Question Number 33944 by rahul 19 last updated on 28/Apr/18
![If the equation (p^2 −4)(p^2 −9)x^3 +[((p−2)/2)]x^2 +(p−4)(p−3)(p−2)x+{2p−1}=0. is satisfied by all values of x in (0,3] then sum of all possible integral values of ′p′ is ? {.} = fractional part function. [.]= greatest integer function.](https://www.tinkutara.com/question/Q33944.png)
$$\boldsymbol{\mathrm{I}}\mathrm{f}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\left(\mathrm{p}^{\mathrm{2}} −\mathrm{4}\right)\left(\mathrm{p}^{\mathrm{2}} −\mathrm{9}\right){x}^{\mathrm{3}} +\left[\frac{\mathrm{p}−\mathrm{2}}{\mathrm{2}}\right]{x}^{\mathrm{2}} +\left(\mathrm{p}−\mathrm{4}\right)\left(\mathrm{p}−\mathrm{3}\right)\left(\mathrm{p}−\mathrm{2}\right){x}+\left\{\mathrm{2p}−\mathrm{1}\right\}=\mathrm{0}. \\ $$$$\mathrm{is}\:\mathrm{satisfied}\:\mathrm{by}\:\mathrm{all}\:\mathrm{values}\:\mathrm{of}\:{x}\:\mathrm{in}\:\left(\mathrm{0},\mathrm{3}\right]\:{then} \\ $$$${sum}\:{of}\:{all}\:{possible}\:{integral}\:{values}\:{of} \\ $$$$'{p}'\:{is}\:? \\ $$$$\left\{.\right\}\:=\:{fractional}\:{part}\:{function}. \\ $$$$\left[.\right]=\:{greatest}\:{integer}\:{function}. \\ $$
Answered by MJS last updated on 28/Apr/18
![p∈Z ⇒ {2p−1}=0 now it′s easy (p^2 −4)(p^2 −9)=0 ⇒ p∈{−3; −2; 2; 3} [((p−2)/2)]=0 ⇒ [(p/2)]=1 ⇒ p∈{2; 3} (p−4)(p−3)(p−2)=0 ⇒ {2; 3; 4} p∈{2; 3} sum({2; 3})=5](https://www.tinkutara.com/question/Q33952.png)
$${p}\in\mathbb{Z}\:\Rightarrow\:\left\{\mathrm{2}{p}−\mathrm{1}\right\}=\mathrm{0} \\ $$$$\mathrm{now}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy} \\ $$$$\left({p}^{\mathrm{2}} −\mathrm{4}\right)\left({p}^{\mathrm{2}} −\mathrm{9}\right)=\mathrm{0}\:\Rightarrow\:{p}\in\left\{−\mathrm{3};\:−\mathrm{2};\:\mathrm{2};\:\mathrm{3}\right\} \\ $$$$\left[\frac{{p}−\mathrm{2}}{\mathrm{2}}\right]=\mathrm{0}\:\Rightarrow\:\left[\frac{{p}}{\mathrm{2}}\right]=\mathrm{1}\:\Rightarrow\:{p}\in\left\{\mathrm{2};\:\mathrm{3}\right\} \\ $$$$\left(\mathrm{p}−\mathrm{4}\right)\left(\mathrm{p}−\mathrm{3}\right)\left(\mathrm{p}−\mathrm{2}\right)=\mathrm{0}\:\Rightarrow\:\left\{\mathrm{2};\:\mathrm{3};\:\mathrm{4}\right\} \\ $$$${p}\in\left\{\mathrm{2};\:\mathrm{3}\right\} \\ $$$${sum}\left(\left\{\mathrm{2};\:\mathrm{3}\right\}\right)=\mathrm{5} \\ $$
Commented by rahul 19 last updated on 28/Apr/18
$$\mathscr{T}{hank}\:{you}\:{sir}\:! \\ $$