117348
IF \(f:[2,3] \rightarrow R\) is defined by \(\mathrm{f}(\mathrm{x})=\mathrm{x}^3+3 \mathrm{x}-2\), then range \(f(x)\) is contained in the interval
1 \([1,12]\)
2 \([12,34]\)
3 \([35,50]\)
4 \([-12,12]\)
Explanation:
B Given that \(\mathrm{f}:[2,3] \rightarrow \mathrm{R}\) And, \(f(x)=x^3+3 x-2\) \(\therefore(\mathrm{x}=2) \quad \mathrm{f}(2)=2^3+3 \times 2-2\) \(\mathrm{f}(2) =8+6-2\) \(\mathrm{f}(2) =12\) And, \((x=3) f(3)=3^3+3 \times 3-2\) \(\mathrm{f}(3)=27+9-2\) \(\mathrm{f}(3)=34\)Hence, the range \(\mathrm{f}(\mathrm{x})\) is contained in the interval \([12,34]\).
AP EAMCET-2009
Sets, Relation and Function
117349
If \(f: R \rightarrow R\) is defined by \(f(x)=[2 x]-2[x]\) for \(x \in R\), where \([x]\) is the greatest integer not exceeding \(x\), then the range of \(f\) is:
1 \(\{x \in R: 0 \leq x \leq 1\}\)
2 \(\{0,1\}\)
3 \(\{x \in R: x>0\}\)
4 \(\{x \in R: x \leq 0\}\)
Explanation:
B Given, If \(f: R \rightarrow R\) Then, for integer valued function \([2 \mathrm{x}]\), If \(x=1\) \({[2 \mathrm{x}]=[\mathrm{x}]+\left[\mathrm{x}+\left(\frac{1}{2}\right)\right]}\) \(\therefore \quad \mathrm{f}(\mathrm{x})=[2 \mathrm{x}]-2[\mathrm{x}]\) \(=[\mathrm{x}]+\left[\mathrm{x}+\left(\frac{1}{2}\right)\right]-2[\mathrm{x}]\) \(=\left[\mathrm{x}+\left(\frac{1}{2}\right)\right]-[\mathrm{x}]\) \(\therefore \mathrm{f}(1)=\left[1+\left(\frac{1}{2}\right)\right]-[1]\) \(=[1]-1\) \(=0\) \(\text { If, } \mathrm{x}=0.25,\) \(\therefore \mathrm{f}(0.25)=\left[0.25+\left(\frac{1}{2}\right)\right]-[0.25]\) \(=0-0=0\) If \(x=0.75\) \(f(0.75)=\left[0.75+\left(\frac{1}{2}\right)\right]-[0.75]\) \(=[1.25]-[0.75]\) \(=1-0\) \(=1\) If \(x=0.5\) \(\therefore \mathrm{f}(0.5) =\left[0.5+\left(\frac{1}{2}\right)\right]-[0.5]\) \(=[1.0]-[0.5]\) \(=1-0\) \(=1\)Therefore, the range of \(\mathrm{flies}\) between \(\{0,1\}\)
AP EAMCET-22.04.2018
Sets, Relation and Function
117350
The solution set of the in equation \(\sqrt{x^2+6 x+5}>(8-x)\) is
117352
The value of \(\left\{x \in R \mid \log \left[(1.6)^{1-x^2}-(0.625)^{6(1+x)}\right] \in \mathbf{R}\right\}\)
1 \((-\infty,-1) \cup(7, \infty)\)
2 \((-1,5)\)
3 \((1,7)\)
4 \((-1,7)\)
Explanation:
D Given, \(x \in R\) and \(\log \left[(1.6)^{1-\mathrm{x}^2}-(0.625)^{6(1+\mathrm{x})}\right]\) \(\because \quad \log _{\mathrm{b}} \mathrm{a}\) is defined when \(\mathrm{b}>0, \mathrm{~b} \neq 1\) and \(\mathrm{x}>0\) Then, \((1.6)^{1-\mathrm{x}^2}>(0.625)^{6(1+\mathrm{x})}\) \(\left(\frac{16}{10}\right)^{1-\mathrm{x}^2}>\left(\frac{625}{1000}\right)^{6(1+\mathrm{x})}\) \(\left(\frac{8}{5}\right)^{1-\mathrm{x}^2}>\left(\frac{8}{5}\right)^{-6(1+\mathrm{x})}\) On comparing the power of both sides, we get - \(1-x^2>-6(1+x)\) \(1-x^2>-6-6 x\) \(x^2-6 x-7\lt 0\) \(x^2-7 x+x-7\lt 0\) \(x(x-7)+1(x-7)\lt 0\) \((x-7)(x+1)\lt 0\) \(x\lt -1 \text { or } x\lt 7\)Hence, \(\mathrm{x} \in(-1,7)\)
AP EAMCET-2013
Sets, Relation and Function
117353
If \(R\) is the set of all real number and \(f: R-\{2\}\) \(\rightarrow R\) is defined by \(f(x)=\frac{2+x}{2-x}\) for \(x \in R-\{2\}\)
1 \(\mathrm{R}-\{-2\}\)
2 \(\mathrm{R}\)
3 \(\mathrm{R}-\{1\}\)
4 \(\mathrm{R}-\{-1\}\)
Explanation:
D Given, \(\mathrm{f}=\mathrm{R}-\{2\} \rightarrow \mathrm{R}\) \(\mathrm{f}(\mathrm{x})=\frac{2+\mathrm{x}}{2-\mathrm{x}},(\mathrm{x} \in \mathrm{R}-\{2\})\) \(\mathrm{f}(\mathrm{x})(2-\mathrm{x})=2+\mathrm{x}\) \(2 \mathrm{f}(\mathrm{x})-\mathrm{xf}(\mathrm{x})=2+\mathrm{x}\) \(\mathrm{x}+\mathrm{xf}(\mathrm{x})=2 \mathrm{f}(\mathrm{x})-2\) \(\mathrm{x}\{1+\mathrm{xf}(\mathrm{x})\}=2 \mathrm{f}(\mathrm{x})-2\) \(\mathrm{x}=\frac{2 \mathrm{f}(\mathrm{x})-2}{\mathrm{f}(\mathrm{x})+1}\) Here, \(\quad \mathrm{f}(\mathrm{x}) \neq-1\) Hence, the range of \(\mathrm{f}\) is \(\mathrm{R}-\{-1\}\)
117348
IF \(f:[2,3] \rightarrow R\) is defined by \(\mathrm{f}(\mathrm{x})=\mathrm{x}^3+3 \mathrm{x}-2\), then range \(f(x)\) is contained in the interval
1 \([1,12]\)
2 \([12,34]\)
3 \([35,50]\)
4 \([-12,12]\)
Explanation:
B Given that \(\mathrm{f}:[2,3] \rightarrow \mathrm{R}\) And, \(f(x)=x^3+3 x-2\) \(\therefore(\mathrm{x}=2) \quad \mathrm{f}(2)=2^3+3 \times 2-2\) \(\mathrm{f}(2) =8+6-2\) \(\mathrm{f}(2) =12\) And, \((x=3) f(3)=3^3+3 \times 3-2\) \(\mathrm{f}(3)=27+9-2\) \(\mathrm{f}(3)=34\)Hence, the range \(\mathrm{f}(\mathrm{x})\) is contained in the interval \([12,34]\).
