TS EAMCET · Maths · Application of Derivatives
The slope of the normal drawn at a point \(P\) to the curve \(y\) \(=x^3-10 x^2+31 x-30\) is \(-\frac{1}{14}\)
- A \(\frac{-11}{7}\)
- B \(22\)
- C \(\frac{11}{7}\)
- D \(-22\)
Answer & Solution
Correct Answer
(C) \(\frac{11}{7}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text y=x^3-10 x^2+31 x-30 \\ & \frac{d y}{d x}=3 x^2-20 x+31 \\ & \therefore \quad \text { Slope of normal }=\frac{-1}{\left(\frac{d y}{d x}\right)}\end{aligned}\)…
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