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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))
$$\:\:\:\blacksquare\sqrt{\mathrm{bemath}}\bigstar \\ $$$$\left(\mathrm{1}\right)\mathrm{If}\:\sqrt{\mathrm{a}}\:−\sqrt{\mathrm{b}}\:=\:\mathrm{20}\:,\:\mathrm{a},\mathrm{b}\in\mathbb{R}\:,\:\mathrm{find}\:\mathrm{maximum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{a}−\mathrm{5b}\:? \\ $$$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\frac{\sqrt{\mathrm{x}}−\sqrt{\mathrm{3}\sqrt{\mathrm{x}}−\mathrm{2}}}{\mathrm{x}^{\mathrm{2}} −\mathrm{16}}\:? \\ $$$$\left(\mathrm{3}\right)\int\:\frac{\mathrm{tan}\:\left(\mathrm{ln}\:\mathrm{x}\right)\:\mathrm{tan}\:\left(\mathrm{ln}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)}{\mathrm{x}}\:\mathrm{dx} \\ $$$$\left(\mathrm{4}\right)\frac{\left(\sqrt{\mathrm{3x}−\mathrm{7}}\right)^{\mathrm{2}} −\mathrm{2}}{\mathrm{x}−\mathrm{3}}\:\leqslant\:\frac{\mathrm{3}−\left(\sqrt{\mathrm{x}}\right)^{\mathrm{2}} }{\mathrm{x}−\mathrm{3}} \\ $$
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))?
$$\boldsymbol{\mathrm{If}}\:\boldsymbol{\mathrm{x}}=\mathrm{0}\:\boldsymbol{\mathrm{then}}\:\sqrt{\mathrm{3}\boldsymbol{\mathrm{x}}−\mathrm{7}}=\sqrt{−\mathrm{7}}? \\ $$
Commented by bemath last updated on 02/Sep/20
(√((3x−7)^2 )) sir
$$\sqrt{\left(\mathrm{3x}−\mathrm{7}\right)^{\mathrm{2}} }\:\mathrm{sir} \\ $$
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
$$\mathrm{typo}\:\mathrm{a}−\mathrm{5b}\:=\:\mathrm{625}−\mathrm{5}×\mathrm{25}=\mathrm{625}−\mathrm{125}=\mathrm{500} \\ $$$$\:\:\:\:\: \\ $$
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))
$$\left.\mathrm{2}\right)\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\frac{{x}−\mathrm{3}\sqrt{{x}}+\mathrm{2}}{{x}−\mathrm{4}}.\frac{\mathrm{1}}{\:\sqrt{{x}}+\sqrt{\mathrm{3}\sqrt{{x}}−\mathrm{2}}}.\frac{\mathrm{1}}{{x}+\mathrm{4}} \\ $$$$\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\frac{{w}^{\mathrm{2}} −\mathrm{3}{w}+\mathrm{2}}{{w}^{\mathrm{2}} −\mathrm{4}}.\frac{\mathrm{1}}{\mathrm{32}}=\frac{\left({w}−\mathrm{1}\right)\left({w}−\mathrm{2}\right)}{\left({w}−\mathrm{2}\right)\left({w}+\mathrm{2}\right)}.\frac{\mathrm{1}}{\mathrm{32}}=\frac{\sqrt{{x}}−\mathrm{1}}{\:\sqrt{{x}}+\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{32}}=\frac{\mathrm{1}}{\mathrm{128}} \\ $$
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)
$$\sqrt{\mathrm{a}}−\sqrt{\mathrm{b}}=\mathrm{20}\Rightarrow\sqrt{\mathrm{b}}=\sqrt{\mathrm{a}}−\mathrm{20}.\mathrm{Hence}, \\ $$$$\mathrm{P}=\mathrm{a}−\mathrm{5b}=\mathrm{a}−\mathrm{5}\left(\sqrt{\mathrm{a}}−\mathrm{20}\right)^{\mathrm{2}} \\ $$$$=\mathrm{a}−\mathrm{5a}+\mathrm{200}\sqrt{\mathrm{a}}−\mathrm{2000} \\ $$$$=−\mathrm{4}\left(\mathrm{a}−\mathrm{50}\sqrt{\mathrm{a}}+\mathrm{625}\right)+\mathrm{500} \\ $$$$=−\mathrm{4}\left(\sqrt{\mathrm{a}}−\mathrm{25}\right)^{\mathrm{2}} +\mathrm{500}\leqslant\mathrm{500} \\ $$$$\mathrm{The}\:\mathrm{equality}\:\mathrm{ocurrs}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if} \\ $$$$\begin{cases}{\sqrt{\mathrm{a}}−\mathrm{25}=\mathrm{0}}\\{\sqrt{\mathrm{a}}−\sqrt{\mathrm{b}}=\mathrm{20}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{a}=\mathrm{625}}\\{\mathrm{b}=\mathrm{25}}\end{cases} \\ $$$$\boldsymbol{\mathrm{Thus}},\boldsymbol{\mathrm{P}}=\boldsymbol{\mathrm{a}}−\mathrm{5}\boldsymbol{\mathrm{b}}\:\boldsymbol{\mathrm{has}}\:\boldsymbol{\mathrm{greatest}}\:\boldsymbol{\mathrm{value}} \\ $$$$\boldsymbol{\mathrm{equal}}\:\boldsymbol{\mathrm{to}}\:\mathrm{500}\:\boldsymbol{\mathrm{when}}\:\left(\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\right)=\left(\mathrm{625},\mathrm{25}\right) \\ $$
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))
