Email  
APPLICATION OF DERIVATIVES
Tags: Engineering Entrance  |  JEE Main  |  CBSE Board  |  JEE Main & Advanced  |  Differential Calculus  |  Class 11  |  Class 12
  « Back to Content











Illustration:
Illustration: Let f(x) = x(x - 1)2, find point at which f(x) have maxima & minima.
Ans: f(x) = x(x - 1)2

    f'(x) = 2x(x - 1) + (x - 1)2

    f'(x) = (x - 1)(3x - 1)

    f'(x) = 0

(x - 1)(3x - 1) = 0

Critical points are x = 1, 1/3

   f''(x) = 6x - 4

f''(1) = 6 - 4 = 2 > 0

f''(-1/3) = - 2 < 0

   f''(1) > 0 so, at n = 1 it has minima

   f''(1/3) < 0 so, maxima at x = 1/3.

Global Maxima & Minima:
(i) Global maxima/minima in [a, b] is a greatest/least value of f(x) in [a, b]

Global maxima/minima in [a, b] would always occur at critical points of f(x) with in [a,b] or end points of interval.
To find global maxima/minima of f(x) in [a, b] find out all critical points of f(x) in [a, b] (i.e. all points at which f'(x) = 0)

Let c1, c2, .......... cn be critical points & f(c1), f(c2), .............. f(cn) be values of function of these points.

Let M1 Global Maxima

     M2 Global Minima

Then M1 = max.{f(a), f(c1), f(c2), ............... f(cn), f(b)}

  &    M2 = min.{f(a), f(c1), f(c2), ................ f(cn), f(b)}

Illustration: If f(x) = 2x3 - 9x2 + 12x + 6. Discuss global maxima and minima of f(x) in (1, 3).
Ans: f(x) = 2x3 - 9x2 + 12x + 6

f'(x) = 6x2 - 18x + 12

f'(x) = 6(x - 1)(x - 2)

f'(x) = 0   x = 1, 2

  f(1) = 11 & 1 - (2) = 10

Since open interval is (1, 3)

Clearly x = 2 is only point in (1, 3) & 1 - (2) = 10

Now,

Dumb Question: How = 11.
Ans: f(x) = 2x2 - 9x2 + 12x + 6

        = f(1 + h)

                  = 2(1 + h)3 - 9(1 + h)2 + 12(1 + h) + 6

                  = 2(1 + h3 + 3h2 + 3h) - 9(1 + h2 + 2h) + 12 + 12h + 6

                  = 11

So, x = 2 is point of global minima in (1, 3) & global maxima doesnot exist in (1, 3) .

Minima of discontinuous function:
For minima at x = a, 4 cases arises.

 

              From figure f(CD) < f(a + h)           From figure f(a) < f(a + h)
                              f(a) < f(a - h)                               f(a)

f(a - h)


              From figure, f(a) < f(a + h)           From figure, f(a)

f(a + h)
                               f(a) < f(a - h)                                f(a) < f(a - h)
From all above case, for minima of discontinuous functions,
f(a)

f(a + h)
2f(a)

f(a - h)
Illustration
: Discuss minima of f(x) = {x}, (where {} is raction part of x) for x = 6.

Ans: For discont functions, maximum & minimum at x = a is attained when
f(a)

f(a + h) & f(a)

f(a - h)
Now, f(x) = {x} is discontinuous function at x = 6
Since f(c) = {6} = 0
     f(6 + h) = {6 + h} = h > 0
&   f(6 - h) = {6 - h} = 1 - h > 0
So, f(6) < f(6 + h) & f(6) < f(6 - h)

f(x) is minimum at x = 6.
Dumb Question
: How f(6) = {6} = 0

Ans: {x} is fractional function. Since6 is integer & no fractional part

f(6) = 0
Maxima of discontinuous function
:


          From figure, f(a) > f(a + h)             From figure, f(a) > f(a + h)
                           f(a) > f(a - h)                                f(a) > f(a - h)


            From figure, f(a)

f(a - h)              From figure, f(a)

f(a + h)
                             f(a) > f(a + h)                               f(a) > f(a - h)
Prove all cases, maxima of discontinuous function,
f(a)

f(a + h)   &   f(a)

f(a - h)
Illustration
: f(x) =

, then for f(x) at x = 1 discuss maxima & minima.

Ans:

    f(x) =



f(1) = 6

f(1 - h) = 6
&  f(1 + h) = 7 - (1 + h) = 6 - h < 6
So, at x = 1 is neither point of maxima nor manima.
Easy Type
:

Q.1. If s = t
3
- 4t, find acceleration at time when velocity is zero.

Ans: s = t
3
- 4t
  

v =

= 3t
2
- 4 ................................................. (i)
       a =

= 6t ....................................................... (ii)
time at which velocity is zero
3t
2
- 4 = 0  

t
2
=


a =


Q.2. If r be radius, s the surface atrea & v the volume of spherical buble, prove that
(i)

   (ii)

Since v =

r
3

(i)

.................................. (i)

(ii) s = 4

r
2



Q.3. On curve x
3
= 12y, find interval at which abscissa changes at a faster rate than ordinate.

Ans: x
3
= 12y
differencing w.r.t. y


Since abscissa change faster than ordinate

> 1   or  

> 1

> 0 where x

0

x
2
- 4 < 0  

(x - 2)(x + 2) < 0

- 2 < x < 2

{0}
So, x

(- 2, 2) - {0} is required interval.
Q.4. Find eq. of tangent to parabola y
2
= 4ax at point (at
2
, 2at)

Ans: y
2
= 4ax ............................................................ (i)
differentiating (i) w.r.t. x,


Eq. of tangent at (at
2
, 2at) is
      (y - 2at) =


  

y - 2at =

(x - at
2
)






   yt = x + at2

Q.5. Find a cute angle b/w curves y = |x
2
- 1| & y = |x
2
- 3| at their points of intersection when x > 0.

Ans: For intersection of given curves
    |x
2
- 1| = |x
2
- 3|  

(x
2
- 1)
2
= (x
2
- 3)
2


2x
2
= 4  

x = ±


Since   x > 0, so, x =

is only point of intersection
y = |x
2
- 1| = (x
2
- 1) since x =


y = |x
2
- 3| = - (x
2
- 3) since x =
     |     

 
  Electricity and Effect of Current      Centre of Mass      Work Power & Energy      Magnetic Effects of Current      AIEEE 2009 Solutions and Analysis      English Core-XII Sample Test Paper      Function Theory      Basis of Inheritance      dasda      Business Studies -XII Sample Test Paper      Chemical Bonding      Group I A      Integration Theory      Organic Chemistry Basic Concepts      Applications of Derivatives      Informatics Practices (065) Sample Question Paper -III      Inorganic Chemistry Group VI A      Properties of Matter      Heat & Thermodynamics      Disha - Path to Success Series     
 



Free Sign Up!
Sponsored Ads

Preparing for JEE?

Kickstart your preparation with new improved study material - Books & Online Test Series for JEE 2014/ 2015


@ INR 5,443/-

For Quick Info

Name

Mobile

E-mail

City

Class