117123
The number of onto mappings from the set \(A=\{1,2 \ldots . . ., 100\}\) to set \(B=\{1,2\}\) is :
1 \(2^{100}-2\)
2 \(2^{100}\)
3 \(2^{99}-2\)
4 \(2^{99}\)
Explanation:
Exp: (A) : Given, \(A=\{1,2 \ldots \ldots, 100\}\) \(B=\{1,2\}\) Then, no. of elements in set \(A=100\) And, no. of elements in set \(B=2\) So, number of possible onto mapping is \(2^{100}\) Since, this also contain the number of elements in \(B\) differently. Hence, the total no of possible onto mapping from the set \(A\) to set \(B\) is \(2^{100}-2\).
[BCECE-2006]
Sets, Relation and Function
117124
If the function \(f:(-\infty, \infty) \rightarrow B\) defined by \(f(x)=-x^2+6 x-8\) is bijective, then \(B=\)
1 \([1, \infty)\)
2 \((-\infty, 1]\)
3 \((-\infty, \infty)\)
4 None of these
Explanation:
Exp: (B) : Since, the function \(\mathrm{f}\) is bijective, therefore \(f\) is onto. Therefore range of \(f=B\). Let, \(y=-x^2+6 x-8\) \(\Rightarrow \mathrm{x}^2-6 \mathrm{x}+(8+\mathrm{y})=0\) \(\Rightarrow x=\frac{6 \pm \sqrt{36-4(8+y)}}{2}=\frac{6 \pm \sqrt{4-4 y}}{2}\) For \(x\) to be real, \(4-4 y \geq 0 \Rightarrow y \leq 1\) \(\therefore \mathrm{B}=\) range of \(\mathrm{f}=(-\infty, 1)\)
[BITSAT-2009]
Sets, Relation and Function
117125
Let \(E=\{1,2,3,4\}\) and \(F=\{1,2\}\). Then the number of onto functions from \(E\) to \(F\) is
1 14
2 16
3 12
4 8
Explanation:
Exp: (A) : If set \(\mathrm{A}\) has \(\mathrm{m}\) elements and set \(\mathrm{B}\) has \(\mathrm{n}\) elements then number of onto functions from \(A\) to \(B\) is \(\sum_{\mathrm{r}=1}^{\mathrm{n}}(-1)^{\mathrm{n}-\mathrm{r} n} \mathrm{C}_{\mathrm{r}} \mathrm{r}^{\mathrm{m}}\) where \(1 \leq \mathrm{n} \leq \mathrm{m}\) Here \(\mathrm{E}=\{1,2,3,4\}, \mathrm{F}=\{1,2\}\) \(\mathrm{m}=4, \mathrm{n}=2\) \(\therefore\) No. of onto functions from \(\mathrm{E}\) to \(\mathrm{F}\) \(=\sum_{\mathrm{r}=1}^2(-1)^{2-\mathrm{r} 2} \mathrm{C}_{\mathrm{r}}(\mathrm{r})^4\) \(=(-1)^1 \times{ }^2 \mathrm{C}_1 \times(1)^4+(-1)^2 \times{ }^2 \mathrm{C}_2 \times(2)^4\) \(=-2+16=14\)
[BITSAT-2010]
Sets, Relation and Function
117126
If \(A=\{x \mid x \in N, x \leq 5\}, B=\left\{x \mid x \in Z, x^2-5 x+6\right.\) \(=0\}\), then the number of onto functions from \(A\) to \(B\) is
1 2
2 30
3 23
4 32
Explanation:
Exp: (B) : Given, \(A=\{x \mid x \in N, x \leq 5\}\) \(B=\left\{x \mid x \in Z, x^2-5 x+6=0\right\}\) Then, \(\mathrm{A}=\{1,2,3,4,5\}\) \(B=x^2-5 x+6\) And, \(B=\{(x-2)(x-3)\}=0 \Rightarrow x=\{2,3\}\) \(B=\{2,3\}\) So, the number of onto functions from \(A\) to \(B\) is - \(=2^5-2\) \(=32-2\) \(=30\)
117123
The number of onto mappings from the set \(A=\{1,2 \ldots . . ., 100\}\) to set \(B=\{1,2\}\) is :
1 \(2^{100}-2\)
2 \(2^{100}\)
3 \(2^{99}-2\)
4 \(2^{99}\)
Explanation:
Exp: (A) : Given, \(A=\{1,2 \ldots \ldots, 100\}\) \(B=\{1,2\}\) Then, no. of elements in set \(A=100\) And, no. of elements in set \(B=2\) So, number of possible onto mapping is \(2^{100}\) Since, this also contain the number of elements in \(B\) differently. Hence, the total no of possible onto mapping from the set \(A\) to set \(B\) is \(2^{100}-2\).