AP EAMCET-2009
Sets, Relation and Function
117349
If \(f: R \rightarrow R\) is defined by \(f(x)=[2 x]-2[x]\) for \(x \in R\), where \([x]\) is the greatest integer not exceeding \(x\), then the range of \(f\) is:
1 \(\{x \in R: 0 \leq x \leq 1\}\)
2 \(\{0,1\}\)
3 \(\{x \in R: x>0\}\)
4 \(\{x \in R: x \leq 0\}\)
Explanation:
B Given, If \(f: R \rightarrow R\) Then, for integer valued function \([2 \mathrm{x}]\), If \(x=1\) \({[2 \mathrm{x}]=[\mathrm{x}]+\left[\mathrm{x}+\left(\frac{1}{2}\right)\right]}\) \(\therefore \quad \mathrm{f}(\mathrm{x})=[2 \mathrm{x}]-2[\mathrm{x}]\) \(=[\mathrm{x}]+\left[\mathrm{x}+\left(\frac{1}{2}\right)\right]-2[\mathrm{x}]\) \(=\left[\mathrm{x}+\left(\frac{1}{2}\right)\right]-[\mathrm{x}]\) \(\therefore \mathrm{f}(1)=\left[1+\left(\frac{1}{2}\right)\right]-[1]\) \(=[1]-1\) \(=0\) \(\text { If, } \mathrm{x}=0.25,\) \(\therefore \mathrm{f}(0.25)=\left[0.25+\left(\frac{1}{2}\right)\right]-[0.25]\) \(=0-0=0\) If \(x=0.75\) \(f(0.75)=\left[0.75+\left(\frac{1}{2}\right)\right]-[0.75]\) \(=[1.25]-[0.75]\) \(=1-0\) \(=1\) If \(x=0.5\) \(\therefore \mathrm{f}(0.5) =\left[0.5+\left(\frac{1}{2}\right)\right]-[0.5]\) \(=[1.0]-[0.5]\) \(=1-0\) \(=1\)Therefore, the range of \(\mathrm{flies}\) between \(\{0,1\}\)
AP EAMCET-22.04.2018
Sets, Relation and Function
117350
The solution set of the in equation \(\sqrt{x^2+6 x+5}>(8-x)\) is
117352
The value of \(\left\{x \in R \mid \log \left[(1.6)^{1-x^2}-(0.625)^{6(1+x)}\right] \in \mathbf{R}\right\}\)
1 \((-\infty,-1) \cup(7, \infty)\)
2 \((-1,5)\)
3 \((1,7)\)
4 \((-1,7)\)
Explanation:
D Given, \(x \in R\) and \(\log \left[(1.6)^{1-\mathrm{x}^2}-(0.625)^{6(1+\mathrm{x})}\right]\) \(\because \quad \log _{\mathrm{b}} \mathrm{a}\) is defined when \(\mathrm{b}>0, \mathrm{~b} \neq 1\) and \(\mathrm{x}>0\) Then, \((1.6)^{1-\mathrm{x}^2}>(0.625)^{6(1+\mathrm{x})}\) \(\left(\frac{16}{10}\right)^{1-\mathrm{x}^2}>\left(\frac{625}{1000}\right)^{6(1+\mathrm{x})}\) \(\left(\frac{8}{5}\right)^{1-\mathrm{x}^2}>\left(\frac{8}{5}\right)^{-6(1+\mathrm{x})}\) On comparing the power of both sides, we get - \(1-x^2>-6(1+x)\) \(1-x^2>-6-6 x\) \(x^2-6 x-7\lt 0\) \(x^2-7 x+x-7\lt 0\) \(x(x-7)+1(x-7)\lt 0\) \((x-7)(x+1)\lt 0\) \(x\lt -1 \text { or } x\lt 7\)Hence, \(\mathrm{x} \in(-1,7)\)
AP EAMCET-2013
Sets, Relation and Function
117353
If \(R\) is the set of all real number and \(f: R-\{2\}\) \(\rightarrow R\) is defined by \(f(x)=\frac{2+x}{2-x}\) for \(x \in R-\{2\}\)
1 \(\mathrm{R}-\{-2\}\)
2 \(\mathrm{R}\)
3 \(\mathrm{R}-\{1\}\)
