116946
The number of non-constant functions f from \(X\) \(=\{0,1,2\}\) to \(Y=\{1,2,3,4,5,6,7,8\}\) such that \(f(i) \leq f(j)\) for \(i, j \in X\) and \(i\lt j\) is
1 120
2 92
3 56
4 112
Explanation:
D Given, Sets \(X=\{0,1,2\}\) and \(Y=\{1,2,3,4,5,6,7,8\}\) and non - constant function \(\mathrm{f}: \mathrm{X} \rightarrow \mathrm{Y}\), such that \(f(i) \leq f(j) i, j \in X\) and \(i\lt j\). Now following two cases are possible Case - I Let range of 'f' are \(a, b, c \in y\), that \(a\lt b\lt c\) means function is strictly increasing. Then number of way of selection 3 distinct number \(a_1 b_2\) and \(\mathrm{c}\) is \({ }^8 \mathrm{C}_3=56\) Case - II If \(\mathrm{a}=\mathrm{b}\lt \mathrm{c}\) or \(\mathrm{a}\lt \mathrm{b}=\mathrm{c}\) Now number of ways of selecting two numbers \(\mathrm{a}=\mathrm{b}, \mathrm{c}\) or \(\mathrm{a}, \mathrm{b}=\mathrm{c}\) is \({ }^8 \mathrm{C}_2\) and since two elements are identical which we can make in \({ }^2 \mathrm{C}_1\) ways. So, number of ways to make such combination is \({ }^8 \mathrm{C}_2 \times{ }^2 \mathrm{C}_1\) \(=56\) So, required number of non - constant functions are \(=56+56=112\).
Shift-II
Sets, Relation and Function
116947
If the function \(f:[a, b] \rightarrow\) defined by \(f(x)=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+\sin x & 1 \\ 1+\cos x & 1 & 1\end{array}\right]\) is one-one and onto, then
A Given. \(f(x)=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin x & 1 \\ 1+\cos x & 1 & 1 \end{array}\right|\) On applying \(\mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_1\) and \(\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1\) we get - \(f(x)=\left|\begin{array}{ccc} 1 & 0 & 0 \\ 1 & \sin x & 0 \\ 1+\cos x & -\cos x & -\cos x \end{array}\right|\) \(f(x)=-\sin x \cdot \cos x\) \(f(x)=-\sin x \cos x\) \(f(x)=\frac{-\sin 2 x}{2}\) \(=\quad \frac{-\sqrt{3}}{4} \leq-\frac{-\sin 2 x}{2} \leq \frac{1}{2}\) \(=\quad-\frac{\sqrt{3}}{2} \leq-\sin 2 \mathrm{x} \leq 1\) \(=-1 \leq \sin 2 x \leq \frac{\sqrt{3}}{2}\) \(=2 \mathrm{x}=-\frac{\pi}{2}\) \(x=\frac{-\pi}{4}\) \(2 \mathrm{x}=\frac{\pi}{3}\) \(\mathrm{x}=\frac{\pi}{6}\) So, \(\quad-\frac{\pi}{4} \leq \mathrm{x} \leq \frac{\pi}{6}\)
Shift-II
Sets, Relation and Function
116948
If \(\mathbf{f}: \mathbf{R} \rightarrow \mathbf{R}\) is defined as \(\mathbf{f}(\mathbf{x})=\mathbf{x}-[\mathbf{x}]+3, \forall \mathbf{x}\) \(\pi R\), then \(f\) is
1 Not a function
2 A periodic function with period \(\pi\)
3 A periodic function with period 1
4 An invertible function
Explanation:
C Given, \(\mathrm{f}(\mathrm{x}) =\mathrm{x}-[\mathrm{x}]+3\) \(=\mathrm{x}-(\mathrm{x}-\{\mathrm{x}\})+3 \quad[\because[\mathrm{x}]=\mathrm{x}-\{\mathrm{x}\}]\) \(\mathrm{f}(\mathrm{x}) =\{\mathrm{x}\}+3\) \(\mathrm{f}(\mathrm{x}) \text { is periodic function with period } 1 .\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is periodic function with period 1 .
