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 $[-2,2]$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now