A) \[6\text{ }L\text{ }HCl(aq)\]
B) \[3\text{ }L\text{ }{{H}_{2}}(g)\]
C) \[33.6\,L\,{{H}_{2}}(g)\]
D) \[Al\]
Correct Answer: B
Solution :
Differentiate w.r.t.\[x\]and it\[f'(x)>0\]for given interval, then the function is increasing. \[\because \]\[f(x)={{\tan }^{-1}}(\sin x+\cos x)\] \[\therefore \]\[f'(x)=\frac{1}{1+{{(\sin x+\cos x)}^{2}}}(\cos x-\sin x)\] \[=\frac{\sqrt{2}\cos \left( x+\frac{\pi }{4} \right)}{1+{{(\sin x+\cos x)}^{2}}}\] \[f(x)\]is increasing, if\[-\frac{\pi }{2}<x+\frac{\pi }{4}<\frac{\pi }{2}\] \[\Rightarrow \]\[-\frac{3\pi }{4}<x<\frac{\pi }{4}\] Hence,\[f(x)\]is increasing when \[x\in \left( -\frac{\pi }{2},\frac{\pi }{4} \right)\]
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