Question Bank(unit-4)

Question-1: Verify Rolle’s theorem for the function f(x) = x(x+3)

in interval [-3 , 0].Sol. First we will differentiate the given function with respect to x, we get

f’(x) = (x²+3x) + (2x + 3)

=

This shows that f’(x) exists for all x, therefore f(x) is continuous for all x.

Now, f(-3) = 0 and f(0) = 0 , so that f(-3) = f(0).

Here f(x) satisfies all the conditions of Rolle’s theorem,

Then,

f’(x) = 0 , which gives

= 0

We get,

X = 3 and x = -2

Here we can see that clearly -3<-2<0 , therefore there exists -2 ∈ (-3,0) such that

f’(-2) = 0

That means the Rolle’s theorem is true for the given function.

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 Rolle’s theorem for sin x in the interval []

Sol. Suppose f(x) = sin x

We know that sin x is continuous for all x.

Now , f’(x) = cos x exists for all x in () and

f(0

f(0

Thus f(x) satisfies all the conditions of Rolle’s theorem.

Now,

f’(x) = 0 that gives , cos x = 0

x =

Here we notice that both intervals lie in (.

There exists, c =

So that, f’(c) = 0

The Rolle’s theorem has been verified.

Question-4: Verify Lagrange’s mean value theorem for f(x) = (x-1)(x-2)(x-3) in [0,4].

Sol. 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.

Question-5: Verify Lagrange’s mean value theorem for f(x) = log x in [1,e].

Sol. We already know that the function which is log x is continuous for all x>0.

So that this is the continuous function In [1,e]

Now,

f’(x) = 1/x

Which is exists for all x in (1,e)

So that f(x) is differentiable in (1,e).

By Lagrange’s mean value theorem, we get

f’(c) = , let x = c,

then ,

f’(c) =

We get,

c = e-1

e-1 will always lies between 1 and e .

Hence the function is verified by Lagrange’s mean value theorem

Question-6: Verify Cauchy’s mean value theorem for the function f(x) = sin x and g(x) = cos x in [ 0 , π/2]

Sol. It is given that ,

f(x) = sin x and g(x) = cos x

now ,

f’(x) = cos x and g’(x) = - sin x

We know that both the functions are continuous in [ 0 , π/2] and differentiable in ( 0 , π/2 )

Also , g’(x) = -sin x ≠ 0 for all x ϵ( 0 , π/2 )

By Cauchy’s mean value theorem , we get

for some c: 0< c <

That means

which gives,

Cot c = 1

C =

Now we see that lies between 0 and

Question-7: Find the Taylor’s series for the function f(x) = 3x² - 6x + 5 in ascending powers of x-3.

Sol. We know that the Taylor’s theorem for the function f(x) in ascending powers of (x – a) is,

f(x) = f(a) + (x – a)f’(a) + + ………. ………….(1)

Here ,

f(x) = 3x² - 6x + 5 and a = 3. So that f(a) = 14

Now , we will find out the derivatives ,

f’(x) = 6x – 6 and f’(3) = 12

f’’(x) = 6 and f’’(3) = 6

f’’’(x) = 0 and f’’’(x) = 14

Put these values in Taylor’s theorem, we get

3x² - 6x + 5 = 14 + 12(x-3) + 6 +…………. Ans.

Question-8: Determine the area under the curves y = 2x² + 10 and y = 4x + 16.

Sol. Here we will find out the intersection, which are the boundaries of the curve,

2x² + 10 = 4x + 16

2x² - 4x – 6 = 0

2 (x + 1) ( x – 3 ) = 0

We get, x = -1 and x = 3,

We know that, Area under the curve,

A =

Question-9: 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?

Sol. 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

Question-10: using slicing method, find out the volume of solid of revolution bounded by f(x) = x² - 4x +5 , x = 1 and x = 4, rotated about the x-axis.

Sol. Intervals are given. [ 1 , 4]

Here the solid is formed by revolving the region about x-axis , the cross – sections are circle .

Area of the cross section is , then is the area of circle, here radius of the circle is given by f(x).

We know that,

Area of the circle= π r²

= π [f(x)]²

= π[x² - 4x +5]²

Volume will be,

V =

=

= which gives,

= π

So the volume is π

Question-11 Find the volume of the solid of revolution generated by rotating the region between the graph of f(x) = √x over the interval [1,4] around x-axis.

Sol. The graph of the function will look like as follow,

On rotation it will make circle ( cross-section)

We know that,

V =

= = , which gives

=

The volume is

Question-12: suppose a wire hanging on two poles follows the curve,

f(x) = a cosh(x/a)

Find the length of the wire.

Sol. We will find the first derivative of f(x),

f’(x) = sinh(x/a)

The curve of f(x) will look like,

Hence the curve os symmetric, we will measure the length of one side first,

The limits on one side will be, 0 to b

We know that

Length of Arc =

Put f’(x) = sinh(x/a), we get

Length of Arc =

Use the identity,

=

=

We get on solving,

a sinh(b/a)

On both sides, the length of the curve will be,

2 a sinh(b/a)

Question-13: find the surface of the surface generated by revolving the graph f(x) = over the interval [1,4] around x-axis.

Sol. The graph of the function will look like

f(x) = then f’(x) = 1/2 (

(f’(x))² = 1/4x

Then,

S =

S =

S = ,

Let u = , then du = dx, when x = 1 , u = 5/4 and when x = 4 , u = 17/4,

This will give,

=

On solving the integral we get,

= 30.84 (approx.)

Question-14: Show that

Solution : =

=

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

=

=

Question-15: Evaluate dx.

Solution : Let dx

X | 0 | |

t | 0 |

Put or ;dx =2t dt .

dt

dt

Question-16: Evaluate dx.

Solution : Let dx

X | 0 | |

t | 0 |

Put or ;dx =2t dt .

dt

dt

Question-17: Evaluate I =

Solution:

= 2 π/3

Question-18: Prove that

Solution : Let

Put or ,

Question-19: Evaluate

Solution :Let

Put or ,,

When,;,

o | ||

1 | 0 |

Also

Question-20: Show that

Solution :

=

(0<p<1)

(by above result)