117339
The set of all real numbers satisfying the in equation \(\mathbf{x}^2-|\mathbf{x}+2|+\mathbf{x}>\mathbf{0}\) is
1 \([-2,-\sqrt{2}) \cup(\sqrt{2}, \infty)\)
2 \((-\infty,-2) \cup(2, \infty)\)
3 \((-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty)\)
4 \((-\infty,-2) \cup(\sqrt{2}, \infty)\)
Explanation:
C Given \(x^2-|x+2|+x>0\) Case I : When \(\mathrm{x}+2\lt 0\) \(x^2+x+2+x>0\) \(x^2+2 x+2>0\) \(x^2+2 x+1+1>0\) \((x+1)^2+1>0\) Which is true for all \(\mathrm{x}\) \(\mathrm{x} \leq-2\), or \(\mathrm{x} \in(-\infty,-2)\) Case II: When \(\mathrm{x}+2 \geq 0\) \(x^2-x-2+x>0\) \(x^2-2>0\) \(x\lt -\sqrt{2} \quad \text { or } \quad x>\sqrt{2}\) \(\mathrm{x} \in[-\infty,-\sqrt{2}] \cup(\sqrt{2}, \infty)\)From equation (ii) and (iii), we get \(x \in(-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty)\)
Shift-I
Sets, Relation and Function
117340
If \(x \in R\), then the range of \(\frac{x}{x^2-5 x+9}\) is
C The range of the function - \(f(x)=y=\frac{x}{x^2-5 x+9}\) \(\mathrm{x}^2 \mathrm{y}-5 \mathrm{xy}+9 \mathrm{y}-\mathrm{x}=0\) \(\mathrm{x}^2 \mathrm{y}-\mathrm{x}(5 \mathrm{y}+1)+9 \mathrm{y}=0\) \(\mathrm{a}=\mathrm{y}, \mathrm{b}=-(5 \mathrm{y}+1), \quad \mathrm{c}=9 \mathrm{y}\) \(\because \quad \mathrm{x}=\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 \mathrm{ac}}}{2 \mathrm{a}}\) \(\therefore \quad x=\frac{(5 y+1) \pm \sqrt{(5 y+1)^2-4 \times y \times 9 y}}{2 \times y}\) \(x=\frac{(5 y+1) \pm \sqrt{(5 y+1)^2-36 y^2}}{2 y}\) \((5 \mathrm{y}+1)^2-(6 \mathrm{y})^2 \geq 0\) \((5 \mathrm{y}+1+6 \mathrm{y})(5 \mathrm{y}+1-6 \mathrm{y}) \geq 0\) \((11 y+1)(1-y) \geq 0\) \(1-\mathrm{y} \geq 0 \quad \text { or } \quad 11 \mathrm{y}+1 \geq 0\) \(y \leq 1 \quad \text { or } \quad \mathrm{y} \geq \frac{-1}{11}\) Hence, range : \(\mathrm{y} \in\left[-\frac{1}{11}, 1\right]\)
Shift-II
Sets, Relation and Function
117342
Find the domain of the real valued function \(f\) \((x)=\left([x]^2-[x]-2\right)^{-1 / 2}\) where \([\cdot]\) is the greatest integer function.
1 \(\mathrm{R}-(-1,3]\)
2 \(\mathrm{R}-[-1,3)\)
3 \(\mathrm{R}-(-1,3)\)
4 \(\mathrm{R}-[-1,3]\)
Explanation:
A Let, \(\mathrm{y}=f(\mathrm{x})=\left([\mathrm{x}]^2-[\mathrm{x}]-2\right)^{-1 / 2}\) \(\Rightarrow \quad y^2=\frac{1}{[x]^2-[x]-2}\) For real valued function - \({[\mathrm{x}]^2-[\mathrm{x}]-2>0}\) \(\Rightarrow \quad \{[\mathrm{x}]-2\}\{[\mathrm{x}]+1\}>0\) \({[\mathrm{x}] \in \mathrm{R}-(-1,2)}\)So, \(\mathrm{x} \in \mathrm{R}-[-1,3]\)
Shift-I
Sets, Relation and Function
117343
The solution set of the equation \(\sin ^{-1} x=2 \tan ^{-1} x\) is
1 \(\{1,2\}\)
2 \(\{-1,2\}\)
3 \(\{-1,1,0\}\)
4 \(\{1,1 / 2,0\}\)
Explanation:
C We have, \(\sin ^{-1} x=2 \tan ^{-1} x\) \(\sin ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\) \(x=\frac{2 x}{1+x^2}\) \(x\left(1+x^2\right)=2 x\) \(x^3-x=0\) \(x\left(x^2-1\right)=0\) \(x=0,1 \text {, or }-1\)So, the solution let have \(\{-1,1,0\}\).
