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:

Option 1 -

[1, e)

Option 2 -

$[l o g_{e} 2, l o g_{e} 3)$

Option 3 -

$[0, l o g_{e} 2)$

Option 4 -

$[0, 1 / e)$

3 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 3

Detailed Solution:

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

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

->t² + t – 2 = 0 Þ (t + 2) (t – 1) = 0

t = -2 $[e^{- x}]$ -> = -2 not possible

$∴ t = [e^{x}] = 1 \Rightarrow 1 \leq e^{x} < 2$

$\Rightarrow 0 \leq x < l n 2$

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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:

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

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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: