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JEE Advanced 2020 · Paper 1, Mathematics question 17

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

S={(x2−1)2(a0+a1x+a2x2+a3x3):a0,a1,a2,a3∈R}.S = \left\{(x^2-1)^2\left(a_0 + a_1x + a_2x^2 + a_3x^3\right) : a_0, a_1, a_2, a_3 \in \mathbb R\right\}.

For a polynomial ff, let f′f' and f′′f'' denote its first and second order derivatives, respectively. Then the minimum possible value of (mf′+mf′′)\left(m_{f'} + m_{f''}\right), where f∈Sf\in S, is ______

Answer and worked solution

Official answer: 5

Lower bound for f′f'. Every f∈Sf\in S has double roots at ±1\pm1, so f′(±1)=0f'(\pm1)=0. By Rolle's theorem f′f' also vanishes somewhere in (−1,1)(-1,1). So mf′≥3m_{f'}\ge3.

Lower bound for f′′f''. Rolle's theorem between consecutive roots of f′f' gives mf′′≥2m_{f''}\ge2.

Achieved. Take f=(x2−1)2f=(x^2-1)^2:

  • f′=4x(x2−1)f'=4x(x^2-1) has 3 roots;
  • f′′=12x2−4f''=12x^2-4 has 2 roots.

The minimum of mf′+mf′′m_{f'}+m_{f''} is 3+2=53+2=5.

Answer: 5

Topics:Rolle's and mean value theoremsTheory of equations (higher degree)
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