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JEE Advanced 2007Paper 2
Exam
JEE Advanced
Year
2007
Paper
Paper 2
Subject
Mathematics
Chapter
Functions

JEE Advanced 2007 · Paper 2 · Mathematics · Functions

Paragraph: If a continuous f defined on the real line R , assume...

Paragraph: If a continuous ff defined on the real line RR, assume positive and negative values in RR, then the equation f(x)=0f(x)=0 has a root in RR. For example, if it is known that a continuous function ff on RR is positive at some point and its minimum values is negative, then the equation f(x)=0f(x)=0 has a root in RR. Consider f(x)=kex−xf(x)=k e^x-x for all real xx, where kk is real constant.Question: The positive value of kk for which kex−x=0k e^x-x=0 has only one root is

Options

  1. A1e\frac{1}{e}
  2. B1
  3. Cee
  4. Dlog⁡e2\log _e 2

Correct answer

A. 1e\frac{1}{e}

Step-by-step solution

Let f(x)=kex−xf(x)=k e^x-xf′(x)=kex−1=0x=−ln⁡k∴f′′(x)=kex Hence, [f′′(x)]x=−ln⁡k=1>0f(−ln⁡k)=1+ln⁡k\begin{aligned} & & f^{\prime}(x) & =k e^x-1=0 \\ & & x & =-\ln k \\ & \therefore & f^{\prime \prime}(x) & =k e^x \\ & \text{ Hence, } & {\left[f^{\prime \prime}(x)\right]_{x=-\ln k} } & =1 \gt 0 \\ & & f(-\ln k) & =1+\ln k\end{aligned}ForonerootofgivenequationFor one root of given equation\begin{aligned}1+\ln k & =0 \\ Hence, k & =\frac{1}{e} . \end{aligned}$$
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