117305
The domain of the function \(f(x)=\frac{\sqrt{4-x^2}}{\cos ^{-1}(2-x)}\) is:
1 \([0,2]\)
2 \([0,2)\)
3 \([1,2)\)
4 \([1,2]\)
Explanation:
C Given, \(f(x)=\frac{\sqrt{4-x^2}}{\cos ^{-1}(2-x)}\) For domain \(\mathrm{f}(\mathrm{x})\) be defined \(4-\mathrm{x}^2 \geq 0,-1 \leq 2-\mathrm{x} \leq 1\) and \(2-\mathrm{x}\) \(\mathrm{x}^2 \leq 4,-1 \leq 2-\mathrm{x} \leq 1\) \(-2 \leq \mathrm{x} \leq 2\) and \(-3 \leq-\mathrm{x} \leq-1\) \(-2 \leq \mathrm{x} \leq 2\) and \(1 \leq \mathrm{x} \leq 3\) Then, So, common interval \(=\) domain of \(f(x)=[1,2)\)
UPSEE-2015
Sets, Relation and Function
117306
Let \(f=\left\{\left(x, \frac{x^2}{1+x^2}\right): x \in R\right\}\) be a function from \(R\) to \(R\). Determine the range of \(\boldsymbol{f}\).
1 \([0,1)\)
2 \([0,-1]\)
3 \([0,2)\)
4 None of these
Explanation:
A Given, \(f(x)=\left\{\left(x, \frac{x^2}{1+x^2}\right): x \in R\right\}\) be a function from \(R\) to \(R\). Consider, \(y=\frac{x^2}{1+x^2}\) \(y+y^{\frac{1}{2}+x^2}=x^2\) \(y=x^2-y x^2\) \(y=x^2(1-y) \Rightarrow x^2=\frac{y}{1-y}\) \(x=\sqrt{\frac{y}{1-y}}\) \(\because \mathrm{x}\) is real, then - \(\frac{y}{1-y} \geq 0\) \(\frac{y(1-y)}{(1-y)^2} \geq 0\) Then, \(y(1-y) \geq 0 \text { and }(1-y)^2>0\) \(0 \leq \mathrm{y} \leq 1\) and \(-\mathrm{y}>-1\) \(0 \leq \mathrm{y} \leq 1\) and \(\mathrm{y}\lt 1\) So, common interval \(=\) range of \(\mathrm{f}=0 \leq \mathrm{y} \leq 1=[0,1)\)
UPSEE-2014
Sets, Relation and Function
117307
The greatest value of \(f(x)=(x+1)^{1 / 3}-(x-1)^{1 / 3}\) on \([0,1]\) is
1 1
2 2
3 3
4 \(\frac{1}{3}\)
Explanation:
B Given, \(f(x)=(x+1)^{\frac{1}{3}}-(x-1)^{\frac{1}{3}}\) Both side difference- Then, \(\mathrm{f}^{\prime}(\mathrm{x})=\left[\frac{1}{3} \times \frac{1}{(\mathrm{x}+1)^{\frac{2}{3}}}-\frac{1}{3} \times \frac{1}{(\mathrm{x}+1)^{\frac{2}{3}}}\right]\) \(\mathrm{f}^{\prime}(\mathrm{x}) =\frac{1}{3}(\mathrm{x}+1)^{\frac{-2}{3}}-\frac{1}{3}(\mathrm{x}-1)^{\frac{-2}{3}}\) \(=\frac{(\mathrm{x}-1)^{\frac{2}{3}}-(\mathrm{x}+1)^{\frac{2}{3}}}{3(\mathrm{x}+1)^{\frac{2}{3}}}\) \(\mathrm{f}^{\prime}(\mathrm{x})\) does exist at \(\mathrm{x}= \pm 1\) \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \((\mathrm{x}-1)^{\frac{2}{3}}=(\mathrm{x}+1)^{\frac{2}{3}}\) \(\mathrm{x}=0\) \(\because \mathrm{f}^{\prime}(\mathrm{x}) \neq 0\) for any other values of \(\mathrm{x} \in[0,1]\). the value of \(f(x)\) at \(x=0\) is 2 . Hence, the greatest value of \(\mathrm{f}(\mathrm{x})\) is 2 .
