JEE Mains · Maths · STD 12 - 6. Application of derivatives
The number of critical points of the function \(f(x)=(x-2)^{2 / 3}(2 x+1)\) is :
- A \(2\)
- B \(0\)
- C \(1\)
- D \(3\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
\( f(x)=(x-2)^{2 / 3}(2 x+1) \) \( f^{\prime}(x)=\frac{2}{3}(x-2)^{-1 / 3}(2 x+1)+(x-2)^{2 / 3} \) \( f^{\prime}(x)=2 \times \frac{(2 x+1)+(x-2)}{3(x-2)^{1 / 3}} \) \( \frac{3 x-1}{(x-2)^{1 / 3}}=0\) Critical points \(\mathrm{x}=\frac{1}{3}\) and \(\mathrm{x}=2\)
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