Question 46: Given the function \(f\left( x \right)\) whose upper derivative satisfies \(\left| {f\left( {x + h} \right) – f\left( {x – h} \right)} \right| \le {h^2}\) for all , h > 0. Set \(g\left( x \right) = {\left[ {x + f’\left( x \right)} \right]^{2019}} + {\left[ {x + f’\left( x \right)} \right]^{29 – m}} – \left( {{m^4} – 29{m^2} + 100} \right). {\sin ^2}x – 1\) with parameter m. Let S be the set of all integer values of m < 27 such that \(g\left( x \right)\) is minimized at x = 0. The sum of the squares of the elements of S is
For all and h > 0, we have
\(\left| {f\left( {x + h} \right) – f\left( {x – h} \right)} \right| \le {h^2} \Leftrightarrow \frac{{\left | {f\left( {x + h} \right) – f\left( x \right) + f\left( x \right) – f\left( {x – h} \right)} \right|} }{h} \le h\)
\(\left| {\frac{{f\left( {x + h} \right) – f\left( x \right)}}{h} + \frac{{f\left( {x – h} \right) – f\left( x \right)}}{{ – h}}} \right| \le h\).
Given \(h \to 0\), we get \(f’\left( x \right) = 0\). Then \(g\left( x \right) = {x^{2019}} + {x^{29 – m}} – \left( {{m^4} – 29{m^2} + 100} \right) {\sin ^2}x – 1\)
\(g’\left( x \right) = 2019{x^{2018}} + \left( {29 – m} \right){x^{28 – m}} – \left( {{m^4) } – 29{m^2} + 100} \right)\sin 2x \Rightarrow g’\left( 0 \right) = 0\)
Consider \(g”\left( x \right) = 2019.2018{x^{2017}} + \left( {29 – m} \right)\left( {28 – m} \right){x^{27 – m}} – 2\left( {{m^4} – 29{m^2} + 100} \right)\cos 2x\)
\( \Rightarrow g”\left( 0 \right) = – 2\left( {{m^4} – 29{m^2} + 100} \right)\)
*) If \(g”\left( 0 \right) = 0 \Leftrightarrow {m^4} – 29{m^2} + 100 = 0 \Leftrightarrow \left[\begin{array}{l}m=\pm5\\m=\pm2\end{array}\right\)[\begin{array}{l}m=\pm5\m=\pm2\end{array}\right\)
+ If m = 5 then \(g’\left( x \right) = 2019{x^{2018}} + 24{x^{23}} = {x^{23}}\left( {2019{x) ^{1995}} + 24} \right)\) change the sign from negative to positive when passing x = 0 so the requirement is satisfied.
+ If m = – 5 then \(g’\left( x \right) = 2019{x^{2018}} + 34{x^{33}} = {x^{33}}\left( {2019{ x^{1985}} + 34} \right)\) changes the sign from negative to positive when passing x = 0 so the requirement is satisfied.
+ If m = 2 then \(g’\left( x \right) = 2019{x^{2018}} + 27{x^{26}} = {x^{26}}\left( {2019{x) ^{1992}} + 27} \right)\) does not change sign when x = 0 so it is eliminated.
+ If m = – 2 then \(g’\left( x \right) = 2019{x^{2018}} + 31{x^{30}} = {x^{30}}\left( {2019{ x^{1988}} + 31} \right)\) does not change sign when x = 0 so it is eliminated.
*) If \(g”\left( x \right) \ne 0\) then \(g\left( x \right)\) is minimized at \(x = 0 \Leftrightarrow g”\left( 0 \ right) > 0 \Leftrightarrow – 2\left( {{m^4} – 29{m^2} + 100} \right) > 0\)
\(\Leftrightarrow {m^4} – 29{m^2} + 100 < 0 \Leftrightarrow 4 < {m^2} < 25\). Since m is integer, \(m \in \left\{ { – 4; – 3;3;4} \right\}\)
So \(S = \left\{ { – 5; – 4; – 3;3;4;5} \right\}\). So the sum of the squares of the elements of S is 100.
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