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If-1-ax-bx-2-1-2x-18-can-be-expanded-using-binomial-theorem-in-ascending-power-of-x-Determine-the-value-of-a-and-b-if-the-coefficient-of-x-3-and-x-4-are-both-zero-

Question Number 40523 by scientist last updated on 23/Jul/18 $${If}\:\left(\mathrm{1}+{ax}+{bx}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{2}{x}\right)^{\mathrm{18}} \:\:{can}\:{be}\:{expanded}\:{using} \\ $$$${binomial}\:{theorem}\:{in}\:{ascending}\:{power}\:{of}\:{x}.{Determine} \\ $$$${the}\:{value}\:{of}\:\:\:{a}\:{and}\:{b},{if}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{3}} \:\:{and}\:{x}^{\mathrm{4}\:} \:\:{are}\:{both}\:{zero}. \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated…

Given-that-ax-b-is-a-factor-of-x-2-2x-2-1-and-a-is-a-root-of-x-2-2x-1-0-show-that-the-value-of-b-is-1-or-3-2-2-

Question Number 40510 by KMA last updated on 23/Jul/18 $${Given}\:{that}\:{ax}+{b}\:{is}\:{a}\:{factor}\:{of}\:{x}^{\mathrm{2}} \\ $$$$+\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\:{and}\:−{a}\:{is}\:{a}\:{root}\:{of}\:{x}^{\mathrm{2}} +\mathrm{2}{x} \\ $$$$−\mathrm{1}=\mathrm{0}\:{show}\:{that}\:{the}\:{value}\:{of}\:{b}\:{is} \\ $$$$−\mathrm{1}\:{or}\:\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}\:.} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated…

if-f-o-g-x-x-and-f-x-1-f-x-2-then-g-2-

Question Number 106039 by 175mohamed last updated on 02/Aug/20 $${if}\:\:\left({f}\:{o}\:{g}\:\right)\left({x}\right)\:=\:{x}\:\:{and}\:{f}'\left({x}\right)=\mathrm{1}\:+\:\left({f}\left({x}\right)\right)^{\mathrm{2}} \\ $$$${then}\:{g}'\left(\mathrm{2}\right)\:=\:…. \\ $$$$ \\ $$ Answered by bemath last updated on 02/Aug/20 $$\frac{\mathrm{d}\left(\mathrm{f}\circ\mathrm{g}\right)\left(\mathrm{x}\right)}{\mathrm{dx}}\:=\:\mathrm{g}'\left(\mathrm{x}\right).\mathrm{f}\:'\left(\mathrm{g}\left(\mathrm{x}\right)\right)=\:\mathrm{1} \\…

Question-171518

Question Number 171518 by Mastermind last updated on 16/Jun/22 Commented by JDamian last updated on 16/Jun/22 $${the}\:{same}\:{ever}\:{tricky}\:{puzzles}\:{again}\:{and}\:{again}… \\ $$$${Both}\:{overall}\:{figures}\:{are}\:{not}\:{triangles} \\ $$$${because}\:{the}\:{red}\:{triangle}\:{and}\:{the}\:{dark} \\ $$$${green}\:{one}\:{are}\:\boldsymbol{{not}}\:{similar}. \\ $$$$…