Question

Let \([x]\) denote the greatest integer less than or equal to x. Then, the values of \(x \in R\) satisfying the equation \([e^x]^2 + [e^x + 1] - 3 = 0\) lie in the interval :

• A. \(\left[0, \frac{1}{e}\right)\)
• B. \([\log_e 2, \log_e 3)\)
• C. \([1, e)\)
• D. \([0, \log_e 2)\)

Hint:

Use \([e^x + 1] = [e^x] + 1\) and solve the resulting quadratic in \([e^x]\).

Answer 1

The Correct Option is D

Since 1 is an integer, \([e^x + 1] = [e^x] + 1\). Let \(t = [e^x]\).
The equation becomes \(t^2 + t + 1 - 3 = 0 \Rightarrow t^2 + t - 2 = 0 \Rightarrow (t + 2)(t - 1) = 0\), so \(t = 1\) or \(t = -2\).
As \(e^x \gt 0\), \([e^x] \ge 0\), so \(t = -2\) is rejected and \(t = 1\).
\([e^x] = 1 \Rightarrow 1 \le e^x \lt 2 \Rightarrow 0 \le x \lt \log_e 2\).
Answer: (4) \([0, \log_e 2)\)