117339
The set of all real numbers satisfying the in equation \(\mathbf{x}^2-|\mathbf{x}+2|+\mathbf{x}>\mathbf{0}\) is
1 \([-2,-\sqrt{2}) \cup(\sqrt{2}, \infty)\)
2 \((-\infty,-2) \cup(2, \infty)\)
3 \((-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty)\)
4 \((-\infty,-2) \cup(\sqrt{2}, \infty)\)
Explanation:
C Given \(x^2-|x+2|+x>0\) Case I : When \(\mathrm{x}+2\lt 0\) \(x^2+x+2+x>0\) \(x^2+2 x+2>0\) \(x^2+2 x+1+1>0\) \((x+1)^2+1>0\) Which is true for all \(\mathrm{x}\) \(\mathrm{x} \leq-2\), or \(\mathrm{x} \in(-\infty,-2)\) Case II: When \(\mathrm{x}+2 \geq 0\) \(x^2-x-2+x>0\) \(x^2-2>0\) \(x\lt -\sqrt{2} \quad \text { or } \quad x>\sqrt{2}\) \(\mathrm{x} \in[-\infty,-\sqrt{2}] \cup(\sqrt{2}, \infty)\)From equation (ii) and (iii), we get \(x \in(-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty)\)
Shift-I
Sets, Relation and Function
117340
If \(x \in R\), then the range of \(\frac{x}{x^2-5 x+9}\) is
C The range of the function - \(f(x)=y=\frac{x}{x^2-5 x+9}\) \(\mathrm{x}^2 \mathrm{y}-5 \mathrm{xy}+9 \mathrm{y}-\mathrm{x}=0\) \(\mathrm{x}^2 \mathrm{y}-\mathrm{x}(5 \mathrm{y}+1)+9 \mathrm{y}=0\) \(\mathrm{a}=\mathrm{y}, \mathrm{b}=-(5 \mathrm{y}+1), \quad \mathrm{c}=9 \mathrm{y}\) \(\because \quad \mathrm{x}=\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 \mathrm{ac}}}{2 \mathrm{a}}\) \(\therefore \quad x=\frac{(5 y+1) \pm \sqrt{(5 y+1)^2-4 \times y \times 9 y}}{2 \times y}\) \(x=\frac{(5 y+1) \pm \sqrt{(5 y+1)^2-36 y^2}}{2 y}\) \((5 \mathrm{y}+1)^2-(6 \mathrm{y})^2 \geq 0\) \((5 \mathrm{y}+1+6 \mathrm{y})(5 \mathrm{y}+1-6 \mathrm{y}) \geq 0\) \((11 y+1)(1-y) \geq 0\) \(1-\mathrm{y} \geq 0 \quad \text { or } \quad 11 \mathrm{y}+1 \geq 0\) \(y \leq 1 \quad \text { or } \quad \mathrm{y} \geq \frac{-1}{11}\) Hence, range : \(\mathrm{y} \in\left[-\frac{1}{11}, 1\right]\)
Shift-II
Sets, Relation and Function
117342
Find the domain of the real valued function \(f\) \((x)=\left([x]^2-[x]-2\right)^{-1 / 2}\) where \([\cdot]\) is the greatest integer function.
1 \(\mathrm{R}-(-1,3]\)
2 \(\mathrm{R}-[-1,3)\)
3 \(\mathrm{R}-(-1,3)\)
4 \(\mathrm{R}-[-1,3]\)
Explanation:
A Let, \(\mathrm{y}=f(\mathrm{x})=\left([\mathrm{x}]^2-[\mathrm{x}]-2\right)^{-1 / 2}\) \(\Rightarrow \quad y^2=\frac{1}{[x]^2-[x]-2}\) For real valued function - \({[\mathrm{x}]^2-[\mathrm{x}]-2>0}\) \(\Rightarrow \quad \{[\mathrm{x}]-2\}\{[\mathrm{x}]+1\}>0\) \({[\mathrm{x}] \in \mathrm{R}-(-1,2)}\)So, \(\mathrm{x} \in \mathrm{R}-[-1,3]\)
Shift-I
Sets, Relation and Function
117343
The solution set of the equation \(\sin ^{-1} x=2 \tan ^{-1} x\) is
1 \(\{1,2\}\)
2 \(\{-1,2\}\)
3 \(\{-1,1,0\}\)
4 \(\{1,1 / 2,0\}\)
Explanation:
C We have, \(\sin ^{-1} x=2 \tan ^{-1} x\) \(\sin ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\) \(x=\frac{2 x}{1+x^2}\) \(x\left(1+x^2\right)=2 x\) \(x^3-x=0\) \(x\left(x^2-1\right)=0\) \(x=0,1 \text {, or }-1\)So, the solution let have \(\{-1,1,0\}\).
NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117339
The set of all real numbers satisfying the in equation \(\mathbf{x}^2-|\mathbf{x}+2|+\mathbf{x}>\mathbf{0}\) is
1 \([-2,-\sqrt{2}) \cup(\sqrt{2}, \infty)\)
2 \((-\infty,-2) \cup(2, \infty)\)
3 \((-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty)\)
4 \((-\infty,-2) \cup(\sqrt{2}, \infty)\)
Explanation:
C Given \(x^2-|x+2|+x>0\) Case I : When \(\mathrm{x}+2\lt 0\) \(x^2+x+2+x>0\) \(x^2+2 x+2>0\) \(x^2+2 x+1+1>0\) \((x+1)^2+1>0\) Which is true for all \(\mathrm{x}\) \(\mathrm{x} \leq-2\), or \(\mathrm{x} \in(-\infty,-2)\) Case II: When \(\mathrm{x}+2 \geq 0\) \(x^2-x-2+x>0\) \(x^2-2>0\) \(x\lt -\sqrt{2} \quad \text { or } \quad x>\sqrt{2}\) \(\mathrm{x} \in[-\infty,-\sqrt{2}] \cup(\sqrt{2}, \infty)\)From equation (ii) and (iii), we get \(x \in(-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty)\)
Shift-I
Sets, Relation and Function
117340
If \(x \in R\), then the range of \(\frac{x}{x^2-5 x+9}\) is
C The range of the function - \(f(x)=y=\frac{x}{x^2-5 x+9}\) \(\mathrm{x}^2 \mathrm{y}-5 \mathrm{xy}+9 \mathrm{y}-\mathrm{x}=0\) \(\mathrm{x}^2 \mathrm{y}-\mathrm{x}(5 \mathrm{y}+1)+9 \mathrm{y}=0\) \(\mathrm{a}=\mathrm{y}, \mathrm{b}=-(5 \mathrm{y}+1), \quad \mathrm{c}=9 \mathrm{y}\) \(\because \quad \mathrm{x}=\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 \mathrm{ac}}}{2 \mathrm{a}}\) \(\therefore \quad x=\frac{(5 y+1) \pm \sqrt{(5 y+1)^2-4 \times y \times 9 y}}{2 \times y}\) \(x=\frac{(5 y+1) \pm \sqrt{(5 y+1)^2-36 y^2}}{2 y}\) \((5 \mathrm{y}+1)^2-(6 \mathrm{y})^2 \geq 0\) \((5 \mathrm{y}+1+6 \mathrm{y})(5 \mathrm{y}+1-6 \mathrm{y}) \geq 0\) \((11 y+1)(1-y) \geq 0\) \(1-\mathrm{y} \geq 0 \quad \text { or } \quad 11 \mathrm{y}+1 \geq 0\) \(y \leq 1 \quad \text { or } \quad \mathrm{y} \geq \frac{-1}{11}\) Hence, range : \(\mathrm{y} \in\left[-\frac{1}{11}, 1\right]\)
Shift-II
Sets, Relation and Function
117342
Find the domain of the real valued function \(f\) \((x)=\left([x]^2-[x]-2\right)^{-1 / 2}\) where \([\cdot]\) is the greatest integer function.
1 \(\mathrm{R}-(-1,3]\)
2 \(\mathrm{R}-[-1,3)\)
3 \(\mathrm{R}-(-1,3)\)
4 \(\mathrm{R}-[-1,3]\)
Explanation:
A Let, \(\mathrm{y}=f(\mathrm{x})=\left([\mathrm{x}]^2-[\mathrm{x}]-2\right)^{-1 / 2}\) \(\Rightarrow \quad y^2=\frac{1}{[x]^2-[x]-2}\) For real valued function - \({[\mathrm{x}]^2-[\mathrm{x}]-2>0}\) \(\Rightarrow \quad \{[\mathrm{x}]-2\}\{[\mathrm{x}]+1\}>0\) \({[\mathrm{x}] \in \mathrm{R}-(-1,2)}\)So, \(\mathrm{x} \in \mathrm{R}-[-1,3]\)
Shift-I
Sets, Relation and Function
117343
The solution set of the equation \(\sin ^{-1} x=2 \tan ^{-1} x\) is
1 \(\{1,2\}\)
2 \(\{-1,2\}\)
3 \(\{-1,1,0\}\)
4 \(\{1,1 / 2,0\}\)
Explanation:
C We have, \(\sin ^{-1} x=2 \tan ^{-1} x\) \(\sin ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\) \(x=\frac{2 x}{1+x^2}\) \(x\left(1+x^2\right)=2 x\) \(x^3-x=0\) \(x\left(x^2-1\right)=0\) \(x=0,1 \text {, or }-1\)So, the solution let have \(\{-1,1,0\}\).
