Equation {e^x} - x - 1 = 0 has

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4-2.Quadratic Equations and Inequations

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

A

Only one real root $x = 0$

B

At least two real roots

C

Exactly two real roots

D

Infinitely many real roots

Solution

(a) ${e^x} = x + 1 \Rightarrow 1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + …… = x + 1$

==> $\frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + …… = 0$

${x^2} = 0,{x^3} = 0,$……${x^n} = 0$

Hence, $x = 0$only one real root.

Trick : Check the equation with options then only option $(a)$ satisfies the equation.

Standard 11

Mathematics

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