Let -x- denote the greatest integer less than or equal to x- Then- the values of x - satisfying the equation -ex-2-ex-1-3-0 lies in the interval-

Given: [e^x]^2+[e^x+1] 3=0 ⇒ [e^x]^2+[e^x] 2=0 ⇒ ([e^x]+2)([e^x] 1)=0 ∴ [e^x]=1, since [e^x]= 2 is not possible. ∴ e^x ∈ [1,2) ∴ x ∈ [0, log_e 2) nbsp; ...

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Standard XIII

Mathematics

Squaring an Inequality

Let [x] denot...

Question

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

[0, \frac{1}{e})

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[0, {log}_{e} 2)

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[{log}_{e} 2, {log}_{e} 3)

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[1, e)

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Solution

The correct option is B $[0, {log}_{e} 2)$
Given: $[e^{x}]^{2} + [e^{x} + 1] - 3 = 0$
$\Rightarrow [e^{x}]^{2} + [e^{x}] - 2 = 0$
$\Rightarrow ([e^{x}] + 2) ([e^{x}] - 1) = 0$
$∴ [e^{x}] = 1$ , since $[e^{x}] = - 2$ is not possible.
$∴ e^{x} \in [1, 2)$
$∴ x \in [0, {log}_{e} 2)$

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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 [ex]2+[ex+1]−3=0 lies in the interval:

The correct option is B [0,loge2)Given: [ex]2+[ex+1]−3=0 ⇒[ex]2+[ex]−2=0 ⇒([ex]+2)([ex]−1)=0 ∴[ex]=1, since [ex]=−2 is not possible. ∴ex∈[1,2) ∴x∈[0,loge2)

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

The correct option is B $[0, {log}_{e} 2)$
Given: $[e^{x}]^{2} + [e^{x} + 1] - 3 = 0$
$\Rightarrow [e^{x}]^{2} + [e^{x}] - 2 = 0$
$\Rightarrow ([e^{x}] + 2) ([e^{x}] - 1) = 0$
$∴ [e^{x}] = 1$ , since $[e^{x}] = - 2$ is not possible.
$∴ e^{x} \in [1, 2)$
$∴ x \in [0, {log}_{e} 2)$