10.4 Mean value Theorems (VSAQs)

Maths-1B | 10.4 Mean value Theorems – VSAQs:
Welcome to VSAQs in Chapter 10: 10.4 Mean value Theorems. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

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VSAQ-1 : State Rolle’s Mean value theorem

If a function f(x) is

(i) continuous on [a,b]

(ii) derivable on (a,b)

(iii) f(a) = f(b) then there exists c ∈ (a,b) such that f'(c) = 0

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VSAQ-2 : State Lagrange’s Mean value theorem

If a function f(x) is

(i) Continuous on [a,b]

(ii) deriivable on (a,b) then there exists c ∈ (a,b) such that

$$[
f'(c) = \frac{f(b) – f(a)}{b – a}
]$$

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VSAQ-3 : Verify Rolle’s theorem for the function y = f(x) = x² + 4 on [-3,3]

$$f(x) = x^2 + 4$$

$$f'(x) = 2x$$

$$\text{(i) continuous on } [-3,3]$$

$$\text{(ii) differentiable in } (-3,3)$$

$$f(-3) = (-3)^2 + 4 = 9 + 4 = 13$$

$$f(3) = 3^2 + 4 = 9 + 4 = 13$$

$$f(-3) = f(3)$$

$$\text{So, from Rolle’s theorem } f'(c) = 0$$

$$2c = 0$$

$$c = 0$$

$$c = 0 \in (-3,3)$$

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VSAQ-4 : Verify Rolle’s theorem for the function x² – 5x + 6 on |-3,8|

$$f(x) = x^2 – 5x + 6$$

$$f'(x) = 2x – 5$$

$$\text{(i) continuous on } [-3,8]$$

$$\text{(ii) differentiable in } (-3,8)$$

$$f(-3) = (-3)^2 – 5(-3) + 6 = 9 + 15 + 6 = 30$$

$$f(8) = 8^2 – 5 \times 8 + 6 = 64 – 40 + 6 = 30$$

$$f(-3) = f(8)$$

$$\text{So, from Rolle’s theorem } f'(c) = 0$$

$$2c – 5 = 0$$

$$2c = 5$$

$$c = \frac{5}{2} = 2.5$$

$$c = 2.5 \in (-3,8)$$

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VSAQ-5 : Verify the conditions of Lagrange’s mean value theorem for the function x² – 1 on |2,3|

$$f(x) = x^2 – 1$$

$$f'(x) = 2x$$

$$\text{(i) continuous on } [2,3]$$

$$\text{(ii) differentiable in } (2,3)$$

$$\text{So, from Lagrange’s theorem }$$

$$f'(c) = \frac{f(3) – f(2)}{3 – 2}$$

$$f'(c) = \frac{(3^2 – 1) – (2^2 – 1)}{3 – 2} = \frac{(9 – 1) – (4 – 1)}{1} = \frac{8 – 3}{1} = 5$$

$$2c = 5$$

$$c = \frac{5}{2} = 2.5$$

$$c = 2.5 \in (2,3)$$

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VSAQ-6 : Verify Lagrange’s mean value theorem for the function f(x) = x² on [2,4]

$$f(x) = x^2$$

$$f'(x) = 2x$$

$$\text{(i) continuous on } [2,4]$$

$$\text{(ii) differentiable in } (2,4)$$

$$\text{So, from Lagrange’s theorem,}$$

$$f'(c) = \frac{f(4) – f(2)}{4 – 2}$$

$$f'(c) = \frac{4^2 – 2^2}{4 – 2} = \frac{16 – 4}{2} = \frac{12}{2} = 6$$

$$2c = 6$$

$$c = 3$$

$$c = 3 \in (2,4)$$