117366
The range of the function \(f(x)=x^2+\frac{1}{x^2+1}\) is
1 \([1, \infty)\)
2 \((2, \infty)\)
3 \(\left(\frac{3}{2}, \infty\right)\)
4 \((0,1)\)
Explanation:
A Given, \(\mathrm{f}(\mathrm{x})=\mathrm{y}=\mathrm{x}^2+\frac{1}{\mathrm{x}^2+1}\) \(\Rightarrow \mathrm{y}=\frac{\mathrm{x}^4+\mathrm{x}^2+1}{\mathrm{x}^2+1}\) \(\Rightarrow \mathrm{x}^4+\mathrm{x}^2+1=\mathrm{y}\left(\mathrm{x}^2+1\right)\) \(\Rightarrow \mathrm{x}^4+\mathrm{x}^2 \times(1-\mathrm{y})+(1-\mathrm{y})=0\) \(\because \mathrm{x} \in \mathrm{R}, \text { So, discriminate } \geq 0\) \(\mathrm{~b}^2-4 \mathrm{ac} \geq 0\) \(\Rightarrow (1-\mathrm{y})^2-4 \times 1 \times(1-\mathrm{y}) \geq 0\) \(\Rightarrow (1-\mathrm{y})(1-\mathrm{y}-4) \geq 0\) \(\Rightarrow (1-\mathrm{y})(-\mathrm{y}-3) \geq 0\) \(\Rightarrow (\mathrm{y}+3)(1-\mathrm{y}) \geq 0\) \(\Rightarrow \mathrm{y} \in(-\infty,-3] \cup[1, \infty)\) \(\text { But, } \mathrm{y}>0,\) \(\text { So, range of function is }[1, \infty)\)So, range of function is \([1, \infty\) )
Shift-I
Sets, Relation and Function
117367
For \(f(x)=\sin \left(\frac{1}{|x| \sqrt{x^2-1}}\right)\) the domain and range of \(\boldsymbol{f}(\mathrm{x})\) in \(\mathrm{R}\) are
1 \(\mathrm{R}-(0, \pm 1)\) and \([-1,1]\), respectively
2 \(\mathrm{R}-(-1,1)\) and \([-1,1]\) respectively
3 \(\mathrm{R}-(0, \pm 1)\) and \([0,1]\), respectively
4 \(\mathrm{R}-[-1,1]\) and \([0,1]\), respectively
Explanation:
B Given, Function \(f(x)=\sin \left(\frac{1}{|x| \sqrt{x^2-1}}\right)\) For domain, \(|x| \neq 0 \Rightarrow x \neq 0\) And, \(x^2-1>0\) \((x-1)(x+1)>0\) Domain \(\mathrm{x} \neq(1,-1)\) Hence, \(x \in R-(-1,1)\) For range \(\because-1 \leq \sin x \leq 1\) Then, the range of sine function is \([-1,1]\).