Shift-I
Sets, Relation and Function
116949
If \(5^x=(0.5)^y=1000\), then \(\frac{1}{x}-\frac{1}{y}\) is equal to
116946
The number of non-constant functions f from \(X\) \(=\{0,1,2\}\) to \(Y=\{1,2,3,4,5,6,7,8\}\) such that \(f(i) \leq f(j)\) for \(i, j \in X\) and \(i\lt j\) is
1 120
2 92
3 56
4 112
Explanation:
D Given, Sets \(X=\{0,1,2\}\) and \(Y=\{1,2,3,4,5,6,7,8\}\) and non - constant function \(\mathrm{f}: \mathrm{X} \rightarrow \mathrm{Y}\), such that \(f(i) \leq f(j) i, j \in X\) and \(i\lt j\). Now following two cases are possible Case - I Let range of 'f' are \(a, b, c \in y\), that \(a\lt b\lt c\) means function is strictly increasing. Then number of way of selection 3 distinct number \(a_1 b_2\) and \(\mathrm{c}\) is \({ }^8 \mathrm{C}_3=56\) Case - II If \(\mathrm{a}=\mathrm{b}\lt \mathrm{c}\) or \(\mathrm{a}\lt \mathrm{b}=\mathrm{c}\) Now number of ways of selecting two numbers \(\mathrm{a}=\mathrm{b}, \mathrm{c}\) or \(\mathrm{a}, \mathrm{b}=\mathrm{c}\) is \({ }^8 \mathrm{C}_2\) and since two elements are identical which we can make in \({ }^2 \mathrm{C}_1\) ways. So, number of ways to make such combination is \({ }^8 \mathrm{C}_2 \times{ }^2 \mathrm{C}_1\) \(=56\) So, required number of non - constant functions are \(=56+56=112\).
Shift-II
Sets, Relation and Function
116947
If the function \(f:[a, b] \rightarrow\) defined by \(f(x)=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+\sin x & 1 \\ 1+\cos x & 1 & 1\end{array}\right]\) is one-one and onto, then
A Given. \(f(x)=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin x & 1 \\ 1+\cos x & 1 & 1 \end{array}\right|\) On applying \(\mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_1\) and \(\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1\) we get - \(f(x)=\left|\begin{array}{ccc} 1 & 0 & 0 \\ 1 & \sin x & 0 \\ 1+\cos x & -\cos x & -\cos x \end{array}\right|\) \(f(x)=-\sin x \cdot \cos x\) \(f(x)=-\sin x \cos x\) \(f(x)=\frac{-\sin 2 x}{2}\) \(=\quad \frac{-\sqrt{3}}{4} \leq-\frac{-\sin 2 x}{2} \leq \frac{1}{2}\) \(=\quad-\frac{\sqrt{3}}{2} \leq-\sin 2 \mathrm{x} \leq 1\) \(=-1 \leq \sin 2 x \leq \frac{\sqrt{3}}{2}\) \(=2 \mathrm{x}=-\frac{\pi}{2}\) \(x=\frac{-\pi}{4}\) \(2 \mathrm{x}=\frac{\pi}{3}\) \(\mathrm{x}=\frac{\pi}{6}\) So, \(\quad-\frac{\pi}{4} \leq \mathrm{x} \leq \frac{\pi}{6}\)
Shift-II
Sets, Relation and Function
116948
If \(\mathbf{f}: \mathbf{R} \rightarrow \mathbf{R}\) is defined as \(\mathbf{f}(\mathbf{x})=\mathbf{x}-[\mathbf{x}]+3, \forall \mathbf{x}\) \(\pi R\), then \(f\) is
1 Not a function
2 A periodic function with period \(\pi\)
3 A periodic function with period 1
4 An invertible function
Explanation:
C Given, \(\mathrm{f}(\mathrm{x}) =\mathrm{x}-[\mathrm{x}]+3\) \(=\mathrm{x}-(\mathrm{x}-\{\mathrm{x}\})+3 \quad[\because[\mathrm{x}]=\mathrm{x}-\{\mathrm{x}\}]\) \(\mathrm{f}(\mathrm{x}) =\{\mathrm{x}\}+3\) \(\mathrm{f}(\mathrm{x}) \text { is periodic function with period } 1 .\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is periodic function with period 1 .
