) = 0 and f( π) = 0 ⇒ f( 0) = f( π)

There must exists at least one point c∈ ] 0 ,π[.

Let ⇒ ( ) ( )0 2sin 2 cos sin log 0 x f x x x x a a′ = ⇒ + =

Sinx = 0 or Tan x2

log a=

⇒ x = ( 0 or π) ∉] 0 ,π[ but x = ( )1 2tan 0,

log aπ

− ⎛ ⎞∈⎜ ⎟

⎝ ⎠

∴ f(x) is applicable for the Rolle’s theorem on [0 ,π ].38 . Verify Rolle’s theorem for the function ( ) ( )

5sin cos on ,

4 4

x f x e x x

π π ⎡ ⎤= − ⎢ ⎥⎣ ⎦

Solution.

( ) ( )sin cos x f x e x x= − is defined on [ π/4 , 5π/4 ].

( i ) f(x) is product of two continuous functions and therefore f(x) is also is a

continuous function in [ π/4 , 5π/4 ].

( ii ) f(x) is differentiable in ] π/4 , 5π/4[.

i.e ( ) ( ) ( )sin cos cos sin x x f x e x x e x x′ = − + +

= 2Sinx e x (iii ) f(π/4) = 0 and f(5π/4 ) = 0

⇒ f(π/4) = f(5π/4 )

There must exists at least one point c∈] π/4 , 5π/4[ such that ( ) f c′ = 0

Let ( ) f x′ = 0 ⇒ 2Sinx e x = 0

⇒ Sinx = 0 but e x > 0 ,∀ x

x = π ∈] π/4 , 5π/4[

∴ f(x) verifies Rolle’s theorem 0n on [ π/4 , 5π/4 ].

39. Verify L.M.V.T for the function ( ) log f x x= on [1 ,2 ].

Solution.

( ) log f x x= is defined on [1 ,2 ].

( i ) f(x) is a logarithmic function and it is continuous on [ 1 , 2 ]( ii ) f’(x ) is differentiable in ] 1 , 2 [





i.e1

( ) f x x

′ =

There must exists at least one point c ∈ ] 1 , 2 [ such that

( ) ( )( )

f b f a f c

b c

−′ =

−

Now( ) ( )2 1

( )2 1

f f f x

−′ =

−

⇒ 1

log 2 log1 x

= −

1log 2

x=

x =log2 e ∈ ] 1 , 2 [

∴ f(x) verifies L.M.V.T on[ 1, 2] at c = log2 e ∈ ] 1 , 2 [.

40. Verify L.M.V.T for the function ( ) 3 22 12 24 16 f x x x x= − + − on [3 , 5 ].

Solution.

( ) 3 22 12 24 16 f x x x x= − + − is defined on [3 , 5 ].

( i ) f(x) is a polynomial function and it is continuous on [ 3 , 5 ].( ii ) f( x) is differentiable on] 3 , 5[ .

i.e ( ) 26 24 24 f x x x′ = − −

There must exists at least one point c∈ ]3 , 5 [ such that

( ) ( ) ( ) f b f a f c

b a−′ =−

Now ( ) ( ) ( )5 3

5 3

f f f x

−′ =

−

2 54 26 24 24

2 x x

−− + =

6x2 – 24x + 24 = 26 ⇒ 6x2 – 24x –2 = 0

⇒ 3x2 – 12 x –1 = 0

12 144 12

6 x

± +

=

12 156 6 39

6 3

+ += = ∈ ]3 , 5[

∴ The function f(x ) verifies L.M.V.T on [ 3 , 5 ] for6 39

3c

+= ∈ ]3 , 5[





41. Verify L.M.V.T for the function on ( ) 1

4 1 f x

x

−=

+ [ 1 , 3 ]

Solution

The function ( ) 1

4 1 f x

x−=

+ is defined on [1,3]

( i ) f( x) is continuous on [1,3].( ii ) f(x) is differentiable on ]1,3[.

i.e ( )( )

2

4

4 1 f x

x′ =

+

There must exists at least one point c ∈] 1 , 3 [ such that

( ) ( ) ( ) f b f a

f cb a

−′ =

−

Now , ( ) ( ) ( )3 13 1

f f f x

−′ = −

( )

( ) ( )2

1/13 1/ 54

24 1 x

− − −=

+

8

130=

( )2 2

4 1 65 8 0 x x x+ = ⇒ + − =

1 33

2 x

− ±=

⇒ 1 33

2 x

− += ∈ ]1 , 3 [

∴ The function f(x) verifies L.M.V.T on [ 1 , 3 ] at1 33

2c

− += ∈ ]1,3[

42. Discuss the applicability of L.M.V.T for the function f(x) = x – 2Sinx

on [- , ].

Solution .

The function f(x) = x – 2Sinx is defined on [-π , π ].( i ) f(x ) is difference of two continuous function and therefore f(x) is also

continuous on[-π , π ].

