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Correct option is B)

$(1,−5)$ lies on the curve

$⇒−5=1+a−b⇒a−b=−6$ .(i)

Also, $y_{′}=3x_{2}+a$

$y_{(1,−5)}=3+a$ (slope of tangent)

$∵$ this tangent is $⊥$ to $−x+y+4=0$

$⇒(3+a)(1)=−1$

$⇒a=−4$ .(ii)

By (i) and (ii): $a=−4$, $b=2$

$∴y=x_{3}−4x−2$

$(2,−2)$ lies on this curve.

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