Question Bank | unit 1 calculus

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Question Bank

UNIT–1

Calculus

Question-1: Verify Rolle’s theorem for the function f(x) = x2 for

Solution:

Here f(x) = x2;

i) Since f(x) is algebraic polynomial which is continuous in [-1, 1]

Ii) Consider f(x) = x2

Diff. w.r.t. x we get

f'(x) = 2x

Clearly f’(x) exists in (-1, 1) and does not becomes infinite.

Iii) Clearly

f(-1) = (-1)2 = 1

f(1) = (1)2 = 1

f(-1) = f(1).

Hence by Rolle’s theorem, there exist such that

f’(c) = 0

i.e. 2c = 0

c = 0

Thus such that

f'(c) = 0

Hence Rolle’s Theorem is verified.

Question-2: 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]

Sol. (1)

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.

Question-3: Verify the Lagrange’s mean value theorem for

Solution:

Here

i) Clearly f(x) = log x is logarithmic function. Hence it is continuous in [1, e]

Ii) Consider f(x) = log x.

Diff. w.r.t. x we get,

Clearly f’(x) exists for each value of & is finite.

Hence all conditions of LMVT are satisfied Hence at least

Such that

i.e.

Since e = 2.7183

Clearly c = 1.7183

Hence LMVT is verified.

Question-4: Verify mean value theorem for f(x) = tan-1x in [0, 1]

Solution:

Here ;

i) Clearly is an inverse trigonometric function and hence it is continuous in [0, 1]

Ii) Consider

Diff. w.r.t. x we get,

Clearly f’(x) is continuous and differentiable in (0, 1) & is finite

Hence all conditions of LMVT are satisfied, Thus there exist

Such that

i.e.

Clearly

Hence LMVT is verified.

Question-5: Verify Cauchy mean value theorems for &in

Solution:

Let &;

i) Clearly f(x) and g(x) both are trigonometric functions. Hence continuous in

Ii) Since &

Diff. w.r.t. x we get,

Clearly both f’(x) and g’(x) exist & finite in . Hence f(x) and g(x) is derivable in and

Iii)

Hence by Cauchy mean value theorem, there exist at least such that

i.e.

i.e. 1 = cot c

i.e.

Clearly

Hence Cauchy mean value theorem is verified.

Question-6: Verify Cauchy’s mean value theorem for the function f(x) = x⁴ and g(x) = x² in the interval [1,2]

Sol. 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.

Question-7: Expand in power of (x – 3)

Solution:

Let

Here a = 3

Now by Taylor’s series expansion,

… (1)

equation (1) becomes.

Question-8: Expand in ascending powers of x.

Solution:

Here

i.e.

Here h = -2

By Taylors series,

… (1)

equation (1) becomes,

Thus

Question-9: Evaluate

Sol. Let f(x) = , then

And

= 0

But if we use L’Hospital rule again, then we get-

Question-10: Evaluate

Sol. 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

Question-11: Find , n>0.

Sol. Let f(x) = log x and g(x) =

These two functions satisfied the theorem that we have discussed above-

So that,

Qustion-12: Evaluate

Sol. Here we find that-

So that this limit is the form of 0.

Now,

Change to obtain the limit-

Now this is the form of 0/0,

Apply L’Hospital’s rule-

Question-13: Find out the maxima and minima of the function

Given …(i)

Partially differentiating (i) with respect to x we get

….(ii)

Partially differentiating (i) with respect to y we get

….(iii)

Now, form the equations

Using (ii) and (iii) we get

using above two equations

Squaring both side we get

This show that

Also we get

Thus we get the pair of value as

Now, we calculate

Putting above values in

At point (0,0) we get

So, the point (0,0) is a saddle point.

At point we get

So the point is the minimum point where

In case

So the point is the maximum point where

Question-14: Evaluate

Sol. Here is an increasing function on [1 , 2]

So that,

…. (1)

We know that-

And

Then equation (1) becomes-

Question-15: Evaluate-

Sol.

Question-16: Determine the area enclosed by the curves-

Sol. We know that the curves are equal at the points of interaction, thus equating the values of y of each curve-

Which gives-

By factorization,

Which means,

x = 2 and x = -3

By determining the intersection points the range the values of x has been found-

-3

-2

-1

And

x	-3	0	2
y = 7 - x	10	7	5

We get the following figure by using above two tables-

Area of shaded region =

= ( 12 – 2 – 8/3 ) – (-18 – 9/2 + 9)

= 125/6 square unit

Question-17: Find the area enclosed by the two functions-

and g(x) = 6 – x

Sol. 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.

Question-18: 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].

Sol. 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-

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