( ii ) f(x) is differentiable on ] -π , π [.i.e ( ) f x′ = 1 – 2Cosx

There must exists at least one point c∈]-π,π[ such that( ) ( )

( ) f b f a

f cb a

−′ =

−





Now ,( ) ( )

( )( )

f f f x

π π

π π

−′ =

− −

1 – 2Cosx = 1 ⇒ Cosx = 0

⇒ x = Cos –1

(0 ) = { -π/2 , π/2 } ∈] -π , π [ .

∴ The function f(x) verifies L.M.V.T on [-π , π ] at c ={ -π/2 , π/2 } ∈] -π , π [.

43. Discuss the applicability of L.M.V.T for the function ( ) 1 2 2 f x x= + −

on [ 3 , 6 ].

Solution.

The function ( ) 1 2 2 f x x= + − is defined on [3 , 6 ].

( i ) f(x ) is a sum of square root function and constant function, therefore f(x)is continuous on[3 , 6 ].( ii) f(x) is differentiable on ] 3 , 6 [.

i.e ( ) 1

2 f x

x′ =

−

There must exists at least one point c∈] 3 , 6[ such that

( ) ( ) ( ) f b f a

f cb a

−′ =

−

Now ( ) ( ) ( )6 36 3

f f f x −′ =−

1 5 3

32 x

−=

−

4x – 8 = 9 ⇒ x =17/4

⇒ c∈] 3 , 6[

∴ The function f(x) verifies L.M.V.T on [ 3 , 6] at c= 17/4 c∈] 3 , 6[.44. Discuss the applicability of L.M.V.T for the function

( ) 2 39 12 6 f x x x x= − + − on [-2 , 0] .

Solution.

The function f(x) = 9 – 12x+ 6x2 – x3 is defined on [-2 , 0].(i ) f(x) is a polynomial function ,therefore f(x) is continuous on [-2 , 0 ]( ii ) f(x) is differentiable on ] –2 , 0 [.

i..e., ( ) 212 12 3 f x x x′ = − + −





There must exists at least one point c∈ ] –2, 0 [ such that

( ) ( ) ( ) f b f a

f cb a

−′ =

−

Now

( ) ( ) ( )0 2

2

f f f x

− −′ =

2 9 6512 12 3

2 x x

−− + − =

2 6 2 213 12 16 0

3 x x x

−− − = ⇒ = ] [2,0−

∴ The function f(x) verifies L.M.V.T on [ -2 , 0 ] at ] [6 2 21

2,03

c −

= −

APPROXIMATION BY DIFFERENTIAL

45. Find the approximate value of ( 0 .024) ½ by differential .

Solution.

Let f(x ) = x ½

⇒ f( x + Δ x) = f(x) + Δ y

Wheredy

y xdx

Δ ≅ Δ

∴ ( ) ( ) dy

f x x f x xdx

+ Δ ≅ + Δ

( )1/2 1/2

1/2( )

2

x x x x

xΔ+ Δ ≅ +

Let x + Δ x = 0.024 ⇒ x = 0.01 and Δ x = 0.014

( ) ( ) ( )

( )

1/2 1/2

1/2

0.0140.024 0.01

2 0.01≅ +

140.1

200≅ +

≅ 0.1 + 0.07

∴ (0.024) ½ ≅ 0.107

46. Find the approximate value of ( 0.049) 1/2 by differential.

Solution.

Let f(x ) = x ½





⇒ f( x + Δx) = f(x) + Δy

wheredy

y xdx

Δ ≅ Δ

∴ ( ) ( ) dy

f x x f x xdx

+ Δ ≅ + Δ

( )1/2 1/2

1/2( )

2

x x x x

x

Δ+ Δ ≅ +

Let x + Δx = 0.049 ⇒ x = 0.04 and Δx = 0.009

( ) ( )( )

1/2 1/2

1/2

0.0090.049 0.04

2 0.04≅ +

≅ 0.2 + 0.0225

∴ (0.049) ½ ≅ 0.2225

47. Find the approximate value of 62 1/6

using differential.

Solution.Let f(x) = x 1/6

⇒ f( x + Δx) = f(x) + Δy

wheredy

y xdx

Δ ≅ Δ

( ) ( ) dy

f x x f x xdx

+ Δ ≅ + Δ

1/6

5/ 66

x x

x

Δ≅ +

( )1/6 1/6

5/ 66

x x x x

xΔ+ Δ ≅ +

Let x + Δx = 62 ⇒ x = 64 and Δx = ‐ 2

( ) ( )( )

1/6 1/6

5/ 6

262 64

6 64≅ −

≅ 2 ‐ 0.0104

∴ 62 1/6 ≅ 1. 9896

48. Find the approximate value of (0.123 ) 1/3 using differential .

Solution.

