117344
If \(f(x)=|\sin x|\) has an inverse, then its domain is
1 \([-\pi / 2, \pi]\)
2 \([0, \pi / 2]\)
3 \([0,2 \pi]\)
4 \([-\pi, \pi]\)
Explanation:
B \(\mathrm{f}(\mathrm{x})=|\sin \mathrm{x}|\) \(\sin x>, 0\) when \(x \in[0, \pi / 2]\) \(|\sin x|=\sin x\) \(f(x)=\sin x, x \in[0, \pi / 2]\) \(\sin x\) increases from increases 0 to \(\pi / 2\), then the value of \(\sin \mathrm{x}\) increases from 0 to 1 . \(f(x)\) is one - one and onto when \(x \in[0, \pi / 2]\) So, \(f(x)\) is invertible if its domain \([0, \pi / 2]\)
AMU-2012
Sets, Relation and Function
117345
The range of the real valued function \(f(x)=\) \(\sqrt{\frac{x^2+2 x+8}{x^2+2 x+4}}\) is
1 \(\left(\sqrt{\frac{7}{3}}, \infty\right)\)
2 \((0, \infty)\)
3 \((1, \infty)\)
4 \(\left(1, \sqrt{\frac{7}{3}}\right]\)
Explanation:
D The range of the real valued function of - \(f(x)=y=\sqrt{\frac{x^2+2 x+8}{x^2+2 x+4}}\) \(y^2=\frac{x^2+2 x+8}{x^2+2 x+4}\) \(x^2 y^2+2 x y^2+4 y^2-x^2-2 x-8=0\) \(\Rightarrow x^2\left(y^2-1\right)+2 x\left(y^2-1\right)+4\left(y^2-2\right)=0\) \(a=\left(y^2-1\right), b=2\left(y^2-1\right), c=4\left(y^2-2\right)\) \(\because x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\) \(=\frac{-2\left(y^2-1\right) \pm \sqrt{\left\{2\left(y^2-1\right)\right\}^2-4 \times\left(y^2-1\right) \times 4\left(y^2-2\right)}}{2\left(y^2-1\right)}\) \(=\frac{-2\left(y^2-1\right) \pm \sqrt{\left(y^2-1\right)\left[4\left(y^2-1\right)-16\left(y^2-2\right)\right)}}{2\left(y^2-1\right)}\) \(=\frac{-2\left(y^2-1\right) \pm \sqrt{\left(y^2-1\right)\left(7-3 y^2\right) \times 4}}{2\left(y^2-1\right)}\) \(\therefore\left(y^2-1\right)\left(7-3 y^2\right) \times 4 \geq 0\) \(\left(y^2-1\right) \geq 0 \quad \text { or } 7-3 y^2 \geq 0\) \(y \geq 1 \text { or } \mathrm{y} \geq \sqrt{\frac{7}{3}}\) \(\mathrm{y} \geq 1 \text { or } \mathrm{y} \geq \sqrt{\frac{7}{3}}\)So, range of \(\mathrm{f}(\mathrm{x})\) is \(\left(1, \sqrt{\frac{7}{3}}\right]\).
Shift-I
Sets, Relation and Function
117346
If \(f: R \rightarrow R\) is defined as \(f(x)=\frac{x^6}{x^6+2020} \cdot \forall x\) \(\in R\), then the range of \(\boldsymbol{f}\) is .....
