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

Which of the followings are true?

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

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

**
C.
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$~af(b)-bf(a)=(a-b)R(0).$

*
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

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

$\therefore~\boxed{\text{TRUE}}$

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

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

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

C.Let $R(x)=px+q.$

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

$\therefore~\boxed{\text{TRUE}}$

From the followings, only

AandCare correct.