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f-x-x-1-3-2f-x-1-f-10-

Question Number 48208 by naka3546 last updated on 20/Nov/18 $${f}\left({x}\right)\:+\:\left({x}\:+\:\mathrm{1}\right)^{\mathrm{3}} \:\:=\:\:\mathrm{2}{f}\left({x}\:+\:\mathrm{1}\right) \\ $$$${f}\left(\mathrm{10}\right)\:\:=\:\:? \\ $$ Answered by MJS last updated on 20/Nov/18 $${f}\left({x}\right)={ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}…

Question-113714

Question Number 113714 by mathdave last updated on 14/Sep/20 Answered by Olaf last updated on 15/Sep/20 $$\mathrm{I}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}} \mathrm{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$\mathrm{I}_{\mathrm{0}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}}…

tanx-dx-

Question Number 113706 by Rasikh last updated on 15/Sep/20 $$\:\:\:\int\sqrt{\mathrm{tanx}}\:\mathrm{dx}\:=?\:\:\:\: \\ $$ Answered by MJS_new last updated on 15/Sep/20 $$\int\sqrt{\mathrm{tan}\:{x}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{tan}\:{x}}\:\rightarrow\:{dx}=\mathrm{2cos}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{tan}\:{x}}\:{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}}…

A-nice-question-lt-3-If-a-quadratic-equation-1-q-p-2-2-x-2-p-1-q-x-q-q-1-p-2-2-0-has-equal-roots-prove-that-p-2-4q-

Question Number 113684 by ZiYangLee last updated on 14/Sep/20 $$\mathrm{A}\:\mathrm{nice}\:\mathrm{question}\:<\mathrm{3} \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{a}\:\mathrm{quadratic}\:\mathrm{equation}\: \\ $$$$\left(\mathrm{1}−{q}+\frac{{p}^{\mathrm{2}} }{\mathrm{2}}\right){x}^{\mathrm{2}} +{p}\left(\mathrm{1}+{q}\right){x}+{q}\left({q}−\mathrm{1}\right)+\frac{{p}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{equal}\:\mathrm{roots},\:\mathrm{prove}\:\mathrm{that}\:{p}^{\mathrm{2}} =\mathrm{4}{q} \\ $$ Commented…

In-ABC-BC-5cm-AC-4cm-cos-A-B-31-32-Find-the-area-of-ABC-

Question Number 113667 by ZiYangLee last updated on 14/Sep/20 $$\mathrm{In}\:\bigtriangleup\mathrm{ABC},\:\mathrm{BC}=\mathrm{5cm}\:\mathrm{AC}=\mathrm{4cm} \\ $$$$\mathrm{cos}\left(\mathrm{A}−\mathrm{B}\right)=\frac{\mathrm{31}}{\mathrm{32}}\:\: \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\bigtriangleup\mathrm{ABC}. \\ $$ Answered by som(math1967) last updated on 14/Sep/20 $$\mathrm{tan}^{\mathrm{2}} \frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}=\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{A}−\mathrm{B}\right)}{\mathrm{1}+\mathrm{cos}\left(\mathrm{A}−\mathrm{B}\right)}…