1 \((0,1)\)
2 \([0, \infty]\)
3 \([0,1]\)
4 \(\left[0, \frac{1}{2020}\right]\)
Explanation:
C Given, \(x \in\) Real value \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^6}{\mathrm{x}^6+2020}\) \(\therefore \mathrm{x}^6+2020>\mathrm{x}^6\) \(\Rightarrow \frac{\mathrm{x}^6}{\mathrm{x}^6+2020}\lt 1\) And, \(x^6\lt 0\) Hence, range of \(f(x)\) is \([0,1]\)
Shift-II
Sets, Relation and Function
117347
\(\left\{x \in R: \frac{14 x}{x+1}-\frac{9 x-30}{x-4}\lt 0\right\}\) is equal to
117344
If \(f(x)=|\sin x|\) has an inverse, then its domain is
1 \([-\pi / 2, \pi]\)
2 \([0, \pi / 2]\)
3 \([0,2 \pi]\)
4 \([-\pi, \pi]\)
Explanation:
B \(\mathrm{f}(\mathrm{x})=|\sin \mathrm{x}|\) \(\sin x>, 0\) when \(x \in[0, \pi / 2]\) \(|\sin x|=\sin x\) \(f(x)=\sin x, x \in[0, \pi / 2]\) \(\sin x\) increases from increases 0 to \(\pi / 2\), then the value of \(\sin \mathrm{x}\) increases from 0 to 1 . \(f(x)\) is one - one and onto when \(x \in[0, \pi / 2]\) So, \(f(x)\) is invertible if its domain \([0, \pi / 2]\)
AMU-2012
Sets, Relation and Function
117345
The range of the real valued function \(f(x)=\) \(\sqrt{\frac{x^2+2 x+8}{x^2+2 x+4}}\) is
1 \(\left(\sqrt{\frac{7}{3}}, \infty\right)\)
2 \((0, \infty)\)
3 \((1, \infty)\)
4 \(\left(1, \sqrt{\frac{7}{3}}\right]\)
Explanation:
D The range of the real valued function of - \(f(x)=y=\sqrt{\frac{x^2+2 x+8}{x^2+2 x+4}}\) \(y^2=\frac{x^2+2 x+8}{x^2+2 x+4}\) \(x^2 y^2+2 x y^2+4 y^2-x^2-2 x-8=0\) \(\Rightarrow x^2\left(y^2-1\right)+2 x\left(y^2-1\right)+4\left(y^2-2\right)=0\) \(a=\left(y^2-1\right), b=2\left(y^2-1\right), c=4\left(y^2-2\right)\) \(\because x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\) \(=\frac{-2\left(y^2-1\right) \pm \sqrt{\left\{2\left(y^2-1\right)\right\}^2-4 \times\left(y^2-1\right) \times 4\left(y^2-2\right)}}{2\left(y^2-1\right)}\) \(=\frac{-2\left(y^2-1\right) \pm \sqrt{\left(y^2-1\right)\left[4\left(y^2-1\right)-16\left(y^2-2\right)\right)}}{2\left(y^2-1\right)}\) \(=\frac{-2\left(y^2-1\right) \pm \sqrt{\left(y^2-1\right)\left(7-3 y^2\right) \times 4}}{2\left(y^2-1\right)}\) \(\therefore\left(y^2-1\right)\left(7-3 y^2\right) \times 4 \geq 0\) \(\left(y^2-1\right) \geq 0 \quad \text { or } 7-3 y^2 \geq 0\) \(y \geq 1 \text { or } \mathrm{y} \geq \sqrt{\frac{7}{3}}\) \(\mathrm{y} \geq 1 \text { or } \mathrm{y} \geq \sqrt{\frac{7}{3}}\)So, range of \(\mathrm{f}(\mathrm{x})\) is \(\left(1, \sqrt{\frac{7}{3}}\right]\).
Shift-I
Sets, Relation and Function
117346
If \(f: R \rightarrow R\) is defined as \(f(x)=\frac{x^6}{x^6+2020} \cdot \forall x\) \(\in R\), then the range of \(\boldsymbol{f}\) is .....
