Question Number 167196 by mnjuly1970 last updated on 09/Mar/22
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:#\:{Question}\:# \\ $$$$\:\:\:\:\:{If}\:,\:{a}\:\notin\:\mathbb{Z}\:\:{and}\:\:{the}\:{function}\:{with}\:{the}\:\:{following}\:\:{rule}\:\: \\ $$$$\:\:\:\:\:\:{is}\:\:{differentiable}\:{at}\:\:\:\:''\:\:{x}\:=\:\mathrm{1}\:''\:\:{then}\:\:{find}\:\:{the}\:{value}\:{of} \\ $$$$\:\:\:\:\:\:\:``\:\:\:{a}\:\:``\:\:\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)\:=\:\left(\lfloor\:\frac{{x}}{\mathrm{2}}\:\rfloor\:+\:\lfloor\:\frac{−{x}}{\mathrm{2}}\:\rfloor\:\right)\mid\:{x}^{\:\mathrm{2}} +{x}\:−\mathrm{2}\mid\:\lfloor{ax}\:\rfloor\:\:\:\:\:\:\:\:\:\:\:\blacksquare\:{m}.{n} \\ $$$$ \\ $$
Commented by MJS_new last updated on 09/Mar/22
$$\mathrm{I}\:\mathrm{think}\:\lfloor{a}\rfloor=\mathrm{0}\:\Rightarrow\:\mathrm{0}<{a}<\mathrm{1} \\ $$
Commented by mnjuly1970 last updated on 09/Mar/22
$${yes}\:{sir}\:\:\:\:{just}\:\:\:\:\:\:\mathrm{0}<{a}<\mathrm{1} \\ $$
Commented by MJS_new last updated on 09/Mar/22
$$\mathrm{yes}\:\mathrm{of}\:\mathrm{course}\:\mathrm{because}\:{a}\notin\mathbb{Z} \\ $$
Commented by mnjuly1970 last updated on 09/Mar/22
$${thanks}\:{alot}\:... \\ $$