117074
Let \(a, b, c\) be real numbers, each greater than 1, such that \(\frac{2}{3} \log _b a+\frac{3}{5} \log _c b+\frac{5}{2} \log _a c=3\). If the value of \(b\) is 9 , then the value of ' \(a\) ' must be
1 \(\sqrt[3]{81}\)
2 \(\frac{27}{2}\)
3 18
4 27
Explanation:
D Given, \(\mathrm{b}=9\) \(\frac{2}{3} \log _b a+\frac{3}{5} \log _c b+\frac{5}{2} \log _a c=3\) \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are real \(\frac{2}{3} \log _9 \mathrm{a}+\frac{3}{5} \log _{\mathrm{c}} 9+\frac{5}{2} \log _{\mathrm{a}} \mathrm{c}=3\) \(\Rightarrow \log _9 \mathrm{a}^{2 / 3}+\log _{\mathrm{c}} 9^{3 / 5}+\log _{\mathrm{a}} \mathrm{c}^{5 / 3}=3\) As there are two unknown variable a \& c, considering every term is equal to 1 . \(\log _9 \mathrm{a}^{2 / 3}=1, \log _{\mathrm{c}} 9^{3 / 5}=1, \log _{\mathrm{a}} \mathrm{c}^{5 / 3}=1\) \(\therefore \frac{\log \mathrm{a}^{2 / 3}}{\log 9}=1 \quad\left(\because \log _{\mathrm{x}} \mathrm{y}=\frac{\log \mathrm{y}}{\log \mathrm{x}}\right)\) \(\Rightarrow \log \mathrm{a}^{2 / 3}=\log 9\) \(\Rightarrow \mathrm{a}^{2 / 3}=9\) \(\Rightarrow \mathrm{a}=9^{3 / 2}\) \(\Rightarrow \mathrm{a}=(3)^{2 \times 3 / 2}\) \(=\mathrm{a}=(3)^3=27 .\)Hence option (D) is correct.
WB JEE-2021
Sets, Relation and Function
117075
\(\mathbf{f}: \mathrm{X} \rightarrow \mathrm{R}, \mathrm{X}=\{\mathrm{x} \mid \mathbf{0}\lt \mathrm{x}\lt 1\}\) is defined as \(\mathrm{f}(\mathrm{x})=\) \(\frac{2 \mathrm{x}}{1-|\mathbf{2 x}-\mathbf{1}|}\). Then
1 \(f\) is only injective
2 \(f\) is only surjective
3 \(\mathrm{f}\) is bijective
4 \(\mathrm{f}\) is neither injective nor surjective
Explanation:
A Given, \(f(x)=\frac{2 x-1}{1-|2 x-1|}\) \(f(x)= \begin{cases}\frac{2 x-1}{1+2 x-1}, 0\lt x\lt \frac{1}{2} \\ \frac{2 x}{1-2 x+1}, \frac{1}{2} \leq x\lt 1\end{cases}\) \(f(x)= \begin{cases}\frac{2 x-1}{2 x}, 0\lt x\lt \frac{1}{2} \\ \frac{2 x-1}{2-2 x}, \frac{1}{2} \leq x\lt 1\end{cases}\) Now, let us draw the graph of above function. From the above graph it is clear that \(\mathrm{f}(\mathrm{x})\) is one-one and since the range of \(f(x)\) is real number, which implies that codomain is equal to range. \(\therefore \mathrm{f}(\mathrm{x})\) is one-one and onto both and hence \(\mathrm{f}(\mathrm{x})\) is bijective.
WB JEE-2022
Sets, Relation and Function
117077
For the mapping f: \(\mathbf{R}-\{1\} \rightarrow \mathbf{R}-\{2\}\), given by \(f(x)=\frac{2 x}{x-1}\). Which of the following is correct ?
1 \(f\) is one- one but not onto
2 \(f\) is onto but not one-one
3 \(f\) is neither one-one nor onto
4 \(f\) is both one-one and onto
Explanation:
D Given, \(f(x)=\frac{2 x}{x-1}\) \(y=\frac{2 x}{x-1}=\frac{2 x-2+2}{x-1}=\frac{2(x-1)+2}{(x-1)}\) \(=\frac{2(x-1)}{(x-1)}+\frac{2}{(x-1)}\) \(\therefore \quad(y-2)(x-1)=2\) \(\text { onto } \rightarrow \text { co-domain }=\text { range }\) Thus, \(\mathrm{f}\) is both one-one and onto.