Shift-I
Sets, Relation and Function
116949
If \(5^x=(0.5)^y=1000\), then \(\frac{1}{x}-\frac{1}{y}\) is equal to
116946
The number of non-constant functions f from \(X\) \(=\{0,1,2\}\) to \(Y=\{1,2,3,4,5,6,7,8\}\) such that \(f(i) \leq f(j)\) for \(i, j \in X\) and \(i\lt j\) is
1 120
2 92
3 56
4 112
Explanation:
D Given, Sets \(X=\{0,1,2\}\) and \(Y=\{1,2,3,4,5,6,7,8\}\) and non - constant function \(\mathrm{f}: \mathrm{X} \rightarrow \mathrm{Y}\), such that \(f(i) \leq f(j) i, j \in X\) and \(i\lt j\). Now following two cases are possible Case - I Let range of 'f' are \(a, b, c \in y\), that \(a\lt b\lt c\) means function is strictly increasing. Then number of way of selection 3 distinct number \(a_1 b_2\) and \(\mathrm{c}\) is \({ }^8 \mathrm{C}_3=56\) Case - II If \(\mathrm{a}=\mathrm{b}\lt \mathrm{c}\) or \(\mathrm{a}\lt \mathrm{b}=\mathrm{c}\) Now number of ways of selecting two numbers \(\mathrm{a}=\mathrm{b}, \mathrm{c}\) or \(\mathrm{a}, \mathrm{b}=\mathrm{c}\) is \({ }^8 \mathrm{C}_2\) and since two elements are identical which we can make in \({ }^2 \mathrm{C}_1\) ways. So, number of ways to make such combination is \({ }^8 \mathrm{C}_2 \times{ }^2 \mathrm{C}_1\) \(=56\) So, required number of non - constant functions are \(=56+56=112\).
Shift-II
Sets, Relation and Function
116947
If the function \(f:[a, b] \rightarrow\) defined by \(f(x)=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+\sin x & 1 \\ 1+\cos x & 1 & 1\end{array}\right]\) is one-one and onto, then
A Given. \(f(x)=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin x & 1 \\ 1+\cos x & 1 & 1 \end{array}\right|\) On applying \(\mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_1\) and \(\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1\) we get - \(f(x)=\left|\begin{array}{ccc} 1 & 0 & 0 \\ 1 & \sin x & 0 \\ 1+\cos x & -\cos x & -\cos x \end{array}\right|\) \(f(x)=-\sin x \cdot \cos x\) \(f(x)=-\sin x \cos x\) \(f(x)=\frac{-\sin 2 x}{2}\) \(=\quad \frac{-\sqrt{3}}{4} \leq-\frac{-\sin 2 x}{2} \leq \frac{1}{2}\) \(=\quad-\frac{\sqrt{3}}{2} \leq-\sin 2 \mathrm{x} \leq 1\) \(=-1 \leq \sin 2 x \leq \frac{\sqrt{3}}{2}\) \(=2 \mathrm{x}=-\frac{\pi}{2}\) \(x=\frac{-\pi}{4}\) \(2 \mathrm{x}=\frac{\pi}{3}\) \(\mathrm{x}=\frac{\pi}{6}\) So, \(\quad-\frac{\pi}{4} \leq \mathrm{x} \leq \frac{\pi}{6}\)
Shift-II
Sets, Relation and Function
116948
If \(\mathbf{f}: \mathbf{R} \rightarrow \mathbf{R}\) is defined as \(\mathbf{f}(\mathbf{x})=\mathbf{x}-[\mathbf{x}]+3, \forall \mathbf{x}\) \(\pi R\), then \(f\) is
1 Not a function
2 A periodic function with period \(\pi\)
3 A periodic function with period 1
4 An invertible function
Explanation:
C Given, \(\mathrm{f}(\mathrm{x}) =\mathrm{x}-[\mathrm{x}]+3\) \(=\mathrm{x}-(\mathrm{x}-\{\mathrm{x}\})+3 \quad[\because[\mathrm{x}]=\mathrm{x}-\{\mathrm{x}\}]\) \(\mathrm{f}(\mathrm{x}) =\{\mathrm{x}\}+3\) \(\mathrm{f}(\mathrm{x}) \text { is periodic function with period } 1 .\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is periodic function with period 1 .
