Question

log√5(√2(sinx-cosx)+3)​

Answer 1

Step-by-step explanation:

$$y = \log_\sqrt{5} (\sqrt{2} (\sin x - \cos x) + 3)$$

We know, −

2

≤sinx−cosx≤

2

Hence domain is x∈R

and Range $$[\log_\sqrt{5} (\sqrt{2} (-\sqrt{2}), \log_\sqrt{5} (\sqrt{2} (\sqrt{2}) + 3)]=[\log_\sqrt{5}1, \log_\sqrt{5} 5] = [0,2]$$