KCET · Maths · Binomial Theorem
If \(\mathrm{f}(\mathrm{x})=1+\mathrm{nx}+\frac{\mathrm{n}(\mathrm{n}-1)}{2} \mathrm{x}^{2}\) \(+\frac{n(n-1)(n-2)}{6} x^{3}+\ldots+x^{n}\)
then \(\mathrm{f}^{\prime \prime}(1)\) is equal to
- A \(\mathrm{n}(\mathrm{n}-1) 2^{\mathrm{n}-1}\)
- B \((n-1) 2^{n-1}\)
- C \(n(n-1) 2^{n-2}\)
- D \(n(n-1) 2^{n}\)
Answer & Solution
Correct Answer
(C) \(n(n-1) 2^{n-2}\)
Step-by-step Solution
Detailed explanation
Given, \(\mathrm{f}(\mathrm{x})=1+\mathrm{nx}+\frac{\mathrm{n}(\mathrm{n}-1)}{2 !} \mathrm{x}^{2}\)
\(+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{3 !} \mathrm{x}^{3}+\ldots+\mathrm{x}^{\mathrm{n}}\)
\[
\begin{array}{ll}
\Rightarrow & \mathrm{f}(\mathrm{x})=(1+\mathrm{x})^{\mathrm{n}} \\
\Rightarrow & \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{n}(1+\mathrm{x})^{\mathrm{n}-1} \\
\Rightarrow & \mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{n}(\mathrm{n}-1)(1+\mathrm{x})^{\mathrm{n}-2} \\
\Rightarrow & \mathrm{f}^{\prime \prime}(1)=\mathrm{n}(\mathrm{n}-1) 2^{\mathrm{n}-2}
\end{array}
\]
\(+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{3 !} \mathrm{x}^{3}+\ldots+\mathrm{x}^{\mathrm{n}}\)
\[
\begin{array}{ll}
\Rightarrow & \mathrm{f}(\mathrm{x})=(1+\mathrm{x})^{\mathrm{n}} \\
\Rightarrow & \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{n}(1+\mathrm{x})^{\mathrm{n}-1} \\
\Rightarrow & \mathrm{f}^{\prime \prime}(\mathrm{x})=\mathrm{n}(\mathrm{n}-1)(1+\mathrm{x})^{\mathrm{n}-2} \\
\Rightarrow & \mathrm{f}^{\prime \prime}(1)=\mathrm{n}(\mathrm{n}-1) 2^{\mathrm{n}-2}
\end{array}
\]
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