The values of f(x)=2
Question:

The values of $f(x)=2 \sin \sqrt{x^{2}+x+1}$ lie in the interval _________________ .

Solution:

$f(x)=2 \sin \sqrt{x^{2}+x+1}$

Since $-1 \leq \sin \theta \leq 1$

i. $\mathrm{e}-1 \leq \sin \sqrt{x^{2}+x+1} \leq 1$

i. $\mathrm{e}-2 \leq 2 \sin \sqrt{x^{2}+x+1} \leq 2$

i. e $2 \sin \sqrt{x^{2}+x+1} \in[-2,2]$

i. e. $2 \sin \sqrt{x^{2}+x+1}$ lies in interval $[-2,2]$

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