1 votes 1 votes For existence of f(x) = , x lies in (take n I) (1) [0, 2nπ] (2) (3) (4) None of these how to solve such question? Pl give detailed explanation. Always find difficult to solve Range and domain of such complex question. Linear Algebra linear-algebra range domain trigonomatric function modulus + – stanchion asked Dec 19, 2017 stanchion 549 views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply joshi_nitish commented Dec 19, 2017 reply Follow Share for existence of f(x), $|\cos x|+\cos x\geq 0$ ,now, $|\cos x|+\cos x= \begin{cases} 2\cos x(+ve) & \text{ if } \cos x\geq 0\\ -\cos x+\cos x=0(+ve)& \text{ if } \cos x< 0 \end{cases}$ now, see above no matter what is value of 'x', f(x) is always defined. so domain of f(x) = $\left ( -\infty ,+\infty \right )$ 0 votes 0 votes stanchion commented Dec 20, 2017 reply Follow Share @ joshi_nitish Veteran I guess, you have stated wrong. Domain is calc for 'x' which is given as option (2) as answer . What you have mentioned in your last line is Range of F(x) = (−∞,+∞) , as range is associated with f(x). So again "how to get the domain of x? " or "x lies in ______ ? " 0 votes 0 votes joshi_nitish commented Dec 20, 2017 reply Follow Share @stanchion, read my answer completely, not just a last line and then comment again. 0 votes 0 votes Xylene commented Dec 20, 2017 i edited by Xylene Dec 20, 2017 reply Follow Share @stanchion, By saying f(x) exists it means that the expression inside the root should be >=0. If that's the case then domain of x is as @nitish said. If by f(x) exists, they mean that expression is strictly greater than 0 (not equal to 0) as roots of negative numbers are complex, then answer should be option b) according to me with the values inside open interval and not closed interval. 0 votes 0 votes hs_yadav commented Dec 20, 2017 i edited by hs_yadav Dec 20, 2017 reply Follow Share $\left | cos x \right | +cos x\geq 0$ or $\left | cos x \right | \geq -cos x $ and this is trivial property of inequality ...true for all value of x ....therefor x might be in the range.. -$\infty < x <\infty$ 0 votes 0 votes joshi_nitish commented Dec 20, 2017 reply Follow Share @hs_yadav $x\geq \infty$ is this even possible ?? 1 votes 1 votes hs_yadav commented Dec 20, 2017 reply Follow Share ...... :) 0 votes 0 votes stanchion commented Dec 21, 2017 reply Follow Share @ joshi_nitish Veteran But given answer is option (B) ie. x lies in { } I don't know, How? 0 votes 0 votes Please log in or register to add a comment.