Shift-I
Sets, Relation and Function
116949
If \(5^x=(0.5)^y=1000\), then \(\frac{1}{x}-\frac{1}{y}\) is equal to
116946
The number of non-constant functions f from \(X\) \(=\{0,1,2\}\) to \(Y=\{1,2,3,4,5,6,7,8\}\) such that \(f(i) \leq f(j)\) for \(i, j \in X\) and \(i\lt j\) is
1 120
2 92
3 56
4 112
Explanation:
D Given, Sets \(X=\{0,1,2\}\) and \(Y=\{1,2,3,4,5,6,7,8\}\) and non - constant function \(\mathrm{f}: \mathrm{X} \rightarrow \mathrm{Y}\), such that \(f(i) \leq f(j) i, j \in X\) and \(i\lt j\). Now following two cases are possible Case - I Let range of 'f' are \(a, b, c \in y\), that \(a\lt b\lt c\) means function is strictly increasing. Then number of way of selection 3 distinct number \(a_1 b_2\) and \(\mathrm{c}\) is \({ }^8 \mathrm{C}_3=56\) Case - II If \(\mathrm{a}=\mathrm{b}\lt \mathrm{c}\) or \(\mathrm{a}\lt \mathrm{b}=\mathrm{c}\) Now number of ways of selecting two numbers \(\mathrm{a}=\mathrm{b}, \mathrm{c}\) or \(\mathrm{a}, \mathrm{b}=\mathrm{c}\) is \({ }^8 \mathrm{C}_2\) and since two elements are identical which we can make in \({ }^2 \mathrm{C}_1\) ways. So, number of ways to make such combination is \({ }^8 \mathrm{C}_2 \times{ }^2 \mathrm{C}_1\) \(=56\) So, required number of non - constant functions are \(=56+56=112\).
Shift-II
Sets, Relation and Function
116947
If the function \(f:[a, b] \rightarrow\) defined by \(f(x)=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+\sin x & 1 \\ 1+\cos x & 1 & 1\end{array}\right]\) is one-one and onto, then
A Given. \(f(x)=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin x & 1 \\ 1+\cos x & 1 & 1 \end{array}\right|\) On applying \(\mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_1\) and \(\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1\) we get - \(f(x)=\left|\begin{array}{ccc} 1 & 0 & 0 \\ 1 & \sin x & 0 \\ 1+\cos x & -\cos x & -\cos x \end{array}\right|\) \(f(x)=-\sin x \cdot \cos x\) \(f(x)=-\sin x \cos x\) \(f(x)=\frac{-\sin 2 x}{2}\) \(=\quad \frac{-\sqrt{3}}{4} \leq-\frac{-\sin 2 x}{2} \leq \frac{1}{2}\) \(=\quad-\frac{\sqrt{3}}{2} \leq-\sin 2 \mathrm{x} \leq 1\) \(=-1 \leq \sin 2 x \leq \frac{\sqrt{3}}{2}\) \(=2 \mathrm{x}=-\frac{\pi}{2}\) \(x=\frac{-\pi}{4}\) \(2 \mathrm{x}=\frac{\pi}{3}\) \(\mathrm{x}=\frac{\pi}{6}\) So, \(\quad-\frac{\pi}{4} \leq \mathrm{x} \leq \frac{\pi}{6}\)
Shift-II
Sets, Relation and Function
116948
If \(\mathbf{f}: \mathbf{R} \rightarrow \mathbf{R}\) is defined as \(\mathbf{f}(\mathbf{x})=\mathbf{x}-[\mathbf{x}]+3, \forall \mathbf{x}\) \(\pi R\), then \(f\) is
1 Not a function
2 A periodic function with period \(\pi\)
3 A periodic function with period 1
4 An invertible function
Explanation:
C Given, \(\mathrm{f}(\mathrm{x}) =\mathrm{x}-[\mathrm{x}]+3\) \(=\mathrm{x}-(\mathrm{x}-\{\mathrm{x}\})+3 \quad[\because[\mathrm{x}]=\mathrm{x}-\{\mathrm{x}\}]\) \(\mathrm{f}(\mathrm{x}) =\{\mathrm{x}\}+3\) \(\mathrm{f}(\mathrm{x}) \text { is periodic function with period } 1 .\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is periodic function with period 1 .
Shift-I
Sets, Relation and Function
116949
If \(5^x=(0.5)^y=1000\), then \(\frac{1}{x}-\frac{1}{y}\) is equal to
