RD Sharma Solutions

RD Sharma Class 12 Solutions Chapter 15 - Mean Value Theorems (Ex 15.1) Exercise 15.1

RD Sharma Class 12 Solutions Chapter 15 - Mean Value Theorems (Ex 15.1) Exercise 15.1

Q: 1. What are the essential steps to verify Rolle's Theorem for a function in RD Sharma Class 12 Exercise 15.1?

To correctly solve a problem verifying Rolle's Theorem from this exercise, you must follow these sequential steps:Step 1: Check Continuity. Verify that the function f(x) is continuous on the closed interval [a, b]. For polynomial, sine, or cosine functions, this is generally true.Step 2: Check Differentiability. Verify that the function f(x) is differentiable on the open interval (a, b). This involves ensuring the derivative f'(x) is defined for all points within the interval.Step 3: Check Endpoint Values. Calculate and confirm that f(a) is equal to f(b).Step 4: Find the Derivative. If the above conditions are met, find the derivative f'(x).Step 5: Solve for 'c'. Set the derivative f'(c) = 0 and solve for the value of 'c'.Step 6: Final Verification. Ensure that the calculated value of 'c' lies strictly between 'a' and 'b', i.e., c ∈ (a, b).

Q: 2. How is Lagrange's Mean Value Theorem (LMVT) applied differently from Rolle's Theorem when solving problems in Chapter 15?

The primary difference lies in the third condition and the final equation. While both theorems require continuity on [a, b] and differentiability on (a, b), the subsequent steps diverge:Rolle's Theorem has an additional condition: you must verify that f(a) = f(b). The goal is to find a point 'c' where the tangent is horizontal, meaning you solve for f'(c) = 0.Lagrange's Mean Value Theorem (LMVT) does not require f(a) = f(b). It is a more general theorem. The goal is to find a point 'c' where the tangent's slope is equal to the slope of the secant line connecting the endpoints. Therefore, you solve the equation f'(c) = (f(b) - f(a)) / (b - a).

Q: 3. Why is it crucial to check the conditions of continuity and differentiability before applying any Mean Value Theorem?

Checking the conditions of continuity on the closed interval [a, b] and differentiability on the open interval (a, b) is a mandatory first step. The Mean Value Theorems guarantee the existence of a point 'c' only if these foundational conditions are met. Skipping this check is a common error that can lead to an incorrect conclusion. If a function fails either of these conditions, the theorem is not applicable, and no such point 'c' is guaranteed to exist. For example, a function with a sharp corner (like |x| at x=0) is not differentiable at that point, and the theorem may not hold if that point is inside the interval.

Q: 4. How can I graphically interpret my solution for a Lagrange's Mean Value Theorem problem from Exercise 15.1?

Graphical interpretation is an excellent way to verify your answer. Once you have calculated the value of 'c', you can follow these steps:1. Sketch the graph of the function f(x) over the interval [a, b].2. Draw a straight line, called the secant line, connecting the two endpoints of the curve, (a, f(a)) and (b, f(b)).3. Locate the point (c, f(c)) on the curve using the value of 'c' you found by solving the LMVT equation.4. Draw the tangent line to the curve at this point.If your solution is correct, the tangent line at x=c will be parallel to the secant line you drew in step 2. This visually confirms that their slopes are equal.

Q: 5. What does it mean if I find more than one value for 'c' when solving a Mean Value Theorem problem?

Finding multiple values for 'c' is possible and can be a correct outcome. The Mean Value Theorems guarantee the existence of at least one point 'c' in the interval (a, b). It does not state that there is only one. For higher-degree polynomial functions or trigonometric functions, the curve may oscillate, creating several points where the tangent is parallel to the secant line. As long as each value of 'c' you find lies within the open interval (a, b), all of them are valid solutions to the problem.

Q: 6. How should I approach a problem from RD Sharma Ex 15.1 where a function appears non-differentiable at a point within the given interval?

If you encounter a function that is not differentiable at a specific point within the open interval (a, b), the Mean Value Theorem cannot be applied. The correct approach is to first identify the point of non-differentiability. For instance, for the function f(x) = |x - 2| on the interval [1, 3], the function is not differentiable at x=2. Since 2 is within the interval (1, 3), the condition of differentiability is violated. Your final answer should state that the theorem is not applicable because the function fails to satisfy all the required conditions on the given interval.

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An overview of RD Sharma Class 12 Solutions Chapter 15 - Mean Value Theorems (Ex 15.1) Exercise 15.1

Free PDF download of RD Sharma Class 12 Solutions Chapter 15 - Mean Value Theorems Exercise 15.1 solved by Expert Mathematics Teachers on Vedantu. All Chapter 15 - Mean Value Theorems Ex 15.1 Questions with Solutions for RD Sharma Class 12 Mathematics to help you to revise complete Syllabus and Score More marks. Register for online coaching for IIT JEE (Mains & Advanced) and other Engineering entrance exams.

Overview of RD Sharma Class 12 Solutions Chapter 15

Let us first discuss what is the mean value theorem? It defines that for any given curve in between two ending points, there should be a point at which the slope of the tangent to the curve is almost same to the slope of the secant through its ending points.

