1.  f(x) = x³ - 6x²+11x-6  in the interval [1,3]

2. f(x) = x²-4x+8  in the interval [1,3]

As we know that every polynomial is continuous and differentiable for all points, so that the given function is continuous and differentiable in the interval [1,3]

Also, f(1) = f(3) = 0

Now we find f’(x) = 0

3x² - 12x +11 = 0

We get, x = 2+ and 2 -

Hence both of them lie in (1,3).

Hence the theorem holds good for the given function in interval [1,3]

(2) As we know that every polynomial is continuous and differentiable for all points, so that the given function is continuous and differentiable in the interval [1,3]

Also,  f(1) = 1 -4 +8 = 5 and f(3) = 9 – 12 + 8 = 5

Hence f(1) = f(3)

 Now the first derivative of the function,

f’(x) = 0

2x – 4 = 0 , gives

X = 2

We can see that 1<2<3, hence there exists 2 between 1 and 3. And f’(2) = 0.

This means that the Rolle’s theorem holds good for the given function and given interval.

As we see that the given function is a polynomial and we know that the polynomial is continuous in [0,4] and differentiable in (0,4).

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

f(x) =  x-6x²+11x-6

now at x = 0, we get

f(0) = -6  and

at x = 4, we get.

f(4) = 6

diff. the function w.r.t.x , we get

f’(x) = 3x²-6x+11

suppose x = c, we get

f’(c) = 3c²-6c+11

by Lagrange’s mean value theorem,

f’(c) =     =     =   = 3

now we get,

3c²-6c+11 = 3

3c²-6c+8 = 0

On solving the quadratic equation, we get

C = 2

Here we see that the value of c lies between 0 and 4

Therefore the given function is verified.

We are given, f(x) = x⁴ and g(x) = x

Derivative of these fucntions ,

      f’(x) = 4x³ and g’(x) = 2x

put these values in Cauchy’s formula, we get

2c² =

c² =

c =

now put the values of a = 1 and b = 2 ,we get

c = = = (approx)

Hence the Cauchy’s theorem is verified.

Let f(x) = sin x

     Then,

                    =

          By using Taylor’s theorem-

+ …….     (1)

Here f(x) = sin x and a = π/2

f’(x) = cosx  , f’’(x) = - sin x    ,     f’’’(x) = - cos x  and so on.

Putting x = π/2 , we get

f(x) = sin x = = 1

f’(x) = cos x = = 0

f’’(x) = -sin x = = -1

f’’’(x) = -cos x = = 0

from equation (1) put a = and substitute these values, we get-

+ …….    

         = ………………………..

Let-

Put these values in Maclaurin’s series we get-

Here we notice that it is an indeterminate form of .

So that , we can apply L’Hospital rule-

Let

   …

By L – Hospital rule

   …

   …

    … (1)

    …

But

From equation (1)

We can see that this is an indeterminate form of type 0/0.

Apply L’Hospital’s rule, we get

But this is again an indeterminate form, so that we will again apply L’Hospital’s rule-

We get

=

Apply L’Hospital rule as we can see that this is the form of

=

Note- In some cases like above example, we can not apply L’Hospital’s rule.

Let

   …

Taking log on both sides,

  …

By L – Hospital rule,

i.e.

Let

Then first derivative-

So that-

Now-

So that-

So that-

Therefore f(x) has a maximum at .

.

Here we notice that  f:x→cos x  is a decreasing function on [a , b],

Therefore by the definition of the definite integrals-

Then

Now,

Here

Thus

=

                                          =

                                          = )  .......................

                                          =

                                          =

Here limits are given a = 1 , b = 4,

We know that, area under the curve,

A =

= 

=

= 

= (16 – 7/4) =  57/4

So that the area of the region is 57/4 unit square.

Here,

F = kx

So that,

1200 = 2k

K = 600 N/cm

In that case,

F = 600x

 We know that,

W =

W = , which gives

W  = 3600 N.cm

We get the following figure by using the equations of three straight lines-

                                         y = 4 – x, y = 3x and 3y = x

Area of shaded region-

and g(x) = 6 – x

We get the following figure by using these two equations

To find the intersection points of two functions f(x) and g(x)-

f(x) = g(x)

On factorizing, we get-

x = 6, -2

Now

Then, area under the curve-

                                        A =

Therefore the area under the curve is  64/3  square unit.

The graph of the function f(x) = 1/x will look like-

The volume of the solid of revolution generated by revolving R(violet region) about the y-axis over the interval [1 , 3]

  Then the volume of the solid will be-