C Given, \(\mathrm{x} \in \mathrm{R}: \frac{\sqrt{6+x-x^2}}{2 \mathrm{x}+5} \geq \frac{\sqrt{6+x-x^2}}{\mathrm{x}+4}\) \(\Rightarrow \quad \sqrt{6+\mathrm{x}-\mathrm{x}^2}\left(\frac{1}{2 \mathrm{x}+5}-\frac{1}{\mathrm{x}+4}\right) \geq 0\) \(\Rightarrow \quad \sqrt{6+\mathrm{x}-\mathrm{x}^2}\left(\frac{\mathrm{x}+4-2 \mathrm{x}-5}{(2 \mathrm{x}+5)(\mathrm{x}+4)}\right) \geq 0\) \(\Rightarrow \quad \sqrt{6+\mathrm{x}-\mathrm{x}^2}\left(\frac{-(\mathrm{x}+1)}{(2 \mathrm{x}+5)(\mathrm{x}+4)}\right) \geq 0\) \(\Rightarrow \quad (\mathrm{x}+1) \leq 0,(2 \mathrm{x}+5)(\mathrm{x}+4)\lt 0\) \(\Rightarrow \quad \mathrm{x} \in(-\infty,-4) \cup\left(\frac{-5}{2},-1\right]\) For expression to be exist - \(6+x-x^2 \geq 0\) \(\Rightarrow x^2-x-6 \leq 0\) \(\Rightarrow x^2-3 x+2 x-6 \leq 0\) \(\Rightarrow x(x-3)+2(x-3) \leq 0\) \(\Rightarrow (x-3)(x+2) \leq 0\) \(\Rightarrow x \in[-2,3]\) From equation (i) and (ii), we get - \(x \in[-2,-1] \cup\{3\}\)
Shift-II
Sets, Relation and Function
117370
Let \(D=\left\{x \in R: f(x)=\sqrt{\frac{x-|x|}{x-[x]}}\right\}\) is defined and \(C\) be the range of the real function \(g(x)=\frac{2 x}{4+x^2}\). Then \(D \cap C=\)
117366
The range of the function \(f(x)=x^2+\frac{1}{x^2+1}\) is
1 \([1, \infty)\)
2 \((2, \infty)\)
3 \(\left(\frac{3}{2}, \infty\right)\)
4 \((0,1)\)
Explanation:
A Given, \(\mathrm{f}(\mathrm{x})=\mathrm{y}=\mathrm{x}^2+\frac{1}{\mathrm{x}^2+1}\) \(\Rightarrow \mathrm{y}=\frac{\mathrm{x}^4+\mathrm{x}^2+1}{\mathrm{x}^2+1}\) \(\Rightarrow \mathrm{x}^4+\mathrm{x}^2+1=\mathrm{y}\left(\mathrm{x}^2+1\right)\) \(\Rightarrow \mathrm{x}^4+\mathrm{x}^2 \times(1-\mathrm{y})+(1-\mathrm{y})=0\) \(\because \mathrm{x} \in \mathrm{R}, \text { So, discriminate } \geq 0\) \(\mathrm{~b}^2-4 \mathrm{ac} \geq 0\) \(\Rightarrow (1-\mathrm{y})^2-4 \times 1 \times(1-\mathrm{y}) \geq 0\) \(\Rightarrow (1-\mathrm{y})(1-\mathrm{y}-4) \geq 0\) \(\Rightarrow (1-\mathrm{y})(-\mathrm{y}-3) \geq 0\) \(\Rightarrow (\mathrm{y}+3)(1-\mathrm{y}) \geq 0\) \(\Rightarrow \mathrm{y} \in(-\infty,-3] \cup[1, \infty)\) \(\text { But, } \mathrm{y}>0,\) \(\text { So, range of function is }[1, \infty)\)So, range of function is \([1, \infty\) )
Shift-I
Sets, Relation and Function
117367
For \(f(x)=\sin \left(\frac{1}{|x| \sqrt{x^2-1}}\right)\) the domain and range of \(\boldsymbol{f}(\mathrm{x})\) in \(\mathrm{R}\) are
1 \(\mathrm{R}-(0, \pm 1)\) and \([-1,1]\), respectively
2 \(\mathrm{R}-(-1,1)\) and \([-1,1]\) respectively
3 \(\mathrm{R}-(0, \pm 1)\) and \([0,1]\), respectively
4 \(\mathrm{R}-[-1,1]\) and \([0,1]\), respectively
Explanation:
B Given, Function \(f(x)=\sin \left(\frac{1}{|x| \sqrt{x^2-1}}\right)\) For domain, \(|x| \neq 0 \Rightarrow x \neq 0\) And, \(x^2-1>0\) \((x-1)(x+1)>0\) Domain \(\mathrm{x} \neq(1,-1)\) Hence, \(x \in R-(-1,1)\) For range \(\because-1 \leq \sin x \leq 1\) Then, the range of sine function is \([-1,1]\).
