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Question Number 110944 by bemath last updated on 01/Sep/20
■(√(bemath))★ (1)If (√a) −(√b) = 20 , a,b∈R , find maximum value of a−5b ? (2)lim_(x→4) (((√x)−(√(3(√x)−2)))/(x^2 −16)) ? (3)∫ ((tan (ln x) tan (ln ((x/2))))/x) dx (4)((((√(3x−7)))^2 −2)/(x−3)) ≤ ((3−((√x))^2 )/(x−3))
◼bemath★(1)Ifa−b=20,a,b∈R,findmaximumvalueofa−5b?(2)limx→4x−3x−2x2−16?(3)∫tan(lnx)tan(ln(x2))xdx(4)(3x−7)2−2x−3⩽3−(x)2x−3
Commented by bemath last updated on 01/Sep/20
Commented by 1549442205PVT last updated on 02/Sep/20
If x=0 then (√(3x−7))=(√(−7))?
Ifx=0then3x−7=−7?
Commented by bemath last updated on 02/Sep/20
(√((3x−7)^2 )) sir
(3x−7)2sir
Answered by john santu last updated on 01/Sep/20
Commented by bemath last updated on 01/Sep/20
typo a−5b = 625−5×25=625−125=500
typoa−5b=625−5×25=625−125=500
Answered by Dwaipayan Shikari last updated on 01/Sep/20
2)lim_(x→4) ((x−3(√x)+2)/(x−4)).(1/( (√x)+(√(3(√x)−2)))).(1/(x+4)) lim_(x→4) ((w^2 −3w+2)/(w^2 −4)).(1/(32))=(((w−1)(w−2))/((w−2)(w+2))).(1/(32))=(((√x)−1)/( (√x)+2)).(1/(32))=(1/(128))
2)limx→4x−3x+2x−4.1x+3x−2.1x+4limx→4w2−3w+2w2−4.132=(w−1)(w−2)(w−2)(w+2).132=x−1x+2.132=1128
Answered by 1549442205PVT last updated on 02/Sep/20
(√a)−(√b)=20⇒(√b)=(√a)−20.Hence, P=a−5b=a−5((√a)−20)^2 =a−5a+200(√a)−2000 =−4(a−50(√a)+625)+500 =−4((√a)−25)^2 +500≤500 The equality ocurrs if and only if { (((√a)−25=0)),(((√a)−(√b)=20)) :}⇔ { ((a=625)),((b=25)) :} Thus,P=a−5b has greatest value equal to 500 when (a,b)=(625,25)
a−b=20⇒b=a−20.Hence,P=a−5b=a−5(a−20)2=a−5a+200a−2000=−4(a−50a+625)+500=−4(a−25)2+500⩽500Theequalityocurrsifandonlyif{a−25=0a−b=20⇔{a=625b=25Thus,P=a−5bhasgreatestvalueequalto500when(a,b)=(625,25)
Answered by 1549442205PVT last updated on 02/Sep/20
2)Set x−4=t(x→4)∼(t→0),x=t+4 (2)lim(((√x)−(√(3(√x)−2)))/(x^2 −16)) =lim_(t→0) (((√(t+4))−(√(3(√(t+4))−2)))/((t+4)^2 −16)) =lim_(t→0) ((t+4−(3(√(t+4))−2))/( [(√(t+4))+(√(3(√(t+4))−2]))(t^2 +8t))) =lim_(t→0) ((t+6−3(√(t+4)))/( [(√(t+4))+(√(3(√(t+4))−2]))(t^2 +8t))) =lim_(t→0) (((t+6)^2 −(3(√(t+4)))^2 )/( [(√(t+4))+(√(3(√(t+4))−2]))(t^2 +8t)(t+6+3(√(t+4))))) =lim_(t→0) ((t^2 +12t+36−9t−36)/( [(√(t+4))+(√(3(√(t+4))−2]))(t^2 +8t)(t+6+3(√(t+4))))) =lim_(t→0) ((t+3)/( [(√(t+4))+(√(3(√(t+4))−2]))(t+6+3(√(t+4)))(t+8))) =(3/((2+2)(6+6).8))=(1/(128))
2)Setx−4=t(x→4)∼(t→0),x=t+4(2)limx−3x−2x2−16=limt→0t+4−3t+4−2(t+4)2−16=limt→0t+4−(3t+4−2)[t+4+3t+4−2](t2+8t)=limt→0t+6−3t+4[t+4+3t+4−2](t2+8t)=limt→0(t+6)2−(3t+4)2[t+4+3t+4−2](t2+8t)(t+6+3t+4)=limt→0t2+12t+36−9t−36[t+4+3t+4−2](t2+8t)(t+6+3t+4)=limt→0t+3[t+4+3t+4−2](t+6+3t+4)(t+8)=3(2+2)(6+6).8=1128
Answered by 1549442205PVT last updated on 02/Sep/20
(4)((((√(3x−7)))^2 −2)/(x−3)) ≤ ((3−((√x))^2 )/(x−3)) ⇔((3x−7−2)/(x−3))≤((3−x)/(x−3)) (1).Under the condition { ((x≥(7/3))),((x≠3)) :} (1)⇔((3x−9)/(x−3))≤−1⇔((x−3)/(x−3))≤−1⇔1≤−1 the inequality has no roots
(4)(3x−7)2−2x−3⩽3−(x)2x−3⇔3x−7−2x−3⩽3−xx−3(1).Underthecondition{x⩾73x≠3(1)⇔3x−9x−3⩽−1⇔x−3x−3⩽−1⇔1⩽−1theinequalityhasnoroots
Commented by bemath last updated on 02/Sep/20
wrong sir. if (√((3x−7)^2 )) = ∣3x−7∣ not (√((3x−7)^2 )) = 3x−7
wrongsir.if(3x−7)2=∣3x−7∣not(3x−7)2=3x−7

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