IIT-JEE 2007 · Paper 2 · Mathematics Q61

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MathematicsSingle correct+4 / -1Question 61 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 line $y=x$ meets $y=ke^x$ for $k\le0$ at

no point

one point

two points

more than two points

Answer and worked solution

Official answer: B

Let $f(x)=ke^x-x$ .

$k=0$ : $f(x)=-x$ has the single root 0.
$k<0$ : $f'(x)=ke^x-1<0$ , so $f$ is strictly decreasing, from $+\infty$ (as $x\to-\infty$ ) to $-\infty$ : exactly one root.

One point of intersection.

(B)

Topics:Monotonicity

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IIT-JEE 2007 · Paper 2, Mathematics question 61