The equation $e^{x} - x - 1 = 0$ has

Only one real root x=0

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At least two real roots

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Exactly two real roots

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Infinitely many real roots

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Solution

The correct option is A Only one real root x=0
$e^{x} = x + 1 \Rightarrow 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + . . . . . . . . = x + 1$
$\Rightarrow \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . . . . = 0$
$x^{2} = 0, x^{3} = 0, . . . . . . . . x^{n} = 0$

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