JEE Advance - Mathematics (2020 - Paper 1 Offline - No. 17)

For a polynomial g(x) with real coefficients, let mg denote the number of distinct real roots of g(x). Suppose S is the set of polynomials with real coefficients defined by

$$S = \{ {({x^2} - 1)^2}({a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3}):{a_0},{a_1},{a_2},{a_3} \in R\} $$;

For a polynomial f, let f' and f'' denote its first and second order derivatives, respectively. Then the minimum possible value of (mf' + mf''), where f $$ \in $$ S, is ..............
Answer
5

Explanation

Given set S of polynomials with real coefficients

$$S = \{ {({x^2} - 1)^2}({a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3}):{a_0},{a_1},{a_2},{a_3} \in R\} $$

and for a polynomial $$f \in S$$, Let

$$f(x) = {({x^2} - 1)^2}({a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3})$$

it have $$-$$1 and 1 as repeated roots twice, so graph of f(x) touches the X-axis at x = $$-$$1 and x = 1, so f'(x) having at least three roots x = $$-$$1, 1 and $$\alpha $$. Where $$\alpha $$$$ \in $$($$-$$1, 1) and f''(x) having at least two roots in interval ($$-$$1, 1)

So, mf' = 3 and mf'' = 2

$$ \therefore $$ Minimum possible value of (mf' + mf'') = 5

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