4 \(\mathrm{R}-\{-1\}\)
Explanation:
D Given, \(\mathrm{f}=\mathrm{R}-\{2\} \rightarrow \mathrm{R}\) \(\mathrm{f}(\mathrm{x})=\frac{2+\mathrm{x}}{2-\mathrm{x}},(\mathrm{x} \in \mathrm{R}-\{2\})\) \(\mathrm{f}(\mathrm{x})(2-\mathrm{x})=2+\mathrm{x}\) \(2 \mathrm{f}(\mathrm{x})-\mathrm{xf}(\mathrm{x})=2+\mathrm{x}\) \(\mathrm{x}+\mathrm{xf}(\mathrm{x})=2 \mathrm{f}(\mathrm{x})-2\) \(\mathrm{x}\{1+\mathrm{xf}(\mathrm{x})\}=2 \mathrm{f}(\mathrm{x})-2\) \(\mathrm{x}=\frac{2 \mathrm{f}(\mathrm{x})-2}{\mathrm{f}(\mathrm{x})+1}\) Here, \(\quad \mathrm{f}(\mathrm{x}) \neq-1\) Hence, the range of \(\mathrm{f}\) is \(\mathrm{R}-\{-1\}\)
117348
IF \(f:[2,3] \rightarrow R\) is defined by \(\mathrm{f}(\mathrm{x})=\mathrm{x}^3+3 \mathrm{x}-2\), then range \(f(x)\) is contained in the interval
1 \([1,12]\)
2 \([12,34]\)
3 \([35,50]\)
4 \([-12,12]\)
Explanation:
B Given that \(\mathrm{f}:[2,3] \rightarrow \mathrm{R}\) And, \(f(x)=x^3+3 x-2\) \(\therefore(\mathrm{x}=2) \quad \mathrm{f}(2)=2^3+3 \times 2-2\) \(\mathrm{f}(2) =8+6-2\) \(\mathrm{f}(2) =12\) And, \((x=3) f(3)=3^3+3 \times 3-2\) \(\mathrm{f}(3)=27+9-2\) \(\mathrm{f}(3)=34\)Hence, the range \(\mathrm{f}(\mathrm{x})\) is contained in the interval \([12,34]\).
AP EAMCET-2009
Sets, Relation and Function
117349
If \(f: R \rightarrow R\) is defined by \(f(x)=[2 x]-2[x]\) for \(x \in R\), where \([x]\) is the greatest integer not exceeding \(x\), then the range of \(f\) is:
1 \(\{x \in R: 0 \leq x \leq 1\}\)
2 \(\{0,1\}\)
3 \(\{x \in R: x>0\}\)
4 \(\{x \in R: x \leq 0\}\)
Explanation:
B Given, If \(f: R \rightarrow R\) Then, for integer valued function \([2 \mathrm{x}]\), If \(x=1\) \({[2 \mathrm{x}]=[\mathrm{x}]+\left[\mathrm{x}+\left(\frac{1}{2}\right)\right]}\) \(\therefore \quad \mathrm{f}(\mathrm{x})=[2 \mathrm{x}]-2[\mathrm{x}]\) \(=[\mathrm{x}]+\left[\mathrm{x}+\left(\frac{1}{2}\right)\right]-2[\mathrm{x}]\) \(=\left[\mathrm{x}+\left(\frac{1}{2}\right)\right]-[\mathrm{x}]\) \(\therefore \mathrm{f}(1)=\left[1+\left(\frac{1}{2}\right)\right]-[1]\) \(=[1]-1\) \(=0\) \(\text { If, } \mathrm{x}=0.25,\) \(\therefore \mathrm{f}(0.25)=\left[0.25+\left(\frac{1}{2}\right)\right]-[0.25]\) \(=0-0=0\) If \(x=0.75\) \(f(0.75)=\left[0.75+\left(\frac{1}{2}\right)\right]-[0.75]\) \(=[1.25]-[0.75]\) \(=1-0\) \(=1\) If \(x=0.5\) \(\therefore \mathrm{f}(0.5) =\left[0.5+\left(\frac{1}{2}\right)\right]-[0.5]\) \(=[1.0]-[0.5]\) \(=1-0\) \(=1\)Therefore, the range of \(\mathrm{flies}\) between \(\{0,1\}\)
AP EAMCET-22.04.2018