[BCECE-2006]
Sets, Relation and Function
117124
If the function \(f:(-\infty, \infty) \rightarrow B\) defined by \(f(x)=-x^2+6 x-8\) is bijective, then \(B=\)
1 \([1, \infty)\)
2 \((-\infty, 1]\)
3 \((-\infty, \infty)\)
4 None of these
Explanation:
Exp: (B) : Since, the function \(\mathrm{f}\) is bijective, therefore \(f\) is onto. Therefore range of \(f=B\). Let, \(y=-x^2+6 x-8\) \(\Rightarrow \mathrm{x}^2-6 \mathrm{x}+(8+\mathrm{y})=0\) \(\Rightarrow x=\frac{6 \pm \sqrt{36-4(8+y)}}{2}=\frac{6 \pm \sqrt{4-4 y}}{2}\) For \(x\) to be real, \(4-4 y \geq 0 \Rightarrow y \leq 1\) \(\therefore \mathrm{B}=\) range of \(\mathrm{f}=(-\infty, 1)\)
[BITSAT-2009]
Sets, Relation and Function
117125
Let \(E=\{1,2,3,4\}\) and \(F=\{1,2\}\). Then the number of onto functions from \(E\) to \(F\) is
1 14
2 16
3 12
4 8
Explanation:
Exp: (A) : If set \(\mathrm{A}\) has \(\mathrm{m}\) elements and set \(\mathrm{B}\) has \(\mathrm{n}\) elements then number of onto functions from \(A\) to \(B\) is \(\sum_{\mathrm{r}=1}^{\mathrm{n}}(-1)^{\mathrm{n}-\mathrm{r} n} \mathrm{C}_{\mathrm{r}} \mathrm{r}^{\mathrm{m}}\) where \(1 \leq \mathrm{n} \leq \mathrm{m}\) Here \(\mathrm{E}=\{1,2,3,4\}, \mathrm{F}=\{1,2\}\) \(\mathrm{m}=4, \mathrm{n}=2\) \(\therefore\) No. of onto functions from \(\mathrm{E}\) to \(\mathrm{F}\) \(=\sum_{\mathrm{r}=1}^2(-1)^{2-\mathrm{r} 2} \mathrm{C}_{\mathrm{r}}(\mathrm{r})^4\) \(=(-1)^1 \times{ }^2 \mathrm{C}_1 \times(1)^4+(-1)^2 \times{ }^2 \mathrm{C}_2 \times(2)^4\) \(=-2+16=14\)
[BITSAT-2010]
Sets, Relation and Function
117126
If \(A=\{x \mid x \in N, x \leq 5\}, B=\left\{x \mid x \in Z, x^2-5 x+6\right.\) \(=0\}\), then the number of onto functions from \(A\) to \(B\) is
1 2
2 30
3 23
4 32
Explanation:
Exp: (B) : Given, \(A=\{x \mid x \in N, x \leq 5\}\) \(B=\left\{x \mid x \in Z, x^2-5 x+6=0\right\}\) Then, \(\mathrm{A}=\{1,2,3,4,5\}\) \(B=x^2-5 x+6\) And, \(B=\{(x-2)(x-3)\}=0 \Rightarrow x=\{2,3\}\) \(B=\{2,3\}\) So, the number of onto functions from \(A\) to \(B\) is - \(=2^5-2\) \(=32-2\) \(=30\)
117123
The number of onto mappings from the set \(A=\{1,2 \ldots . . ., 100\}\) to set \(B=\{1,2\}\) is :
1 \(2^{100}-2\)
2 \(2^{100}\)
3 \(2^{99}-2\)
4 \(2^{99}\)
Explanation:
Exp: (A) : Given, \(A=\{1,2 \ldots \ldots, 100\}\) \(B=\{1,2\}\) Then, no. of elements in set \(A=100\) And, no. of elements in set \(B=2\) So, number of possible onto mapping is \(2^{100}\) Since, this also contain the number of elements in \(B\) differently. Hence, the total no of possible onto mapping from the set \(A\) to set \(B\) is \(2^{100}-2\).
