0 votes 0 votes If $f(x) = \dfrac{\sqrt{3}\sin x}{2+\cos x}$, then the range of $f(x)$ is the interval $[-1, \sqrt{3}/2]$ the interval $[- \sqrt{3}/2, 1]$ the interval $[-1, 1]$ none of the above Calculus isi2015-mma calculus functions range trigonometry non-gate + – Arjun asked Sep 23, 2019 recategorized Nov 16, 2019 by Lakshman Bhaiya Arjun 571 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply `JEET commented Sep 24, 2019 reply Follow Share is $C$ the answer. 0 votes 0 votes Lakshman Bhaiya commented Sep 24, 2019 i edited by Lakshman Bhaiya Sep 24, 2019 reply Follow Share Yes, you are right. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Note that $y=f(x)=\dfrac{\sqrt 3 \sin x}{2+\cos x}$. Note that $y^2=\dfrac{3 \sin ^2 x}{(2+\cos ^2 x)^2}=\dfrac{3 (1- \cos ^2 x)}{(2+\cos ^2 x)^2}$ Put $\cos x =u$, we have \begin{align} y^2(2+u)^2 =3(1-u^2) \\ \implies (3+y^2)u^2+4y^2u+4y^2-3=0 \end{align} For this to have a solution, we need discriminant of the quadratic to be non-negative. Hence, we get $$(4y)^2-4(3+y^2)(4y^2-3) \ge 0 \\ \implies y^2 \le 1\\ \implies-1 \le y \le 1$$ So, the correct answer is $(C)$. Source from: https://math.stackexchange.com/questions/382136/a-problem-on-range-of-a-trigonometric-function-what-is-the-range-of-frac-sqr/383172#383172 Lakshman Bhaiya answered Sep 24, 2019 Lakshman Bhaiya comment Share Follow See all 0 reply Please log in or register to add a comment.