Sets, Relation and Function
117350
The solution set of the in equation \(\sqrt{x^2+6 x+5}>(8-x)\) is
117352
The value of \(\left\{x \in R \mid \log \left[(1.6)^{1-x^2}-(0.625)^{6(1+x)}\right] \in \mathbf{R}\right\}\)
1 \((-\infty,-1) \cup(7, \infty)\)
2 \((-1,5)\)
3 \((1,7)\)
4 \((-1,7)\)
Explanation:
D Given, \(x \in R\) and \(\log \left[(1.6)^{1-\mathrm{x}^2}-(0.625)^{6(1+\mathrm{x})}\right]\) \(\because \quad \log _{\mathrm{b}} \mathrm{a}\) is defined when \(\mathrm{b}>0, \mathrm{~b} \neq 1\) and \(\mathrm{x}>0\) Then, \((1.6)^{1-\mathrm{x}^2}>(0.625)^{6(1+\mathrm{x})}\) \(\left(\frac{16}{10}\right)^{1-\mathrm{x}^2}>\left(\frac{625}{1000}\right)^{6(1+\mathrm{x})}\) \(\left(\frac{8}{5}\right)^{1-\mathrm{x}^2}>\left(\frac{8}{5}\right)^{-6(1+\mathrm{x})}\) On comparing the power of both sides, we get - \(1-x^2>-6(1+x)\) \(1-x^2>-6-6 x\) \(x^2-6 x-7\lt 0\) \(x^2-7 x+x-7\lt 0\) \(x(x-7)+1(x-7)\lt 0\) \((x-7)(x+1)\lt 0\) \(x\lt -1 \text { or } x\lt 7\)Hence, \(\mathrm{x} \in(-1,7)\)
AP EAMCET-2013
Sets, Relation and Function
117353
If \(R\) is the set of all real number and \(f: R-\{2\}\) \(\rightarrow R\) is defined by \(f(x)=\frac{2+x}{2-x}\) for \(x \in R-\{2\}\)
1 \(\mathrm{R}-\{-2\}\)
2 \(\mathrm{R}\)
3 \(\mathrm{R}-\{1\}\)
4 \(\mathrm{R}-\{-1\}\)
Explanation:
D Given, \(\mathrm{f}=\mathrm{R}-\{2\} \rightarrow \mathrm{R}\) \(\mathrm{f}(\mathrm{x})=\frac{2+\mathrm{x}}{2-\mathrm{x}},(\mathrm{x} \in \mathrm{R}-\{2\})\) \(\mathrm{f}(\mathrm{x})(2-\mathrm{x})=2+\mathrm{x}\) \(2 \mathrm{f}(\mathrm{x})-\mathrm{xf}(\mathrm{x})=2+\mathrm{x}\) \(\mathrm{x}+\mathrm{xf}(\mathrm{x})=2 \mathrm{f}(\mathrm{x})-2\) \(\mathrm{x}\{1+\mathrm{xf}(\mathrm{x})\}=2 \mathrm{f}(\mathrm{x})-2\) \(\mathrm{x}=\frac{2 \mathrm{f}(\mathrm{x})-2}{\mathrm{f}(\mathrm{x})+1}\) Here, \(\quad \mathrm{f}(\mathrm{x}) \neq-1\) Hence, the range of \(\mathrm{f}\) is \(\mathrm{R}-\{-1\}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
117348
IF \(f:[2,3] \rightarrow R\) is defined by \(\mathrm{f}(\mathrm{x})=\mathrm{x}^3+3 \mathrm{x}-2\), then range \(f(x)\) is contained in the interval
1 \([1,12]\)
2 \([12,34]\)
3 \([35,50]\)
4 \([-12,12]\)
Explanation:
B Given that \(\mathrm{f}:[2,3] \rightarrow \mathrm{R}\) And, \(f(x)=x^3+3 x-2\) \(\therefore(\mathrm{x}=2) \quad \mathrm{f}(2)=2^3+3 \times 2-2\) \(\mathrm{f}(2) =8+6-2\) \(\mathrm{f}(2) =12\) And, \((x=3) f(3)=3^3+3 \times 3-2\) \(\mathrm{f}(3)=27+9-2\) \(\mathrm{f}(3)=34\)Hence, the range \(\mathrm{f}(\mathrm{x})\) is contained in the interval \([12,34]\).
