117333
The range of the function \(f(x)=\tan \sqrt{\frac{\pi^2}{9}-x^2}\) is
1 \([0, \sqrt{3}]\)
2 \((0, \sqrt{3})\)
3 \([0, \sqrt{3})\)
4 \((0, \sqrt{3}]\)
Explanation:
A Find, the domain of function - \(f(x)=\tan \sqrt{\frac{\pi^2}{9}-x^2}\) \(\because\) The terms under the square root, it must be \(>0\) \(=\frac{\pi^2}{9}-x^2 \geq 0\) \(=x^2-\frac{\pi^2}{9} \leq 0=x^2 \leq \frac{\pi^2}{9}\) \(=-\frac{\pi}{3} \leq x \leq \frac{\pi}{3}\) Domain of \(\mathrm{x} \in\left[-\frac{\pi}{3}, \frac{\pi}{3}\right]\) \(\frac{-\pi}{3} \leq \mathrm{x} \leq \frac{\pi}{3}\) \(=0 \leq \mathrm{x}^2 \leq \frac{\pi^2}{9}\) Multiplying both sides by ( -1 ) \(=\frac{-\pi^2}{9} \leq-\mathrm{x}^2 \leq 0\) Adding both sides by \(\frac{\pi^2}{9}\) \(0 \leq \frac{\pi^2}{9}-\mathrm{x}^2 \leq \frac{\pi^2}{9}\) \(=0 \leq \sqrt{\frac{\pi^2}{9}-\mathrm{x}^2} \leq \frac{\pi}{3}\) Tanking tan on both the sides \(0 \leq \tan \left(\sqrt{\frac{\pi^2}{9}-x^2}\right) \leq \sqrt{3}\)So, the range of the function is \([0, \sqrt{3}]\)
AMU-2021
Sets, Relation and Function
117336
Consider the following relations in the real number \(\mathbf{R}_1=\left\{(x, y) \mid x^2+y^2 \leq 25\right\}, \mathbf{R}_2=\left\{(x, y) \left\lvert\, y \geq \frac{4 x^2}{9}\right.\right\}\) , then the range of \(R_1 \cap R_2\) is
1 \([0,5]\)
2 \([-3,3]\)
3 \([-5,5]\)
4 \([-3,5]\)
Explanation:
A Given, \(\mathrm{R}_1=\left\{(\mathrm{x}, \mathrm{y}) \mid \mathrm{x}^2+\mathrm{y}^2 \leq 25\right\}\) \(-5 \leq \mathrm{x} \leq 5,-5 \leq \mathrm{y} \leq 5\) Range of \(R_1=[-5,5]\) And, \(\mathrm{R}_2=\left\{(\mathrm{x}, \mathrm{y}) \left\lvert\, \mathrm{y} \geq \frac{4 \mathrm{x}^2}{9}\right.\right\}\) Range of \(R_2=[0, \infty)\) So, range of \(R_1 \cap R_2 \Rightarrow[0,5]\)
AMU-2018
Sets, Relation and Function
117337
The domain of the real valued function \(f(x)=\frac{\log _2(x+3)}{\sqrt{x^2+3 x+2}}\) is
1 \((-3, \infty)\)
2 \((-3,1) \cup(1-, \infty)\)
3 \((-3,-2) \cup(-2,-1) \cup(-1, \infty)\)
4 \((-3,-2) \cup(-1, \infty)\)
Explanation:
D The domain of the function \(\log _2(\mathrm{x}+3)\) \(\mathrm{x}+3>0\) \(\mathrm{x}>-3\) And \(x^2+3 x+2>0\) \(x^2+2 x+x+2>0\) \(x(x+2)+1(x+2)>0\) \((x+2)(x+1)>0\) \(x \in(-\infty,-2) \cup(-1, \infty)\) From (i) and (ii) we get - \(x \in(-3,-2) \cup(-1, \infty)\)
Shift-II
Sets, Relation and Function
117338
The domain of the real valued function \(f(x)=\frac{\sqrt{2-x}+\sqrt{1+x}}{\sqrt{x+3}}\) is
1 \([-1,2]\)
2 \((-1,2)\)
3 \([-1, \infty)\)
4 \([2, \infty)\)
Explanation:
A The domain of the function - \(2-\mathrm{x} \geq 0\) \(\mathrm{x} \leq 2\) \(1+\mathrm{x} \geq 0\) \(\mathrm{x} \geq-1\) \(\mathrm{x}+3 \geq 0\) \(\mathrm{x} \geq-3\) \(\therefore \mathrm{x} \in[-1,2]\)
117333
