Question-45164

Question Number 45164 by Tinkutara last updated on 09/Oct/18

Commented by maxmathsup by imad last updated on 09/Oct/18

$${let}\:\varphi\left({x}\right)=\:\left(\frac{{x}+{a}}{{x}+{b}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:=\left(\frac{{x}+{b}\:+{a}−{b}}{{x}+{b}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:=\left(\mathrm{1}+\frac{{a}−{b}}{{x}+{b}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:\Rightarrow \\ $$$$\varphi^{'} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(−\frac{{a}−{b}}{\left({x}+{b}\right)^{\mathrm{2}} }\right)\left(\frac{{x}+{a}}{{x}+{b}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} =\frac{{b}−{a}}{\mathrm{4}}\left({x}+{b}\right)^{−\mathrm{2}} \:\frac{\left({x}+{a}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} }{\left({x}+{b}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} } \\ $$$$=\frac{{b}−{a}}{\mathrm{4}}\:\left({x}+{b}\right)^{−\mathrm{2}+\frac{\mathrm{3}}{\mathrm{4}}} \:\left({x}+{a}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} \:=\frac{{b}−{a}}{\mathrm{4}}\left({x}+{a}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} \:\left({x}+{b}\right)^{−\frac{\mathrm{5}}{\mathrm{4}}} \:\Rightarrow \\ $$$$\frac{\mathrm{4}}{{b}−{a}}\:\varphi^{'} \left({x}\right)\:=\:\left\{\:\left({x}+{a}\right)^{−\mathrm{3}} \left({x}+{b}\right)^{−\mathrm{5}} \right\}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\Rightarrow \\ $$$$\int\:\:\left\{\left({x}+{a}\right)^{−\mathrm{3}} \left({x}+{b}\right)^{−\mathrm{5}} \right\}^{\frac{\mathrm{1}}{\mathrm{4}}} \:=\frac{\mathrm{4}}{{b}−{a}}\:\left(\frac{{x}+{a}}{{x}+{b}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:+{c}\:\:\:{the}\:{correct}\:{answer}\:{is}\:\left({C}\right). \\ $$

Commented by Tinkutara last updated on 09/Oct/18

But Sir it is by option checking, isn't there a method without seeing options?

Commented by MJS last updated on 10/Oct/18

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{trust}\:\mathrm{in}\:\left(\mathrm{b}\right)\:\mathrm{and}\:\left(\mathrm{d}\right)\:\mathrm{because}\:\mathrm{why}\:\mathrm{should} \\ $$$$\mathrm{there}\:\mathrm{be}\:\mathrm{a}\:\mathrm{change}\:\mathrm{of}\:\mathrm{sign}? \\ $$$$\int{t}^{\frac{\mathrm{1}}{\mathrm{4}}} {dt}=\frac{\mathrm{4}}{\mathrm{5}}{t}^{\frac{\mathrm{5}}{\mathrm{4}}} \\ $$$$\mathrm{so}\:\mathrm{by}\:\mathrm{instinct}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{the}\:\mathrm{one}\:\mathrm{with}\:\mathrm{the}\:\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{I}'\mathrm{d}\:\mathrm{vote}\:\mathrm{for}\:\left(\mathrm{c}\right) \\ $$$$\frac{{d}}{{dx}}\left[\frac{\mathrm{4}}{{b}−{a}}\left(\frac{{x}+{a}}{{x}+{b}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \right]=\frac{\mathrm{4}}{{b}−{a}}×\frac{{d}}{{dx}}\left[\left(\frac{{x}+{a}}{{x}+{b}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \right]= \\ $$$$=\frac{\mathrm{4}}{{b}−{a}}×\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{x}+{a}}{{x}+{b}}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} ×\frac{\mathrm{1}×\left({x}+{b}\right)−\mathrm{1}×\left({x}+{a}\right)}{\left({x}+{b}\right)^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{1}}{{b}−{a}}\left(\frac{{x}+{a}}{{x}+{b}}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} ×\frac{{b}−{a}}{\left({x}+{b}\right)^{\mathrm{2}} }=\left(\frac{{x}+{b}}{{x}+{a}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} ×\frac{\mathrm{1}}{\left({x}+{b}\right)^{\mathrm{2}} }= \\ $$$$=\frac{\left({x}+{b}\right)^{\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{2}} }{\left({x}+{a}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} }=\frac{\left({x}+{b}\right)^{−\frac{\mathrm{5}}{\mathrm{4}}} }{\left({x}+{a}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} }=\left({x}+{b}\right)^{−\frac{\mathrm{5}}{\mathrm{4}}} \left({x}+{a}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} = \\ $$$$=\left(\left({x}+{a}\right)^{−\mathrm{3}} \left({x}+{b}\right)^{−\mathrm{5}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$

Commented by math khazana by abdo last updated on 10/Oct/18

$${sir}\:{you}\:{can}\:{try}\:{the}\:{changement}\:\frac{{x}+{a}}{{x}+{b}}={t}^{\mathrm{4}} \:… \\ $$

Commented by MJS last updated on 10/Oct/18

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$

Commented by Tinkutara last updated on 10/Oct/18

Thank you very much Sir! I got the answer. ��������

Terms of Service
Privacy Policy
Contact: info@tinkutara.com