AP EAMCET-2009
Sets, Relation and Function
117349
If \(f: R \rightarrow R\) is defined by \(f(x)=[2 x]-2[x]\) for \(x \in R\), where \([x]\) is the greatest integer not exceeding \(x\), then the range of \(f\) is:
1 \(\{x \in R: 0 \leq x \leq 1\}\)
2 \(\{0,1\}\)
3 \(\{x \in R: x>0\}\)
4 \(\{x \in R: x \leq 0\}\)
Explanation:
B Given, If \(f: R \rightarrow R\) Then, for integer valued function \([2 \mathrm{x}]\), If \(x=1\) \({[2 \mathrm{x}]=[\mathrm{x}]+\left[\mathrm{x}+\left(\frac{1}{2}\right)\right]}\) \(\therefore \quad \mathrm{f}(\mathrm{x})=[2 \mathrm{x}]-2[\mathrm{x}]\) \(=[\mathrm{x}]+\left[\mathrm{x}+\left(\frac{1}{2}\right)\right]-2[\mathrm{x}]\) \(=\left[\mathrm{x}+\left(\frac{1}{2}\right)\right]-[\mathrm{x}]\) \(\therefore \mathrm{f}(1)=\left[1+\left(\frac{1}{2}\right)\right]-[1]\) \(=[1]-1\) \(=0\) \(\text { If, } \mathrm{x}=0.25,\) \(\therefore \mathrm{f}(0.25)=\left[0.25+\left(\frac{1}{2}\right)\right]-[0.25]\) \(=0-0=0\) If \(x=0.75\) \(f(0.75)=\left[0.75+\left(\frac{1}{2}\right)\right]-[0.75]\) \(=[1.25]-[0.75]\) \(=1-0\) \(=1\) If \(x=0.5\) \(\therefore \mathrm{f}(0.5) =\left[0.5+\left(\frac{1}{2}\right)\right]-[0.5]\) \(=[1.0]-[0.5]\) \(=1-0\) \(=1\)Therefore, the range of \(\mathrm{flies}\) between \(\{0,1\}\)
AP EAMCET-22.04.2018
Sets, Relation and Function
117350
The solution set of the in equation \(\sqrt{x^2+6 x+5}>(8-x)\) is
117352
The value of \(\left\{x \in R \mid \log \left[(1.6)^{1-x^2}-(0.625)^{6(1+x)}\right] \in \mathbf{R}\right\}\)
1 \((-\infty,-1) \cup(7, \infty)\)
2 \((-1,5)\)
3 \((1,7)\)
4 \((-1,7)\)
Explanation:
D Given, \(x \in R\) and \(\log \left[(1.6)^{1-\mathrm{x}^2}-(0.625)^{6(1+\mathrm{x})}\right]\) \(\because \quad \log _{\mathrm{b}} \mathrm{a}\) is defined when \(\mathrm{b}>0, \mathrm{~b} \neq 1\) and \(\mathrm{x}>0\) Then, \((1.6)^{1-\mathrm{x}^2}>(0.625)^{6(1+\mathrm{x})}\) \(\left(\frac{16}{10}\right)^{1-\mathrm{x}^2}>\left(\frac{625}{1000}\right)^{6(1+\mathrm{x})}\) \(\left(\frac{8}{5}\right)^{1-\mathrm{x}^2}>\left(\frac{8}{5}\right)^{-6(1+\mathrm{x})}\) On comparing the power of both sides, we get - \(1-x^2>-6(1+x)\) \(1-x^2>-6-6 x\) \(x^2-6 x-7\lt 0\) \(x^2-7 x+x-7\lt 0\) \(x(x-7)+1(x-7)\lt 0\) \((x-7)(x+1)\lt 0\) \(x\lt -1 \text { or } x\lt 7\)Hence, \(\mathrm{x} \in(-1,7)\)
AP EAMCET-2013
Sets, Relation and Function