UPSEE-2010
Sets, Relation and Function
117308
The set of points where the function \(f(x)=x|x|\) is differentiable is
117305
The domain of the function \(f(x)=\frac{\sqrt{4-x^2}}{\cos ^{-1}(2-x)}\) is:
1 \([0,2]\)
2 \([0,2)\)
3 \([1,2)\)
4 \([1,2]\)
Explanation:
C Given, \(f(x)=\frac{\sqrt{4-x^2}}{\cos ^{-1}(2-x)}\) For domain \(\mathrm{f}(\mathrm{x})\) be defined \(4-\mathrm{x}^2 \geq 0,-1 \leq 2-\mathrm{x} \leq 1\) and \(2-\mathrm{x}\) \(\mathrm{x}^2 \leq 4,-1 \leq 2-\mathrm{x} \leq 1\) \(-2 \leq \mathrm{x} \leq 2\) and \(-3 \leq-\mathrm{x} \leq-1\) \(-2 \leq \mathrm{x} \leq 2\) and \(1 \leq \mathrm{x} \leq 3\) Then, So, common interval \(=\) domain of \(f(x)=[1,2)\)
UPSEE-2015
Sets, Relation and Function
117306
Let \(f=\left\{\left(x, \frac{x^2}{1+x^2}\right): x \in R\right\}\) be a function from \(R\) to \(R\). Determine the range of \(\boldsymbol{f}\).
1 \([0,1)\)
2 \([0,-1]\)
3 \([0,2)\)
4 None of these
Explanation:
A Given, \(f(x)=\left\{\left(x, \frac{x^2}{1+x^2}\right): x \in R\right\}\) be a function from \(R\) to \(R\). Consider, \(y=\frac{x^2}{1+x^2}\) \(y+y^{\frac{1}{2}+x^2}=x^2\) \(y=x^2-y x^2\) \(y=x^2(1-y) \Rightarrow x^2=\frac{y}{1-y}\) \(x=\sqrt{\frac{y}{1-y}}\) \(\because \mathrm{x}\) is real, then - \(\frac{y}{1-y} \geq 0\) \(\frac{y(1-y)}{(1-y)^2} \geq 0\) Then, \(y(1-y) \geq 0 \text { and }(1-y)^2>0\) \(0 \leq \mathrm{y} \leq 1\) and \(-\mathrm{y}>-1\) \(0 \leq \mathrm{y} \leq 1\) and \(\mathrm{y}\lt 1\) So, common interval \(=\) range of \(\mathrm{f}=0 \leq \mathrm{y} \leq 1=[0,1)\)
UPSEE-2014
Sets, Relation and Function
117307
The greatest value of \(f(x)=(x+1)^{1 / 3}-(x-1)^{1 / 3}\) on \([0,1]\) is
1 1
2 2
3 3
4 \(\frac{1}{3}\)
Explanation:
B Given, \(f(x)=(x+1)^{\frac{1}{3}}-(x-1)^{\frac{1}{3}}\) Both side difference- Then, \(\mathrm{f}^{\prime}(\mathrm{x})=\left[\frac{1}{3} \times \frac{1}{(\mathrm{x}+1)^{\frac{2}{3}}}-\frac{1}{3} \times \frac{1}{(\mathrm{x}+1)^{\frac{2}{3}}}\right]\) \(\mathrm{f}^{\prime}(\mathrm{x}) =\frac{1}{3}(\mathrm{x}+1)^{\frac{-2}{3}}-\frac{1}{3}(\mathrm{x}-1)^{\frac{-2}{3}}\) \(=\frac{(\mathrm{x}-1)^{\frac{2}{3}}-(\mathrm{x}+1)^{\frac{2}{3}}}{3(\mathrm{x}+1)^{\frac{2}{3}}}\) \(\mathrm{f}^{\prime}(\mathrm{x})\) does exist at \(\mathrm{x}= \pm 1\) \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \((\mathrm{x}-1)^{\frac{2}{3}}=(\mathrm{x}+1)^{\frac{2}{3}}\) \(\mathrm{x}=0\) \(\because \mathrm{f}^{\prime}(\mathrm{x}) \neq 0\) for any other values of \(\mathrm{x} \in[0,1]\). the value of \(f(x)\) at \(x=0\) is 2 . Hence, the greatest value of \(\mathrm{f}(\mathrm{x})\) is 2 .