Let f(x) = x 1/3

⇒ f( x + Δ x) = f(x) + Δ y

wheredy

y xdx

Δ ≅ Δ





∴ ( ) ( ) dy

f x x f x xdx

+ Δ ≅ + Δ

( )1/3 1/3

2/ 33

x x x x

x

Δ+ Δ ≅ +

Let x + Δx = 0.123 ⇒ x = 0.125 and Δx =‐ 0. 002

( ) ( )( )

1/3 1/3

2/ 3

0.0020.123 0.125

3 0.125≅ −

0.0020.5

0.75≅ −

≅ 0.5 ‐ 0.0026

∴(0.123) 1/3 ≅ 0.4984

49. Find the approximate value of log 16.34 , given that log 4 = 0. 6021and log10e =0.4343 .

Solution.

Let f(x ) = log10 x ⇒ f(x) = 0.4343 log e x

⇒ f( x + Δ x) = f(x) + Δ y

wheredy

y xdx

Δ ≅ Δ

∴ ( ) ( ) dy

f x x f x xdx

+ Δ ≅ + Δ

log (x + Δ x)0.4343

log x

x x

Δ≅ +

Let x + Δ x = 16.34 ⇒ x = 16 and Δ x = 0.34

log 16.34( )( )0.4343 0.34

log16.34 log1616

≅ ≅ +

≅ 2 log 4 +0.00922

≅ 2 (0.6021) +0.00922

≅ 1.2042 +0.00922

log 16.34 ≅ 1. 2134.

50. Find the approximate value of log 37 , given that log6 =0.7782 andloge = 0.4343 .

Solution.


⇒ f( x + Δ x) = f(x) + Δ y

wheredy

y xdx

Δ ≅ Δ





( ) ( ) dy

f x x f x xdx

+ Δ ≅ + Δ

log (x + Δx)0.4343

log x

x x

Δ≅ +

Let x + Δx = 37 ⇒ x = 36 and Δx = 1

( )0.4343 .1log37 log36

36≅ +

≅ 2 log 6 +0.01206

≅ 2(0.7782) +0.01206

≅ 1.5564 +0.01206

∴ log 37 ≅ 1.568551. Find the approximate value of log 0.256 , Given that log 0.5 =

1.6990 and log e = 0.4343 .

Solution.


⇒ f( x + Δx) = f(x) + Δy

wheredy

y xdx

Δ ≅ Δ

∴ ( ) ( ) dy

f x x f x xdx

+ Δ ≅ + Δ

log (x + Δx)0.4343

log x x x

≅ + Δ

Let x + Δx = 0.256 ⇒ x = 0.25 and Δx = 0.006

Log 0.256 ( ) ( )( )

0.4343 0.006log0.250.25

≅ +

≅ 2 log 0.5 +0.0104

≅ 1.3984 +0.0104

∴ Log 0.256 ≅ 1.4084

52. Find the approximate value of Sin 50°30’ , Given that Sin 50° = 0.7660 and

Cos50°=0.6228 . ( 1° = 0.01745 radian )

Solution.

Let 50°30’ = (50 .5)° and f(x) = Six x° ⇒ f(x) = Sin (π x/180)

⇒ f( x + Δ x) = f(x) + Δ y





wheredy

y xdx

Δ ≅ Δ

∴ ( ) ( ) dy

f x x f x xdx

+ Δ ≅ + Δ

Sin (x + Δ x)° ≅ Sin x° + Cos(π x/180) (π/180) Δ x

Let (x + Δ x ) = 50.5 ⇒ x = 50 and Δ x = 0.5 , (π/180) = 0.01745 radian

Sin 50 .5° ≅ Sin 50° +Cos 50° (0.01745) (0.5)

≅ 0.7660 + (0.6228) (0.01745) (0.5)

≅ 0.7660 + 0.0054

Sin 50 .5° ≅ 0.7714

53. Find the approximate value of Cos 60°45’ , Given that Cos 60° =0.5 ,

Sin60° =0.8660. (1° = 0.01745 radian )

Solution

Let 60° 45’ = ( 60.75) ° and f(x) = Cos x° ⇒ f( x) = Cos(π x/180)

⇒ f( x + Δ x) = f(x) + Δ y

wheredy

y xdx

Δ ≅ Δ

∴ ( ) ( ) dy

f x x f x xdx

+ Δ ≅ + Δ

Cos (x + Δ x)° ≅ Cosx° - Sin(π x/180) (π/180) Δ x

Let (x + Δ x ) = 60.75 ⇒ x = 60 and Δ x = 0.75 , (π/180) = 0.01745radian

Cos 60.45° ≅ Cos 60° - Sin 60° (0.01745) (0.75)

≅ 0.5 - (0.8660) (0.01745) (0.75)

≅ 0.5 - 0.0113

∴ Cos 60.45° ≅ 0.4887

54. Find the approximate value of Tan 30°15’ ,given that Tan30°

=0.5774 and 1° = 0.01745 radian.