117353
If \(R\) is the set of all real number and \(f: R-\{2\}\) \(\rightarrow R\) is defined by \(f(x)=\frac{2+x}{2-x}\) for \(x \in R-\{2\}\)
1 \(\mathrm{R}-\{-2\}\)
2 \(\mathrm{R}\)
3 \(\mathrm{R}-\{1\}\)
4 \(\mathrm{R}-\{-1\}\)
Explanation:
D Given, \(\mathrm{f}=\mathrm{R}-\{2\} \rightarrow \mathrm{R}\) \(\mathrm{f}(\mathrm{x})=\frac{2+\mathrm{x}}{2-\mathrm{x}},(\mathrm{x} \in \mathrm{R}-\{2\})\) \(\mathrm{f}(\mathrm{x})(2-\mathrm{x})=2+\mathrm{x}\) \(2 \mathrm{f}(\mathrm{x})-\mathrm{xf}(\mathrm{x})=2+\mathrm{x}\) \(\mathrm{x}+\mathrm{xf}(\mathrm{x})=2 \mathrm{f}(\mathrm{x})-2\) \(\mathrm{x}\{1+\mathrm{xf}(\mathrm{x})\}=2 \mathrm{f}(\mathrm{x})-2\) \(\mathrm{x}=\frac{2 \mathrm{f}(\mathrm{x})-2}{\mathrm{f}(\mathrm{x})+1}\) Here, \(\quad \mathrm{f}(\mathrm{x}) \neq-1\) Hence, the range of \(\mathrm{f}\) is \(\mathrm{R}-\{-1\}\)
117348
IF \(f:[2,3] \rightarrow R\) is defined by \(\mathrm{f}(\mathrm{x})=\mathrm{x}^3+3 \mathrm{x}-2\), then range \(f(x)\) is contained in the interval
1 \([1,12]\)
2 \([12,34]\)
3 \([35,50]\)
4 \([-12,12]\)
Explanation:
B Given that \(\mathrm{f}:[2,3] \rightarrow \mathrm{R}\) And, \(f(x)=x^3+3 x-2\) \(\therefore(\mathrm{x}=2) \quad \mathrm{f}(2)=2^3+3 \times 2-2\) \(\mathrm{f}(2) =8+6-2\) \(\mathrm{f}(2) =12\) And, \((x=3) f(3)=3^3+3 \times 3-2\) \(\mathrm{f}(3)=27+9-2\) \(\mathrm{f}(3)=34\)Hence, the range \(\mathrm{f}(\mathrm{x})\) is contained in the interval \([12,34]\).
AP EAMCET-2009
Sets, Relation and Function
117349
If \(f: R \rightarrow R\) is defined by \(f(x)=[2 x]-2[x]\) for \(x \in R\), where \([x]\) is the greatest integer not exceeding \(x\), then the range of \(f\) is:
1 \(\{x \in R: 0 \leq x \leq 1\}\)
2 \(\{0,1\}\)
3 \(\{x \in R: x>0\}\)
4 \(\{x \in R: x \leq 0\}\)
Explanation:
B Given, If \(f: R \rightarrow R\) Then, for integer valued function \([2 \mathrm{x}]\), If \(x=1\) \({[2 \mathrm{x}]=[\mathrm{x}]+\left[\mathrm{x}+\left(\frac{1}{2}\right)\right]}\) \(\therefore \quad \mathrm{f}(\mathrm{x})=[2 \mathrm{x}]-2[\mathrm{x}]\) \(=[\mathrm{x}]+\left[\mathrm{x}+\left(\frac{1}{2}\right)\right]-2[\mathrm{x}]\) \(=\left[\mathrm{x}+\left(\frac{1}{2}\right)\right]-[\mathrm{x}]\) \(\therefore \mathrm{f}(1)=\left[1+\left(\frac{1}{2}\right)\right]-[1]\) \(=[1]-1\) \(=0\) \(\text { If, } \mathrm{x}=0.25,\) \(\therefore \mathrm{f}(0.25)=\left[0.25+\left(\frac{1}{2}\right)\right]-[0.25]\) \(=0-0=0\) If \(x=0.75\) \(f(0.75)=\left[0.75+\left(\frac{1}{2}\right)\right]-[0.75]\) \(=[1.25]-[0.75]\) \(=1-0\) \(=1\) If \(x=0.5\) \(\therefore \mathrm{f}(0.5) =\left[0.5+\left(\frac{1}{2}\right)\right]-[0.5]\) \(=[1.0]-[0.5]\) \(=1-0\) \(=1\)Therefore, the range of \(\mathrm{flies}\) between \(\{0,1\}\)
AP EAMCET-22.04.2018
Sets, Relation and Function
117350
The solution set of the in equation \(\sqrt{x^2+6 x+5}>(8-x)\) is
117352
The value of \(\left\{x \in R \mid \log \left[(1.6)^{1-x^2}-(0.625)^{6(1+x)}\right] \in \mathbf{R}\right\}\)
1 \((-\infty,-1) \cup(7, \infty)\)
2 \((-1,5)\)
3 \((1,7)\)
4 \((-1,7)\)
Explanation:
D Given, \(x \in R\) and \(\log \left[(1.6)^{1-\mathrm{x}^2}-(0.625)^{6(1+\mathrm{x})}\right]\) \(\because \quad \log _{\mathrm{b}} \mathrm{a}\) is defined when \(\mathrm{b}>0, \mathrm{~b} \neq 1\) and \(\mathrm{x}>0\) Then, \((1.6)^{1-\mathrm{x}^2}>(0.625)^{6(1+\mathrm{x})}\) \(\left(\frac{16}{10}\right)^{1-\mathrm{x}^2}>\left(\frac{625}{1000}\right)^{6(1+\mathrm{x})}\) \(\left(\frac{8}{5}\right)^{1-\mathrm{x}^2}>\left(\frac{8}{5}\right)^{-6(1+\mathrm{x})}\) On comparing the power of both sides, we get - \(1-x^2>-6(1+x)\) \(1-x^2>-6-6 x\) \(x^2-6 x-7\lt 0\) \(x^2-7 x+x-7\lt 0\) \(x(x-7)+1(x-7)\lt 0\) \((x-7)(x+1)\lt 0\) \(x\lt -1 \text { or } x\lt 7\)Hence, \(\mathrm{x} \in(-1,7)\)
AP EAMCET-2013
Sets, Relation and Function
117353
If \(R\) is the set of all real number and \(f: R-\{2\}\) \(\rightarrow R\) is defined by \(f(x)=\frac{2+x}{2-x}\) for \(x \in R-\{2\}\)
1 \(\mathrm{R}-\{-2\}\)
2 \(\mathrm{R}\)
3 \(\mathrm{R}-\{1\}\)
4 \(\mathrm{R}-\{-1\}\)
Explanation:
D Given, \(\mathrm{f}=\mathrm{R}-\{2\} \rightarrow \mathrm{R}\) \(\mathrm{f}(\mathrm{x})=\frac{2+\mathrm{x}}{2-\mathrm{x}},(\mathrm{x} \in \mathrm{R}-\{2\})\) \(\mathrm{f}(\mathrm{x})(2-\mathrm{x})=2+\mathrm{x}\) \(2 \mathrm{f}(\mathrm{x})-\mathrm{xf}(\mathrm{x})=2+\mathrm{x}\) \(\mathrm{x}+\mathrm{xf}(\mathrm{x})=2 \mathrm{f}(\mathrm{x})-2\) \(\mathrm{x}\{1+\mathrm{xf}(\mathrm{x})\}=2 \mathrm{f}(\mathrm{x})-2\) \(\mathrm{x}=\frac{2 \mathrm{f}(\mathrm{x})-2}{\mathrm{f}(\mathrm{x})+1}\) Here, \(\quad \mathrm{f}(\mathrm{x}) \neq-1\) Hence, the range of \(\mathrm{f}\) is \(\mathrm{R}-\{-1\}\)