The range of the function \(f(x)=\tan \sqrt{\frac{\pi^2}{9}-x^2}\) is
1 \([0, \sqrt{3}]\)
2 \((0, \sqrt{3})\)
3 \([0, \sqrt{3})\)
4 \((0, \sqrt{3}]\)
Explanation:
A Find, the domain of function - \(f(x)=\tan \sqrt{\frac{\pi^2}{9}-x^2}\) \(\because\) The terms under the square root, it must be \(>0\) \(=\frac{\pi^2}{9}-x^2 \geq 0\) \(=x^2-\frac{\pi^2}{9} \leq 0=x^2 \leq \frac{\pi^2}{9}\) \(=-\frac{\pi}{3} \leq x \leq \frac{\pi}{3}\) Domain of \(\mathrm{x} \in\left[-\frac{\pi}{3}, \frac{\pi}{3}\right]\) \(\frac{-\pi}{3} \leq \mathrm{x} \leq \frac{\pi}{3}\) \(=0 \leq \mathrm{x}^2 \leq \frac{\pi^2}{9}\) Multiplying both sides by ( -1 ) \(=\frac{-\pi^2}{9} \leq-\mathrm{x}^2 \leq 0\) Adding both sides by \(\frac{\pi^2}{9}\) \(0 \leq \frac{\pi^2}{9}-\mathrm{x}^2 \leq \frac{\pi^2}{9}\) \(=0 \leq \sqrt{\frac{\pi^2}{9}-\mathrm{x}^2} \leq \frac{\pi}{3}\) Tanking tan on both the sides \(0 \leq \tan \left(\sqrt{\frac{\pi^2}{9}-x^2}\right) \leq \sqrt{3}\)So, the range of the function is \([0, \sqrt{3}]\)
AMU-2021
Sets, Relation and Function
117336
Consider the following relations in the real number \(\mathbf{R}_1=\left\{(x, y) \mid x^2+y^2 \leq 25\right\}, \mathbf{R}_2=\left\{(x, y) \left\lvert\, y \geq \frac{4 x^2}{9}\right.\right\}\) , then the range of \(R_1 \cap R_2\) is
1 \([0,5]\)
2 \([-3,3]\)
3 \([-5,5]\)
4 \([-3,5]\)
Explanation:
A Given, \(\mathrm{R}_1=\left\{(\mathrm{x}, \mathrm{y}) \mid \mathrm{x}^2+\mathrm{y}^2 \leq 25\right\}\) \(-5 \leq \mathrm{x} \leq 5,-5 \leq \mathrm{y} \leq 5\) Range of \(R_1=[-5,5]\) And, \(\mathrm{R}_2=\left\{(\mathrm{x}, \mathrm{y}) \left\lvert\, \mathrm{y} \geq \frac{4 \mathrm{x}^2}{9}\right.\right\}\) Range of \(R_2=[0, \infty)\) So, range of \(R_1 \cap R_2 \Rightarrow[0,5]\)
AMU-2018
Sets, Relation and Function
117337
The domain of the real valued function \(f(x)=\frac{\log _2(x+3)}{\sqrt{x^2+3 x+2}}\) is
1 \((-3, \infty)\)
2 \((-3,1) \cup(1-, \infty)\)
3 \((-3,-2) \cup(-2,-1) \cup(-1, \infty)\)
4 \((-3,-2) \cup(-1, \infty)\)
Explanation:
D The domain of the function \(\log _2(\mathrm{x}+3)\) \(\mathrm{x}+3>0\) \(\mathrm{x}>-3\) And \(x^2+3 x+2>0\) \(x^2+2 x+x+2>0\) \(x(x+2)+1(x+2)>0\) \((x+2)(x+1)>0\) \(x \in(-\infty,-2) \cup(-1, \infty)\) From (i) and (ii) we get - \(x \in(-3,-2) \cup(-1, \infty)\)
Shift-II
Sets, Relation and Function
117338
The domain of the real valued function \(f(x)=\frac{\sqrt{2-x}+\sqrt{1+x}}{\sqrt{x+3}}\) is
1 \([-1,2]\)
2 \((-1,2)\)
3 \([-1, \infty)\)
4 \([2, \infty)\)
Explanation:
A The domain of the function - \(2-\mathrm{x} \geq 0\) \(\mathrm{x} \leq 2\) \(1+\mathrm{x} \geq 0\) \(\mathrm{x} \geq-1\) \(\mathrm{x}+3 \geq 0\) \(\mathrm{x} \geq-3\) \(\therefore \mathrm{x} \in[-1,2]\)
117333
