If $f (x) = | x |$ then find the number of solution of $f (x) + 1 = e^{x}$ .

0

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1

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2

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3

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Solution

The correct option is B $1$
Graphs of $| x |$ and $e^{x} - 1$ can be drawn as shown. $e^{x}$ increases faster than $x$ . Hence only one point of intersection.

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