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4-2.Quadratic Equations and Inequations
medium
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