The range of the function \(f(x)=\tan \sqrt{\frac{\pi^2}{9}-x^2}\) is
1 \([0, \sqrt{3}]\)
2 \((0, \sqrt{3})\)
3 \([0, \sqrt{3})\)
4 \((0, \sqrt{3}]\)
Explanation:
A Find, the domain of function - \(f(x)=\tan \sqrt{\frac{\pi^2}{9}-x^2}\) \(\because\) The terms under the square root, it must be \(>0\) \(=\frac{\pi^2}{9}-x^2 \geq 0\) \(=x^2-\frac{\pi^2}{9} \leq 0=x^2 \leq \frac{\pi^2}{9}\) \(=-\frac{\pi}{3} \leq x \leq \frac{\pi}{3}\) Domain of \(\mathrm{x} \in\left[-\frac{\pi}{3}, \frac{\pi}{3}\right]\) \(\frac{-\pi}{3} \leq \mathrm{x} \leq \frac{\pi}{3}\) \(=0 \leq \mathrm{x}^2 \leq \frac{\pi^2}{9}\) Multiplying both sides by ( -1 ) \(=\frac{-\pi^2}{9} \leq-\mathrm{x}^2 \leq 0\) Adding both sides by \(\frac{\pi^2}{9}\) \(0 \leq \frac{\pi^2}{9}-\mathrm{x}^2 \leq \frac{\pi^2}{9}\) \(=0 \leq \sqrt{\frac{\pi^2}{9}-\mathrm{x}^2} \leq \frac{\pi}{3}\) Tanking tan on both the sides \(0 \leq \tan \left(\sqrt{\frac{\pi^2}{9}-x^2}\right) \leq \sqrt{3}\)So, the range of the function is \([0, \sqrt{3}]\)
AMU-2021
Sets, Relation and Function
117336
Consider the following relations in the real number \(\mathbf{R}_1=\left\{(x, y) \mid x^2+y^2 \leq 25\right\}, \mathbf{R}_2=\left\{(x, y) \left\lvert\, y \geq \frac{4 x^2}{9}\right.\right\}\) , then the range of \(R_1 \cap R_2\) is
1 \([0,5]\)
2 \([-3,3]\)
3 \([-5,5]\)
4 \([-3,5]\)
Explanation:
A Given, \(\mathrm{R}_1=\left\{(\mathrm{x}, \mathrm{y}) \mid \mathrm{x}^2+\mathrm{y}^2 \leq 25\right\}\) \(-5 \leq \mathrm{x} \leq 5,-5 \leq \mathrm{y} \leq 5\) Range of \(R_1=[-5,5]\) And, \(\mathrm{R}_2=\left\{(\mathrm{x}, \mathrm{y}) \left\lvert\, \mathrm{y} \geq \frac{4 \mathrm{x}^2}{9}\right.\right\}\) Range of \(R_2=[0, \infty)\) So, range of \(R_1 \cap R_2 \Rightarrow[0,5]\)
AMU-2018
Sets, Relation and Function
117337
The domain of the real valued function \(f(x)=\frac{\log _2(x+3)}{\sqrt{x^2+3 x+2}}\) is
1 \((-3, \infty)\)
2 \((-3,1) \cup(1-, \infty)\)
3 \((-3,-2) \cup(-2,-1) \cup(-1, \infty)\)
4 \((-3,-2) \cup(-1, \infty)\)
Explanation:
D The domain of the function \(\log _2(\mathrm{x}+3)\) \(\mathrm{x}+3>0\) \(\mathrm{x}>-3\) And \(x^2+3 x+2>0\) \(x^2+2 x+x+2>0\) \(x(x+2)+1(x+2)>0\) \((x+2)(x+1)>0\) \(x \in(-\infty,-2) \cup(-1, \infty)\) From (i) and (ii) we get - \(x \in(-3,-2) \cup(-1, \infty)\)
Shift-II
Sets, Relation and Function
117338
The domain of the real valued function \(f(x)=\frac{\sqrt{2-x}+\sqrt{1+x}}{\sqrt{x+3}}\) is
1 \([-1,2]\)
2 \((-1,2)\)
3 \([-1, \infty)\)
4 \([2, \infty)\)
Explanation:
A The domain of the function - \(2-\mathrm{x} \geq 0\) \(\mathrm{x} \leq 2\) \(1+\mathrm{x} \geq 0\) \(\mathrm{x} \geq-1\) \(\mathrm{x}+3 \geq 0\) \(\mathrm{x} \geq-3\) \(\therefore \mathrm{x} \in[-1,2]\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
