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Unit-1Calculus Q1) Verify Rolle’s theorem for the given functions below-1. f(x) = x³ - 6x²+11x-6 in the interval [1,3]2. f(x) = x²-4x+8 in the interval [1,3]A1)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. Q2) Verify Lagrange’s mean value theorem for f(x) = (x-1)(x-2)(x-3) in [0,4].A2)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 4Therefore the given function is verified. Q3) Verify Cauchy’s mean value theorem for the function f(x) = x⁴ and g(x) = x² in the interval [1,2]A3)We are given, f(x) = x⁴ and g(x) = xDerivative 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.

Q4) Expand sin x in powers of

A4)

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-

+ …….

= ………………………..

Q5) By using Maclaurin’s series expand tan x.A5)Let-

Put these values in Maclaurin’s series we get-

Q6) Evaluate

A6)Here we notice that it is an indeterminate form of

.So that , we can apply L’Hospital rule-

Q7) Find the value of a, b if

A7)Let

…

By L – Hospital rule

…

… (1)

…

But

From equation (1)

Q8) Evaluate

A8)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

Q9) Evaluate

A9:

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.

Q10) Evaluate

A10)Let

…

Taking log on both sides,

…

By L – Hospital rule,

i.e.

Q11) Show that sin x (1 + cos x) is a maximum when

.A11)Let

Then first derivative-

So that-

Now-

So that-

Therefore f(x) has a maximum at .

Q12) Evaluate

.A12)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

Q13)

A13)

Q14) Show that

A14)

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

Q15) find the area under the curves where f(x) = x+4 and g(x) = 3 – x/2 over the interval [1,4]A15)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.

Q16) A force of 1200 N compresses a spring from its natural length of 18 cm to a length of 16 cm. How much work is done in compressing it from 16 cm to 14 cm?A16)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

Q17) Determine the area bounded by three straight lines y = 4 – x, y = 3x and 3y = xA17)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-

Q18) Find the area enclosed by the two functions-

and g(x) = 6 – xA18)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.

Q19) Find the volume of the solid of revolution formed by revolving R around y-axis of the function f(x) = 1/x over the interval [1 , 3].A19)

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-

Q20) check whether the series

is convergent or divergent. Find its value in case of convergent.A20)The general formula for this series is given by,

Sn = = )

We get,

) = 3/2

Hence the series is convergent and its values is 3/2.

Example-3: check whether the series is convergent or divergent.

Sol. The general formula can be written as,

We get on applying limits,

= 3/4

This is the convergent series and its value is 3 / 4

Q21) Test the convergence of the following series:

A21)

Here we take,

Which not zero and finite ,

So by comparison test , and both converges or diverges, but by p-series test

Is convergent. so that is convergent.

Q22) Test for the convergence of the n’th term of the series given below-

A22)

We have ,

Now , by D’Almbert ratio test converges if and diverges if

At x = 1 , this test fails.

Now , when x = 1

The limit is finite and not zero.Then by comparison test,

converges or diverges together.Since

is the form of

, in which

Hence

diverges then

will also diverge.Therefore in the given series

converges if x<1 and diverges if x≥1 Q23) If the series

converges, then find the value of x.A23)

Here

Then,

By D’Almbert’s ratio test the series is convergent for |x|<1 and divergent if |x|>1.

So at x = 1

The series becomes-

At x = -1

This is an alternately convergent series.This is also convergent series, p = 2Here, the interval of convergence is

Q24) Expand

as far as the term in

.A24)We know that the power series for

is-

Here we have to find-

So that-

On solving, we get-

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