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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 of the following function on the indicated interval
$\text{f}(\text{x})=\sin^4\text{x}+\cos^4\text{x}\text{ on }\Big[0,\frac{\pi}{2}\Big]$
✓

Answer

The given function is $\text{f}(\text{x})=\sin^4\text{x}+\cos^4\text{x}$
Since $\sin\text{x}$ and $\cos\text{x}$ are everywhere continuous and differentiable $\text{f}(\text{x})=\sin^4\text{x}+\cos^4\text{x}$ is continuous on $\Big[0,\frac{\pi}{2}\Big]$ and differentiable on $\Big(0,\frac{\pi}{2}\Big).$
Also,
$\text{f}\Big(\frac{\pi}{2}\Big)=\text{f}(0)=1$
Thus, f(x) satisfies all the conditionss of Rolle's theorem.
Now, we have to show that there exists $\text{c}\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c) = 0.
We have
$\text{f}(\text{x})=\sin^4\text{x}+\cos^4\text{x}$
$\Rightarrow\text{f}'(\text{x})=4\sin^3\text{x}\cos\text{x}-4\cos^3\text{x}\sin\text{x}$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow4\sin^3\text{x}\cos\text{x}-4\cos^3\text{x}\sin\text{x}=0$
$\Rightarrow\sin^3\text{x}\cos\text{x}-\cos^3\text{x}\sin\text{x}=0$
$\Rightarrow\tan^3\text{x}-\tan\text{x}=0$
$\Rightarrow\tan\text{x}(\tan^2\text{x}-1)=0$
$\Rightarrow\tan\text{x}=0,\tan^2\text{x}=1$
$\Rightarrow\tan\text{x}=0,\tan\text{x}=\pm1$
$\Rightarrow\text{x}=0,\text{x}=\frac{\pi}{4},\frac{3\pi}{4}$
Since $\text{c}=\frac{\pi}{4}\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c)=0
Hence, Rolle's theorem is verified.

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