MCQ
The equation${e^x} - x - 1 = 0$ has
- ✓Only one real root $x = 0$
- BAt least two real roots
- CExactly two real roots
- DInfinitely many real roots
==> $\frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ...... = 0$
${x^2} = 0,{x^3} = 0,$......${x^n} = 0$
Hence, $x = 0$only one real root.
Trick : Check the equation with options then only option $(a)$ satisfies the equation.
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