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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\frac{\sin\text{x}}{\text{e}^{\text{x}}}\text{ on }0\leq\text{x}\leq\pi$
✓

Answer

The given function is $\text{f}(\text{x})=\frac{\sin\text{x}}{\text{e}^{\text{x}}}.$
Since $\cos\text{x}$ and $\text{e}^\text{x}$ are everywhere continuous and differentiable, being a quotient of these two, f(x) is continuous on $[0,\pi]$ and differentiable on $(0,\pi).$
Also,
$\text{f}(\pi)=\text{f}(0)=0$
Thus, f(x) satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists $\text{c}\in(0,\pi)$ such that f'(c) = 0.
We have
$\text{f}(\text{x})=\frac{\sin\text{x}}{\text{e}^{\text{x}}}$
$\Rightarrow \text{f}'(\text{x})=\frac{\cos\text{x}-\sin\text{x}}{\text{e}^{\text{x}}}$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow\frac{\cos\text{x}-\sin\text{x}}{\text{e}^{\text{x}}}=0$
$\Rightarrow\cos\text{x}-\sin\text{x}=0$
$\Rightarrow\tan\text{x}=1$
$\Rightarrow\text{x}=\frac{\pi}{4}$
Thus, $\text{c}=\frac{\pi}{4}\in(0,\pi)$ such that f'(c) = 0
Hence, Rolle's theorem is verified.

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