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Question Number 2705 by Syaka last updated on 25/Nov/15

if function f satisfy form : f(x) = (∫_0 ^1 f(x) dx)x^2 + (∫_0 ^2 f(x) dx)x + (∫_0 ^3 f(x) dx) + 1 then the value of f(4) is..... ?

$${if}\:{function}\:{f}\:{satisfy}\:{form}\:: \\ $$$${f}\left({x}\right)\:=\:\left(\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\left({x}\right)\:{dx}\right){x}^{\mathrm{2}} \:+\:\left(\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right)\:{dx}\right){x}\:+\:\left(\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}{f}\left({x}\right)\:{dx}\right)\:+\:\mathrm{1} \\ $$$${then}\:{the}\:{value}\:{of}\:{f}\left(\mathrm{4}\right)\:{is}.....\:? \\ $$

Answered by prakash jain last updated on 25/Nov/15

f(x)=ax^2 +bx+c ....(1) ∫f(x)=a(x^3 /3)+b(x^2 /2)+cx+C ∫_0 ^1 f(x)=(a/3)+(b/2)+c ∫_0 ^2 f(x)=((8a)/3)+2b+2c ∫_0 ^3 f(x)=9a+(9/2)b+3c f(x)= ((a/3)+(b/2)+c)x^2 +(((8a)/3)+2b+2c)x+(9a+(9/2)b+3c)+1 ..(2) Equating coefficients in (1) and (2) (x^2 coeffiient) (a/3)+(b/2)+c=a⇒((2a)/3)=(b/2)+c⇒a=(3/4)b+(3/2)c (x coefficient) ((8a)/3)+2b+2c=b⇒2b+4c+2b+2c=b⇒ 5b=−6c⇒b=((−6c)/5) a=((3b)/4)+((3c)/2)⇒a=−(3/4)×((−6c)/5)+((3c)/2)=((12c)/(20))=((3c)/5) (constant term) (9a+(9/2)b+3c)+1=c To be solved for a, b and c ((27c)/5)−(9/2)×((6c)/5)+2c=−1 c=((−1)/2), a=((−3)/(10)), b=(3/5) f(x)=−(3/(10))x^2 +(3/5)x−(1/2) f(4)=((−3)/(10))×16+(3/5)×4−(1/2)=((−48+24−5)/(10))=((−29)/(10))