Solution.





Let 30°15’ =( 30. 25)° and f(x) = Tan x° ⇒ f(x) = Tan(π x/180)

⇒ f( x + Δ x) = f(x) + Δ y

where

dy

y xdxΔ ≅ Δ

∴ ( ) ( ) dy

f x x f x xdx

+ Δ ≅ + Δ

Tan( x + Δ x)° ≅ Tan x° + Sec2 (π x/180) (π/180) (Δ x)

Let x + Δ x = 30.25 ⇒ x = 30 and Δ x = 0.25

Tan (30.25)° ≅ Tan 30° + Sec 2 30° (0.01745) (0.25)

≅ 0.5774 +(1.33) (0.01745) (0.25)

≅ 0.5774 +0.0057632

∴ Tan (30.25)° ≅ 0.5832

55. Find the approximate value of Cot 46° , Given that 1°= 0.01745 .

Solution.

Let f(x) = Cot x° ⇒ f(x) = Cot(π x/180)

⇒ f( x + Δ x) = f(x) + Δ y

wheredy

y xdx

Δ ≅ Δ

( ) ( ) dy

f x x f x xdx

+ Δ ≅ + Δ

Cot( x + Δ x)° ≅ Cotx° -Cosec2 (π x/180) (π/180) (Δ x)

Let x + Δ x = 46 ⇒ x = 45 and Δ x =1

Cot 46° ≅ Cot 45° - Cosec 2 45° (0.01745) .1

≅ 1 -(2) (0.01745) .1

≅ 1 - 0.0349

∴ Cot 46° ≅ 0.9651





MAXIMUM AND MINIMUM (SIX MARKS PROBLEMS )

56. Find the value of local maximum and local minimum for the function

( ) 3 22 21 60 12 f x x x x= − + +

Solution.

Equation of the curve : ( ) 3 22 21 60 12 f x x x x= − + +

⇒ ( ) 26 42 60 f x x x′ = − +

= 6 (x –5 ) ( x – 2)

If ( ) f x′ = 0 ⇒ 6 (x –5 ) ( x – 2) = 0

⇒ x = 5 or x = 2

Now , ( ) f x′ = 12x – 42

When x= 5 , ( )5 f ′ > 0 ⇒ x = 5 is a point of local minimum and f(5 ) is the

local minimum value.i.e local minimum value f(5 ) = 37.

When x = 2 , ( )2 f ′ < 0 ⇒ x= 2 is a point of local maximum and f(2) is the local

maximum value.i.e local maximum value f( 2) = 64.

57. Show that the semi –vertical angle of the cone of the maximum volume and the

given slant height is 1tan 2−

Solution

h

r

θ

A

B CD

l

Let ‘ r’ and ‘h’ be respectively radius andheight of a right circular cone.

Let ‘ θ’ be the semi –vertical angle. And ‘l’ isthe given slant height.

i.e r = l Sinθ , and h= l Cosθ .

Volume : 21

3V r hπ =

( )3

2sin cos3

lV

π θ θ =





⇒ ( )3

2 2sin 2 cos sin3

dV l

d

π θ θ θ

θ = −

32 2sin cos 2

3

dV lTan

d

π θ θ θ

θ ⎡ ⎤= −⎣ ⎦

If 0dV

d θ = Then 2‐ Tan2θ= 0 But Sinθ ≠ 0

and cos 0θ ≠ , Since 02

π θ < <

⇒ tan 2θ = (or) ( )1tan 2θ −=

58. Show that semi-vertical angle of a right circular cone of given surfacearea and maximum volume is Sin –1 ( 1/3) .Solution.

Let ‘r’ , ‘h’ and ‘l’ be respectively radius , height

h

r

θ

A

B CD

l

and slant height of a right circular cone.

Let ‘ θ’ be the semi-vertical angle of the cone.

i.e sin , cosr l h lθ θ = =

Surface area 2S rl r π π = +

⇒ ( )2 2sin sinS lπ θ θ = +

⇒ ( )33 3 6 3

sin 1 sinS lπ θ θ = +

⇒ ( )

36

33 3sin 1 sin

S l

π θ θ

=+

………….(1)

Volume 21

3V r hπ =

( )3 2sin cos3

V lπ

θ θ =

( )2

2 6 4 2sin cos9

V lπ

θ θ =

From ( i ) ,

( ) ( )

32 2 4 2

33 3sin cos

sin 1 sin

S V π θ θ

π θ θ

⎛ ⎞= ⎜ ⎟

⎜ ⎟+⎝ ⎠





let V2 = f(θ) and V is maxima or minima as f(θ) is.

