JEE Advanced · Mathematics · 22. Functions
Paragraph:
If a continuous \(f\) defined on the real line \(R\), assume positive and negative values in \(R\), then the equation \(f(x)=0\) has a root in \(R\). For example, if it is known that a continuous function \(f\) on \(R\) is positive at some point and its minimum values is negative, then the equation \(f(x)=0\) has a root in \(R\).
Consider \(f(x)=k e^x-x\) for all real \(x\), where \(k\) is real constant.Question:
The positive value of \(k\) for which \(k e^x-x=0\) has only one root is
- A
\(\frac{1}{e}\)
- B
1
- C
\(e\)
- D
\(\log _e 2\)
Answer & Solution
Correct Answer
(A)
\(\frac{1}{e}\)
Step-by-step Solution
Detailed explanation
Let \(f(x)=k e^x-x\)
\(\begin{aligned} & & f^{\prime}(x) & =k e^x-1=0 \\ & & x & =-\ln k \\ & \therefore & f^{\prime \prime}(x) & =k e^x \\ & \text { Hence, } & {\left[f^{\prime \prime}(x)\right]_{x=-\ln k} } & =1>0 \\ & & f(-\ln k) & =1+\ln k\end{aligned}\)
For one root of given equation
\[
\begin{aligned}
1+\ln k & =0 \\
Hence,
k & =\frac{1}{e} .
\end{aligned}
\]
\(\begin{aligned} & & f^{\prime}(x) & =k e^x-1=0 \\ & & x & =-\ln k \\ & \therefore & f^{\prime \prime}(x) & =k e^x \\ & \text { Hence, } & {\left[f^{\prime \prime}(x)\right]_{x=-\ln k} } & =1>0 \\ & & f(-\ln k) & =1+\ln k\end{aligned}\)
For one root of given equation
\[
\begin{aligned}
1+\ln k & =0 \\
Hence,
k & =\frac{1}{e} .
\end{aligned}
\]
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