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JEE Mains · Maths · STD 12 - 6. Application of derivatives
Statement \(- 1:\) The function \(x^2 (e^x + e^{-x})\) is increasing for all \(x > 0.\) Statement \(-2:\) The functions \(x^2e^x\) and \(x^2e^{-x}\) are increasing for all \(x > 0\) and the sum of two increasing functions in any interval \((a, b)\) is an increasing function in \((a, b).\)
- A Statement \(-1\) is false; Statement \(-2\) is true.
- B Statement \(-1\) is true; Statement \(- 2\) is true;Statement \(-2\) is not a correct explanation for Statement \(- 1.\)
- C Statement \(-1\) is true; Statement \(-2\) is false.
- D Statement \(- 1\) is true; Statement \(-2\) is true; Statement \(-2\) is a correct explanation for statement \(-1.\)
Answer & Solution
Correct Answer
(C) Statement \(-1\) is true; Statement \(-2\) is false.
Step-by-step Solution
Detailed explanation
Let \(y=x^{2} \cdot e^{-x}\) For increasing function, \(\frac{d y}{d x}>0 \Rightarrow x\left[(2-x) e^{-x}\right]>0\) \(\because x>0, \therefore(2-x) e^{-x}>0\) \(\Rightarrow(2-x) \frac{1}{e^{x}}>0\) For \(0 < x < 2,\,\,\,(2 - x) < 0\) \(\therefore \frac{1}{e^{x}}<0,\) but it is…
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