UPSEE-2010
Sets, Relation and Function
117308
The set of points where the function \(f(x)=x|x|\) is differentiable is
117305
The domain of the function \(f(x)=\frac{\sqrt{4-x^2}}{\cos ^{-1}(2-x)}\) is:
1 \([0,2]\)
2 \([0,2)\)
3 \([1,2)\)
4 \([1,2]\)
Explanation:
C Given, \(f(x)=\frac{\sqrt{4-x^2}}{\cos ^{-1}(2-x)}\) For domain \(\mathrm{f}(\mathrm{x})\) be defined \(4-\mathrm{x}^2 \geq 0,-1 \leq 2-\mathrm{x} \leq 1\) and \(2-\mathrm{x}\) \(\mathrm{x}^2 \leq 4,-1 \leq 2-\mathrm{x} \leq 1\) \(-2 \leq \mathrm{x} \leq 2\) and \(-3 \leq-\mathrm{x} \leq-1\) \(-2 \leq \mathrm{x} \leq 2\) and \(1 \leq \mathrm{x} \leq 3\) Then, So, common interval \(=\) domain of \(f(x)=[1,2)\)
UPSEE-2015
Sets, Relation and Function
117306
Let \(f=\left\{\left(x, \frac{x^2}{1+x^2}\right): x \in R\right\}\) be a function from \(R\) to \(R\). Determine the range of \(\boldsymbol{f}\).
1 \([0,1)\)
2 \([0,-1]\)
3 \([0,2)\)
4 None of these
Explanation:
A Given, \(f(x)=\left\{\left(x, \frac{x^2}{1+x^2}\right): x \in R\right\}\) be a function from \(R\) to \(R\). Consider, \(y=\frac{x^2}{1+x^2}\) \(y+y^{\frac{1}{2}+x^2}=x^2\) \(y=x^2-y x^2\) \(y=x^2(1-y) \Rightarrow x^2=\frac{y}{1-y}\) \(x=\sqrt{\frac{y}{1-y}}\) \(\because \mathrm{x}\) is real, then - \(\frac{y}{1-y} \geq 0\) \(\frac{y(1-y)}{(1-y)^2} \geq 0\) Then, \(y(1-y) \geq 0 \text { and }(1-y)^2>0\) \(0 \leq \mathrm{y} \leq 1\) and \(-\mathrm{y}>-1\) \(0 \leq \mathrm{y} \leq 1\) and \(\mathrm{y}\lt 1\) So, common interval \(=\) range of \(\mathrm{f}=0 \leq \mathrm{y} \leq 1=[0,1)\)
UPSEE-2014
Sets, Relation and Function
117307
The greatest value of \(f(x)=(x+1)^{1 / 3}-(x-1)^{1 / 3}\) on \([0,1]\) is
1 1
2 2
3 3
4 \(\frac{1}{3}\)
Explanation:
B Given, \(f(x)=(x+1)^{\frac{1}{3}}-(x-1)^{\frac{1}{3}}\) Both side difference- Then, \(\mathrm{f}^{\prime}(\mathrm{x})=\left[\frac{1}{3} \times \frac{1}{(\mathrm{x}+1)^{\frac{2}{3}}}-\frac{1}{3} \times \frac{1}{(\mathrm{x}+1)^{\frac{2}{3}}}\right]\) \(\mathrm{f}^{\prime}(\mathrm{x}) =\frac{1}{3}(\mathrm{x}+1)^{\frac{-2}{3}}-\frac{1}{3}(\mathrm{x}-1)^{\frac{-2}{3}}\) \(=\frac{(\mathrm{x}-1)^{\frac{2}{3}}-(\mathrm{x}+1)^{\frac{2}{3}}}{3(\mathrm{x}+1)^{\frac{2}{3}}}\) \(\mathrm{f}^{\prime}(\mathrm{x})\) does exist at \(\mathrm{x}= \pm 1\) \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \((\mathrm{x}-1)^{\frac{2}{3}}=(\mathrm{x}+1)^{\frac{2}{3}}\) \(\mathrm{x}=0\) \(\because \mathrm{f}^{\prime}(\mathrm{x}) \neq 0\) for any other values of \(\mathrm{x} \in[0,1]\). the value of \(f(x)\) at \(x=0\) is 2 . Hence, the greatest value of \(\mathrm{f}(\mathrm{x})\) is 2 .
