• # question_answer If the tangent to the curve, $y={{x}^{3}}+axb$ at the point (1, -5) is perpendicular to the line, $x+y+4=0,$then which one of the following points lies on the curve? [JEE Main 9-4-2019 Morning] A) (-2, 2)             B) (2, -2)C) (2, -1)             D) (-2, 1)

$y={{x}^{3}}+ax-b$ (1, -5) lies on the curve $\Rightarrow -5=1+a-b\Rightarrow a-b=-6$                         ... (i)             Also,$y'=3{{x}^{2}}+a$ $y{{'}_{\left( 1,5 \right)}}=3+a$               (slope of tangent) $\because$this tangent is $\bot$to $x+y+4=0$ $\Rightarrow (3+a)$(1) = -1 $\Rightarrow a=-4$                                        ....(ii) By (i) and (ii) : $a=-4,b=2$ $\therefore$$y={{x}^{3}}-4x-2.$ (2,-2) lies on this curve.