JEE MAIN - Mathematics (2021 - 22th July Evening Shift - No. 9)
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 :
$$\left[ {0,{1 \over e}} \right)$$
[loge2, loge3)
[1, e)
[0, loge2)
Explanation
$${[{e^x}]^2} + [{e^x} + 1] - 3 = 0$$
$$ \Rightarrow {[{e^x}]^2} + [{e^x}] + 1 - 3 = 0$$
Let $$[{e^x}] = t$$
$$ \Rightarrow {t^2} + t - 2 = 0$$
$$ \Rightarrow t = - 2,1$$
$$[{e^x}] = - 2$$ (Not possible)
or $$[{e^x}] = 1$$ $$\therefore$$ $$1 \le {e^x} < 2$$
$$ \Rightarrow \ln (1) \le x < \ln (2)$$
$$ \Rightarrow 0 \le x < \ln (2)$$
$$ \Rightarrow x \in [0,\ln 2)$$
$$ \Rightarrow {[{e^x}]^2} + [{e^x}] + 1 - 3 = 0$$
Let $$[{e^x}] = t$$
$$ \Rightarrow {t^2} + t - 2 = 0$$
$$ \Rightarrow t = - 2,1$$
$$[{e^x}] = - 2$$ (Not possible)
or $$[{e^x}] = 1$$ $$\therefore$$ $$1 \le {e^x} < 2$$
$$ \Rightarrow \ln (1) \le x < \ln (2)$$
$$ \Rightarrow 0 \le x < \ln (2)$$
$$ \Rightarrow x \in [0,\ln 2)$$
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