Question Number 39142 by rahul 19 last updated on 03/Jul/18
![F(x) = x^3 −9x^2 +24x+c=0 has three real and distinct roots α , β & γ . Q.1 → Possible value of c is : Q.2 → If [α]+[β]+[γ]= 8 then c is : Q.3 → If [α]+[β]+[γ]=7 then c is : Options for the above 3 Q. → a) (−20,−16) b) (−20,−18) c) (−18,−16) d) none of these. [.] = greatest integer function.](https://www.tinkutara.com/question/Q39142.png)
$$\mathrm{F}\left({x}\right)\:=\:{x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{24}{x}+{c}=\mathrm{0}\:{has}\:\mathrm{three} \\ $$$$\mathrm{real}\:\mathrm{and}\:\mathrm{distinct}\:\mathrm{roots}\:\alpha\:,\:\beta\:\&\:\gamma\:. \\ $$$$\mathrm{Q}.\mathrm{1}\:\rightarrow\:\mathrm{Possible}\:\mathrm{value}\:\mathrm{of}\:\mathrm{c}\:\mathrm{is}\:: \\ $$$$\mathrm{Q}.\mathrm{2}\:\rightarrow\:\mathrm{If}\:\left[\alpha\right]+\left[\beta\right]+\left[\gamma\right]=\:\mathrm{8}\:\mathrm{then}\:\mathrm{c}\:\mathrm{is}\:: \\ $$$$\mathrm{Q}.\mathrm{3}\:\rightarrow\:\mathrm{If}\:\left[\alpha\right]+\left[\beta\right]+\left[\gamma\right]=\mathrm{7}\:\mathrm{then}\:\mathrm{c}\:\mathrm{is}\:: \\ $$$$ \\ $$$$\mathrm{Options}\:\mathrm{for}\:\mathrm{the}\:\mathrm{above}\:\mathrm{3}\:\mathrm{Q}.\:\rightarrow \\ $$$$\left.\mathrm{a}\left.\right)\:\left(−\mathrm{20},−\mathrm{16}\right)\:\:\:\:\:\:\:\:\mathrm{b}\right)\:\left(−\mathrm{20},−\mathrm{18}\right) \\ $$$$\left.\mathrm{c}\left.\right)\:\left(−\mathrm{18},−\mathrm{16}\right)\:\:\:\:\:\:\:\:\:\mathrm{d}\right)\:\mathrm{none}\:\mathrm{of}\:\mathrm{these}. \\ $$$$ \\ $$$$\left[.\right]\:=\:\mathrm{greatest}\:\mathrm{integer}\:\mathrm{function}. \\ $$
Answered by MJS last updated on 03/Jul/18
![f(x)=x^3 −9x^2 +24x+c f′(x)=3x^2 −18x+24 x^2 −6x+8=0 x_1 =2; x_2 =4 f(2)=20+c [local max] f(4)=16+c [local min] ⇒ f(x) has 3 real and distinct roots with −20<c<−16 ⇒ if c∈Z: c∈{−19, −18, −17} for Q2 & Q3 we have to solve the three equations x^3 −9x^2 +24x−17=0 x^3 −9x^2 +24x−18=0 x^3 −9x^2 +24x−19=0 with x=z+3 we get z^3 −3z+1=0 z^3 −3z=0 z^3 +3z−1=0 the 2^(nd) one is easy to solve for the 1^(st) & 3^(rd) we use the trigonometric formula z=2(√(−(p/3)))sin((1/3)(arcsin(((9q)/(2p^2 ))(√(−(p/3))))+2kπ)) with k=0, 1, 2 z^3 −3z+1=0 z={−2cos (π/9); 2sin (π/(18)); 2cos ((2π)/9)} x={3−2cos (π/9); 3+2sin (π/(18)); 3+2cos ((2π)/9)} [x]={1; 3; 4} ⇒ sum([x])=8 z^3 −3z=0 z={−(√3); 0; (√3)} x={3−(√3); 3; 3+(√3)} [x]={1; 3; 4} ⇒ sum([x])=8 z^3 −3z−1=0 z={−2cos ((2π)/9); −2sin (π/(18)); 2cos (π/9)} x={3−2cos ((2π)/9); 3−2sin (π/(18)); 3+2cos (π/9)} [x]={1; 2; 4} ⇒ sum([x])=7 Q1: c∈{−19; −18; −17} Q2: c=−18 ∨ c=−17 Q3: c=−19](https://www.tinkutara.com/question/Q39146.png)
$${f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{24}{x}+{c} \\ $$$${f}'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{18}{x}+\mathrm{24} \\ $$$${x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{8}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{2};\:{x}_{\mathrm{2}} =\mathrm{4} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{20}+{c}\:\left[\mathrm{local}\:\mathrm{max}\right] \\ $$$${f}\left(\mathrm{4}\right)=\mathrm{16}+{c}\:\left[\mathrm{local}\:\mathrm{min}\right] \\ $$$$\Rightarrow\:{f}\left({x}\right)\:\mathrm{has}\:\mathrm{3}\:\mathrm{real}\:\mathrm{and}\:\mathrm{distinct}\:\mathrm{roots}\:\mathrm{with} \\ $$$$−\mathrm{20}<{c}<−\mathrm{16}\:\Rightarrow\:\mathrm{if}\:{c}\in\mathbb{Z}:\:{c}\in\left\{−\mathrm{19},\:−\mathrm{18},\:−\mathrm{17}\right\} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{Q2}\:\&\:\mathrm{Q3}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{three} \\ $$$$\mathrm{equations} \\ $$$${x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{24}{x}−\mathrm{17}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{24}{x}−\mathrm{18}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{24}{x}−\mathrm{19}=\mathrm{0} \\ $$$$\mathrm{with}\:{x}={z}+\mathrm{3}\:\mathrm{we}\:\mathrm{get} \\ $$$${z}^{\mathrm{3}} −\mathrm{3}{z}+\mathrm{1}=\mathrm{0} \\ $$$${z}^{\mathrm{3}} −\mathrm{3}{z}=\mathrm{0} \\ $$$${z}^{\mathrm{3}} +\mathrm{3}{z}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{one}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\&\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{we}\:\mathrm{use}\:\mathrm{the}\:\mathrm{trigonometric} \\ $$$$\mathrm{formula} \\ $$$${z}=\mathrm{2}\sqrt{−\frac{{p}}{\mathrm{3}}}\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{arcsin}\left(\frac{\mathrm{9}{q}}{\mathrm{2}{p}^{\mathrm{2}} }\sqrt{−\frac{{p}}{\mathrm{3}}}\right)+\mathrm{2}{k}\pi\right)\right)\:\mathrm{with}\:{k}=\mathrm{0},\:\mathrm{1},\:\mathrm{2} \\ $$$$ \\ $$$${z}^{\mathrm{3}} −\mathrm{3}{z}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:{z}=\left\{−\mathrm{2cos}\:\frac{\pi}{\mathrm{9}};\:\mathrm{2sin}\:\frac{\pi}{\mathrm{18}};\:\mathrm{2cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\right\} \\ $$$$\:\:\:\:\:{x}=\left\{\mathrm{3}−\mathrm{2cos}\:\frac{\pi}{\mathrm{9}};\:\mathrm{3}+\mathrm{2sin}\:\frac{\pi}{\mathrm{18}};\:\mathrm{3}+\mathrm{2cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\right\} \\ $$$$\:\:\:\:\:\left[{x}\right]=\left\{\mathrm{1};\:\mathrm{3};\:\mathrm{4}\right\}\:\Rightarrow\:\mathrm{sum}\left(\left[{x}\right]\right)=\mathrm{8} \\ $$$${z}^{\mathrm{3}} −\mathrm{3}{z}=\mathrm{0} \\ $$$$\:\:\:\:\:{z}=\left\{−\sqrt{\mathrm{3}};\:\mathrm{0};\:\sqrt{\mathrm{3}}\right\} \\ $$$$\:\:\:\:\:{x}=\left\{\mathrm{3}−\sqrt{\mathrm{3}};\:\mathrm{3};\:\mathrm{3}+\sqrt{\mathrm{3}}\right\} \\ $$$$\:\:\:\:\:\left[{x}\right]=\left\{\mathrm{1};\:\mathrm{3};\:\mathrm{4}\right\}\:\Rightarrow\:\mathrm{sum}\left(\left[{x}\right]\right)=\mathrm{8} \\ $$$${z}^{\mathrm{3}} −\mathrm{3}{z}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:{z}=\left\{−\mathrm{2cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}};\:−\mathrm{2sin}\:\frac{\pi}{\mathrm{18}};\:\mathrm{2cos}\:\frac{\pi}{\mathrm{9}}\right\} \\ $$$$\:\:\:\:\:{x}=\left\{\mathrm{3}−\mathrm{2cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}};\:\mathrm{3}−\mathrm{2sin}\:\frac{\pi}{\mathrm{18}};\:\mathrm{3}+\mathrm{2cos}\:\frac{\pi}{\mathrm{9}}\right\} \\ $$$$\:\:\:\:\:\left[{x}\right]=\left\{\mathrm{1};\:\mathrm{2};\:\mathrm{4}\right\}\:\Rightarrow\:\mathrm{sum}\left(\left[{x}\right]\right)=\mathrm{7} \\ $$$$ \\ $$$$\mathrm{Q1}:\:{c}\in\left\{−\mathrm{19};\:−\mathrm{18};\:−\mathrm{17}\right\} \\ $$$$\mathrm{Q2}:\:{c}=−\mathrm{18}\:\vee\:{c}=−\mathrm{17} \\ $$$$\mathrm{Q3}:\:{c}=−\mathrm{19} \\ $$
Commented by rahul 19 last updated on 03/Jul/18
Thank You Sir ! ����