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Differentiation and its Application - JEE Mathematics question (JEE Advanced 2007 Paper 2)

Comprehension · Questions 1–1

Read the passage given below and answer the questions that follow.

If a continuous function $f$ defined on the real line $\mathbf R$, assumes positive and negative values in $\mathbf R$ then the equation $f(x)=0$ has a root in $\mathbf R$. For example, if it is known that a continuous function $f$ on $\mathbf R$ is positive at some point and its minimum value is negative then the equation $f(x)=0$ has a root in $\mathbf R$. Consider $f(x)=ke^x-x$ for all real $x$ where $k$ is a real constant.
JEE Advanced 2007 Paper 2Question 1
The positive value of $k$ for which $ke^x-x=0$ has only one root is
Differentiation and its Application
Options for question 1, choose one

Solution

Concept. Here $f(x)=ke^x-x$; its roots are the meeting points of $y=ke^x$ and $y=x$. For $k>0$ there is exactly one root when $y=ke^x$ touches $y=x$. Formulas used. (1) Touching: equal values and equal slopesStep 1 - Equations. By (1): $ke^x=x$ and $ke^x=1$. Step 2 - Solve. So $x=1$ and $k=\frac1e$. Answer: (A)

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