( )

3 2

2

sin sin( )

9 1 sin

S f

θ θ θ

π θ

⎡ ⎤−⎣ ⎦=+

………………[2]

[ ]( )

3

3

cos 3sin cos( )

9 1 sin

S f

θ θ θ θ

π θ

−′ =+

[ ]

( )

3

3

cos 1 3sin( )

9 1 sin

S f

θ θ θ

π θ

−′ =

+ ………………….. [3]

When f’(θ) = 0 , Sinθ = 1/3 ,but Cosθ >0 Since 0 < θ <π/2 .

If θ < Sin –1 (1/3) , Then 3Sinθ ‐ 1 < 0 ⇒ 1 – 3 Sinθ > 0 ,∴ f’(θ) > 0

If θ > Sin –1 (1/3) , Then 3Sinθ ‐ 1 > 0 ⇒ 1 – 3 Sinθ < 0 ,∴ f ’(θ) < 0

⇒ f ’(θ) changes from =+ve into –ive when θ = Sin –1 (1/3).

i.e f(θ) is maximum when θ = Sin –1 (1/3). And so the volume of the cone is maximum for

the semi‐vertical angle θ = Sin –1 (1/3). Another method

2S r rlπ π = +

2s r

lr

π

π

−⇒ =

And21

3V r hπ =

⇒ ( )2 2 4 2 21

9V r l r π = −

22

2 2 4 21

9

s r V r r

r

π π

π

⎛ ⎞⎛ ⎞−⎜ ⎟= −⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

2 2 4 2 2 42 2 4

2 2

1 2

9

s r s r r V r

r

π π π π

π

⎛ ⎞+ − −= ⎜ ⎟

⎝ ⎠

( )2 2 2 212

9V r s s r π = −

Let ( )2 2 21( ) 2

9 f r r s s r π = − ∵ V is maximum whenever ( ) f r is maximum

( )2 31( ) 2 8

9 f r s r s r π ′ = −

If ( ) 0 f r ′ = then

2

4S r π = and

( ) 0 , f r r ′′ < ∀

Let24S r π =

⇒ 2 24r rl r π π π + =

1

3

r

l=

1sin

3θ =

1 1sin

3θ

− ⎛ ⎞⇒ = ⎜ ⎟

⎝ ⎠.





59.A point on the hypotenuse of s triangle is at distances ‘a’ and ‘b’ fromthe sides of the triangle Show that the minimum length of thehypotenuse ( a 2/3 + b 2/3 ) 3/2 .

A

B C

P

a

bθ

θ

Solution.

Let ABC be a right triangle and right angle at ‘B’Let ‘P’ be a point on the hypotenuse and PL = a , PM =b .From the Fig.

AP = b Secθ and PC = a Cosecθ

∴ Hypotenuse h = AP + PC

⇒ h = bSecθ + a Cosecθ ………( 1 )

⇒ sec tan cos cotdh

b a ecd

θ θ θ θ θ

= −

3 3

2 2

sin cos

cos sin

dh b a

d

θ θ

θ θ θ

−=

3

2

cos tan

sin

b adh

d

θ θ

θ θ

⎡ ⎤−⎣ ⎦= ………….(2)

When 30, tan 0dh b ad

θ θ

= − = but sin 0 and cos 0θ θ > >

( 0 < θ < π/2 )

⇒ Tan3θ = a/b (or)

1/3

1tan a

bθ

− ⎛ ⎞= ⎜ ⎟

⎝ ⎠

If

1/3

1tan ,a

bθ

− ⎛ ⎞< ⎜ ⎟

⎝ ⎠ T hen 3

tan 0b aθ − <

0dh

d θ ⇒ <

If

1/3

1tan ab

θ − ⎛ ⎞> ⎜ ⎟⎝ ⎠

, then 3tan 0b aθ − >

0dh

d θ ⇒ >

i.e The sign ofdh

d θ changes from ‐ive into +ive when 3tan

b

aθ =





∴ Hypotenuse ‘h’ is minimum for Tan3θ = b/a .

When( )

1/22/3 2/31/3

1/3tan ,sec

a bb

a bθ θ

+⎛ ⎞= =⎜ ⎟

⎝ ⎠ and

( )1/2

2/3 2/3

1/3sec

a bco

aθ

+=

The hypotenuse h = bSecθ + a Cosecθ

= b 2/3(a 2/3 + b 2/3) ½ + a 2/3( a 2/3 + b 2/3) ½

h =(a 2/3 + b 2/3) 3/2 .

60.Find the maximum area of an isosceles triangle inscribed in the ellipse

2 2

2 2 1

x y

a b+ = with its vertex one end of the major axis .

θ

θ − A

B

C

X

Y

OD

solution:

Let ABC be an isosceles triangle inscribed

In an ellipse2 2

2 2 1

x y

a b+ = , and its vertices

are A(- a , 0) ,B(aCosθ,bSinθ ) and C( aCosθ,-bsinθ)From fig.

