MHT CET · Maths · Application of Derivatives
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?
- A (2,-2)
- B (-2,2)
- C (-2,1)
- D (2,-1)
Answer & Solution
Correct Answer
(A) (2,-2)
Step-by-step Solution
Detailed explanation
\(y=x^3+a x-b\)
\(\text { slope of tangent }=\frac{\mathrm{d} y}{\mathrm{~d} x}=3 x^2+a\)
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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