A) Only one real root \[x=0\]
B) At least two real roots
C) Exactly two real roots
D) Infinitely many real roots
Correct Answer: A
Solution :
\[{{e}^{x}}=x+1\Rightarrow 1+\frac{x}{1!}+\frac{{{x}^{2}}}{2!}+......=x+1\] Þ \[\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+......=0\] \[{{x}^{2}}=0,{{x}^{3}}=0,\]......\[{{x}^{n}}=0\] Hence, \[x=0\]only one real root. Trick: Check the equation with options then only option (a) satisfies the equation.
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