Area of ( ) ( )1

2 BC ADΔ =

( ) ( )1

2 sin cos2

A b a aθ θ = +

⇒ A = ab Sinθ ( 1 + Cosθ ) …….(1 ).

⇒

dA

d θ = ab [Cosθ(1+Cos

θ)- Sin

2θ ]

dA

d θ = ab [2Cos2θ +cosθ - 1 ]

= ab (Cosθ +1 ) (2Cosθ - 1 ) ……(2 )





WhendA

d θ = 0 , 2Cosθ ‐ 1 = 0 , but Cosθ +1 > 0 ,Since 0 < θ < π/2 .

⇒ Cosθ = ½ ⇒ θ = π/3

Now ,2

2

d A

d θ = ab [‐2 CosθSinθ ‐ Sinθ ]

= ‐ab Sinθ[2Cosθ + 1]

When θ =π/3 ,2

2 0

d A

d θ <

i.e Area is maximum when θ =π/3 .

∴ Maximum area = ab Sin(π/3) [ 1 + Cos(π/3) ]

=3 3

4

ab sq. units

61. Show that height of the cylinder of greatest volume which can be inscribed in a

right

circular

cone

of

height

‘h’

and

having

semi-

vertical

angle θ

is

one

third

that

of

the cone and greatest volume of cylinder is 3 24tan

27hπ θ

h

h1

r

r 1

A

B CD

E

Solution.

In a right circular cone, θ : Semi – vertical angle, h : Height of the cone, r :Radius of the cone.Let r1 , h1 be respectively radius and height of the inscribed cylinder .

From the Fig.

1

1

tan tanr r

r h

h h h

θ θ = = ⇒ =

−

And( )

22

12

1 2

r h hr

h

−=

⇒ ( )

22 2

12

1 2

tanh h hr

h

θ −=

r1 2 = Tan 2θ (h –h1)2





Volume of the inscribed cylinder

V = π r1 2 h1

= π Tan2 θ (h –h1)2 h1 .

= π Tan2 θ[ h2h1 –2hh12+ h1

3 ]

⇒ 1

dV

dh = π Tan2 θ[h2 -4h h1 + 3h1 2 ]

= π Tan2 θ[h1 – h] [3h1 – h ]

When1

0dV

dh= , h1 – h = 0 ( or) 3h1 – h = 0

⇒ h1 = h/3 , but h1 ≠ h

Now ,2

2

1

d V

dh = π Tan2 θ[ -4h + 6h1 ]

When2

1 2

1

, 03

h d V h

dh= <

∴Volume is maximum when h1 = h/3

⇒ Maximum volume : 3 24tan

27

V hπ θ =

62. Show that the height of the cylinder of maximum volume that can be

inscribed in a sphere of radius R is2

3

R. Also find the maximum volume.

R

r

Rh

Solution.

In a sphere of radius ‘R’ , a cylinder is inscribed





Let r be the radius of the cylinder.Let h be the height of the cylinder.From the fig.

AC2 = AB 2 + BC 2

4R2 = 4r 2 + h2

⇒ 2 2

2 4

4

R hr

−=

Now , Volume of cylinder :

V = π r 2 h2 24

4

R hV h

⎡ ⎤−= ⎢ ⎥

⎣ ⎦

= ( )2 344

R h hπ

−

⇒ 2 2

4 34

dV R hdh

π

⎡ ⎤= −⎣ ⎦

When 0dV

dh= , 4 R 2 – 3 h 2 = 0 ⇒

2

3

Rh =

Now ,2

2

3

2

d V h

dh

π = −

When2

2

2, 0

3

R d V h

dh= <

∴Volume of the inscribed cylinder is maximum when 23

Rh =

Maximum volume = ( ) ( )3

24 2 / 3 2 / 34

R R Rπ ⎡ ⎤−

⎢ ⎥⎣ ⎦

34

3 3

RV

π =

63.A window is in the form of rectangle surmounted by a semi-circularopening. The total perimeter of the window is 40m. Find the dimension ofthe window to admit maximum light through the opening.

Solution





x

A

B

Cx x

y y

xE D

Solution.

In a rectangular window with semi- circular opening top,2x : Width of the window y : Height of the window.

i.e Perimeter = 2x + 2y + π x40 =x( π+ 2) +2y

⇒ ( )40 2

2

x y

π − +=

Area of the window =2

22

x xy

π +

( )2

40 22

x A x x

π π = − + +⎡ ⎤⎣ ⎦

2240 2

2

x A x x

π = − −

40 4dA

x xdx

π = − −

When 0dA

dx= , 40– x(π+ 4 ) = 0 ⇒

40

4 x

π =

+

Now , ( )2

2 4 0

d A

dxπ = − + <

This shows that the area of the window is maximum for40

4 x

π =

+

( )40 2

2

x y

π − +⇒ =

=( )

( )

40 220

2 4

π

π

+−

+

∴Height( )

40

4 y

π =

+ and Width

( )80

24

xπ

=+





64. Show that a right circular cone of least curved surface area and given

volume has an altitude equal to 2 times the radius of the base.Solution

Solution.