If f(x) is a function, so that f(x) is continuous on the closed interval p,q and also differentiable on the open interval (p, q), then there is point r in (p, q) that is, p < r < q in such a way that f’(r) = f(q) -f(p)/ q-p

Lagrange’s Mean Value Theorem or first mean value theorem is a synonym for the mean value theorem. This article discuss about Mean Value Theorem for Integrals, Mean ValueTheorem for Integrals problems and Cauchy Mean Value Theorem

Geometrical Representation of Mean Value Theorem

The mean value theorem graph shows the graph of the function f(x).

Let us look at the point A as (a,f(a)) and point B as = (b,f(b))]

The point C in the graph from where the tangent passes through the curve is (cf(c)).

The slope of the tangent line is almost the same as the secant line i.e. both the tangent line and the secant line are parallel to each other.

FAQs on RD Sharma Class 12 Solutions Chapter 15 - Mean Value Theorems (Ex 15.1) Exercise 15.1

1. What are the essential steps to verify Rolle's Theorem for a function in RD Sharma Class 12 Exercise 15.1?

To correctly solve a problem verifying Rolle's Theorem from this exercise, you must follow these sequential steps:

Step 1: Check Continuity. Verify that the function f(x) is continuous on the closed interval [a, b]. For polynomial, sine, or cosine functions, this is generally true.
Step 2: Check Differentiability. Verify that the function f(x) is differentiable on the open interval (a, b). This involves ensuring the derivative f'(x) is defined for all points within the interval.
Step 3: Check Endpoint Values. Calculate and confirm that f(a) is equal to f(b).
Step 4: Find the Derivative. If the above conditions are met, find the derivative f'(x).
Step 5: Solve for 'c'. Set the derivative f'(c) = 0 and solve for the value of 'c'.
Step 6: Final Verification. Ensure that the calculated value of 'c' lies strictly between 'a' and 'b', i.e., c ∈ (a, b).

2. How is Lagrange's Mean Value Theorem (LMVT) applied differently from Rolle's Theorem when solving problems in Chapter 15?

The primary difference lies in the third condition and the final equation. While both theorems require continuity on [a, b] and differentiability on (a, b), the subsequent steps diverge:

Rolle's Theorem has an additional condition: you must verify that f(a) = f(b). The goal is to find a point 'c' where the tangent is horizontal, meaning you solve for f'(c) = 0.
Lagrange's Mean Value Theorem (LMVT) does not require f(a) = f(b). It is a more general theorem. The goal is to find a point 'c' where the tangent's slope is equal to the slope of the secant line connecting the endpoints. Therefore, you solve the equation f'(c) = (f(b) - f(a)) / (b - a).

3. Why is it crucial to check the conditions of continuity and differentiability before applying any Mean Value Theorem?

Checking the conditions of continuity on the closed interval [a, b] and differentiability on the open interval (a, b) is a mandatory first step. The Mean Value Theorems guarantee the existence of a point 'c' only if these foundational conditions are met. Skipping this check is a common error that can lead to an incorrect conclusion. If a function fails either of these conditions, the theorem is not applicable, and no such point 'c' is guaranteed to exist. For example, a function with a sharp corner (like |x| at x=0) is not differentiable at that point, and the theorem may not hold if that point is inside the interval.

4. How can I graphically interpret my solution for a Lagrange's Mean Value Theorem problem from Exercise 15.1?

Graphical interpretation is an excellent way to verify your answer. Once you have calculated the value of 'c', you can follow these steps:
1. Sketch the graph of the function f(x) over the interval [a, b].
2. Draw a straight line, called the secant line, connecting the two endpoints of the curve, (a, f(a)) and (b, f(b)).
3. Locate the point (c, f(c)) on the curve using the value of 'c' you found by solving the LMVT equation.
4. Draw the tangent line to the curve at this point.
If your solution is correct, the tangent line at x=c will be parallel to the secant line you drew in step 2. This visually confirms that their slopes are equal.

5. What does it mean if I find more than one value for 'c' when solving a Mean Value Theorem problem?

Finding multiple values for 'c' is possible and can be a correct outcome. The Mean Value Theorems guarantee the existence of at least one point 'c' in the interval (a, b). It does not state that there is only one. For higher-degree polynomial functions or trigonometric functions, the curve may oscillate, creating several points where the tangent is parallel to the secant line. As long as each value of 'c' you find lies within the open interval (a, b), all of them are valid solutions to the problem.

6. How should I approach a problem from RD Sharma Ex 15.1 where a function appears non-differentiable at a point within the given interval?

If you encounter a function that is not differentiable at a specific point within the open interval (a, b), the Mean Value Theorem cannot be applied. The correct approach is to first identify the point of non-differentiability. For instance, for the function f(x) = |x - 2| on the interval [1, 3], the function is not differentiable at x=2. Since 2 is within the interval (1, 3), the condition of differentiability is violated. Your final answer should state that the theorem is not applicable because the function fails to satisfy all the required conditions on the given interval.