117339
The set of all real numbers satisfying the in equation \(\mathbf{x}^2-|\mathbf{x}+2|+\mathbf{x}>\mathbf{0}\) is
1 \([-2,-\sqrt{2}) \cup(\sqrt{2}, \infty)\)
2 \((-\infty,-2) \cup(2, \infty)\)
3 \((-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty)\)
4 \((-\infty,-2) \cup(\sqrt{2}, \infty)\)
Explanation:
C Given \(x^2-|x+2|+x>0\) Case I : When \(\mathrm{x}+2\lt 0\) \(x^2+x+2+x>0\) \(x^2+2 x+2>0\) \(x^2+2 x+1+1>0\) \((x+1)^2+1>0\) Which is true for all \(\mathrm{x}\) \(\mathrm{x} \leq-2\), or \(\mathrm{x} \in(-\infty,-2)\) Case II: When \(\mathrm{x}+2 \geq 0\) \(x^2-x-2+x>0\) \(x^2-2>0\) \(x\lt -\sqrt{2} \quad \text { or } \quad x>\sqrt{2}\) \(\mathrm{x} \in[-\infty,-\sqrt{2}] \cup(\sqrt{2}, \infty)\)From equation (ii) and (iii), we get \(x \in(-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty)\)
Shift-I
Sets, Relation and Function
117340
If \(x \in R\), then the range of \(\frac{x}{x^2-5 x+9}\) is
C The range of the function - \(f(x)=y=\frac{x}{x^2-5 x+9}\) \(\mathrm{x}^2 \mathrm{y}-5 \mathrm{xy}+9 \mathrm{y}-\mathrm{x}=0\) \(\mathrm{x}^2 \mathrm{y}-\mathrm{x}(5 \mathrm{y}+1)+9 \mathrm{y}=0\) \(\mathrm{a}=\mathrm{y}, \mathrm{b}=-(5 \mathrm{y}+1), \quad \mathrm{c}=9 \mathrm{y}\) \(\because \quad \mathrm{x}=\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 \mathrm{ac}}}{2 \mathrm{a}}\) \(\therefore \quad x=\frac{(5 y+1) \pm \sqrt{(5 y+1)^2-4 \times y \times 9 y}}{2 \times y}\) \(x=\frac{(5 y+1) \pm \sqrt{(5 y+1)^2-36 y^2}}{2 y}\) \((5 \mathrm{y}+1)^2-(6 \mathrm{y})^2 \geq 0\) \((5 \mathrm{y}+1+6 \mathrm{y})(5 \mathrm{y}+1-6 \mathrm{y}) \geq 0\) \((11 y+1)(1-y) \geq 0\) \(1-\mathrm{y} \geq 0 \quad \text { or } \quad 11 \mathrm{y}+1 \geq 0\) \(y \leq 1 \quad \text { or } \quad \mathrm{y} \geq \frac{-1}{11}\) Hence, range : \(\mathrm{y} \in\left[-\frac{1}{11}, 1\right]\)
Shift-II
Sets, Relation and Function
117342
Find the domain of the real valued function \(f\) \((x)=\left([x]^2-[x]-2\right)^{-1 / 2}\) where \([\cdot]\) is the greatest integer function.
1 \(\mathrm{R}-(-1,3]\)
2 \(\mathrm{R}-[-1,3)\)
3 \(\mathrm{R}-(-1,3)\)
4 \(\mathrm{R}-[-1,3]\)
Explanation:
A Let, \(\mathrm{y}=f(\mathrm{x})=\left([\mathrm{x}]^2-[\mathrm{x}]-2\right)^{-1 / 2}\) \(\Rightarrow \quad y^2=\frac{1}{[x]^2-[x]-2}\) For real valued function - \({[\mathrm{x}]^2-[\mathrm{x}]-2>0}\) \(\Rightarrow \quad \{[\mathrm{x}]-2\}\{[\mathrm{x}]+1\}>0\) \({[\mathrm{x}] \in \mathrm{R}-(-1,2)}\)So, \(\mathrm{x} \in \mathrm{R}-[-1,3]\)
Shift-I
Sets, Relation and Function
117343
The solution set of the equation \(\sin ^{-1} x=2 \tan ^{-1} x\) is
1 \(\{1,2\}\)
2 \(\{-1,2\}\)
3 \(\{-1,1,0\}\)
4 \(\{1,1 / 2,0\}\)
Explanation:
C We have, \(\sin ^{-1} x=2 \tan ^{-1} x\) \(\sin ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\) \(x=\frac{2 x}{1+x^2}\) \(x\left(1+x^2\right)=2 x\) \(x^3-x=0\) \(x\left(x^2-1\right)=0\) \(x=0,1 \text {, or }-1\)So, the solution let have \(\{-1,1,0\}\).