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?