AP EAMCET · Maths · Application of Derivatives
The points on the curve \(y=\frac{2}{3} x^3+\frac{1}{2} x^2\), where the tangents make equal angles with coordinate axes are
- A \(\left(\frac{1}{2}, \frac{5}{24}\right)\) and \(\left(-1, \frac{-1}{6}\right)\)
- B \(\left(1, \frac{7}{6}\right)\) and \(\left(-1, \frac{-1}{6}\right)\)
- C \(\left(2, \frac{22}{3}\right)\) and \(\left(\frac{1}{2}, \frac{5}{24}\right)\)
- D \(\left(1, \frac{7}{6}\right)\) and \(\left(2, \frac{22}{3}\right)\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{1}{2}, \frac{5}{24}\right)\) and \(\left(-1, \frac{-1}{6}\right)\)
Step-by-step Solution
Detailed explanation
\(y=\frac{2}{3} x^3+\frac{1}{2} x^2\) \[ \begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{2}{3}\left(3 x^2\right)+\frac{1}{2}(2 x) \\ & =2 x^2+x \end{aligned} \] \(\because\) Tangents make equal angles with coordinate axis. Hence, \(\frac{d y}{d x}=1\)…
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