117333
The range of the function \(f(x)=\tan \sqrt{\frac{\pi^2}{9}-x^2}\) is
1 \([0, \sqrt{3}]\)
2 \((0, \sqrt{3})\)
3 \([0, \sqrt{3})\)
4 \((0, \sqrt{3}]\)
Explanation:
A Find, the domain of function - \(f(x)=\tan \sqrt{\frac{\pi^2}{9}-x^2}\) \(\because\) The terms under the square root, it must be \(>0\) \(=\frac{\pi^2}{9}-x^2 \geq 0\) \(=x^2-\frac{\pi^2}{9} \leq 0=x^2 \leq \frac{\pi^2}{9}\) \(=-\frac{\pi}{3} \leq x \leq \frac{\pi}{3}\) Domain of \(\mathrm{x} \in\left[-\frac{\pi}{3}, \frac{\pi}{3}\right]\) \(\frac{-\pi}{3} \leq \mathrm{x} \leq \frac{\pi}{3}\) \(=0 \leq \mathrm{x}^2 \leq \frac{\pi^2}{9}\) Multiplying both sides by ( -1 ) \(=\frac{-\pi^2}{9} \leq-\mathrm{x}^2 \leq 0\) Adding both sides by \(\frac{\pi^2}{9}\) \(0 \leq \frac{\pi^2}{9}-\mathrm{x}^2 \leq \frac{\pi^2}{9}\) \(=0 \leq \sqrt{\frac{\pi^2}{9}-\mathrm{x}^2} \leq \frac{\pi}{3}\) Tanking tan on both the sides \(0 \leq \tan \left(\sqrt{\frac{\pi^2}{9}-x^2}\right) \leq \sqrt{3}\)So, the range of the function is \([0, \sqrt{3}]\)
AMU-2021
Sets, Relation and Function
117336
Consider the following relations in the real number \(\mathbf{R}_1=\left\{(x, y) \mid x^2+y^2 \leq 25\right\}, \mathbf{R}_2=\left\{(x, y) \left\lvert\, y \geq \frac{4 x^2}{9}\right.\right\}\) , then the range of \(R_1 \cap R_2\) is
1 \([0,5]\)
2 \([-3,3]\)
3 \([-5,5]\)
4 \([-3,5]\)
Explanation:
A Given, \(\mathrm{R}_1=\left\{(\mathrm{x}, \mathrm{y}) \mid \mathrm{x}^2+\mathrm{y}^2 \leq 25\right\}\) \(-5 \leq \mathrm{x} \leq 5,-5 \leq \mathrm{y} \leq 5\) Range of \(R_1=[-5,5]\) And, \(\mathrm{R}_2=\left\{(\mathrm{x}, \mathrm{y}) \left\lvert\, \mathrm{y} \geq \frac{4 \mathrm{x}^2}{9}\right.\right\}\) Range of \(R_2=[0, \infty)\) So, range of \(R_1 \cap R_2 \Rightarrow[0,5]\)
AMU-2018
Sets, Relation and Function
117337
The domain of the real valued function \(f(x)=\frac{\log _2(x+3)}{\sqrt{x^2+3 x+2}}\) is
1 \((-3, \infty)\)
2 \((-3,1) \cup(1-, \infty)\)
3 \((-3,-2) \cup(-2,-1) \cup(-1, \infty)\)
4 \((-3,-2) \cup(-1, \infty)\)
Explanation:
D The domain of the function \(\log _2(\mathrm{x}+3)\) \(\mathrm{x}+3>0\) \(\mathrm{x}>-3\) And \(x^2+3 x+2>0\) \(x^2+2 x+x+2>0\) \(x(x+2)+1(x+2)>0\) \((x+2)(x+1)>0\) \(x \in(-\infty,-2) \cup(-1, \infty)\) From (i) and (ii) we get - \(x \in(-3,-2) \cup(-1, \infty)\)
Shift-II
Sets, Relation and Function
117338
The domain of the real valued function \(f(x)=\frac{\sqrt{2-x}+\sqrt{1+x}}{\sqrt{x+3}}\) is
1 \([-1,2]\)
2 \((-1,2)\)
3 \([-1, \infty)\)
4 \([2, \infty)\)
Explanation:
A The domain of the function - \(2-\mathrm{x} \geq 0\) \(\mathrm{x} \leq 2\) \(1+\mathrm{x} \geq 0\) \(\mathrm{x} \geq-1\) \(\mathrm{x}+3 \geq 0\) \(\mathrm{x} \geq-3\) \(\therefore \mathrm{x} \in[-1,2]\)