C Given, \(\mathrm{x} \in \mathrm{R}: \frac{\sqrt{6+x-x^2}}{2 \mathrm{x}+5} \geq \frac{\sqrt{6+x-x^2}}{\mathrm{x}+4}\) \(\Rightarrow \quad \sqrt{6+\mathrm{x}-\mathrm{x}^2}\left(\frac{1}{2 \mathrm{x}+5}-\frac{1}{\mathrm{x}+4}\right) \geq 0\) \(\Rightarrow \quad \sqrt{6+\mathrm{x}-\mathrm{x}^2}\left(\frac{\mathrm{x}+4-2 \mathrm{x}-5}{(2 \mathrm{x}+5)(\mathrm{x}+4)}\right) \geq 0\) \(\Rightarrow \quad \sqrt{6+\mathrm{x}-\mathrm{x}^2}\left(\frac{-(\mathrm{x}+1)}{(2 \mathrm{x}+5)(\mathrm{x}+4)}\right) \geq 0\) \(\Rightarrow \quad (\mathrm{x}+1) \leq 0,(2 \mathrm{x}+5)(\mathrm{x}+4)\lt 0\) \(\Rightarrow \quad \mathrm{x} \in(-\infty,-4) \cup\left(\frac{-5}{2},-1\right]\) For expression to be exist - \(6+x-x^2 \geq 0\) \(\Rightarrow x^2-x-6 \leq 0\) \(\Rightarrow x^2-3 x+2 x-6 \leq 0\) \(\Rightarrow x(x-3)+2(x-3) \leq 0\) \(\Rightarrow (x-3)(x+2) \leq 0\) \(\Rightarrow x \in[-2,3]\) From equation (i) and (ii), we get - \(x \in[-2,-1] \cup\{3\}\)
Shift-II
Sets, Relation and Function
117370
Let \(D=\left\{x \in R: f(x)=\sqrt{\frac{x-|x|}{x-[x]}}\right\}\) is defined and \(C\) be the range of the real function \(g(x)=\frac{2 x}{4+x^2}\). Then \(D \cap C=\)
117366
The range of the function \(f(x)=x^2+\frac{1}{x^2+1}\) is
1 \([1, \infty)\)
2 \((2, \infty)\)
3 \(\left(\frac{3}{2}, \infty\right)\)
4 \((0,1)\)
Explanation:
A Given, \(\mathrm{f}(\mathrm{x})=\mathrm{y}=\mathrm{x}^2+\frac{1}{\mathrm{x}^2+1}\) \(\Rightarrow \mathrm{y}=\frac{\mathrm{x}^4+\mathrm{x}^2+1}{\mathrm{x}^2+1}\) \(\Rightarrow \mathrm{x}^4+\mathrm{x}^2+1=\mathrm{y}\left(\mathrm{x}^2+1\right)\) \(\Rightarrow \mathrm{x}^4+\mathrm{x}^2 \times(1-\mathrm{y})+(1-\mathrm{y})=0\) \(\because \mathrm{x} \in \mathrm{R}, \text { So, discriminate } \geq 0\) \(\mathrm{~b}^2-4 \mathrm{ac} \geq 0\) \(\Rightarrow (1-\mathrm{y})^2-4 \times 1 \times(1-\mathrm{y}) \geq 0\) \(\Rightarrow (1-\mathrm{y})(1-\mathrm{y}-4) \geq 0\) \(\Rightarrow (1-\mathrm{y})(-\mathrm{y}-3) \geq 0\) \(\Rightarrow (\mathrm{y}+3)(1-\mathrm{y}) \geq 0\) \(\Rightarrow \mathrm{y} \in(-\infty,-3] \cup[1, \infty)\) \(\text { But, } \mathrm{y}>0,\) \(\text { So, range of function is }[1, \infty)\)So, range of function is \([1, \infty\) )
Shift-I
Sets, Relation and Function
117367
For \(f(x)=\sin \left(\frac{1}{|x| \sqrt{x^2-1}}\right)\) the domain and range of \(\boldsymbol{f}(\mathrm{x})\) in \(\mathrm{R}\) are
1 \(\mathrm{R}-(0, \pm 1)\) and \([-1,1]\), respectively
2 \(\mathrm{R}-(-1,1)\) and \([-1,1]\) respectively
3 \(\mathrm{R}-(0, \pm 1)\) and \([0,1]\), respectively
4 \(\mathrm{R}-[-1,1]\) and \([0,1]\), respectively
Explanation:
B Given, Function \(f(x)=\sin \left(\frac{1}{|x| \sqrt{x^2-1}}\right)\) For domain, \(|x| \neq 0 \Rightarrow x \neq 0\) And, \(x^2-1>0\) \((x-1)(x+1)>0\) Domain \(\mathrm{x} \neq(1,-1)\) Hence, \(x \in R-(-1,1)\) For range \(\because-1 \leq \sin x \leq 1\) Then, the range of sine function is \([-1,1]\).
