117453
The domain of the function $(x)=\frac{\log _2(x+3)}{x^2+3 x+2}$ is
1 $\mathrm{R}-\{-1,-2\}$
2 $\mathrm{R}-\{-1,-2,0\}$
3 $(-3,-1) \cup(-1, \infty)$
4 $(-3, \infty)-\{-1,-2\}$
5 $(0, \infty)$
Explanation:
D Given function, $f(x)=\frac{\log _2(x+3)}{\left(x^2+3 x+2\right)}$ Define the above function as, $\lvert\, \begin{aligned} & x^2+3 x+2 \neq 0 \\ & (x+1)(x+2) \neq 0 \\ & x \neq-1,-2 \end{aligned}.$ Then, \(\begin{aligned} & =x+3>0 \\ & =x>-3 \end{aligned}\) Since, the domain of \(f(x)\) is - \((-3, \infty)-\{-1,-2\}\)
Kerala CEE-2013
Sets, Relation and Function
117454
The domain of the function \(f(x)=\sin ^{-1}\left(\frac{x+5}{2}\right)\) is
1 \([-1,1]\)
2 \([2,3]\)
3 \([3,7]\)
4 \([-7,-3]\)
5 \((-\infty, \infty)\)
Explanation:
D Given, \(f(x)=\sin ^{-1}\left(\frac{x+5}{2}\right)\) \(\left[\because\right.\) Domain of \(\sin ^{-1} \mathrm{x}\) is \(\mathrm{x} \in[-1,1]\) \(\therefore\) For \(f(x)\) to be defined is \(\begin{aligned} & \Rightarrow-1 \leq \frac{\mathrm{x}+5}{2} \leq 1 \\ & \Rightarrow-2 \leq \mathrm{x}+5 \leq 2 \\ & \Rightarrow-2-5 \leq \mathrm{x} \leq 2-5 \\ & \Rightarrow-7 \leq \mathrm{x} \leq-3 \\ & \Rightarrow \mathrm{x} \in[-7,-3] \end{aligned}\)
Kerala CEE-2012
Sets, Relation and Function
117455
The range of the function \(f(x)=\frac{x^2+8}{x^2+4}, x \in R\) is
1 \(\left[-1, \frac{3}{2}\right]\)
2 \((1,2]\)
3 \((1,2)\)
4 \([1,2]\)
5 \(\left[\frac{3}{2}, 2\right]\)
Explanation:
B Let, \(\mathrm{y}=\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^2+8}{\mathrm{x}^2+4}\) \(\Rightarrow \mathrm{x}^2(\mathrm{y}-1)+4 \mathrm{y}=8\) \(x^2=\frac{8-4 y}{y-1}\) \(x=\sqrt{\frac{8-4 y}{y-1}}\) Since, x is real, \(\therefore \mathrm{y}-1>0\) and \(8-4 y \geq 0\) \(y>1\) and \(y \leq 2\) \(\mathrm{y} \in(1,2]\)
117453
The domain of the function $(x)=\frac{\log _2(x+3)}{x^2+3 x+2}$ is
1 $\mathrm{R}-\{-1,-2\}$
2 $\mathrm{R}-\{-1,-2,0\}$
3 $(-3,-1) \cup(-1, \infty)$
4 $(-3, \infty)-\{-1,-2\}$
5 $(0, \infty)$
Explanation:
D Given function, $f(x)=\frac{\log _2(x+3)}{\left(x^2+3 x+2\right)}$ Define the above function as, $\lvert\, \begin{aligned} & x^2+3 x+2 \neq 0 \\ & (x+1)(x+2) \neq 0 \\ & x \neq-1,-2 \end{aligned}.$ Then, \(\begin{aligned} & =x+3>0 \\ & =x>-3 \end{aligned}\) Since, the domain of \(f(x)\) is - \((-3, \infty)-\{-1,-2\}\)
Kerala CEE-2013
Sets, Relation and Function
117454