$$\left.\mathrm{2}\right)\mathrm{Set}\:\mathrm{x}−\mathrm{4}=\mathrm{t}\left(\mathrm{x}\rightarrow\mathrm{4}\right)\sim\left(\mathrm{t}\rightarrow\mathrm{0}\right),\mathrm{x}=\mathrm{t}+\mathrm{4} \\ $$$$\left(\mathrm{2}\right)\mathrm{lim}\frac{\sqrt{\mathrm{x}}−\sqrt{\mathrm{3}\sqrt{\mathrm{x}}−\mathrm{2}}}{\mathrm{x}^{\mathrm{2}} −\mathrm{16}}\:=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{t}+\mathrm{4}}−\sqrt{\mathrm{3}\sqrt{\mathrm{t}+\mathrm{4}}−\mathrm{2}}}{\left(\mathrm{t}+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{16}} \\ $$$$=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{t}+\mathrm{4}−\left(\mathrm{3}\sqrt{\mathrm{t}+\mathrm{4}}−\mathrm{2}\right)}{\:\left[\sqrt{\mathrm{t}+\mathrm{4}}+\sqrt{\left.\mathrm{3}\sqrt{\mathrm{t}+\mathrm{4}}−\mathrm{2}\right]}\left(\mathrm{t}^{\mathrm{2}} +\mathrm{8t}\right)\right.} \\ $$$$=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{t}+\mathrm{6}−\mathrm{3}\sqrt{\mathrm{t}+\mathrm{4}}}{\:\left[\sqrt{\mathrm{t}+\mathrm{4}}+\sqrt{\left.\mathrm{3}\sqrt{\mathrm{t}+\mathrm{4}}−\mathrm{2}\right]}\left(\mathrm{t}^{\mathrm{2}} +\mathrm{8t}\right)\right.} \\ $$$$=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{t}+\mathrm{6}\right)^{\mathrm{2}} −\left(\mathrm{3}\sqrt{\mathrm{t}+\mathrm{4}}\right)^{\mathrm{2}} }{\:\left[\sqrt{\mathrm{t}+\mathrm{4}}+\sqrt{\left.\mathrm{3}\sqrt{\mathrm{t}+\mathrm{4}}−\mathrm{2}\right]}\left(\mathrm{t}^{\mathrm{2}} +\mathrm{8t}\right)\left(\mathrm{t}+\mathrm{6}+\mathrm{3}\sqrt{\mathrm{t}+\mathrm{4}}\right)\right.} \\ $$$$=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{12t}+\mathrm{36}−\mathrm{9t}−\mathrm{36}}{\:\left[\sqrt{\mathrm{t}+\mathrm{4}}+\sqrt{\left.\mathrm{3}\sqrt{\mathrm{t}+\mathrm{4}}−\mathrm{2}\right]}\left(\mathrm{t}^{\mathrm{2}} +\mathrm{8t}\right)\left(\mathrm{t}+\mathrm{6}+\mathrm{3}\sqrt{\mathrm{t}+\mathrm{4}}\right)\right.} \\ $$$$=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{t}+\mathrm{3}}{\:\left[\sqrt{\mathrm{t}+\mathrm{4}}+\sqrt{\left.\mathrm{3}\sqrt{\mathrm{t}+\mathrm{4}}−\mathrm{2}\right]}\left(\mathrm{t}+\mathrm{6}+\mathrm{3}\sqrt{\mathrm{t}+\mathrm{4}}\right)\left(\mathrm{t}+\mathrm{8}\right)\right.} \\ $$$$=\frac{\mathrm{3}}{\left(\mathrm{2}+\mathrm{2}\right)\left(\mathrm{6}+\mathrm{6}\right).\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{128}} \\ $$
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
$$\left(\mathrm{4}\right)\frac{\left(\sqrt{\mathrm{3x}−\mathrm{7}}\right)^{\mathrm{2}} −\mathrm{2}}{\mathrm{x}−\mathrm{3}}\:\leqslant\:\frac{\mathrm{3}−\left(\sqrt{\mathrm{x}}\right)^{\mathrm{2}} }{\mathrm{x}−\mathrm{3}} \\ $$$$\Leftrightarrow\frac{\mathrm{3x}−\mathrm{7}−\mathrm{2}}{\mathrm{x}−\mathrm{3}}\leqslant\frac{\mathrm{3}−\mathrm{x}}{\mathrm{x}−\mathrm{3}}\:\left(\mathrm{1}\right).\mathrm{Under}\:\mathrm{the} \\ $$$$\mathrm{condition\begin{cases}{\mathrm{x}\geqslant\frac{\mathrm{7}}{\mathrm{3}}}\\{\mathrm{x}\neq\mathrm{3}}\end{cases}} \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\frac{\mathrm{3x}−\mathrm{9}}{\mathrm{x}−\mathrm{3}}\leqslant−\mathrm{1}\Leftrightarrow\frac{\mathrm{x}−\mathrm{3}}{\mathrm{x}−\mathrm{3}}\leqslant−\mathrm{1}\Leftrightarrow\mathrm{1}\leqslant−\mathrm{1} \\ $$$$\mathrm{the}\:\mathrm{inequality}\:\mathrm{has}\:\mathrm{no}\:\mathrm{roots} \\ $$
Commented by bemath last updated on 02/Sep/20
wrong sir. if (√((3x−7)^2 )) = ∣3x−7∣ not (√((3x−7)^2 )) = 3x−7
$$\mathrm{wrong}\:\mathrm{sir}.\:\mathrm{if}\:\sqrt{\left(\mathrm{3x}−\mathrm{7}\right)^{\mathrm{2}} }\:=\:\mid\mathrm{3x}−\mathrm{7}\mid \\ $$$$\mathrm{not}\:\sqrt{\left(\mathrm{3x}−\mathrm{7}\right)^{\mathrm{2}} }\:=\:\mathrm{3x}−\mathrm{7} \\ $$

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