WB JEE-2022
Sets, Relation and Function
117078
The function \(f: X \rightarrow Y\) defined by \(f(x)=\sin x\) is one-one but not onto, if \(X\) and \(Y\) are respectively equal to
1 \(\mathrm{R}\) and \(\mathrm{R}\)
2 \([0, \pi]\) and \([0,1]\)
3 \(\left[0, \frac{\pi}{2}\right]\) and \([-1,1]\)
4 \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) and \([-1,1]\)
Explanation:
C One-one or injective function- Basically denotes the mapping of two sets. Since, \(\mathrm{f}: \mathrm{X} \rightarrow \mathrm{Y}\) Then \(\mathrm{f}(\mathrm{x})=\sin \mathrm{x}\) Now check option \(\mathrm{C}\) Domain \(=\left[0, \frac{\pi}{2}\right]\) Range \(=[-1,1]\) For every value of \(x\), we get unique value of \(y\). But the value of \(y\) in \([-1,0)\) does not have any pre-image \(\therefore\) Function is one-one but not onto. Hence, option (c) is correct.
117074
Let \(a, b, c\) be real numbers, each greater than 1, such that \(\frac{2}{3} \log _b a+\frac{3}{5} \log _c b+\frac{5}{2} \log _a c=3\). If the value of \(b\) is 9 , then the value of ' \(a\) ' must be
1 \(\sqrt[3]{81}\)
2 \(\frac{27}{2}\)
3 18
4 27
Explanation:
D Given, \(\mathrm{b}=9\) \(\frac{2}{3} \log _b a+\frac{3}{5} \log _c b+\frac{5}{2} \log _a c=3\) \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are real \(\frac{2}{3} \log _9 \mathrm{a}+\frac{3}{5} \log _{\mathrm{c}} 9+\frac{5}{2} \log _{\mathrm{a}} \mathrm{c}=3\) \(\Rightarrow \log _9 \mathrm{a}^{2 / 3}+\log _{\mathrm{c}} 9^{3 / 5}+\log _{\mathrm{a}} \mathrm{c}^{5 / 3}=3\) As there are two unknown variable a \& c, considering every term is equal to 1 . \(\log _9 \mathrm{a}^{2 / 3}=1, \log _{\mathrm{c}} 9^{3 / 5}=1, \log _{\mathrm{a}} \mathrm{c}^{5 / 3}=1\) \(\therefore \frac{\log \mathrm{a}^{2 / 3}}{\log 9}=1 \quad\left(\because \log _{\mathrm{x}} \mathrm{y}=\frac{\log \mathrm{y}}{\log \mathrm{x}}\right)\) \(\Rightarrow \log \mathrm{a}^{2 / 3}=\log 9\) \(\Rightarrow \mathrm{a}^{2 / 3}=9\) \(\Rightarrow \mathrm{a}=9^{3 / 2}\) \(\Rightarrow \mathrm{a}=(3)^{2 \times 3 / 2}\) \(=\mathrm{a}=(3)^3=27 .\)Hence option (D) is correct.
WB JEE-2021
Sets, Relation and Function
117075
\(\mathbf{f}: \mathrm{X} \rightarrow \mathrm{R}, \mathrm{X}=\{\mathrm{x} \mid \mathbf{0}\lt \mathrm{x}\lt 1\}\) is defined as \(\mathrm{f}(\mathrm{x})=\) \(\frac{2 \mathrm{x}}{1-|\mathbf{2 x}-\mathbf{1}|}\). Then
1 \(f\) is only injective
2 \(f\) is only surjective
3 \(\mathrm{f}\) is bijective
4 \(\mathrm{f}\) is neither injective nor surjective
Explanation:
A Given, \(f(x)=\frac{2 x-1}{1-|2 x-1|}\) \(f(x)= \begin{cases}\frac{2 x-1}{1+2 x-1}, 0\lt x\lt \frac{1}{2} \\ \frac{2 x}{1-2 x+1}, \frac{1}{2} \leq x\lt 1\end{cases}\) \(f(x)= \begin{cases}\frac{2 x-1}{2 x}, 0\lt x\lt \frac{1}{2} \\ \frac{2 x-1}{2-2 x}, \frac{1}{2} \leq x\lt 1\end{cases}\) Now, let us draw the graph of above function. From the above graph it is clear that \(\mathrm{f}(\mathrm{x})\) is one-one and since the range of \(f(x)\) is real number, which implies that codomain is equal to range. \(\therefore \mathrm{f}(\mathrm{x})\) is one-one and onto both and hence \(\mathrm{f}(\mathrm{x})\) is bijective.
