Differentiation and its Application: For k>0 , the set of all values of k for which ke^x-x=0 has two… | 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

For $k>0$, the set of all values of $k$ for which $ke^x-x=0$ has two distinct roots is

Differentiation and its Application

$\left(0,\frac1e\right)$✓ Correct answer$\left(\frac1e,1\right)$$\left(\frac1e,\infty\right)$$(0,1)$

Solution

Concept. Here $f(x)=ke^x-x$; its roots are the meeting points of $y=ke^x$ and $y=x$. Separate $k$: the roots are the meeting points of the level $y=k$ with $y=xe^{-x}$. Formulas used. (1) $\dfrac{d}{dx}\,xe^{-x}=(1-x)e^{-x}$Step 1 - Rewrite. $ke^x=x\iff k=xe^{-x}$. Step 2 - Shape. By (1), $xe^{-x}$ increases for $x<1$ and decreases for $x>1$; maximum $\frac1e$ at $x=1$; it tends to $0^+$ as $x\to\infty$. Step 3 - Count (vertical scale stretched):For $0<k<\frac1e$ the level cuts the curve twice; for $k=\frac1e$ once; for $k>\frac1e$ never. So $k\in\left(0,\frac1e\right)$. Answer: (A)

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

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