The correct option is C a=1
Let f(x)=ax4+bx3−x2+2x+3
Using remainder theorem,
f(x)=(x2+x−2)Q(x)+4x+3
f(x)=(x+2)(x−1)Q(x)+4x+3
∴f(−2)=4(−2)+3=−5
⇒a(−2)4+b(−2)3−(−2)2+2(−2)+3=−8+3
⇒16a−8b−4−4+3=−5
⇒16a−8b−5=−5
⇒16a−8b=0
⇒2a=b .....(i)
And f(1)=4(1)+3
⇒a(1)4+b(1)3−(1)2+2(1)+3=4+3
⇒a+b−1+2+3=7
⇒a+b=3 .....(ii)
On solving (i) and (ii),
2a=b and a+b=3,
we get a=1,b=2