WB JEE-2022
Sets, Relation and Function
117077
For the mapping f: \(\mathbf{R}-\{1\} \rightarrow \mathbf{R}-\{2\}\), given by \(f(x)=\frac{2 x}{x-1}\). Which of the following is correct ?
1 \(f\) is one- one but not onto
2 \(f\) is onto but not one-one
3 \(f\) is neither one-one nor onto
4 \(f\) is both one-one and onto
Explanation:
D Given, \(f(x)=\frac{2 x}{x-1}\) \(y=\frac{2 x}{x-1}=\frac{2 x-2+2}{x-1}=\frac{2(x-1)+2}{(x-1)}\) \(=\frac{2(x-1)}{(x-1)}+\frac{2}{(x-1)}\) \(\therefore \quad(y-2)(x-1)=2\) \(\text { onto } \rightarrow \text { co-domain }=\text { range }\) Thus, \(\mathrm{f}\) is both one-one and onto.
WB JEE-2022
Sets, Relation and Function
117078
The function \(f: X \rightarrow Y\) defined by \(f(x)=\sin x\) is one-one but not onto, if \(X\) and \(Y\) are respectively equal to
1 \(\mathrm{R}\) and \(\mathrm{R}\)
2 \([0, \pi]\) and \([0,1]\)
3 \(\left[0, \frac{\pi}{2}\right]\) and \([-1,1]\)
4 \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) and \([-1,1]\)
Explanation:
C One-one or injective function- Basically denotes the mapping of two sets. Since, \(\mathrm{f}: \mathrm{X} \rightarrow \mathrm{Y}\) Then \(\mathrm{f}(\mathrm{x})=\sin \mathrm{x}\) Now check option \(\mathrm{C}\) Domain \(=\left[0, \frac{\pi}{2}\right]\) Range \(=[-1,1]\) For every value of \(x\), we get unique value of \(y\). But the value of \(y\) in \([-1,0)\) does not have any pre-image \(\therefore\) Function is one-one but not onto. Hence, option (c) is correct.
117074
Let \(a, b, c\) be real numbers, each greater than 1, such that \(\frac{2}{3} \log _b a+\frac{3}{5} \log _c b+\frac{5}{2} \log _a c=3\). If the value of \(b\) is 9 , then the value of ' \(a\) ' must be
1 \(\sqrt[3]{81}\)
2 \(\frac{27}{2}\)
3 18
4 27
Explanation:
D Given, \(\mathrm{b}=9\) \(\frac{2}{3} \log _b a+\frac{3}{5} \log _c b+\frac{5}{2} \log _a c=3\) \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are real \(\frac{2}{3} \log _9 \mathrm{a}+\frac{3}{5} \log _{\mathrm{c}} 9+\frac{5}{2} \log _{\mathrm{a}} \mathrm{c}=3\) \(\Rightarrow \log _9 \mathrm{a}^{2 / 3}+\log _{\mathrm{c}} 9^{3 / 5}+\log _{\mathrm{a}} \mathrm{c}^{5 / 3}=3\) As there are two unknown variable a \& c, considering every term is equal to 1 . \(\log _9 \mathrm{a}^{2 / 3}=1, \log _{\mathrm{c}} 9^{3 / 5}=1, \log _{\mathrm{a}} \mathrm{c}^{5 / 3}=1\) \(\therefore \frac{\log \mathrm{a}^{2 / 3}}{\log 9}=1 \quad\left(\because \log _{\mathrm{x}} \mathrm{y}=\frac{\log \mathrm{y}}{\log \mathrm{x}}\right)\) \(\Rightarrow \log \mathrm{a}^{2 / 3}=\log 9\) \(\Rightarrow \mathrm{a}^{2 / 3}=9\) \(\Rightarrow \mathrm{a}=9^{3 / 2}\) \(\Rightarrow \mathrm{a}=(3)^{2 \times 3 / 2}\) \(=\mathrm{a}=(3)^3=27 .\)Hence option (D) is correct.
