#19 of Sept 2015 Grade 10 CSAT (Korean SAT) Mock test

Algebra Level 4

f ( x ) f(x) is a quartic, and R ( x ) R(x) is the remainder of the division f ( x ) ÷ ( x − a ) ( x − b ) . f(x)\div (x-a)(x-b).

Which of the followings are true?

A. f ( a ) − R ( a ) = 0. ~f(a)-R(a)=0.

B. f ( a ) − R ( b ) = f ( b ) − R ( a ) . ~f(a)-R(b)=f(b)-R(a).

C. a f ( b ) − b f ( a ) = ( a − b ) R ( 0 ) . ~af(b)-bf(a)=(a-b)R(0).

This problem is a part of <Grade 10 CSAT Mock test> series .

Only A and B All of them Only B Only B and C Only C Only A None of them Only C and A

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Boi (보이)
Jul 20, 2017

A.

f ( x ) = ( x − a ) ( x − b ) Q ( x ) + R ( x ) , f(x)=(x-a)(x-b)Q(x)+R(x), and therefore, f ( a ) = R ( a ) . f(a)=R(a).

∴ TRUE \therefore~\boxed{\text{TRUE}}

B.

f ( a ) = R ( a ) , f ( b ) = R ( b ) . f(a)=R(a),~f(b)=R(b).

The equation holds when f ( a ) = f ( b ) , f(a)=f(b), which does not make any sense.

∴ FALSE \therefore~\boxed{\text{FALSE}}

C.

Let R ( x ) = p x + q . R(x)=px+q.

Note that R ( 0 ) = q , R(0)=q, and we see that a f ( b ) − b f ( a ) = a R ( b ) − b R ( a ) = a b p + a q − a b p − b q = a q − b q = ( a − b ) q = ( a − b ) R ( 0 ) . af(b)-bf(a)=aR(b)-bR(a)=abp+aq-abp-bq=aq-bq=(a-b)q=(a-b)R(0).

∴ TRUE \therefore~\boxed{\text{TRUE}}

From the followings, only A and C are correct.

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