Q38 / 90 · KVPY 2015 (SB/SX) · Mathematics
Let $\mathrm{p}(\mathrm{x})$ be a polynomial such that $p(x)-p^{\prime}(x)=x^{n}$, where $\mathrm{n}$ is a positive integer. Then $p(0)$ equals.
- Correct answerA$\mathrm n !$
- B$\mathrm(n-1) !$
- C$\frac{1}{n !}$
- D$\frac{1}{(n-1) !}$
Solution
Let $\mathrm{P}(\mathrm{x})=\mathrm{a}_{0} \mathrm{x}^{\mathrm{n}}+\mathrm{a}_{1} \mathrm{x}^{\mathrm{n}-1}+\mathrm{a}_{2} \mathrm{x}^{\mathrm{n}-2} \ldots \ldots . \mathrm{a}_{\mathrm{n}}=\sum_{\mathrm{r}=0}^{\mathrm{n}} \mathrm{a}_{\mathrm{r}} \mathrm{x}^{\mathrm{n}-\mathrm{r}}$
$$
\begin{array}{l}
P^{1}(x)=\sum_{r=0}^{n-1} a_{r}(n-r) x^{n-r-1} \\
P(x)-P^{1}(x)=a_{0} x^{n}+\sum_{r=1}^{n}\left\{a_{r}-a_{r+}(n-r+1)\right\} x^{n-r}=x^{n}
\end{array}
$$
So, $a_{0}=1$ and $a_{r}=a_{r-1}(n-r+1)$
$$
\frac{\mathrm{a}_{\mathrm{r}}}{\mathrm{a}_{\mathrm{r}-1}}=\mathrm{n}-\mathrm{r}+1
$$
Now, $P(0)=a_{n}=\frac{a_{n}}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \ldots \ldots \frac{a_{2}}{a_{1}} \frac{a_{1}}{a_{0}} \cdot a_{0}$ $=(1 \times 2 \ldots . ., \mathrm{n})=\mathrm{n} !$
Hence, the correct answer is option: 'A'.