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F-x-x-3-9x-2-24x-c-0-has-three-real-and-distinct-roots-amp-Q-1-Possible-value-of-c-is-Q-2-If-8-then-c-is-Q-3-If-7-then-c-is-Options-for-the-above

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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.
F(x)=x3−9x2+24x+c=0hasthreerealanddistinctrootsα,β&γ.Q.1→Possiblevalueofcis:Q.2→If[α]+[β]+[γ]=8thencis:Q.3→If[α]+[β]+[γ]=7thencis:Optionsfortheabove3Q.→a)(−20,−16)b)(−20,−18)c)(−18,−16)d)noneofthese.[.]=greatestintegerfunction.
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
f(x)=x3−9x2+24x+cf′(x)=3x2−18x+24x2−6x+8=0x1=2;x2=4f(2)=20+c[localmax]f(4)=16+c[localmin]⇒f(x)has3realanddistinctrootswith−20<c<−16⇒ifc∈Z:c∈{−19,−18,−17}forQ2&Q3wehavetosolvethethreeequationsx3−9x2+24x−17=0x3−9x2+24x−18=0x3−9x2+24x−19=0withx=z+3wegetz3−3z+1=0z3−3z=0z3+3z−1=0the2ndoneiseasytosolveforthe1st&3rdweusethetrigonometricformulaz=2−p3sin(13(arcsin(9q2p2−p3)+2kπ))withk=0,1,2z3−3z+1=0z={−2cosπ9;2sinπ18;2cos2π9}x={3−2cosπ9;3+2sinπ18;3+2cos2π9}[x]={1;3;4}⇒sum([x])=8z3−3z=0z={−3;0;3}x={3−3;3;3+3}[x]={1;3;4}⇒sum([x])=8z3−3z−1=0z={−2cos2π9;−2sinπ18;2cosπ9}x={3−2cos2π9;3−2sinπ18;3+2cosπ9}[x]={1;2;4}⇒sum([x])=7Q1:c∈{−19;−18;−17}Q2:c=−18∨c=−17Q3:c=−19
Commented by rahul 19 last updated on 03/Jul/18
Thank You Sir ! ����

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