C Given, \(\mathrm{x} \in \mathrm{R}: \frac{\sqrt{6+x-x^2}}{2 \mathrm{x}+5} \geq \frac{\sqrt{6+x-x^2}}{\mathrm{x}+4}\) \(\Rightarrow \quad \sqrt{6+\mathrm{x}-\mathrm{x}^2}\left(\frac{1}{2 \mathrm{x}+5}-\frac{1}{\mathrm{x}+4}\right) \geq 0\) \(\Rightarrow \quad \sqrt{6+\mathrm{x}-\mathrm{x}^2}\left(\frac{\mathrm{x}+4-2 \mathrm{x}-5}{(2 \mathrm{x}+5)(\mathrm{x}+4)}\right) \geq 0\) \(\Rightarrow \quad \sqrt{6+\mathrm{x}-\mathrm{x}^2}\left(\frac{-(\mathrm{x}+1)}{(2 \mathrm{x}+5)(\mathrm{x}+4)}\right) \geq 0\) \(\Rightarrow \quad (\mathrm{x}+1) \leq 0,(2 \mathrm{x}+5)(\mathrm{x}+4)\lt 0\) \(\Rightarrow \quad \mathrm{x} \in(-\infty,-4) \cup\left(\frac{-5}{2},-1\right]\) For expression to be exist - \(6+x-x^2 \geq 0\) \(\Rightarrow x^2-x-6 \leq 0\) \(\Rightarrow x^2-3 x+2 x-6 \leq 0\) \(\Rightarrow x(x-3)+2(x-3) \leq 0\) \(\Rightarrow (x-3)(x+2) \leq 0\) \(\Rightarrow x \in[-2,3]\) From equation (i) and (ii), we get - \(x \in[-2,-1] \cup\{3\}\)
Shift-II
Sets, Relation and Function
117370
Let \(D=\left\{x \in R: f(x)=\sqrt{\frac{x-|x|}{x-[x]}}\right\}\) is defined and \(C\) be the range of the real function \(g(x)=\frac{2 x}{4+x^2}\). Then \(D \cap C=\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117366
The range of the function \(f(x)=x^2+\frac{1}{x^2+1}\) is
1 \([1, \infty)\)
2 \((2, \infty)\)
3 \(\left(\frac{3}{2}, \infty\right)\)
4 \((0,1)\)
Explanation:
A Given, \(\mathrm{f}(\mathrm{x})=\mathrm{y}=\mathrm{x}^2+\frac{1}{\mathrm{x}^2+1}\) \(\Rightarrow \mathrm{y}=\frac{\mathrm{x}^4+\mathrm{x}^2+1}{\mathrm{x}^2+1}\) \(\Rightarrow \mathrm{x}^4+\mathrm{x}^2+1=\mathrm{y}\left(\mathrm{x}^2+1\right)\) \(\Rightarrow \mathrm{x}^4+\mathrm{x}^2 \times(1-\mathrm{y})+(1-\mathrm{y})=0\) \(\because \mathrm{x} \in \mathrm{R}, \text { So, discriminate } \geq 0\) \(\mathrm{~b}^2-4 \mathrm{ac} \geq 0\) \(\Rightarrow (1-\mathrm{y})^2-4 \times 1 \times(1-\mathrm{y}) \geq 0\) \(\Rightarrow (1-\mathrm{y})(1-\mathrm{y}-4) \geq 0\) \(\Rightarrow (1-\mathrm{y})(-\mathrm{y}-3) \geq 0\) \(\Rightarrow (\mathrm{y}+3)(1-\mathrm{y}) \geq 0\) \(\Rightarrow \mathrm{y} \in(-\infty,-3] \cup[1, \infty)\) \(\text { But, } \mathrm{y}>0,\) \(\text { So, range of function is }[1, \infty)\)So, range of function is \([1, \infty\) )
