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LIMITS AND DERIVATIVES — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLIMITS AND DERIVATIVES4 Marks
Question
For the function$\text{f}(\text{x})=\frac{\text{x}^{100}}{100}+\frac{\text{x}^{99}}{99}+...+\frac{\text{x}^2}{2}+\text{x}+1.$
Prove that $\text{f}'(1)=100$$\text{f}'(0)$.
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Answer

The given function is $\text{f}(\text{x})=\frac{\text{x}^{100}}{100}+\frac{\text{x}^{99}}{99}+...+\frac{\text{x}^2}{2}+\text{x}+1$ $\frac{\text{d}}{\text{dx}}\text{f}(\text{x})=\frac{\text{d}}{\text{dx}}\bigg[\frac{\text{x}^{100}}{100}+\frac{\text{x}^{99}}{99}+...+\frac{\text{x}^2}{2}+\text{x}+1\bigg]$ $\frac{\text{d}}{\text{dx}}\text{f}(\text{x})=\frac{\text{d}}{\text{dx}}\bigg(\frac{\text{x}^{100}}{100}\bigg)+\frac{\text{d}}{\text{dx}}\bigg(\frac{\text{x}^{99}}{99}\bigg)+...+\frac{\text{d}}{\text{dx}}\bigg(\frac{\text{x}^2}{2}\bigg)+\frac{\text{d}}{\text{dx}}\bigg(\text{x}\bigg)+\frac{\text{d}}{\text{dx}}\bigg(1\bigg)$ On using theorem $\frac{\text{d}}{\text{dx}}(\text{x})^\text{n}=\text{nx}^{\text{n}-1}$, we obtain $\frac{\text{d}}{\text{dx}}\text{f}(\text{x})=\frac{100\text{x}^{99}}{100}+\frac{99\text{x}^{98}}{99}+...+\frac{2\text{x}}{2}+1+0$ $=\text{x}^{99}+\text{x}^{98}+...+\text{x}+1$ $\therefore\text{f}'(\text{x})=\text{x}^{99}+\text{x}^{98}+...+\text{x}+1$ At $\text{x}=0$, $\text{f}'(0)=1$ At $\text{x}=1$,$\text{f}'(1)=1^{99}+1^{98}+...+1+1=[1+1+...+1+1]_{100\text{terms}}=1\times100=100$ Thus,$\text{f}'(1)=100\times\text{f}'(0)$

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