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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-




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..... ?
iffunctionfsatisfyform:f(x)=(10f(x)dx)x2+(20f(x)dx)x+(30f(x)dx)+1thenthevalueoff(4)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(x)=ax2+bx+c.(1)f(x)=ax33+bx22+cx+C01f(x)=a3+b2+c02f(x)=8a3+2b+2c03f(x)=9a+92b+3cf(x)=(a3+b2+c)x2+(8a3+2b+2c)x+(9a+92b+3c)+1..(2)Equatingcoefficientsin(1)and(2)(x2coeffiient)a3+b2+c=a2a3=b2+ca=34b+32c(xcoefficient)8a3+2b+2c=b2b+4c+2b+2c=b5b=6cb=6c5a=3b4+3c2a=34×6c5+3c2=12c20=3c5(constantterm)(9a+92b+3c)+1=cTobesolvedfora,bandc27c592×6c5+2c=1c=12,a=310,b=35f(x)=310x2+35x12f(4)=310×16+35×412=48+24510=2910
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.
Youcanalsostartwithf(x)=ax2+bx+(c+1)Aslongasyoutreat(c+1)asconstantterm.
Commented by Syaka last updated on 25/Nov/15
Is f(x) are f(x) = ax^2  + bx + c + 1, isn′t?
Isf(x)aref(x)=ax2+bx+c+1,isnt?
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.
Forapolynomialcand1bothareconstantssoweuseax2+bx+candwhilecomparingcomparewith03f(x)dx+1.

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