0,\INFTY )\] \[f(x)={{E}^{x}}-x\log x\] \[f'(x)={{e}^{x}}-(\log \,x+1)\] From the FIGURE it is clear that \[f'(x)>0\] on \[(1,\infty )\]. So both statements (1) & (2) are correct.

"> 0,\INFTY )\] \[f(x)={{E}^{x}}-x\log x\] \[f'(x)={{e}^{x}}-(\log \,x+1)\] From the FIGURE it is clear that \[f'(x)>0\] on \[(1,\infty )\]. So both statements (1) & (2) are correct.

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Consider the following statements: 1. \[f(x)=\] ln x is an increasing function on \[\left( 0,\infty \right).\] 2. \[f(x)={{e}^{x}}-x(ln\,\,x)\] is an increasing function on \[\left( 1,\,\infty \right)\]. Which of the above statements is/are correct?

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) . 11 months ago

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[c] \[f(x)=\log \,x\] Clearly \[f(x)\] is increasing on \[(0,\INFTY )\] \[f(x)={{E}^{x}}-x\log x\] \[f'(x)={{e}^{x}}-(\log \,x+1)\] From the FIGURE it is clear that \[f'(x)>0\] on \[(1,\infty )\]. So both statements (1) & (2) are correct.

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Posted on 04 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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