Assertion :Statement 1: A polynomial of least degree that has a maximum equal to $6$ at $x = 1$ minimum equal to $2$ at $x = 3$ is $x^{3} - 6 x^{2} + 9 x + 2$ Reason: The polynomial is everywhere differentiable and the points of extremum can only be roots of derivative

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

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Assertion is correct but Reason is incorrect

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Both Assertion and Reason are incorrect

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Solution

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
Statement 2 is true clearly So applying this we have $P^{'} (x) = a (x - 1) (x - 3)$
Since at $x = 1,$ we must have $P (1) = 6$ , we have
$P (x) = \int_{1}^{x} P^{'} (x) d x + 6 = a \int_{1}^{x} (x^{2} - 4 x + 3) d x + 6$
$= a (\frac{x^{3}}{3} - 2 x^{2} + 3 x - \frac{4}{3}) + 6$
Also $P (3) = 2$ when $a = 3$
Hence $P (x) = x^{3} - 6 x^{2} + 9 x + 2$

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Assertion :Statement 1: A polynomial of least degree that has a maximum equal to 6 at x=1 minimum equal to 2 at x=3 is x3−6x2+9x+2 Reason: The polynomial is everywhere differentiable and the points of extremum can only be roots of derivative

Assertion :Statement 1: A polynomial of least degree that has a maximum equal to $6$ at $x = 1$ minimum equal to $2$ at $x = 3$ is $x^{3} - 6 x^{2} + 9 x + 2$ Reason: The polynomial is everywhere differentiable and the points of extremum can only be roots of derivative