The domain of the function \(f(x)=\sin ^{-1}\left(\frac{x+5}{2}\right)\) is
1 \([-1,1]\)
2 \([2,3]\)
3 \([3,7]\)
4 \([-7,-3]\)
5 \((-\infty, \infty)\)
Explanation:
D Given, \(f(x)=\sin ^{-1}\left(\frac{x+5}{2}\right)\) \(\left[\because\right.\) Domain of \(\sin ^{-1} \mathrm{x}\) is \(\mathrm{x} \in[-1,1]\) \(\therefore\) For \(f(x)\) to be defined is \(\begin{aligned} & \Rightarrow-1 \leq \frac{\mathrm{x}+5}{2} \leq 1 \\ & \Rightarrow-2 \leq \mathrm{x}+5 \leq 2 \\ & \Rightarrow-2-5 \leq \mathrm{x} \leq 2-5 \\ & \Rightarrow-7 \leq \mathrm{x} \leq-3 \\ & \Rightarrow \mathrm{x} \in[-7,-3] \end{aligned}\)
Kerala CEE-2012
Sets, Relation and Function
117455
The range of the function \(f(x)=\frac{x^2+8}{x^2+4}, x \in R\) is
1 \(\left[-1, \frac{3}{2}\right]\)
2 \((1,2]\)
3 \((1,2)\)
4 \([1,2]\)
5 \(\left[\frac{3}{2}, 2\right]\)
Explanation:
B Let, \(\mathrm{y}=\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^2+8}{\mathrm{x}^2+4}\) \(\Rightarrow \mathrm{x}^2(\mathrm{y}-1)+4 \mathrm{y}=8\) \(x^2=\frac{8-4 y}{y-1}\) \(x=\sqrt{\frac{8-4 y}{y-1}}\) Since, x is real, \(\therefore \mathrm{y}-1>0\) and \(8-4 y \geq 0\) \(y>1\) and \(y \leq 2\) \(\mathrm{y} \in(1,2]\)
117453
The domain of the function $(x)=\frac{\log _2(x+3)}{x^2+3 x+2}$ is
1 $\mathrm{R}-\{-1,-2\}$
2 $\mathrm{R}-\{-1,-2,0\}$
3 $(-3,-1) \cup(-1, \infty)$
4 $(-3, \infty)-\{-1,-2\}$
5 $(0, \infty)$
Explanation:
D Given function, $f(x)=\frac{\log _2(x+3)}{\left(x^2+3 x+2\right)}$ Define the above function as, $\lvert\, \begin{aligned} & x^2+3 x+2 \neq 0 \\ & (x+1)(x+2) \neq 0 \\ & x \neq-1,-2 \end{aligned}.$ Then, \(\begin{aligned} & =x+3>0 \\ & =x>-3 \end{aligned}\) Since, the domain of \(f(x)\) is - \((-3, \infty)-\{-1,-2\}\)
Kerala CEE-2013
Sets, Relation and Function
117454
The domain of the function \(f(x)=\sin ^{-1}\left(\frac{x+5}{2}\right)\) is
1 \([-1,1]\)
2 \([2,3]\)
3 \([3,7]\)
4 \([-7,-3]\)
5 \((-\infty, \infty)\)
Explanation:
D Given, \(f(x)=\sin ^{-1}\left(\frac{x+5}{2}\right)\) \(\left[\because\right.\) Domain of \(\sin ^{-1} \mathrm{x}\) is \(\mathrm{x} \in[-1,1]\) \(\therefore\) For \(f(x)\) to be defined is \(\begin{aligned} & \Rightarrow-1 \leq \frac{\mathrm{x}+5}{2} \leq 1 \\ & \Rightarrow-2 \leq \mathrm{x}+5 \leq 2 \\ & \Rightarrow-2-5 \leq \mathrm{x} \leq 2-5 \\ & \Rightarrow-7 \leq \mathrm{x} \leq-3 \\ & \Rightarrow \mathrm{x} \in[-7,-3] \end{aligned}\)
