KCET · Maths · Application of Derivatives
If \(f(x)=x e^{x(1-x)}\), then \(f(x)\) is
- A increasing in \(R\)
- B decreasing in \(R\)
- C decreasing in \(\left[-\frac{1}{2}, 1\right]\)
- D increasing in \(\left[-\frac{1}{2}, 1\right]\)
Answer & Solution
Correct Answer
(D) increasing in \(\left[-\frac{1}{2}, 1\right]\)
Step-by-step Solution
Detailed explanation
Given, \(f(x)=x e^{x(1-x)}\)
\(f^{\prime}(x)=e^{x(1-x)}+x e^{x(1-x)}(1-2 x)\)
\(=e^{x(1-x)}(1+x(1-2 x))\)
\(=-e^{x(1-x)}\left(2 x^2-x-1\right)\)
\(=-e^{x(1-x)}\left(2 x^2-2 x+x-1\right)\)
\(=-e^{x(1-x)}(x-1)(2 x+1)\)
So, \(f(x)\) is increasing when \(f^{\prime}(x) \geq 0\) and decreasing when \(f^{\prime}(x) \leq 0\).
Hence, \(f\) is increasing in \(\left[\frac{-1}{2}, 1\right]\).
\(f^{\prime}(x)=e^{x(1-x)}+x e^{x(1-x)}(1-2 x)\)
\(=e^{x(1-x)}(1+x(1-2 x))\)
\(=-e^{x(1-x)}\left(2 x^2-x-1\right)\)
\(=-e^{x(1-x)}\left(2 x^2-2 x+x-1\right)\)
\(=-e^{x(1-x)}(x-1)(2 x+1)\)
So, \(f(x)\) is increasing when \(f^{\prime}(x) \geq 0\) and decreasing when \(f^{\prime}(x) \leq 0\).
Hence, \(f\) is increasing in \(\left[\frac{-1}{2}, 1\right]\).
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