Search any question & find its solution
Question:
Answered & Verified by Expert
Let $n>2$ be an integer and define a polynomial $p(x)=x^{n}+a_{n-1} x^{n-1}+\ldots \ldots+a_{1} x+a_{0}$
where $\mathrm{a}_{0}, \mathrm{a}_{1}, \ldots \ldots \ldots \mathrm{a}_{\mathrm{n}-1}$ are integers. Suppose we know that $\mathrm{np}(\mathrm{x})=(1+\mathrm{x}) \mathrm{p}^{\prime}(\mathrm{x})$. If $\mathrm{b}=\mathrm{p}(1)$, then
Options:
where $\mathrm{a}_{0}, \mathrm{a}_{1}, \ldots \ldots \ldots \mathrm{a}_{\mathrm{n}-1}$ are integers. Suppose we know that $\mathrm{np}(\mathrm{x})=(1+\mathrm{x}) \mathrm{p}^{\prime}(\mathrm{x})$. If $\mathrm{b}=\mathrm{p}(1)$, then
Solution:
1118 Upvotes
Verified Answer
The correct answer is:
b is a power of 2
$\begin{array}{l}
\mathrm{n}\left[\mathrm{x}^{\mathrm{n}}+\mathrm{a}_{\mathrm{n}-1} \mathrm{x}^{\mathrm{n}-1}+\mathrm{a}_{\mathrm{n}-2} \mathrm{x}^{\mathrm{n}-2}-\mathrm{a}_{1} \mathrm{x}+\mathrm{a}_{0}\right] \\
=(1+\mathrm{x})\left(\mathrm{n} \mathrm{x}^{\mathrm{n}-1}+\mathrm{a}_{\mathrm{n}-1}(\mathrm{n}-1) \mathrm{x}^{\mathrm{n}-2}\right. \\
+\mathrm{a}_{\mathrm{n}-2} \mathrm{x}^{\mathrm{n}-3}(\mathrm{n}-2) \\
\left.\quad+\mathrm{a}_{n-3}(\mathrm{n}-3) \mathrm{x}^{\mathrm{n}-4}+\ldots \ldots\right)
\end{array}$
compare coefficient of $\mathrm{x}^{\mathrm{n}-1}$
$\mathrm{na}_{\mathrm{n}-1}=(\mathrm{n}-1) \mathrm{a}_{\mathrm{n}-1}+\mathrm{n}$
Solve $a_{n-1}=n$ or ${ }^{n} C_{1}$
compare coefficient of $\mathrm{x}^{\mathrm{n}-2}$
$\begin{aligned}
\mathrm{na}_{\mathrm{n}-2} &=(\mathrm{n}-2) \mathrm{a}_{\mathrm{n}-2}+(\mathrm{n}-1) \mathrm{a}_{\mathrm{n}-1} \\
\mathrm{a}_{\mathrm{n}-2} &=\frac{\mathrm{n}(\mathrm{n}-1)}{2}={ }^{\mathrm{n}} \mathrm{C}_{2}
\end{aligned}$
similarly $a_{n-3}={ }^{n} C_{3} \&$ So $\ldots .$ on
$\begin{aligned}
b=P(1) &=1+a_{n-1}+a_{n-2}+\ldots \ldots a_{1}+a_{0} \\
&={ }^{n} C_{0+}{ }^{n} C_{1}+{ }^{n} C_{2}+\ldots \ldots .{ }^{n} C_{n}=2^{n}
\end{aligned}$
\mathrm{n}\left[\mathrm{x}^{\mathrm{n}}+\mathrm{a}_{\mathrm{n}-1} \mathrm{x}^{\mathrm{n}-1}+\mathrm{a}_{\mathrm{n}-2} \mathrm{x}^{\mathrm{n}-2}-\mathrm{a}_{1} \mathrm{x}+\mathrm{a}_{0}\right] \\
=(1+\mathrm{x})\left(\mathrm{n} \mathrm{x}^{\mathrm{n}-1}+\mathrm{a}_{\mathrm{n}-1}(\mathrm{n}-1) \mathrm{x}^{\mathrm{n}-2}\right. \\
+\mathrm{a}_{\mathrm{n}-2} \mathrm{x}^{\mathrm{n}-3}(\mathrm{n}-2) \\
\left.\quad+\mathrm{a}_{n-3}(\mathrm{n}-3) \mathrm{x}^{\mathrm{n}-4}+\ldots \ldots\right)
\end{array}$
compare coefficient of $\mathrm{x}^{\mathrm{n}-1}$
$\mathrm{na}_{\mathrm{n}-1}=(\mathrm{n}-1) \mathrm{a}_{\mathrm{n}-1}+\mathrm{n}$
Solve $a_{n-1}=n$ or ${ }^{n} C_{1}$
compare coefficient of $\mathrm{x}^{\mathrm{n}-2}$
$\begin{aligned}
\mathrm{na}_{\mathrm{n}-2} &=(\mathrm{n}-2) \mathrm{a}_{\mathrm{n}-2}+(\mathrm{n}-1) \mathrm{a}_{\mathrm{n}-1} \\
\mathrm{a}_{\mathrm{n}-2} &=\frac{\mathrm{n}(\mathrm{n}-1)}{2}={ }^{\mathrm{n}} \mathrm{C}_{2}
\end{aligned}$
similarly $a_{n-3}={ }^{n} C_{3} \&$ So $\ldots .$ on
$\begin{aligned}
b=P(1) &=1+a_{n-1}+a_{n-2}+\ldots \ldots a_{1}+a_{0} \\
&={ }^{n} C_{0+}{ }^{n} C_{1}+{ }^{n} C_{2}+\ldots \ldots .{ }^{n} C_{n}=2^{n}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.