1 \((0,1)\)
2 \([0, \infty]\)
3 \([0,1]\)
4 \(\left[0, \frac{1}{2020}\right]\)
Explanation:
C Given, \(x \in\) Real value \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^6}{\mathrm{x}^6+2020}\) \(\therefore \mathrm{x}^6+2020>\mathrm{x}^6\) \(\Rightarrow \frac{\mathrm{x}^6}{\mathrm{x}^6+2020}\lt 1\) And, \(x^6\lt 0\) Hence, range of \(f(x)\) is \([0,1]\)
Shift-II
Sets, Relation and Function
117347
\(\left\{x \in R: \frac{14 x}{x+1}-\frac{9 x-30}{x-4}\lt 0\right\}\) is equal to
117344
If \(f(x)=|\sin x|\) has an inverse, then its domain is
1 \([-\pi / 2, \pi]\)
2 \([0, \pi / 2]\)
3 \([0,2 \pi]\)
4 \([-\pi, \pi]\)
Explanation:
B \(\mathrm{f}(\mathrm{x})=|\sin \mathrm{x}|\) \(\sin x>, 0\) when \(x \in[0, \pi / 2]\) \(|\sin x|=\sin x\) \(f(x)=\sin x, x \in[0, \pi / 2]\) \(\sin x\) increases from increases 0 to \(\pi / 2\), then the value of \(\sin \mathrm{x}\) increases from 0 to 1 . \(f(x)\) is one - one and onto when \(x \in[0, \pi / 2]\) So, \(f(x)\) is invertible if its domain \([0, \pi / 2]\)
AMU-2012
Sets, Relation and Function
117345
The range of the real valued function \(f(x)=\) \(\sqrt{\frac{x^2+2 x+8}{x^2+2 x+4}}\) is
1 \(\left(\sqrt{\frac{7}{3}}, \infty\right)\)
2 \((0, \infty)\)
3 \((1, \infty)\)
4 \(\left(1, \sqrt{\frac{7}{3}}\right]\)
Explanation:
D The range of the real valued function of - \(f(x)=y=\sqrt{\frac{x^2+2 x+8}{x^2+2 x+4}}\) \(y^2=\frac{x^2+2 x+8}{x^2+2 x+4}\) \(x^2 y^2+2 x y^2+4 y^2-x^2-2 x-8=0\) \(\Rightarrow x^2\left(y^2-1\right)+2 x\left(y^2-1\right)+4\left(y^2-2\right)=0\) \(a=\left(y^2-1\right), b=2\left(y^2-1\right), c=4\left(y^2-2\right)\) \(\because x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\) \(=\frac{-2\left(y^2-1\right) \pm \sqrt{\left\{2\left(y^2-1\right)\right\}^2-4 \times\left(y^2-1\right) \times 4\left(y^2-2\right)}}{2\left(y^2-1\right)}\) \(=\frac{-2\left(y^2-1\right) \pm \sqrt{\left(y^2-1\right)\left[4\left(y^2-1\right)-16\left(y^2-2\right)\right)}}{2\left(y^2-1\right)}\) \(=\frac{-2\left(y^2-1\right) \pm \sqrt{\left(y^2-1\right)\left(7-3 y^2\right) \times 4}}{2\left(y^2-1\right)}\) \(\therefore\left(y^2-1\right)\left(7-3 y^2\right) \times 4 \geq 0\) \(\left(y^2-1\right) \geq 0 \quad \text { or } 7-3 y^2 \geq 0\) \(y \geq 1 \text { or } \mathrm{y} \geq \sqrt{\frac{7}{3}}\) \(\mathrm{y} \geq 1 \text { or } \mathrm{y} \geq \sqrt{\frac{7}{3}}\)So, range of \(\mathrm{f}(\mathrm{x})\) is \(\left(1, \sqrt{\frac{7}{3}}\right]\).
Shift-I
Sets, Relation and Function
117346
If \(f: R \rightarrow R\) is defined as \(f(x)=\frac{x^6}{x^6+2020} \cdot \forall x\) \(\in R\), then the range of \(\boldsymbol{f}\) is .....
