Test: Mean Value Theorem - Civil Engineering (CE) MCQ

In order to find the average value, one must integrate the function by using the Fundamental Theorem of calculus and divide the answer by the given length of the interval.

So, the average (or the mean) value of f (x) on [a, b] is defined by

The Mean Value Theorem for Definite Integrals:

Let y = f (x) be a continuous function on the closed interval [a, b]. The mean value theorem for integrals states that there exists a point c in the interval such that

In other words, the mean value theorem for integrals states that there is at least one point c in the interval [a, b] where f (x) attains its average value :

Geometrically, this means that there is a rectangle whose area exactly represents the area of the region under the curve y = f (x). The value of f (c) represents the height of the rectangle and the difference (b - a) represents the width.

Since given function is continuous and differentiable then by Lagrange’s Mean-Value Theorem.

3c² - 3 = 0

c² = 1

∴ c = ±1

From mean value theorem,

we know that 

 Here, a = 0, b = 2

 and 

 Substituting the values 

 f(2) = 2 × f'(x) + f(0) 

lower bound of f(2) =

 = 1 + 4 = 5

Given:

f(x) = e^x, g(x) = e^-x

Now,

derivative of f(x) i.e. f^’(x) = e^x

derivative of g(x) i.e. g^’(x) = - e^x

range is given as [2, 3], that is, a = 2 and b = 3

here, f(x) and g(x) are differentiable and continuous and derivative of g(x) is not equal to 0.

They are satisfying the conditions of Cauchy’s mean value theorem.

Now,

Consider it is in interval of [a, b].

c = ½ (2 + 3) = 5/2 = 2.5

f(x) and g(x) are continuous in [4, 6]

f(x) and g(x) are differentiable in [4, 6]

g′(x) = −2 / x³

g’(x) ≠ 0 in (4, 6)

Now, according to Cauchy’s mean value theorem

∴ c = 4.8

Both the functions are continuous in [1, 2] and differentiable in (1, 2).

By Cauchy's Mean Value theorem we should have

Therefore c can be any value between 1 and 2. 

Rolle’s mean value theorem:

Suppose f(x) is a function that satisfies all of the following

1. f(x) is continuous on the closed interval [a, b]

2. f(x) is differentiable on the closed interval (a, b)

3. f(a) = f(b)

Then there is a number c such that a < c < b and f'(c) = 0

Let h(x) = f(x) – 2g(x)

Given that f(x) and g(x) are continuous as well as differentiable

Hence h(x) is also continuous as well as differentiable.

h(0) = f(0) – 2g(0) = 2 – 2(0) = 2

h(1) = f(1) – 2g(1) = 6 – 2(2) = 2

h(0) = h(1)

hence h(x) satisfies all the conditions of Rolle’s theorem

Now, h'(c) = 0

⇒ f'(c) – 2g'(c) = 0

⇒ f'(c) = 2g'(c)

We know that, (Intermediate value theorem)

If f(a)f(b) < 0 then f(x) has at least one root in (a, b)

Since there is no root in [a, b] this implies f(a) and f(b) are of same sign 

i.e. either they both are positive or they both are negative

In both cases f(a)f(b) > 0

Let f(x) = ax + b

∴ f′(x) = a

Given that, 

6f′(c) + f(−3) = f(3).

⇒ 6a = 3a + b – (−3a + b)

⇒ a = a

∴ C can be any value (−3,3)

So, no condition on c

To find points such that f'(c) = 1
We need to check points on graph where slope remains the same (45 degrees)
In every interval of the form [(n – 1)π, nπ] we must have 2n – 1 points
Because sine curve there has frequency 2n and the graph is going to meet the graph y = x at 2n points.
Hence, in the interval [0, 16π] we have
= 1 + 3 + 5…….(16terms)
=(16)² = 256.