UPSEE-2010
Sets, Relation and Function
117308
The set of points where the function \(f(x)=x|x|\) is differentiable is
117305
The domain of the function \(f(x)=\frac{\sqrt{4-x^2}}{\cos ^{-1}(2-x)}\) is:
1 \([0,2]\)
2 \([0,2)\)
3 \([1,2)\)
4 \([1,2]\)
Explanation:
C Given, \(f(x)=\frac{\sqrt{4-x^2}}{\cos ^{-1}(2-x)}\) For domain \(\mathrm{f}(\mathrm{x})\) be defined \(4-\mathrm{x}^2 \geq 0,-1 \leq 2-\mathrm{x} \leq 1\) and \(2-\mathrm{x}\) \(\mathrm{x}^2 \leq 4,-1 \leq 2-\mathrm{x} \leq 1\) \(-2 \leq \mathrm{x} \leq 2\) and \(-3 \leq-\mathrm{x} \leq-1\) \(-2 \leq \mathrm{x} \leq 2\) and \(1 \leq \mathrm{x} \leq 3\) Then, So, common interval \(=\) domain of \(f(x)=[1,2)\)
UPSEE-2015
Sets, Relation and Function
117306
Let \(f=\left\{\left(x, \frac{x^2}{1+x^2}\right): x \in R\right\}\) be a function from \(R\) to \(R\). Determine the range of \(\boldsymbol{f}\).
1 \([0,1)\)
2 \([0,-1]\)
3 \([0,2)\)
4 None of these
Explanation:
A Given, \(f(x)=\left\{\left(x, \frac{x^2}{1+x^2}\right): x \in R\right\}\) be a function from \(R\) to \(R\). Consider, \(y=\frac{x^2}{1+x^2}\) \(y+y^{\frac{1}{2}+x^2}=x^2\) \(y=x^2-y x^2\) \(y=x^2(1-y) \Rightarrow x^2=\frac{y}{1-y}\) \(x=\sqrt{\frac{y}{1-y}}\) \(\because \mathrm{x}\) is real, then - \(\frac{y}{1-y} \geq 0\) \(\frac{y(1-y)}{(1-y)^2} \geq 0\) Then, \(y(1-y) \geq 0 \text { and }(1-y)^2>0\) \(0 \leq \mathrm{y} \leq 1\) and \(-\mathrm{y}>-1\) \(0 \leq \mathrm{y} \leq 1\) and \(\mathrm{y}\lt 1\) So, common interval \(=\) range of \(\mathrm{f}=0 \leq \mathrm{y} \leq 1=[0,1)\)
UPSEE-2014
Sets, Relation and Function
117307
The greatest value of \(f(x)=(x+1)^{1 / 3}-(x-1)^{1 / 3}\) on \([0,1]\) is
1 1
2 2
3 3
4 \(\frac{1}{3}\)
Explanation:
B Given, \(f(x)=(x+1)^{\frac{1}{3}}-(x-1)^{\frac{1}{3}}\) Both side difference- Then, \(\mathrm{f}^{\prime}(\mathrm{x})=\left[\frac{1}{3} \times \frac{1}{(\mathrm{x}+1)^{\frac{2}{3}}}-\frac{1}{3} \times \frac{1}{(\mathrm{x}+1)^{\frac{2}{3}}}\right]\) \(\mathrm{f}^{\prime}(\mathrm{x}) =\frac{1}{3}(\mathrm{x}+1)^{\frac{-2}{3}}-\frac{1}{3}(\mathrm{x}-1)^{\frac{-2}{3}}\) \(=\frac{(\mathrm{x}-1)^{\frac{2}{3}}-(\mathrm{x}+1)^{\frac{2}{3}}}{3(\mathrm{x}+1)^{\frac{2}{3}}}\) \(\mathrm{f}^{\prime}(\mathrm{x})\) does exist at \(\mathrm{x}= \pm 1\) \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \((\mathrm{x}-1)^{\frac{2}{3}}=(\mathrm{x}+1)^{\frac{2}{3}}\) \(\mathrm{x}=0\) \(\because \mathrm{f}^{\prime}(\mathrm{x}) \neq 0\) for any other values of \(\mathrm{x} \in[0,1]\). the value of \(f(x)\) at \(x=0\) is 2 . Hence, the greatest value of \(\mathrm{f}(\mathrm{x})\) is 2 .
UPSEE-2010
Sets, Relation and Function
117308
The set of points where the function \(f(x)=x|x|\) is differentiable is