WB JEE-2021
Sets, Relation and Function
117075
\(\mathbf{f}: \mathrm{X} \rightarrow \mathrm{R}, \mathrm{X}=\{\mathrm{x} \mid \mathbf{0}\lt \mathrm{x}\lt 1\}\) is defined as \(\mathrm{f}(\mathrm{x})=\) \(\frac{2 \mathrm{x}}{1-|\mathbf{2 x}-\mathbf{1}|}\). Then
1 \(f\) is only injective
2 \(f\) is only surjective
3 \(\mathrm{f}\) is bijective
4 \(\mathrm{f}\) is neither injective nor surjective
Explanation:
A Given, \(f(x)=\frac{2 x-1}{1-|2 x-1|}\) \(f(x)= \begin{cases}\frac{2 x-1}{1+2 x-1}, 0\lt x\lt \frac{1}{2} \\ \frac{2 x}{1-2 x+1}, \frac{1}{2} \leq x\lt 1\end{cases}\) \(f(x)= \begin{cases}\frac{2 x-1}{2 x}, 0\lt x\lt \frac{1}{2} \\ \frac{2 x-1}{2-2 x}, \frac{1}{2} \leq x\lt 1\end{cases}\) Now, let us draw the graph of above function. From the above graph it is clear that \(\mathrm{f}(\mathrm{x})\) is one-one and since the range of \(f(x)\) is real number, which implies that codomain is equal to range. \(\therefore \mathrm{f}(\mathrm{x})\) is one-one and onto both and hence \(\mathrm{f}(\mathrm{x})\) is bijective.
WB JEE-2022
Sets, Relation and Function
117077
For the mapping f: \(\mathbf{R}-\{1\} \rightarrow \mathbf{R}-\{2\}\), given by \(f(x)=\frac{2 x}{x-1}\). Which of the following is correct ?
1 \(f\) is one- one but not onto
2 \(f\) is onto but not one-one
3 \(f\) is neither one-one nor onto
4 \(f\) is both one-one and onto
Explanation:
D Given, \(f(x)=\frac{2 x}{x-1}\) \(y=\frac{2 x}{x-1}=\frac{2 x-2+2}{x-1}=\frac{2(x-1)+2}{(x-1)}\) \(=\frac{2(x-1)}{(x-1)}+\frac{2}{(x-1)}\) \(\therefore \quad(y-2)(x-1)=2\) \(\text { onto } \rightarrow \text { co-domain }=\text { range }\) Thus, \(\mathrm{f}\) is both one-one and onto.
WB JEE-2022
Sets, Relation and Function
117078
The function \(f: X \rightarrow Y\) defined by \(f(x)=\sin x\) is one-one but not onto, if \(X\) and \(Y\) are respectively equal to
1 \(\mathrm{R}\) and \(\mathrm{R}\)
2 \([0, \pi]\) and \([0,1]\)
3 \(\left[0, \frac{\pi}{2}\right]\) and \([-1,1]\)
4 \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) and \([-1,1]\)
Explanation:
C One-one or injective function- Basically denotes the mapping of two sets. Since, \(\mathrm{f}: \mathrm{X} \rightarrow \mathrm{Y}\) Then \(\mathrm{f}(\mathrm{x})=\sin \mathrm{x}\) Now check option \(\mathrm{C}\) Domain \(=\left[0, \frac{\pi}{2}\right]\) Range \(=[-1,1]\) For every value of \(x\), we get unique value of \(y\). But the value of \(y\) in \([-1,0)\) does not have any pre-image \(\therefore\) Function is one-one but not onto. Hence, option (c) is correct.
117074
Let \(a, b, c\) be real numbers, each greater than 1, such that \(\frac{2}{3} \log _b a+\frac{3}{5} \log _c b+\frac{5}{2} \log _a c=3\). If the value of \(b\) is 9 , then the value of ' \(a\) ' must be
1 \(\sqrt[3]{81}\)
2 \(\frac{27}{2}\)
3 18
4 27
Explanation:
D Given, \(\mathrm{b}=9\) \(\frac{2}{3} \log _b a+\frac{3}{5} \log _c b+\frac{5}{2} \log _a c=3\) \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are real \(\frac{2}{3} \log _9 \mathrm{a}+\frac{3}{5} \log _{\mathrm{c}} 9+\frac{5}{2} \log _{\mathrm{a}} \mathrm{c}=3\) \(\Rightarrow \log _9 \mathrm{a}^{2 / 3}+\log _{\mathrm{c}} 9^{3 / 5}+\log _{\mathrm{a}} \mathrm{c}^{5 / 3}=3\) As there are two unknown variable a \& c, considering every term is equal to 1 . \(\log _9 \mathrm{a}^{2 / 3}=1, \log _{\mathrm{c}} 9^{3 / 5}=1, \log _{\mathrm{a}} \mathrm{c}^{5 / 3}=1\) \(\therefore \frac{\log \mathrm{a}^{2 / 3}}{\log 9}=1 \quad\left(\because \log _{\mathrm{x}} \mathrm{y}=\frac{\log \mathrm{y}}{\log \mathrm{x}}\right)\) \(\Rightarrow \log \mathrm{a}^{2 / 3}=\log 9\) \(\Rightarrow \mathrm{a}^{2 / 3}=9\) \(\Rightarrow \mathrm{a}=9^{3 / 2}\) \(\Rightarrow \mathrm{a}=(3)^{2 \times 3 / 2}\) \(=\mathrm{a}=(3)^3=27 .\)Hence option (D) is correct.
