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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Verify the Rolle’s theorem for each of the functions:
$\text{f(x)}=\text{x}(\text{x}+3)\text{e}^{-\frac{\text{x}}{2}}\text{ in }[-3,0].$
✓

Answer

We have, $\text{f(x)}=\text{x}(\text{x}+3)\text{e}^{-\frac{\text{x}}{2}}$
Since polynomial function x(x + 3) and exponential funxtion $\text{e}^{-\frac{\text{x}}{2}}$ are continuous and diffferentiable in R, given function f(x) is also continuous and differentiable in R
Also f(0) = f(-3) = 0
So, conditions of Rolle's theorem are satisfied.
Hence, there exists a real number $\text{c}\in(-3,0)$ such that f'(c) = 0
Now $\text{f(x)}=(\text{x}^2+3\text{x})\text{e}^{-\frac{\text{x}}{2}}$
$\therefore\ \text{f}'(\text{x})=(2\text{x}+3)\text{e}^{-\frac{\text{x}}{2}}-\frac{1}{2}\text{e}^{-\frac{\text{x}}{2}}(\text{x}^2+3\text{x})$
$=-\frac{1}{2}\text{e}^{-\frac{\text{x}}{2}}(\text{x}^2+3\text{x}-4\text{x}-6)=-\frac{1}{2}\text{e}^{-\frac{\text{x}}{2}}(\text{x}^2-\text{x}-6)$
So, f'(c) = 0
$\Rightarrow\ -\frac{1}{2}\text{e}^{-\frac{\text{x}}{2}}(\text{c}+2)(\text{c}-3)=0$
$\Rightarrow\ \text{c}=-2\in(-3,0)$
Therefore, Rolle's theorem has been verified.

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