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.


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Question Number 39142 by rahul 19 last updated on 03/Jul/18

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