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The tangent to the curve $y=x^3+a x-b$ at the point $(1,-5)$ is perpendicular to the line $y-x+4=0$, then which one of the following points lies on the curve?
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Verified Answer
The correct answer is:
(2,-2)
$y=x^3+a x-b$
$\text { slope of tangent }=\frac{\mathrm{d} y}{\mathrm{~d} x}=3 x^2+a$
slope of the line $y-x+4=0$ is 1
$\begin{aligned}
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}_{a t(1-5)}=-1 \\
& \Rightarrow 3 \times 1^2+a=-1 \\
& \Rightarrow a=-4
\end{aligned}$
also $(1,-5)$ lies on the curve $y=x^3+a x-b$
$\begin{aligned}
& \Rightarrow-5=1^3+a \times 1-b=1+(-4) \times 1-b \\
& \Rightarrow b=2
\end{aligned}$
Hence, the curve is $y=x^3-4 x-2$ which is satisfied by $(2,-2)$
$\text { slope of tangent }=\frac{\mathrm{d} y}{\mathrm{~d} x}=3 x^2+a$
slope of the line $y-x+4=0$ is 1
$\begin{aligned}
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}_{a t(1-5)}=-1 \\
& \Rightarrow 3 \times 1^2+a=-1 \\
& \Rightarrow a=-4
\end{aligned}$
also $(1,-5)$ lies on the curve $y=x^3+a x-b$
$\begin{aligned}
& \Rightarrow-5=1^3+a \times 1-b=1+(-4) \times 1-b \\
& \Rightarrow b=2
\end{aligned}$
Hence, the curve is $y=x^3-4 x-2$ which is satisfied by $(2,-2)$
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