117274
The domain of definition of the function \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)}\) is
1 \((1,4)\)
2 \([1,4]\)
3 \((0,5)\)
4 \([0,5]\)
Explanation:
B Given, \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)}\) Then, \(f(x)\) exists only when \(\log _{10}\left(\frac{5 x-x^2}{4}\right) \geq 0\) \(\frac{5 x-x^2}{4} \geq 1\) \(x^2-5 x+4 \leq 0\) \((x-1)(x-4) \leq 0\)So, \(x \in[1,4]\)
SRMJEEE-2010
Sets, Relation and Function
117280
The domain of the function \(f(x)=\sqrt{x}\) is
1 \(\mathrm{R}-\{0\}\)
2 \(\mathrm{R}^{+} \cup\{0\}\)
3 \(\mathrm{R}\)
4 \(\mathrm{R}^{+}\)
Explanation:
B Given, For \(f(x)\) to be defined, the term under the square root should be greater than or equal to zero, i.e. \(x \geq 0\) So, domain is \([0, \infty)\) i.e. \(\mathrm{R}^{+} \cup\{0\}\)
MHT-CET 20
Sets, Relation and Function
117281
The domain of a function \(f(y)=\frac{\cos ^{-1}(y-5)}{\sqrt{25-y^2}}\) is
1 \((4,5]\)
2 \((4,6]\)
3 \((-5,5)\)
4 \([4,5)\)
Explanation:
D Given, We have, \(f(y)=\frac{\cos ^{-1}(y-5)}{\sqrt{25-y^2}}\) Here, \(\quad-1 \leq y-5 \leq 1\) and \(25-y^2>0\) \(\therefore \quad 4 \leq \mathrm{y} \leq 6 \quad\) and \(\quad-5\lt \mathrm{y}\lt 5\) Hence, domain of \(\mathrm{f}(\mathrm{y})\) is[4,5)
MHT-CET 20
Sets, Relation and Function
117282
Given \(A=\{1,2,3,4,5\}, B=\{1,4,5\}\). If \(R\) is a relation from \(A\) to \(B\) such that \((x, y) \in R\) with \(x>y\), then range of \(R\) is
1 \(\{1,4,5\}\)
2 \(\{4,5\}\)
3 \(\{2,4\}\)
4 \(\{1,4\}\)
Explanation:
:Given, From given data, we write \(\mathrm{R}=\{(2,1),(3,1),(4,1),(5,1),(5,4)\}\)So, range of \(\mathrm{R}\) is \(\{1,4\}\)
117274
The domain of definition of the function \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)}\) is
1 \((1,4)\)
2 \([1,4]\)
3 \((0,5)\)
4 \([0,5]\)
Explanation:
B Given, \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)}\) Then, \(f(x)\) exists only when \(\log _{10}\left(\frac{5 x-x^2}{4}\right) \geq 0\) \(\frac{5 x-x^2}{4} \geq 1\) \(x^2-5 x+4 \leq 0\) \((x-1)(x-4) \leq 0\)So, \(x \in[1,4]\)
SRMJEEE-2010
Sets, Relation and Function
117280
The domain of the function \(f(x)=\sqrt{x}\) is
1 \(\mathrm{R}-\{0\}\)
2 \(\mathrm{R}^{+} \cup\{0\}\)
3 \(\mathrm{R}\)
4 \(\mathrm{R}^{+}\)
Explanation:
B Given, For \(f(x)\) to be defined, the term under the square root should be greater than or equal to zero, i.e. \(x \geq 0\) So, domain is \([0, \infty)\) i.e. \(\mathrm{R}^{+} \cup\{0\}\)
MHT-CET 20
Sets, Relation and Function
117281
The domain of a function \(f(y)=\frac{\cos ^{-1}(y-5)}{\sqrt{25-y^2}}\) is
1 \((4,5]\)
2 \((4,6]\)
3 \((-5,5)\)
4 \([4,5)\)
Explanation:
D Given, We have, \(f(y)=\frac{\cos ^{-1}(y-5)}{\sqrt{25-y^2}}\) Here, \(\quad-1 \leq y-5 \leq 1\) and \(25-y^2>0\) \(\therefore \quad 4 \leq \mathrm{y} \leq 6 \quad\) and \(\quad-5\lt \mathrm{y}\lt 5\) Hence, domain of \(\mathrm{f}(\mathrm{y})\) is[4,5)
MHT-CET 20
Sets, Relation and Function
117282
Given \(A=\{1,2,3,4,5\}, B=\{1,4,5\}\). If \(R\) is a relation from \(A\) to \(B\) such that \((x, y) \in R\) with \(x>y\), then range of \(R\) is
1 \(\{1,4,5\}\)
2 \(\{4,5\}\)
3 \(\{2,4\}\)
4 \(\{1,4\}\)
Explanation:
:Given, From given data, we write \(\mathrm{R}=\{(2,1),(3,1),(4,1),(5,1),(5,4)\}\)So, range of \(\mathrm{R}\) is \(\{1,4\}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117274
