JEE Advanced · Mathematics · 25. AOD
Let \(p(x)\) be a real polynomial of least degree which has a local maximum at \(x=1\) and a local minimum at \(x=3\). If \(p(1)=6\) and \(p(3)=2\), then \(p^{\prime}(0)\) is
- A 18
- B 9
- C 20
- D 24
Answer & Solution
Correct Answer
(B) 9
Step-by-step Solution
Detailed explanation
Since, \(p(x)\) has a local maximum at \(x=1\) and a local minimum at \(x=3\) and \(p(x)\) is a real polynomial of least degree.
Hence, let \(p^{\prime}(x)=k(x-1)(x-3)=k\left(x^{2}-4 x+3\right)\)
\(\Rightarrow p(x)=k\left(\frac{x^{3}}{3}-2 x^{2}+3 x\right)+c\)
Now, \(p(1)=6\) and \(p(3)=2\)
\(\begin{array}{l}
\Rightarrow \frac{4}{3} k+C=6 \text { and } 0+C=2 \Rightarrow k=3 \\
\therefore \quad p^{\prime}(x)=3(x-1)(x-3) \Rightarrow p^{\prime}(0)=9
\end{array}\)
Hence, let \(p^{\prime}(x)=k(x-1)(x-3)=k\left(x^{2}-4 x+3\right)\)
\(\Rightarrow p(x)=k\left(\frac{x^{3}}{3}-2 x^{2}+3 x\right)+c\)
Now, \(p(1)=6\) and \(p(3)=2\)
\(\begin{array}{l}
\Rightarrow \frac{4}{3} k+C=6 \text { and } 0+C=2 \Rightarrow k=3 \\
\therefore \quad p^{\prime}(x)=3(x-1)(x-3) \Rightarrow p^{\prime}(0)=9
\end{array}\)
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