Q82 / 125 · BITSAT 2012 · Mathematics
For the function
$f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots \frac{x^{2}}{2}+x+1$
$\mathrm{f}^{\prime}(1)=\mathrm{mf}^{\prime}(0),$ where $\mathrm{m}$ is equal to
- A50
- B0
- Correct answerC100
- D200
Solution
Given, $f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^{2}}{2}+x+1$
$\Rightarrow \quad f^{\prime}(x)=\frac{100 x^{99}}{100}+\frac{99 x^{98}}{99}+\ldots+\frac{2 x}{2}+1+0$
$\left[\because \mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{n}} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{nx}^{\mathrm{n}-1}\right]$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{99}+\mathrm{x}^{98}+\ldots+\mathrm{x}+1$ ...(i)
Putting $\mathrm{x}=1$, we get $f^{\prime}(1)=\underbrace{(1)^{99}+1^{98}+\ldots+1+1}_{100 \text { times }}=\underbrace{1+1+1 \ldots+1+1}_{100 \text { times }}$
$\Rightarrow \quad \mathrm{f}^{\prime}(1)=100$ ...(ii)
Again, putting $\mathrm{x}=0$, we get $\mathrm{f}^{\prime}(0)=0+0+\ldots+0+1 \quad \Rightarrow \mathrm{f}^{\prime}(0)=1$ ...(iii)
From eqs. (ii) and(iii), we get; $\mathrm{f}^{\prime}(1)=100 \mathrm{f}^{\prime}(0)$ Hence, $m=100$
Hence, the correct answer is option: 'C'.