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Question-120008




Question Number 120008 by help last updated on 28/Oct/20
Commented by TANMAY PANACEA last updated on 28/Oct/20
is the first[brackdt term contains  x^2
$${is}\:{the}\:{first}\left[{brackdt}\:{term}\:{contains}\:\:\boldsymbol{{x}}^{\mathrm{2}} \right. \\ $$$$ \\ $$
Commented by help last updated on 28/Oct/20
yes...thinking option B should be right
$${yes}…{thinking}\:{option}\:{B}\:{should}\:{be}\:{right} \\ $$
Answered by TANMAY PANACEA last updated on 28/Oct/20
y=(x^2 +p)(x+q)  y=0  either (x^2 +p)=0  if only if p is negdtive  or (x+q)=0  when q is negetive  so option B and D seem correct
$${y}=\left({x}^{\mathrm{2}} +{p}\right)\left({x}+{q}\right) \\ $$$${y}=\mathrm{0}\:\:\boldsymbol{{either}}\:\left({x}^{\mathrm{2}} +{p}\right)=\mathrm{0}\:\:{if}\:{only}\:{if}\:{p}\:{is}\:{negdtive} \\ $$$$\boldsymbol{{or}}\:\left(\boldsymbol{{x}}+\boldsymbol{{q}}\right)=\mathrm{0}\:\:\boldsymbol{{when}}\:\boldsymbol{{q}}\:\boldsymbol{{is}}\:\boldsymbol{{negetive}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{option}}\:\boldsymbol{{B}}\:{and}\:{D}\:{seem}\:{correct} \\ $$
Commented by help last updated on 28/Oct/20
only one option must be picked
$${only}\:{one}\:{option}\:{must}\:{be}\:{picked} \\ $$
Answered by Don08q last updated on 28/Oct/20
B   B is the option where the equation will   give exactly one real root when y=0   thus intersecting the x−axis at only   one point and also has a positive    coefficient for x^3 , thus, from the left,   giving a maximum point first before   minimum.
$$\boldsymbol{\mathrm{B}} \\ $$$$\:\mathrm{B}\:\mathrm{is}\:\mathrm{the}\:\mathrm{option}\:\mathrm{where}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{will} \\ $$$$\:\mathrm{give}\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{real}\:\mathrm{root}\:\mathrm{when}\:{y}=\mathrm{0} \\ $$$$\:\mathrm{thus}\:\mathrm{intersecting}\:\mathrm{the}\:{x}−{axis}\:\mathrm{at}\:\mathrm{only} \\ $$$$\:\mathrm{one}\:\mathrm{point}\:\mathrm{and}\:\mathrm{also}\:\mathrm{has}\:\mathrm{a}\:{positive}\: \\ $$$$\:\mathrm{coefficient}\:\mathrm{for}\:{x}^{\mathrm{3}} ,\:\mathrm{thus},\:\mathrm{from}\:\mathrm{the}\:\mathrm{left}, \\ $$$$\:\mathrm{giving}\:\mathrm{a}\:\mathrm{maximum}\:\mathrm{point}\:\mathrm{first}\:\mathrm{before} \\ $$$$\:\mathrm{minimum}. \\ $$

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