Mathematics · Application of Derivatives

JEE Advanced 2020 — Paper 1 — Question 44

For a polynomial g(x)g(x) with real coefficients, let mg\mathrm{m}_{\mathrm{g}} denote the number of distinct real roots of g(x)\mathrm{g}(\mathrm{x}). Suppose S in the set of polynomials with real coefficients defined by S={(x2−1)2(a0+a1x+a2x2+a3x3):a0,a1,a2,a3∈R}S=\left\{\left(x^{2}-1\right)^{2}\left(a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}\right): a_{0}, a_{1}, a_{2}, a_{3} \in R\right\} For a polynomial ff, let f′f^{\prime} and f′′f^{\prime \prime} denote its first and second order derivatives, respectively. Then the minimum possible value of (mf′+mf′′)\left(\mathrm{m}_{\mathrm{f}^{\prime}}+\mathrm{m}_{\mathrm{f}^{\prime \prime}}\right), where f∈S\mathrm{f} \in \mathrm{S}, is ____\_\_\_\_

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

f=(x2−1)2(a0+a1x+a2x2+a3x3)f=\left(x^{2}-1\right)^{2}\left(a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}\right)

=(x+1)2(x−1)2(a0+a1x+a2x2+a3x3)=(x+1)^{2}(x-1)^{2}\left(a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}\right)

For min value of mf+mf′,a0+a1x+a2x2+a3x3m_{f}+m_{f^{\prime}}, a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3} must have only 1 real root which is 1 or -1 , then

mf=3m_{f}=3 & mf′=2⇒\mathrm{m}_{\mathrm{f}^{\prime}}=2 \Rightarrow Answer =5=5

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 1
Subject
Mathematics
Chapter
Application of Derivatives
Topic
Maxima and Minima