In a right circular cone,l :Slant height.h:Height of the cone.r :Radius of the base.From the Fig.

sinr l θ = and cosh l θ =

Volume : 21

3V r hπ =

3

sin 2 cos3

lV

π θ θ =

3 3.........(1)

sin 2 cos

V l

π θ θ =

Curved surface area: S = π rl

S = π l2 SinθCosθ

⇒ S 3 = π3 l6 Sin3θCos3θ

( )

23 3 3

2

9sin cos

sin 2 cos

V π θ θ

π θ θ

⎛ ⎞= ⎜ ⎟

⎜ ⎟⎝ ⎠

Let ( ) ( )2

3

3

9,

sin sin

vS f f

π θ θ

θ θ

= =

−

⇒ ( )( )

2 2

23

9 cos 3sin 1

sin sin

v f

π θ θ θ

θ θ

⎡ ⎤−⎣ ⎦′ =−

When f ’(θ) = 0 , 3Sin2θ -1 = 0 but Sinθ> 0 and Cosθ >0 Since 0 < θ <π/2





⇒ Sin2θ = 1/3 ⇒ sinθ =1

3(or) θ = Sin –1

1

3

When θ < Sin –1 1

3⇒ 3Sin2θ -1 < 0 ⇒ f ’(θ) < 0

When θ > Sin –1 13

⇒ 3Sin2θ -1 > 0 ⇒ f’(θ) < 0

⇒f ’(θ) changes from –ive into +ive for θ = Sin –1 1

3

f(θ) is minimum for θ = Sin –1 1

3

⇒ Curved surface area is also minimum for θ = Sin –1 1

3

If Sinθ = 1

3

, then Cosθ =2

3

and h /r = Cotθ

⇒ h/r = 2 ⇒ h = 2 r

65. Prove that the perimeter of a right angled triangle of the givenhypotenuse is maximum when the triangle is isosceles.Solution

h

A

B

C

θ

Solution.

In a right triangle ABC ,BC = x and AB = y and AC = h

Let ∠B = 90° and ∠C =θ From the Fig.

x= hCosθ and y = hSinθ Perimeter : P = x + y + h

= hCosθ + hSinθ + h

P = h ( Cosθ + Sinθ + 1 )

( )sin cosdP

hd

θ θ θ

= − +





When 0dP

d θ = , -Sinθ +Cosθ = 0⇒ Tanθ = 0

⇒ θ = π/4

Now , ( )2

2 cos sin

d Ph

d θ θ

θ = − +

When θ = π/4 ,2

2 0

d P

d θ <

i.e Perimeter is maximum when θ = π/4

x = hCosπ/4 ⇒ 2

h x = and y = h Sinπ/4 ⇒

2

h y =

This shows that the triangle is isosceles.

66.A wire of length 20 cm. is cut into two pieces, one of the pieces isformed into a square and the other is in the form of an equilateraltriangle. Find the lengths of two pieces so that the combined area isminimum .

Solution.

A wire of 20 cm. long is cut into two pieces x cm. and( 20-x) cm.A piece of x cm. is folded into a square.

⇒Perimeter : 4a = x⇒ a= x/4 cm.

∴ Area of square =2

16

x

A piece of( 20-x) cm is formed into an equilateral triangle

⇒Perimeter : 3b =(20 –x) ⇒ side (b) = ( )20

3

x−

⇒Area of equilateral Triangle =( )

2

203

4 3

x−⎡ ⎤⎢ ⎥⎣ ⎦

∴ Area of Triangle = ( )23

2036

x−

The combined area

A = ( )2

23 2016 36 x x+ −

⇒ ( )3 20

8 18

xdA x

dx

−= −

When9 4 3 10 3

072 9

dA x

dx

+= ⇒ =





80

4 3 3 x =

+

Now2

2

1 30

8 18

d A

dx= + >

⇒Combined area is minimum when 2− 804 3 3

x =+

m and ( ) 60 3204 3 3

x− =+

m.

67. Prove that volume of the largest cone that can be inscribed in asphere of radius ‘a’ is 8/27 of the volume of the sphere.

Solution

a

r

h

A

B CD

O

In a sphere of radius ‘a’ , a cone is inscribedr : Radius of the cone.From the Fig.