The domain of definition of the function \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)}\) is
1 \((1,4)\)
2 \([1,4]\)
3 \((0,5)\)
4 \([0,5]\)
Explanation:
B Given, \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)}\) Then, \(f(x)\) exists only when \(\log _{10}\left(\frac{5 x-x^2}{4}\right) \geq 0\) \(\frac{5 x-x^2}{4} \geq 1\) \(x^2-5 x+4 \leq 0\) \((x-1)(x-4) \leq 0\)So, \(x \in[1,4]\)
SRMJEEE-2010
Sets, Relation and Function
117280
The domain of the function \(f(x)=\sqrt{x}\) is
1 \(\mathrm{R}-\{0\}\)
2 \(\mathrm{R}^{+} \cup\{0\}\)
3 \(\mathrm{R}\)
4 \(\mathrm{R}^{+}\)
Explanation:
B Given, For \(f(x)\) to be defined, the term under the square root should be greater than or equal to zero, i.e. \(x \geq 0\) So, domain is \([0, \infty)\) i.e. \(\mathrm{R}^{+} \cup\{0\}\)
MHT-CET 20
Sets, Relation and Function
117281
The domain of a function \(f(y)=\frac{\cos ^{-1}(y-5)}{\sqrt{25-y^2}}\) is
1 \((4,5]\)
2 \((4,6]\)
3 \((-5,5)\)
4 \([4,5)\)
Explanation:
D Given, We have, \(f(y)=\frac{\cos ^{-1}(y-5)}{\sqrt{25-y^2}}\) Here, \(\quad-1 \leq y-5 \leq 1\) and \(25-y^2>0\) \(\therefore \quad 4 \leq \mathrm{y} \leq 6 \quad\) and \(\quad-5\lt \mathrm{y}\lt 5\) Hence, domain of \(\mathrm{f}(\mathrm{y})\) is[4,5)
MHT-CET 20
Sets, Relation and Function
117282
Given \(A=\{1,2,3,4,5\}, B=\{1,4,5\}\). If \(R\) is a relation from \(A\) to \(B\) such that \((x, y) \in R\) with \(x>y\), then range of \(R\) is
1 \(\{1,4,5\}\)
2 \(\{4,5\}\)
3 \(\{2,4\}\)
4 \(\{1,4\}\)
Explanation:
:Given, From given data, we write \(\mathrm{R}=\{(2,1),(3,1),(4,1),(5,1),(5,4)\}\)So, range of \(\mathrm{R}\) is \(\{1,4\}\)
117274
The domain of definition of the function \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)}\) is
1 \((1,4)\)
2 \([1,4]\)
3 \((0,5)\)
4 \([0,5]\)
Explanation:
B Given, \(f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)}\) Then, \(f(x)\) exists only when \(\log _{10}\left(\frac{5 x-x^2}{4}\right) \geq 0\) \(\frac{5 x-x^2}{4} \geq 1\) \(x^2-5 x+4 \leq 0\) \((x-1)(x-4) \leq 0\)So, \(x \in[1,4]\)
SRMJEEE-2010
Sets, Relation and Function
117280
The domain of the function \(f(x)=\sqrt{x}\) is
1 \(\mathrm{R}-\{0\}\)
2 \(\mathrm{R}^{+} \cup\{0\}\)
3 \(\mathrm{R}\)
4 \(\mathrm{R}^{+}\)
Explanation:
B Given, For \(f(x)\) to be defined, the term under the square root should be greater than or equal to zero, i.e. \(x \geq 0\) So, domain is \([0, \infty)\) i.e. \(\mathrm{R}^{+} \cup\{0\}\)
MHT-CET 20
Sets, Relation and Function
117281
The domain of a function \(f(y)=\frac{\cos ^{-1}(y-5)}{\sqrt{25-y^2}}\) is
1 \((4,5]\)
2 \((4,6]\)
3 \((-5,5)\)
4 \([4,5)\)
Explanation:
D Given, We have, \(f(y)=\frac{\cos ^{-1}(y-5)}{\sqrt{25-y^2}}\) Here, \(\quad-1 \leq y-5 \leq 1\) and \(25-y^2>0\) \(\therefore \quad 4 \leq \mathrm{y} \leq 6 \quad\) and \(\quad-5\lt \mathrm{y}\lt 5\) Hence, domain of \(\mathrm{f}(\mathrm{y})\) is[4,5)
MHT-CET 20
Sets, Relation and Function
117282
Given \(A=\{1,2,3,4,5\}, B=\{1,4,5\}\). If \(R\) is a relation from \(A\) to \(B\) such that \((x, y) \in R\) with \(x>y\), then range of \(R\) is
1 \(\{1,4,5\}\)
2 \(\{4,5\}\)
3 \(\{2,4\}\)
4 \(\{1,4\}\)
Explanation:
:Given, From given data, we write \(\mathrm{R}=\{(2,1),(3,1),(4,1),(5,1),(5,4)\}\)So, range of \(\mathrm{R}\) is \(\{1,4\}\)