WB JEE-2021
Sets, Relation and Function
117075
\(\mathbf{f}: \mathrm{X} \rightarrow \mathrm{R}, \mathrm{X}=\{\mathrm{x} \mid \mathbf{0}\lt \mathrm{x}\lt 1\}\) is defined as \(\mathrm{f}(\mathrm{x})=\) \(\frac{2 \mathrm{x}}{1-|\mathbf{2 x}-\mathbf{1}|}\). Then
1 \(f\) is only injective
2 \(f\) is only surjective
3 \(\mathrm{f}\) is bijective
4 \(\mathrm{f}\) is neither injective nor surjective
Explanation:
A Given, \(f(x)=\frac{2 x-1}{1-|2 x-1|}\) \(f(x)= \begin{cases}\frac{2 x-1}{1+2 x-1}, 0\lt x\lt \frac{1}{2} \\ \frac{2 x}{1-2 x+1}, \frac{1}{2} \leq x\lt 1\end{cases}\) \(f(x)= \begin{cases}\frac{2 x-1}{2 x}, 0\lt x\lt \frac{1}{2} \\ \frac{2 x-1}{2-2 x}, \frac{1}{2} \leq x\lt 1\end{cases}\) Now, let us draw the graph of above function. From the above graph it is clear that \(\mathrm{f}(\mathrm{x})\) is one-one and since the range of \(f(x)\) is real number, which implies that codomain is equal to range. \(\therefore \mathrm{f}(\mathrm{x})\) is one-one and onto both and hence \(\mathrm{f}(\mathrm{x})\) is bijective.
WB JEE-2022
Sets, Relation and Function
117077
For the mapping f: \(\mathbf{R}-\{1\} \rightarrow \mathbf{R}-\{2\}\), given by \(f(x)=\frac{2 x}{x-1}\). Which of the following is correct ?
1 \(f\) is one- one but not onto
2 \(f\) is onto but not one-one
3 \(f\) is neither one-one nor onto
4 \(f\) is both one-one and onto
Explanation:
D Given, \(f(x)=\frac{2 x}{x-1}\) \(y=\frac{2 x}{x-1}=\frac{2 x-2+2}{x-1}=\frac{2(x-1)+2}{(x-1)}\) \(=\frac{2(x-1)}{(x-1)}+\frac{2}{(x-1)}\) \(\therefore \quad(y-2)(x-1)=2\) \(\text { onto } \rightarrow \text { co-domain }=\text { range }\) Thus, \(\mathrm{f}\) is both one-one and onto.
WB JEE-2022
Sets, Relation and Function
117078
The function \(f: X \rightarrow Y\) defined by \(f(x)=\sin x\) is one-one but not onto, if \(X\) and \(Y\) are respectively equal to
1 \(\mathrm{R}\) and \(\mathrm{R}\)
2 \([0, \pi]\) and \([0,1]\)
3 \(\left[0, \frac{\pi}{2}\right]\) and \([-1,1]\)
4 \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) and \([-1,1]\)
Explanation:
C One-one or injective function- Basically denotes the mapping of two sets. Since, \(\mathrm{f}: \mathrm{X} \rightarrow \mathrm{Y}\) Then \(\mathrm{f}(\mathrm{x})=\sin \mathrm{x}\) Now check option \(\mathrm{C}\) Domain \(=\left[0, \frac{\pi}{2}\right]\) Range \(=[-1,1]\) For every value of \(x\), we get unique value of \(y\). But the value of \(y\) in \([-1,0)\) does not have any pre-image \(\therefore\) Function is one-one but not onto. Hence, option (c) is correct.
