differentiation-8

As the polynomial functions are continuous and differentiable every where f {f(x)} is continuous and differentiable in [0, 3] except perhaps at the turning points of definition, namely, x = 1, 2 denote f {f(x)} by g(x).

Now g (1+0) = 2-1=1

         g (1-0) = 2+1 = 3

So g(x) is not continuous at x=1.

g (2+0) = 4-2 =2

g (2-0) = 2-2 = 0

So, g(x) is not continuous at x=2.

Hence g(x) is also not differentiable at x = 1, 2.

\The points of discontinuity and non-differentiability are x=1, 2.

Now we have to draw the graph for y = f {f(x)} where

y = 2+x   in [0, 1],

y = 2-x    in [1, 2]

y = 4-x    in [2, 3]

The graph is discontinuous at x = 1, 2.

                        Fig (6)

Dumb Question:

1)      In the above solution, there is a condition 0 £1+x £2 if 0 £ x £ 2 from that it is inferred that 0 £ x£ 1 How?

Ans:

0 £1+x £2       if  0 £ x £ 2

Þ -1 £ x £ 1 if  0 £ x £ 2

Now clearly both the conditions need to be satisfied simultaneously. So the intersection of these two conditions will give the desired results.

i.e. 0 £ x £ 1

2) If then find from the first principle.

Solution:

Clearly the definition of the function f(x) is where h is a very small positive quantity, remains the same, so  is not a turning point of definition.

Now,

 

(Using definition of mod)

Now

(Using L-Hospital Rule)

 

 

 

3) Let a+b =1, 2a²+2b² =1 and f(x) be a continuous function such that f (2+x) +f(x) = 2 for all xÎ(0, 2) and p = then find the least positive integral values of ‘a’ for which the equation ax²²-bx+c = 0 has both roots lies between p and q;

Where a, b, c ÎN.

Solution:

Given   a+b =1 --------- (1)

            2a²+2b² =1 -------- (2)

Solving equation (1) and (2) we get,

a = b = .

Þ ----------- (3)

Given f(x+2) +f(x) = 2 for all x Î[0, 2] -------- (4)

Now p =

(let x = t+2 for second integration}

Then p =0, q=1

Let the roots of equation ax²-bx+c = 0 be a and b

\f (x) = ax²-bx+c= a(x-a) (x-b) -------- (5)

Since equation f(x) = 0 has both roots between 0 and 1.

\f (0).f (1) > 0 ------- (6)

But f (0).f (1) = c (a-b+c) = an integer ------- (7)

Therefore least value of f (0). f (1) = 1 -------- (8)

Now from equation (5)

f (0).f (1) = a a b a (1- a) (1-b)

             = a²ab (1- a) (1- b) --------- (9)

As we know

a (1- a) has greatest values and b (1- b) has greatest value

But b ¹ a

Thus from equation (8) greatest value of

f (0). f (1) < ------ (10)

From (8) and (10)

{As a Î Natural number}

Dumb Question:

1)      Why c (a-b+c) is an integer?

Ans: It is given in the question that a, b, c ÎN

So, cÎN and a-b+cÎI

So, c (a-b+c) ÎI

Hence c (a-b+c) is an integer.

 

 

 

 

 

 

                                              KEYWORDS

ü      Derivative

ü      First Principle

ü     

ü      Right Hand Derivative

ü      Left Hand Derivative

ü      Parametric

ü      Logarithmic Differentiation

ü      Implicit function

ü      Chain Rule