A) \[\left[ \frac{1}{3},3 \right]\]
B) R
C) \[R-\left( \frac{1}{3},3 \right)\]
D) None of these
Correct Answer: A
Solution :
[a] \[f\left( x \right)=\frac{1+{{\tan }^{2}}x-\tan x}{1+{{\tan }^{2}}x+\tan x}=\frac{{{t}^{2}}-t+1}{{{t}^{2}}+t+1}\] Where \[t=tan\text{ }x\] can assume any real x. \[\therefore y=\frac{{{t}^{2}}-t+1}{{{t}^{2}}+t+1}\] \[\Rightarrow (y-1){{t}^{2}}+(y+1)t+y-1=0,t\in R\] \[\Rightarrow {{(y+1)}^{2}}-4{{(y-1)}^{2}}\ge 0\] \[\Rightarrow 3{{y}^{2}}-10y+3\le 0\] \[\Rightarrow 3\left( y-\frac{1}{3} \right)(y-3)\le 0\] \[\Rightarrow \frac{1}{3}\le y\le 3\]
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