TS EAMCET · Maths · Application of Derivatives
The acute angle between the curves \(y=3 x^2-2 x-1\) and \(y=x^3-1\) at their point of intersection which lies in the first quadrant is
- A \(\operatorname{Tan}^{-1}\left(\frac{2}{121}\right)\)
- B \(\operatorname{Tan}^{-1}(2)\)
- C \(\operatorname{Tan}^{-1}\left(\frac{1}{13}\right)\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(A) \(\operatorname{Tan}^{-1}\left(\frac{2}{121}\right)\)
Step-by-step Solution
Detailed explanation
\(3x^2 - 2x - 1 = x^3 - 1\) \(x^3 - 3x^2 + 2x = 0 \Rightarrow x(x-1)(x-2) = 0\) Point of intersection in first quadrant: \((2, 7)\) \(m_1 = \frac{d}{dx}(3x^2 - 2x - 1) = 6x - 2\) \(m_1 = 6(2) - 2 = 10\) \(m_2 = \frac{d}{dx}(x^3 - 1) = 3x^2\) \(m_2 = 3(2)^2 = 12\)…
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