Kerala CEE-2012
Sets, Relation and Function
117455
The range of the function \(f(x)=\frac{x^2+8}{x^2+4}, x \in R\) is
1 \(\left[-1, \frac{3}{2}\right]\)
2 \((1,2]\)
3 \((1,2)\)
4 \([1,2]\)
5 \(\left[\frac{3}{2}, 2\right]\)
Explanation:
B Let, \(\mathrm{y}=\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^2+8}{\mathrm{x}^2+4}\) \(\Rightarrow \mathrm{x}^2(\mathrm{y}-1)+4 \mathrm{y}=8\) \(x^2=\frac{8-4 y}{y-1}\) \(x=\sqrt{\frac{8-4 y}{y-1}}\) Since, x is real, \(\therefore \mathrm{y}-1>0\) and \(8-4 y \geq 0\) \(y>1\) and \(y \leq 2\) \(\mathrm{y} \in(1,2]\)
117453
The domain of the function $(x)=\frac{\log _2(x+3)}{x^2+3 x+2}$ is
1 $\mathrm{R}-\{-1,-2\}$
2 $\mathrm{R}-\{-1,-2,0\}$
3 $(-3,-1) \cup(-1, \infty)$
4 $(-3, \infty)-\{-1,-2\}$
5 $(0, \infty)$
Explanation:
D Given function, $f(x)=\frac{\log _2(x+3)}{\left(x^2+3 x+2\right)}$ Define the above function as, $\lvert\, \begin{aligned} & x^2+3 x+2 \neq 0 \\ & (x+1)(x+2) \neq 0 \\ & x \neq-1,-2 \end{aligned}.$ Then, \(\begin{aligned} & =x+3>0 \\ & =x>-3 \end{aligned}\) Since, the domain of \(f(x)\) is - \((-3, \infty)-\{-1,-2\}\)
Kerala CEE-2013
Sets, Relation and Function
117454
The domain of the function \(f(x)=\sin ^{-1}\left(\frac{x+5}{2}\right)\) is
1 \([-1,1]\)
2 \([2,3]\)
3 \([3,7]\)
4 \([-7,-3]\)
5 \((-\infty, \infty)\)
Explanation:
D Given, \(f(x)=\sin ^{-1}\left(\frac{x+5}{2}\right)\) \(\left[\because\right.\) Domain of \(\sin ^{-1} \mathrm{x}\) is \(\mathrm{x} \in[-1,1]\) \(\therefore\) For \(f(x)\) to be defined is \(\begin{aligned} & \Rightarrow-1 \leq \frac{\mathrm{x}+5}{2} \leq 1 \\ & \Rightarrow-2 \leq \mathrm{x}+5 \leq 2 \\ & \Rightarrow-2-5 \leq \mathrm{x} \leq 2-5 \\ & \Rightarrow-7 \leq \mathrm{x} \leq-3 \\ & \Rightarrow \mathrm{x} \in[-7,-3] \end{aligned}\)
Kerala CEE-2012
Sets, Relation and Function
117455
The range of the function \(f(x)=\frac{x^2+8}{x^2+4}, x \in R\) is
1 \(\left[-1, \frac{3}{2}\right]\)
2 \((1,2]\)
3 \((1,2)\)
4 \([1,2]\)
5 \(\left[\frac{3}{2}, 2\right]\)
Explanation:
B Let, \(\mathrm{y}=\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^2+8}{\mathrm{x}^2+4}\) \(\Rightarrow \mathrm{x}^2(\mathrm{y}-1)+4 \mathrm{y}=8\) \(x^2=\frac{8-4 y}{y-1}\) \(x=\sqrt{\frac{8-4 y}{y-1}}\) Since, x is real, \(\therefore \mathrm{y}-1>0\) and \(8-4 y \geq 0\) \(y>1\) and \(y \leq 2\) \(\mathrm{y} \in(1,2]\)