Shift-I
Sets, Relation and Function
117367
For \(f(x)=\sin \left(\frac{1}{|x| \sqrt{x^2-1}}\right)\) the domain and range of \(\boldsymbol{f}(\mathrm{x})\) in \(\mathrm{R}\) are
1 \(\mathrm{R}-(0, \pm 1)\) and \([-1,1]\), respectively
2 \(\mathrm{R}-(-1,1)\) and \([-1,1]\) respectively
3 \(\mathrm{R}-(0, \pm 1)\) and \([0,1]\), respectively
4 \(\mathrm{R}-[-1,1]\) and \([0,1]\), respectively
Explanation:
B Given, Function \(f(x)=\sin \left(\frac{1}{|x| \sqrt{x^2-1}}\right)\) For domain, \(|x| \neq 0 \Rightarrow x \neq 0\) And, \(x^2-1>0\) \((x-1)(x+1)>0\) Domain \(\mathrm{x} \neq(1,-1)\) Hence, \(x \in R-(-1,1)\) For range \(\because-1 \leq \sin x \leq 1\) Then, the range of sine function is \([-1,1]\).
C Given, \(\mathrm{x} \in \mathrm{R}: \frac{\sqrt{6+x-x^2}}{2 \mathrm{x}+5} \geq \frac{\sqrt{6+x-x^2}}{\mathrm{x}+4}\) \(\Rightarrow \quad \sqrt{6+\mathrm{x}-\mathrm{x}^2}\left(\frac{1}{2 \mathrm{x}+5}-\frac{1}{\mathrm{x}+4}\right) \geq 0\) \(\Rightarrow \quad \sqrt{6+\mathrm{x}-\mathrm{x}^2}\left(\frac{\mathrm{x}+4-2 \mathrm{x}-5}{(2 \mathrm{x}+5)(\mathrm{x}+4)}\right) \geq 0\) \(\Rightarrow \quad \sqrt{6+\mathrm{x}-\mathrm{x}^2}\left(\frac{-(\mathrm{x}+1)}{(2 \mathrm{x}+5)(\mathrm{x}+4)}\right) \geq 0\) \(\Rightarrow \quad (\mathrm{x}+1) \leq 0,(2 \mathrm{x}+5)(\mathrm{x}+4)\lt 0\) \(\Rightarrow \quad \mathrm{x} \in(-\infty,-4) \cup\left(\frac{-5}{2},-1\right]\) For expression to be exist - \(6+x-x^2 \geq 0\) \(\Rightarrow x^2-x-6 \leq 0\) \(\Rightarrow x^2-3 x+2 x-6 \leq 0\) \(\Rightarrow x(x-3)+2(x-3) \leq 0\) \(\Rightarrow (x-3)(x+2) \leq 0\) \(\Rightarrow x \in[-2,3]\) From equation (i) and (ii), we get - \(x \in[-2,-1] \cup\{3\}\)
Shift-II
Sets, Relation and Function
117370
Let \(D=\left\{x \in R: f(x)=\sqrt{\frac{x-|x|}{x-[x]}}\right\}\) is defined and \(C\) be the range of the real function \(g(x)=\frac{2 x}{4+x^2}\). Then \(D \cap C=\)
