Menu Close

Question-196595

Tinku Tara 0 Comments
Pin




Question Number 196595 by sonukgindia last updated on 27/Aug/23
Answered by Rasheed.Sindhi last updated on 28/Aug/23
f(x)=((ax^2 +bx+c)/(dx+e)) f(x)+f(−x)=((ax^2 +bx+c)/(dx+e))+((ax^2 −bx+c)/(−dx+e)) =(((ax^2 +bx+c)(e−dx)+(ax^2 −bx+c)(e+dx))/((e+dx)(e−dx))) =((aex^2 +bex+ce−adx^3 −bdx^2 −cdx+aex^2 −bex+ce+adx^3 −bdx^2 +cdx)/(e^2 −d^2 x^2 )) =((2(ae−bd)x^2 +2ce)/(e^2 −d^2 x^2 )) determinant (((f(x)+f(−x)=((2(ae−bd)x^2 +2ce)/(e^2 −d^2 x^2 ))))) f(0)+f(−0)=((2ce)/e^2 )=((2c)/e) f(0)=(c/e) f(1)+f(−1)=((2(ae−bd)(1)^2 +2ce)/(e^2 −d^2 (1)^2 ))=(2/3)+2 =((2(ae−bd)+2ce)/(e^2 −d^2 ))=(8/3) 6(ae−bd)+6ce−8e^2 +8d^2 =0 determinant (((3(ae−bd)+3ce−4e^2 +4d^2 =0))) ae−bd=((−3ce+4e^2 +4d^2 )/3)....(i) f(2)+f(−2)=((2(ae−bd)(2)^2 +2ce)/(e^2 −d^2 (2)^2 ))=((11)/(13))−1 =((4(ae−bd)+ce)/(e^2 −4d^2 ))=−(1/(13)) 5(ae−bd)+13ce=−e^2 +4d^2 5(ae−bd)+13ce+e^2 −4d^2 =0 determinant (((5(ae−bd)+13ce+e^2 −4d^2 =0))) ae−bd=((−13ce−e^2 +4d^2 )/5)...(ii) f(3)=p =((9a+3b+c)/(3d+e)) f(3)+f(−3)=((2(ae−bd)(3)^2 +2ce)/(e^2 −d^2 (3)^2 ))=p−(6/7) =((18(ae−bd)+2ce)/(e^2 −9d^2 ))=((7p−6)/7) 126(ae−bd)+14ce=(7p−6)e^2 −9(7p−6)d^2 Where determinant (((126(ae−bd)+14ce−(7p−6)e^2 +9(7p−6)d^2 =0_(Where p =((9a−3b+c)/(−3d+e)) ) ))) ae−bd=((−14ce+(7p−6)e^2 −9(7p−6)d^2 )/(126)) (i),(ii) & (iii)⇒ ((−3ce+4e^2 +4d^2 )/3)=((−13ce−e^2 +4d^2 )/5)=((−14ce+(7p−6)e^2 −9(7p−6)d^2 )/(126)) 210(−3ce+4e^2 +4d^2 )=126(−13ce−e^2 +4d^2 )=5(−14ce+(7p−6)e^2 −9(7p−6)d^2 ) ...... f(1)=((a+b+c)/(d+e))=(2/3)⇒3a+3b+3c=2d+2e f(−1)=((a−b+c)/(−d+e))=2 f(2)=((4a+2b+c)/(2d+e))=((11)/(13)) f(−2)=((4a−2b+c)/(−2d+e))=−1 f(−3)=((9a−3b+c)/(−3d+e))=−(6/7)
$${f}\left({x}\right)=\frac{{ax}^{\mathrm{2}} +{bx}+{c}}{{dx}+{e}} \\ $$$${f}\left({x}\right)+{f}\left(−{x}\right)=\frac{{ax}^{\mathrm{2}} +{bx}+{c}}{{dx}+{e}}+\frac{{ax}^{\mathrm{2}} −{bx}+{c}}{−{dx}+{e}} \\ $$$$=\frac{\left({ax}^{\mathrm{2}} +{bx}+{c}\right)\left({e}−{dx}\right)+\left({ax}^{\mathrm{2}} −{bx}+{c}\right)\left({e}+{dx}\right)}{\left({e}+{dx}\right)\left({e}−{dx}\right)} \\ $$$$=\frac{{aex}^{\mathrm{2}} +{bex}+{ce}−\cancel{{adx}^{\mathrm{3}} }−{bdx}^{\mathrm{2}} −{cdx}+{aex}^{\mathrm{2}} −{bex}+{ce}+\cancel{{adx}^{\mathrm{3}} }−{bdx}^{\mathrm{2}} +{cdx}}{{e}^{\mathrm{2}} −{d}^{\mathrm{2}} {x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}\left({ae}−{bd}\right){x}^{\mathrm{2}} +\mathrm{2}{ce}}{{e}^{\mathrm{2}} −{d}^{\mathrm{2}} {x}^{\mathrm{2}} } \\ $$$$\begin{array}{|c|}{{f}\left({x}\right)+{f}\left(−{x}\right)=\frac{\mathrm{2}\left({ae}−{bd}\right){x}^{\mathrm{2}} +\mathrm{2}{ce}}{{e}^{\mathrm{2}} −{d}^{\mathrm{2}} {x}^{\mathrm{2}} }}\\\hline\end{array}\: \\ $$$${f}\left(\mathrm{0}\right)+{f}\left(−\mathrm{0}\right)=\frac{\mathrm{2}{ce}}{{e}^{\mathrm{2}} }=\frac{\mathrm{2}{c}}{{e}} \\ $$$${f}\left(\mathrm{0}\right)=\frac{{c}}{{e}} \\ $$$${f}\left(\mathrm{1}\right)+{f}\left(−\mathrm{1}\right)=\frac{\mathrm{2}\left({ae}−{bd}\right)\left(\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}{ce}}{{e}^{\mathrm{2}} −{d}^{\mathrm{2}} \left(\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{2}}{\mathrm{3}}+\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}\left({ae}−{bd}\right)+\mathrm{2}{ce}}{{e}^{\mathrm{2}} −{d}^{\mathrm{2}} }=\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\mathrm{6}\left({ae}−{bd}\right)+\mathrm{6}{ce}−\mathrm{8}{e}^{\mathrm{2}} +\mathrm{8}{d}^{\mathrm{2}} =\mathrm{0} \\ $$$$\begin{array}{|c|}{\mathrm{3}\left({ae}−{bd}\right)+\mathrm{3}{ce}−\mathrm{4}{e}^{\mathrm{2}} +\mathrm{4}{d}^{\mathrm{2}} =\mathrm{0}}\\\hline\end{array} \\ $$$${ae}−{bd}=\frac{−\mathrm{3}{ce}+\mathrm{4}{e}^{\mathrm{2}} +\mathrm{4}{d}^{\mathrm{2}} }{\mathrm{3}}….\left({i}\right) \\ $$$${f}\left(\mathrm{2}\right)+{f}\left(−\mathrm{2}\right)=\frac{\mathrm{2}\left({ae}−{bd}\right)\left(\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}{ce}}{{e}^{\mathrm{2}} −{d}^{\mathrm{2}} \left(\mathrm{2}\right)^{\mathrm{2}} }=\frac{\mathrm{11}}{\mathrm{13}}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}\left({ae}−{bd}\right)+{ce}}{{e}^{\mathrm{2}} −\mathrm{4}{d}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{13}} \\ $$$$\mathrm{5}\left({ae}−{bd}\right)+\mathrm{13}{ce}=−{e}^{\mathrm{2}} +\mathrm{4}{d}^{\mathrm{2}} \\ $$$$\mathrm{5}\left({ae}−{bd}\right)+\mathrm{13}{ce}+{e}^{\mathrm{2}} −\mathrm{4}{d}^{\mathrm{2}} =\mathrm{0} \\ $$$$\begin{array}{|c|}{\mathrm{5}\left({ae}−{bd}\right)+\mathrm{13}{ce}+{e}^{\mathrm{2}} −\mathrm{4}{d}^{\mathrm{2}} =\mathrm{0}}\\\hline\end{array} \\ $$$${ae}−{bd}=\frac{−\mathrm{13}{ce}−{e}^{\mathrm{2}} +\mathrm{4}{d}^{\mathrm{2}} }{\mathrm{5}}…\left({ii}\right) \\ $$$${f}\left(\mathrm{3}\right)={p}\:=\frac{\mathrm{9}{a}+\mathrm{3}{b}+{c}}{\mathrm{3}{d}+{e}} \\ $$$${f}\left(\mathrm{3}\right)+{f}\left(−\mathrm{3}\right)=\frac{\mathrm{2}\left({ae}−{bd}\right)\left(\mathrm{3}\right)^{\mathrm{2}} +\mathrm{2}{ce}}{{e}^{\mathrm{2}} −{d}^{\mathrm{2}} \left(\mathrm{3}\right)^{\mathrm{2}} }={p}−\frac{\mathrm{6}}{\mathrm{7}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{18}\left({ae}−{bd}\right)+\mathrm{2}{ce}}{{e}^{\mathrm{2}} −\mathrm{9}{d}^{\mathrm{2}} }=\frac{\mathrm{7}{p}−\mathrm{6}}{\mathrm{7}} \\ $$$$\:\:\:\:\mathrm{126}\left({ae}−{bd}\right)+\mathrm{14}{ce}=\left(\mathrm{7}{p}−\mathrm{6}\right){e}^{\mathrm{2}} −\mathrm{9}\left(\mathrm{7}{p}−\mathrm{6}\right){d}^{\mathrm{2}} \:{Where} \\ $$$$\begin{array}{|c|}{\underset{{Where}\:{p}\:=\frac{\mathrm{9}{a}−\mathrm{3}{b}+{c}}{−\mathrm{3}{d}+{e}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:} {\mathrm{126}\left({ae}−{bd}\right)+\mathrm{14}{ce}−\left(\mathrm{7}{p}−\mathrm{6}\right){e}^{\mathrm{2}} +\mathrm{9}\left(\mathrm{7}{p}−\mathrm{6}\right){d}^{\mathrm{2}} =\mathrm{0}}}\\\hline\end{array} \\ $$$${ae}−{bd}=\frac{−\mathrm{14}{ce}+\left(\mathrm{7}{p}−\mathrm{6}\right){e}^{\mathrm{2}} −\mathrm{9}\left(\mathrm{7}{p}−\mathrm{6}\right){d}^{\mathrm{2}} }{\mathrm{126}} \\ $$$$\left({i}\right),\left({ii}\right)\:\&\:\left({iii}\right)\Rightarrow \\ $$$$\frac{−\mathrm{3}{ce}+\mathrm{4}{e}^{\mathrm{2}} +\mathrm{4}{d}^{\mathrm{2}} }{\mathrm{3}}=\frac{−\mathrm{13}{ce}−{e}^{\mathrm{2}} +\mathrm{4}{d}^{\mathrm{2}} }{\mathrm{5}}=\frac{−\mathrm{14}{ce}+\left(\mathrm{7}{p}−\mathrm{6}\right){e}^{\mathrm{2}} −\mathrm{9}\left(\mathrm{7}{p}−\mathrm{6}\right){d}^{\mathrm{2}} }{\mathrm{126}} \\ $$$$\mathrm{210}\left(−\mathrm{3}{ce}+\mathrm{4}{e}^{\mathrm{2}} +\mathrm{4}{d}^{\mathrm{2}} \right)=\mathrm{126}\left(−\mathrm{13}{ce}−{e}^{\mathrm{2}} +\mathrm{4}{d}^{\mathrm{2}} \right)=\mathrm{5}\left(−\mathrm{14}{ce}+\left(\mathrm{7}{p}−\mathrm{6}\right){e}^{\mathrm{2}} −\mathrm{9}\left(\mathrm{7}{p}−\mathrm{6}\right){d}^{\mathrm{2}} \right) \\ $$$$…… \\ $$$${f}\left(\mathrm{1}\right)=\frac{{a}+{b}+{c}}{{d}+{e}}=\frac{\mathrm{2}}{\mathrm{3}}\Rightarrow\mathrm{3}{a}+\mathrm{3}{b}+\mathrm{3}{c}=\mathrm{2}{d}+\mathrm{2}{e} \\ $$$${f}\left(−\mathrm{1}\right)=\frac{{a}−{b}+{c}}{−{d}+{e}}=\mathrm{2} \\ $$$${f}\left(\mathrm{2}\right)=\frac{\mathrm{4}{a}+\mathrm{2}{b}+{c}}{\mathrm{2}{d}+{e}}=\frac{\mathrm{11}}{\mathrm{13}} \\ $$$${f}\left(−\mathrm{2}\right)=\frac{\mathrm{4}{a}−\mathrm{2}{b}+{c}}{−\mathrm{2}{d}+{e}}=−\mathrm{1} \\ $$$${f}\left(−\mathrm{3}\right)=\frac{\mathrm{9}{a}−\mathrm{3}{b}+{c}}{−\mathrm{3}{d}+{e}}=−\frac{\mathrm{6}}{\mathrm{7}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *