IIT-JEE 2007 · Paper 2 · Mathematics Q62

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MathematicsSingle correct+4 / -1Question 62 of 66

Passage

Paragraph for Question Nos. 61 to 63

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.

The positive value of $k$ for which $ke^x-x=0$ has only one root is

\frac1e

e

\log_e2

Answer and worked solution

Official answer: A

Tangency. For $k>0$ , $f(x)=ke^x-x$ is convex with minimum where $ke^x=1$ , i.e. $x=-\ln k$ , and the minimum value is $1+\ln k$ .

One root exactly when the minimum is zero: $1+\ln k=0$ , $k=\dfrac1e$ (the line $y=x$ then touches $y=\frac1ee^x$ at $x=1$ ).

(A)

Topics:Maxima and minima Tangents and normals

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IIT-JEE 2007 · Paper 2, Mathematics question 62