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Mean Value Theorems — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMean Value Theorems5 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. $f(x) = x^3 + 4x - 4x^2 - x^2 - 4 + 4x$
i.e. $f(x) = x^3- 5x^2+ 8x - 4$
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,
$f(x) = x^{3 }+ - 5x^2-^{ }8x - 4$
$\Rightarrow f'(x) = 3x^2 + 8 - 10x$
$\therefore f'(x) = 0 $
$\Rightarrow 3x^{2 }- 10x + 8 = 0$
$\Rightarrow 3x^{2 }- 6x - 4x + 8 = 0$
$\Rightarrow 3x(x - 2) - 4(x - 2) = 0$
$\Rightarrow (x - 2)(3x - 4)$
$\Rightarrow\text{x}=2,\frac{4}{3}$
Thus, $\text{c}\in(1,2)$ such that $f'(c) = 0.$
Hence, Rolle's theorem is verified.

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