OD = x and a2

= x2

+ r2

⇒

r2

= a2

– x2

Volume of inscribed cone

21

3V r hπ =

( )( )2 2

3

a x a xV

π − +=

= 3 2 2 3

3a a x ax x

π ⎡ ⎤+ − −⎣ ⎦

2 22 33

dV a ax x

dx

π ⎡ ⎤⇒ = − −⎣ ⎦

When 0dV

dx= , a2 –2ax – 3x2 = 0 ⇒(3x –a ) (x + a) = 0

⇒ x = a/3 ,but x ≠ -a

Now , [ ]2

2 2 63

d V a x

dx

π −= +





When x = a/3 ,2

2 0

d V

dx<

i.e Volume of inscribed cone is maximum when a = a/3.

∴Maximum volume of cone =2

2

3 9 3

a aa a

π ⎛ ⎞⎛ ⎞− +⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

=38 4

27 3

aπ ⎛ ⎞⎜ ⎟⎝ ⎠

=8

27 Volume of sphere.

68. An open Tank with a Square base and vertical sides is to beconstructed from a metal sheet so as to hold a given quantity of water.Show that the cost of the material will be least when the depth of thetank is half of its width.

Solution .

In a open box with the square base, x : Side of the square base. y : Height of the open box.Given, Volume(V) = x2 y

2

V y

x=

Surface area : S = x 2 + 4xy

⇒ S = x 2 + 4x( V/x2)

2 4V S x

x= +

2

42

dS V x

dx x= − ⇒

When3

2

2 40, 0

dS x V

dx x

−= = ⇒ x3 = 2V

Now2

2 3

82

d S V

dx x= +

When x3 = 2V ,2

2 0

d S

dx> ⇒ Surface area is least and so the cost of the material

is least.

x3 = 2V ⇒ x3 = 2 x2 y x = 2y

⇒ Depth of the tank (y ) = Half of the width (x/2)





69.A square piece of tin of side 20 cm. is to made into a box without topby cutting a square from each corner and folding up the flaps to form abox .What should be the side of the square to be cut so that the volumeof the box is maximum? Also find the maximum volume.

Solution.

x x

xx

xx

x x

20-2x

20-2x

In a square piece of tin sheet, a square of x cm is cut off from each corner. x : Side of the square from each corner.20 –2x : Side of square base of the open box.Volume of the open box.

V = (Base area) (Height)= (20 –2x) 2 x

V = 400x – 80 x2 +4x3

⇒ 2

400 160 12

dV

x xdx = − +

= 4(x –10 ) (3x – 10 )

When 0dV

dx= , 4(x –10 ) (3x – 10 ) = 0⇒ x = 10/3 , but x ≠ 10

Now2

2

d V

dx = -160 +24x

When x = 10/3 ,2

2 0

d V

dx<

∴ Volume is maximum when x= 10/3 .

Maximum volume =

2

10 104 103 3

⎛ ⎞−⎜ ⎟⎝ ⎠

=16000

27 cm.3





70. Show that the height of the cylinder , open at the top of given surfacearea and greatest volume is equal to the radius of its base.

Solution

In a right circular cylinder ,r : Radius of the cylinder.h : Height of the cylinder.

Surface area : S =πr2 + 2πrh2

2

S r h

r

π

π

−=

Volume : V = πr2h

=2

2

2

S r r

r

π π

π

⎛ ⎞−⎜ ⎟⎝ ⎠

3

2

Sr r V

π −=

23

2

dV S r

dr

π −=

When 0dV dr = = 0 ⇒ S - 3πr2 = 0⇒

1/2

3S r π

⎛ ⎞= ⎜ ⎟⎝ ⎠

Now ,2

2 3

d V r

dr π = − and

2

2 0

d V

dr < when

1/2

3

S r

π

⎛ ⎞= ⎜ ⎟

⎝ ⎠

∴ Volume is maximum when S = 3π r2

⇒ πr2 + 2πrh = 3π r2 ⇒ h = r .

71 .Find the local minimum or maximum for the function f(x) = Sin x -

Cos x 0n [ 0 , 2 ] .

Solution.

Equation of the function: f(x) = Sin x – Cos x .

⇒ f’(x) = Cos x + Sin xWhen f’( x) = 0 , Cos x + Sin x = 0

⇒ Tan x = -1

Tan x = Tan ( -π/4)





x = nπ + (-π/4)

⇒ x = 3π/4 , 7π/4 ∈ [ 0 , 2π]Now , f’’(x) = -Sin x + Cos x

When x=3π/4 , f’’(3π/4 ) = -Sin(3π/4 ) + Cos(3π/4 ) = 2− <0

i.e x=3π/4 is a point of local maximum

and the local maximum f(3π/4 ) = 2− .

When x=7π/4 , f’’(7π/4 ) = -Sin(7π/4 ) + Cos(7π/4 ) = 2 >0

i.e x=7π/4 is a point of local minimum and

the local minimum f(7π/4 ) = 2− .

Application of Derivatives -C.B.S

Documents

Transcript of Application of Derivatives -C.B.S