Complex Numbers And Quadratic Equations question 746
Question: The equation $ e^{x}-x-1=0 $ has [Kurukshetra CEE 1998]
Options:
A) Only one real root at $ x=0 $
B) At least two real roots
C) Exactly two real roots
D) Infinitely many real roots
Show Answer
Answer:
Correct Answer: A
Solution:
$ e^{x}=1+\frac{x}{1!}+\frac{x^{2}}{2!}+……\Rightarrow 1+\frac{x}{1!}+\frac{x^{2}}{2!}+……=e^{x}$ Þ $ \frac{x^{2}}{2!}+\frac{x^{3}}{3!}+……=0 $ $ x^{2}=0,x^{3}=0, $ …… $ x^{n}=0 $ Hence, $ x=0 $ is the only real root. Trick: Check the equation with options then only option (a) satisfies the equation.
BETA
