Learn Mean Value Theorem for JEE Main

Differentiability and continuity is an important topic for the JEE Main exam. Students can expect 3-4 questions from this topic for any entrance exam.

Students are recommended to revise and learn previous years’ questions on differentiability and continuity so that they can understand the question pattern and difficulty level of the questions asked from this topic. Mean Value Theorem (MVT) is a tool that is very useful in both differential and integral calculus. It helps to study the identical behaviour of different functions. This theorem is also called Legrange’s MVT. This is also an extension of Rolle’s MVT. This is used in solving many problems in differentiability and continuity.

MVT Theorem

It states that for any two points on the curve, there exists a point on the curve so that the tangent drawn at that point is parallel to the secant through the two points on the curve.

MVT Statement

Let f(x) is a function such that  f(x) is continuous in [a, b] and differentiable in (a, b). Then, there exists a number c, such that  a < c < b.

f(b) – f(a) = f ‘(c) (b – a).

f’(c) is the first derivative at c.

If f(a) = f(b) then, there exists at least one c with a < c < b such that f'(c) = 0. This is called

Rolle’s Theorem.

Using the mean value theorem, we can approximate the derivative of any function. It can build a relationship between the slope of a tangent line and the secant line on a curve.

Let us have a look at some examples.

Example 1: Given f(x) = x² – 8x + 12. Check for Rolle’s theorem on (2, 6).

Solution:

f(x) = x² – 8x + 12

f(x) is differentiable and continuous since it is a polynomial function.

f(2) = 4 – 16 + 12 = 0

f(6) = 36 – 48 + 12 = 0

f(2) = f(6)

Differentiate f(x) w.r.t.x

=> 2x – 8

f’(c) = 2c-8

f’(c) = 0 => 2c -8 = 0

=> 2c = 8

=> c = 8/2 = 4

4 belongs to (2, 6).

Hence Rolle’s theorem is verified.

Example 2: Find the value of c in the MVT for f(x) = e^x, a = 0 and b = 1.

Solution:

Given f(x) = e^x

a = 0 and b = 1

f’(x) = e^x

f’(c) = (f(b)-f(a))/(b-a)

e^c = (e¹ – 1)/(1-0)

e^c = e-1

=> c = log (e-1)

Hence the value of c is log(e-1).

Central Limit Theorem (CLT)

CLT is a theory in statistics. It states that whenever a random sample of size n is taken from any distribution with mean and variance, then the sample mean will be approximately normally distributed with mean and variance. Central limit theorem has great importance in statistics. The higher the value of the sample size, the better the approximation to the normal.

Students are recommended to revise and practise previous years’ questions so that they can be familiar with the type of questions asked from statistics, differentiability and continuity. Visit BYJU’S for important formulas pdf on differentiability and continuity and previous years’ JEE solved question papers.