$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}\:\:\:\:\:\:\:\:....\left(\mathrm{1}\right) \\ $$$$\int{f}\left({x}\right)={a}\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+{b}\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{cx}+{C} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right)=\frac{{a}}{\mathrm{3}}+\frac{{b}}{\mathrm{2}}+{c} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} {f}\left({x}\right)=\frac{\mathrm{8}{a}}{\mathrm{3}}+\mathrm{2}{b}+\mathrm{2}{c} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{3}} {f}\left({x}\right)=\mathrm{9}{a}+\frac{\mathrm{9}}{\mathrm{2}}{b}+\mathrm{3}{c} \\ $$$${f}\left({x}\right)= \\ $$$$\left(\frac{{a}}{\mathrm{3}}+\frac{{b}}{\mathrm{2}}+{c}\right){x}^{\mathrm{2}} +\left(\frac{\mathrm{8}{a}}{\mathrm{3}}+\mathrm{2}{b}+\mathrm{2}{c}\right){x}+\left(\mathrm{9}{a}+\frac{\mathrm{9}}{\mathrm{2}}{b}+\mathrm{3}{c}\right)+\mathrm{1}\:..\left(\mathrm{2}\right) \\ $$$${E}\mathrm{quating}\:\mathrm{coefficients}\:\mathrm{in}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right) \\ $$$$\left({x}^{\mathrm{2}} \:{coeffiient}\right) \\ $$$$\frac{{a}}{\mathrm{3}}+\frac{{b}}{\mathrm{2}}+{c}={a}\Rightarrow\frac{\mathrm{2}{a}}{\mathrm{3}}=\frac{{b}}{\mathrm{2}}+{c}\Rightarrow{a}=\frac{\mathrm{3}}{\mathrm{4}}{b}+\frac{\mathrm{3}}{\mathrm{2}}{c} \\ $$$$\left({x}\:{coefficient}\right) \\ $$$$\frac{\mathrm{8}{a}}{\mathrm{3}}+\mathrm{2}{b}+\mathrm{2}{c}={b}\Rightarrow\mathrm{2}{b}+\mathrm{4}{c}+\mathrm{2}{b}+\mathrm{2}{c}={b}\Rightarrow \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}{b}=−\mathrm{6}{c}\Rightarrow{b}=\frac{−\mathrm{6}{c}}{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}=\frac{\mathrm{3}{b}}{\mathrm{4}}+\frac{\mathrm{3}{c}}{\mathrm{2}}\Rightarrow{a}=−\frac{\mathrm{3}}{\mathrm{4}}×\frac{−\mathrm{6}{c}}{\mathrm{5}}+\frac{\mathrm{3}{c}}{\mathrm{2}}=\frac{\mathrm{12}{c}}{\mathrm{20}}=\frac{\mathrm{3}{c}}{\mathrm{5}} \\ $$$$\left({constant}\:{term}\right) \\ $$$$\left(\mathrm{9}{a}+\frac{\mathrm{9}}{\mathrm{2}}{b}+\mathrm{3}{c}\right)+\mathrm{1}={c} \\ $$$$\mathrm{To}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{for}\:{a},\:{b}\:\mathrm{and}\:{c} \\ $$$$\frac{\mathrm{27}{c}}{\mathrm{5}}−\frac{\mathrm{9}}{\mathrm{2}}×\frac{\mathrm{6}{c}}{\mathrm{5}}+\mathrm{2}{c}=−\mathrm{1} \\ $$$${c}=\frac{−\mathrm{1}}{\mathrm{2}},\:{a}=\frac{−\mathrm{3}}{\mathrm{10}},\:{b}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${f}\left({x}\right)=−\frac{\mathrm{3}}{\mathrm{10}}{x}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{5}}{x}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${f}\left(\mathrm{4}\right)=\frac{−\mathrm{3}}{\mathrm{10}}×\mathrm{16}+\frac{\mathrm{3}}{\mathrm{5}}×\mathrm{4}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{−\mathrm{48}+\mathrm{24}−\mathrm{5}}{\mathrm{10}}=\frac{−\mathrm{29}}{\mathrm{10}} \\ $$

Commented by prakash jain last updated on 25/Nov/15

You can also start with f(x)=ax^2 +bx+(c+1) As long as you treat (c+1) as constant term.

$$\mathrm{You}\:\mathrm{can}\:\mathrm{also}\:\mathrm{start}\:\mathrm{with} \\ $$$${f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+\left({c}+\mathrm{1}\right) \\ $$$$\mathrm{As}\:\mathrm{long}\:\mathrm{as}\:\mathrm{you}\:\mathrm{treat}\:\left(\mathrm{c}+\mathrm{1}\right)\:\mathrm{as}\:\mathrm{constant}\:\mathrm{term}. \\ $$

Commented by Syaka last updated on 25/Nov/15

Is f(x) are f(x) = ax^2 + bx + c + 1, isn′t?

$${Is}\:{f}\left({x}\right)\:{are}\:{f}\left({x}\right)\:=\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:+\:\mathrm{1},\:{isn}'{t}? \\ $$

Commented by prakash jain last updated on 25/Nov/15

For a polynomial c and 1 both are constants so we use ax^2 +bx+c and while comparing compare with ∫_0 ^3 f(x)dx+1.

$$\mathrm{For}\:\mathrm{a}\:\mathrm{polynomial}\:{c}\:{and}\:\mathrm{1}\:{both}\:{are}\:{constants} \\ $$$${so}\:{we}\:{use}\:{ax}^{\mathrm{2}} +{bx}+{c}\:{and}\:{while}\:{comparing} \\ $$$${compare}\:{with}\:\int_{\mathrm{0}} ^{\mathrm{3}} {f}\left({x}\right){dx}+\mathrm{1}. \\ $$

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