Let [x] denote the greatest integer less than or equal to x.

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 :

Accepted Answer

$${[{e^x}]^2} + [{e^x} + 1] - 3 = 0$$

$$ \Rightarrow {[{e^x}]^2} + [{e^x}] + 1 - 3 = 0$$

Let $$[{e^x}] = t$$

$$ \Rightarrow {t^2} + t - 2 = 0$$

$$ \Rightarrow t =  - 2,1$$

$$[{e^x}] =  - 2$$ (Not possible)

or $$[{e^x}] = 1$$ $$\therefore$$ $$1 \le {e^x} < 2$$

$$ \Rightarrow \ln (1) \le x < \ln (2)$$

$$ \Rightarrow 0 \le x < \ln (2)$$

$$ \Rightarrow x \in [0,\ln 2)$$

Solutions · 1