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Mean Value Theorems — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMean Value Theorems4 Marks
Question
Verify Rolle's theorem for the following function on the indicated intervals

f(x) = (x - 1) (x - 2)2 on [1, 2]

✓

Answer

Given:

f(x) = (x - 1)(x - 2)2

i.e. f(x) = x3 + 4x - 4x2 - x2 - 4 + 4x

i.e. f(x) = x3 - 5x2 + 8x - 4

We know that a polynominal function is everywhere derivable and hence continuous.

So, being a polynomial function, f(x) is continuous and derivable on [1, 2]

Also,

f(1) = (1)3 - 5(1)2 + 8(1) - 4 = 0

f(2) = (2)3 - 5(2)2 + 8(2) - 4 = 0

$\therefore$ f(1) = f(2) = 0

Thus, all the continuous of Rolle's theorem are satisfied.

Now, we have to show that there exists $\text{c}\in(1,2)$ such that f'(c) = 0.

We have,

f(x) = x3 + - 5x2 - 8x - 4

⇒ f'(x) = 3x2 + 8 - 10x

$\therefore$ f'(x) = 0 ⇒ 3x2 - 10x + 8 = 0

⇒ 3x2 - 6x - 4x + 8 = 0

⇒ 3x(x - 2) - 4(x - 2) = 0

⇒ (x - 2)(3x - 4)

$\Rightarrow\text{x}=2,\frac{4}{3}$

Thus, $\text{c}\in(1,2)$ such that f'(c) = 0.

Hence, Rolle's theorem is verified.

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