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

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Question

(a) ${e^x} = x + 1 \Rightarrow 1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + ...... = x + 1$

==> $\frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ..

Vedclass · Accepted Answer

Only one real root $x = 0$

Vedclass · Answer

(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.

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

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