1 \((0,1)\)
2 \([0, \infty]\)
3 \([0,1]\)
4 \(\left[0, \frac{1}{2020}\right]\)
Explanation:
C Given, \(x \in\) Real value \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^6}{\mathrm{x}^6+2020}\) \(\therefore \mathrm{x}^6+2020>\mathrm{x}^6\) \(\Rightarrow \frac{\mathrm{x}^6}{\mathrm{x}^6+2020}\lt 1\) And, \(x^6\lt 0\) Hence, range of \(f(x)\) is \([0,1]\)
Shift-II
Sets, Relation and Function
117347
\(\left\{x \in R: \frac{14 x}{x+1}-\frac{9 x-30}{x-4}\lt 0\right\}\) is equal to
117344
If \(f(x)=|\sin x|\) has an inverse, then its domain is
1 \([-\pi / 2, \pi]\)
2 \([0, \pi / 2]\)
3 \([0,2 \pi]\)
4 \([-\pi, \pi]\)
Explanation:
B \(\mathrm{f}(\mathrm{x})=|\sin \mathrm{x}|\) \(\sin x>, 0\) when \(x \in[0, \pi / 2]\) \(|\sin x|=\sin x\) \(f(x)=\sin x, x \in[0, \pi / 2]\) \(\sin x\) increases from increases 0 to \(\pi / 2\), then the value of \(\sin \mathrm{x}\) increases from 0 to 1 . \(f(x)\) is one - one and onto when \(x \in[0, \pi / 2]\) So, \(f(x)\) is invertible if its domain \([0, \pi / 2]\)
AMU-2012
Sets, Relation and Function
117345
The range of the real valued function \(f(x)=\) \(\sqrt{\frac{x^2+2 x+8}{x^2+2 x+4}}\) is
1 \(\left(\sqrt{\frac{7}{3}}, \infty\right)\)
2 \((0, \infty)\)
3 \((1, \infty)\)
4 \(\left(1, \sqrt{\frac{7}{3}}\right]\)
Explanation:
D The range of the real valued function of - \(f(x)=y=\sqrt{\frac{x^2+2 x+8}{x^2+2 x+4}}\) \(y^2=\frac{x^2+2 x+8}{x^2+2 x+4}\) \(x^2 y^2+2 x y^2+4 y^2-x^2-2 x-8=0\) \(\Rightarrow x^2\left(y^2-1\right)+2 x\left(y^2-1\right)+4\left(y^2-2\right)=0\) \(a=\left(y^2-1\right), b=2\left(y^2-1\right), c=4\left(y^2-2\right)\) \(\because x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\) \(=\frac{-2\left(y^2-1\right) \pm \sqrt{\left\{2\left(y^2-1\right)\right\}^2-4 \times\left(y^2-1\right) \times 4\left(y^2-2\right)}}{2\left(y^2-1\right)}\) \(=\frac{-2\left(y^2-1\right) \pm \sqrt{\left(y^2-1\right)\left[4\left(y^2-1\right)-16\left(y^2-2\right)\right)}}{2\left(y^2-1\right)}\) \(=\frac{-2\left(y^2-1\right) \pm \sqrt{\left(y^2-1\right)\left(7-3 y^2\right) \times 4}}{2\left(y^2-1\right)}\) \(\therefore\left(y^2-1\right)\left(7-3 y^2\right) \times 4 \geq 0\) \(\left(y^2-1\right) \geq 0 \quad \text { or } 7-3 y^2 \geq 0\) \(y \geq 1 \text { or } \mathrm{y} \geq \sqrt{\frac{7}{3}}\) \(\mathrm{y} \geq 1 \text { or } \mathrm{y} \geq \sqrt{\frac{7}{3}}\)So, range of \(\mathrm{f}(\mathrm{x})\) is \(\left(1, \sqrt{\frac{7}{3}}\right]\).
Shift-I
Sets, Relation and Function
117346
If \(f: R \rightarrow R\) is defined as \(f(x)=\frac{x^6}{x^6+2020} \cdot \forall x\) \(\in R\), then the range of \(\boldsymbol{f}\) is .....
1 \((0,1)\)
2 \([0, \infty]\)
3 \([0,1]\)
4 \(\left[0, \frac{1}{2020}\right]\)
Explanation:
C Given, \(x \in\) Real value \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^6}{\mathrm{x}^6+2020}\) \(\therefore \mathrm{x}^6+2020>\mathrm{x}^6\) \(\Rightarrow \frac{\mathrm{x}^6}{\mathrm{x}^6+2020}\lt 1\) And, \(x^6\lt 0\) Hence, range of \(f(x)\) is \([0,1]\)
Shift-II
Sets, Relation and Function
117347
\(\left\{x \in R: \frac{14 x}{x+1}-\frac{9 x-30}{x-4}\lt 0\right\}\) is equal to
