Question Number 103492 by abony1303 last updated on 15/Jul/20

Q1:  Evaluate ∫_(−1) ^( 1) ∣x∣∙(x^3 +1)dx    Q2: Find the sum of all integers k for   which the equation 2x^3 −6x^2 +k=0  has more than one solution.    Q3: Find the shortest distance from a   point on the curve y=x^2 −x to the line  y=x−3

Commented byabony1303 last updated on 15/Jul/20

pls help

Answered by mr W last updated on 15/Jul/20

Q2  2x^2 (x−3)+k=0  k<0 ⇒one root: x>3  k=0 ⇒two roots: x=0 or 3  k>0:  (1/x^3 )−(6/(kx))+(2/k)=0  such that three roots:  ((1/k))^2 +(−(2/k))^3 <0  ((1/k))^2 (1−(8/k))<0  1−(8/k)<0  k<8  ⇒k=1,2,3,...,7  Σk=((7(1+7))/2)=28

Commented byabony1303 last updated on 15/Jul/20

thank you sir. but I drew a graph and k=8  also correct? And pls can you explain the  raw such that three roots... and below it.

Commented bymr W last updated on 15/Jul/20

i misread the question as: more than  two roots.  for more than one root: the answer  is Σk=0+1+2+3+...+8=36    we have:  (1/x^3 )−(6/(kx))+(2/k)=0  let t=(1/x), then  t^3 −(6/k)t+(2/k)=0  three roots for x ⇒ three roots for t.    let f(t)=t^3 +bt+c  if f(t)=0 has three roots, it means  f′(t)=0 ⇒3t^2 +b=0⇒t_(1,2) =±(√(−(b/3)))  f(t_1 )f(t_2 )<0, see diagram.  f(t_1 )=t_1 ^3 +bt_1 +c=(1/3)t_1 (3t_1 ^2 +b)+((2bt_1 )/3)+c=((2bt_1 )/3)+c  f(t_2 )=((2bt_2 )/3)+c  (((2bt_1 )/3)+c)(((2bt_2 )/3)+c)<0  ((4b^2 )/9)t_1 t_2 +((2b)/3)(t_1 +t_2 )+c^2 <0  ((4b^2 )/9)×(b/3)+((2b)/3)×0+c^2 <0  ⇒((b/3))^3 +((c/2))^2 <0  with b=−(6/k), c=(2/k)  ⇒(−(2/k))^3 +((1/k))^2 <0

Commented bymr W last updated on 15/Jul/20

Commented bymr W last updated on 15/Jul/20

Commented bymr W last updated on 15/Jul/20

Commented bymr W last updated on 15/Jul/20

Commented bymr W last updated on 15/Jul/20

Commented bymr W last updated on 15/Jul/20

Commented bymr W last updated on 15/Jul/20

three roots only for 1≤k≤7.  two roots only for k=0 or 8.  one root only for k≤−1 or k≥9.

Answered by mr W last updated on 15/Jul/20

Q3  y=x^2 −x  y′=2x−1=1 ⇒x=1, y=0  d_(min) =((∣1×1−1×0−3∣)/(√(1^2 +(−1)^2 )))=(√2)

Answered by mr W last updated on 15/Jul/20

Q1  ∫_(−1) ^( 1) ∣x∣∙(x^3 +1)dx  =∫_(−1) ^( 1) ∣x∣∙x^3 dx+∫_(−1) ^( 1) ∣x∣dx  =0+2∫_0 ^( 1) xdx  =1

Answered by OlafThorendsen last updated on 15/Jul/20

Q1.  I = ∫_(−1) ^1 ∣x∣(x^3 +1)dx  I = ∫_(−1) ^0 −x(x^3 +1)dx+∫_0 ^1 x(x^3 +1)dx  I = [−(x^5 /5)−(x^2 /2)]_(−1) ^0 +[(x^5 /5)+(x^2 /2)]_0 ^1   I = (−(1/5)+(1/2))+((1/5)+(1/2))  I = 1  Q2.  Δ′ = (−3)^2 −(2)(k) = 9−2k  More than one solution ⇔ Δ′>0  ⇔ k<(9/2)  ⇔ k∈{0;1;2;3;4}  ⇒ sum = 0+1+2+3+4 = 10  Q3.  A is a point of the curve  B is a point of the line  A ((a),((a^2 −a)) ) and B ((b),((b−3)) )  AB^2  = (b−a)^2 +(b−3−a^2 +a)^2   for a given a, AB^2  = f(b)  f(b) = (b−a)^2 +(b−3−a^2 +a)^2   f′(b) = 2(b−a)+2(b−3−a^2 +a)  f′(b) = 4b−2a^2 −6  f′(b) = 0 ⇔ b = ((a^2 +3)/2)  Then :  AB^2  = (((a^2 +3)/2)−a)^2 +(((a^2 +3)/2)−3−a^2 +a)^2   AB^2  = ((a^2 /2)−a+(3/2))^2 +(−(a^2 /2)+a−(3/2))^2   AB^2  = 2((a^2 /2)−a+(3/2))^2   AB = (√2)∣(a^2 /2)−a+(3/2)∣ and then :  A ((a),((a^2 −a)) )  and B ((((a^2 +3)/2)),(((a^2 −3)/2)) )

Commented byabony1303 last updated on 15/Jul/20

no, no it′s ok. :)

Commented byOlafThorendsen last updated on 15/Jul/20

sorry ! I need to change my  glasses ! :−)

Commented byabony1303 last updated on 15/Jul/20

Thank you ser.Everything is clear, but   can you pls explain what is △′ in Q2?

Commented bybemath last updated on 15/Jul/20

Δ = b^2 −4ac   discriminant

Commented byabony1303 last updated on 15/Jul/20

but it′s not the quadratic equation?

Commented byOlafThorendsen last updated on 15/Jul/20

Δ′ : reduced discriminant  (sorry my english is not fluent)  x = ((−b±(√Δ))/(2a)) with Δ = b^2 −4ac  but when b is even you can use :  x = ((−b′±(√(Δ′)))/a)  with b′ = (b/2) and Δ′ = b′^2 −ac  (the calculation is faster)

Commented byabony1303 last updated on 15/Jul/20

ser but it′s not the quadratic equation?