[BCECE-2006]
Sets, Relation and Function
117124
If the function \(f:(-\infty, \infty) \rightarrow B\) defined by \(f(x)=-x^2+6 x-8\) is bijective, then \(B=\)
1 \([1, \infty)\)
2 \((-\infty, 1]\)
3 \((-\infty, \infty)\)
4 None of these
Explanation:
Exp: (B) : Since, the function \(\mathrm{f}\) is bijective, therefore \(f\) is onto. Therefore range of \(f=B\). Let, \(y=-x^2+6 x-8\) \(\Rightarrow \mathrm{x}^2-6 \mathrm{x}+(8+\mathrm{y})=0\) \(\Rightarrow x=\frac{6 \pm \sqrt{36-4(8+y)}}{2}=\frac{6 \pm \sqrt{4-4 y}}{2}\) For \(x\) to be real, \(4-4 y \geq 0 \Rightarrow y \leq 1\) \(\therefore \mathrm{B}=\) range of \(\mathrm{f}=(-\infty, 1)\)
[BITSAT-2009]
Sets, Relation and Function
117125
Let \(E=\{1,2,3,4\}\) and \(F=\{1,2\}\). Then the number of onto functions from \(E\) to \(F\) is
1 14
2 16
3 12
4 8
Explanation:
Exp: (A) : If set \(\mathrm{A}\) has \(\mathrm{m}\) elements and set \(\mathrm{B}\) has \(\mathrm{n}\) elements then number of onto functions from \(A\) to \(B\) is \(\sum_{\mathrm{r}=1}^{\mathrm{n}}(-1)^{\mathrm{n}-\mathrm{r} n} \mathrm{C}_{\mathrm{r}} \mathrm{r}^{\mathrm{m}}\) where \(1 \leq \mathrm{n} \leq \mathrm{m}\) Here \(\mathrm{E}=\{1,2,3,4\}, \mathrm{F}=\{1,2\}\) \(\mathrm{m}=4, \mathrm{n}=2\) \(\therefore\) No. of onto functions from \(\mathrm{E}\) to \(\mathrm{F}\) \(=\sum_{\mathrm{r}=1}^2(-1)^{2-\mathrm{r} 2} \mathrm{C}_{\mathrm{r}}(\mathrm{r})^4\) \(=(-1)^1 \times{ }^2 \mathrm{C}_1 \times(1)^4+(-1)^2 \times{ }^2 \mathrm{C}_2 \times(2)^4\) \(=-2+16=14\)
[BITSAT-2010]
Sets, Relation and Function
117126
If \(A=\{x \mid x \in N, x \leq 5\}, B=\left\{x \mid x \in Z, x^2-5 x+6\right.\) \(=0\}\), then the number of onto functions from \(A\) to \(B\) is
1 2
2 30
3 23
4 32
Explanation:
Exp: (B) : Given, \(A=\{x \mid x \in N, x \leq 5\}\) \(B=\left\{x \mid x \in Z, x^2-5 x+6=0\right\}\) Then, \(\mathrm{A}=\{1,2,3,4,5\}\) \(B=x^2-5 x+6\) And, \(B=\{(x-2)(x-3)\}=0 \Rightarrow x=\{2,3\}\) \(B=\{2,3\}\) So, the number of onto functions from \(A\) to \(B\) is - \(=2^5-2\) \(=32-2\) \(=30\)
117123
The number of onto mappings from the set \(A=\{1,2 \ldots . . ., 100\}\) to set \(B=\{1,2\}\) is :
1 \(2^{100}-2\)
2 \(2^{100}\)
3 \(2^{99}-2\)
4 \(2^{99}\)
Explanation:
Exp: (A) : Given, \(A=\{1,2 \ldots \ldots, 100\}\) \(B=\{1,2\}\) Then, no. of elements in set \(A=100\) And, no. of elements in set \(B=2\) So, number of possible onto mapping is \(2^{100}\) Since, this also contain the number of elements in \(B\) differently. Hence, the total no of possible onto mapping from the set \(A\) to set \(B\) is \(2^{100}-2\).
[BCECE-2006]
Sets, Relation and Function
117124
If the function \(f:(-\infty, \infty) \rightarrow B\) defined by \(f(x)=-x^2+6 x-8\) is bijective, then \(B=\)
1 \([1, \infty)\)
2 \((-\infty, 1]\)
3 \((-\infty, \infty)\)
4 None of these
Explanation:
Exp: (B) : Since, the function \(\mathrm{f}\) is bijective, therefore \(f\) is onto. Therefore range of \(f=B\). Let, \(y=-x^2+6 x-8\) \(\Rightarrow \mathrm{x}^2-6 \mathrm{x}+(8+\mathrm{y})=0\) \(\Rightarrow x=\frac{6 \pm \sqrt{36-4(8+y)}}{2}=\frac{6 \pm \sqrt{4-4 y}}{2}\) For \(x\) to be real, \(4-4 y \geq 0 \Rightarrow y \leq 1\) \(\therefore \mathrm{B}=\) range of \(\mathrm{f}=(-\infty, 1)\)
[BITSAT-2009]
Sets, Relation and Function
117125
Let \(E=\{1,2,3,4\}\) and \(F=\{1,2\}\). Then the number of onto functions from \(E\) to \(F\) is
1 14
2 16
3 12
4 8
Explanation:
Exp: (A) : If set \(\mathrm{A}\) has \(\mathrm{m}\) elements and set \(\mathrm{B}\) has \(\mathrm{n}\) elements then number of onto functions from \(A\) to \(B\) is \(\sum_{\mathrm{r}=1}^{\mathrm{n}}(-1)^{\mathrm{n}-\mathrm{r} n} \mathrm{C}_{\mathrm{r}} \mathrm{r}^{\mathrm{m}}\) where \(1 \leq \mathrm{n} \leq \mathrm{m}\) Here \(\mathrm{E}=\{1,2,3,4\}, \mathrm{F}=\{1,2\}\) \(\mathrm{m}=4, \mathrm{n}=2\) \(\therefore\) No. of onto functions from \(\mathrm{E}\) to \(\mathrm{F}\) \(=\sum_{\mathrm{r}=1}^2(-1)^{2-\mathrm{r} 2} \mathrm{C}_{\mathrm{r}}(\mathrm{r})^4\) \(=(-1)^1 \times{ }^2 \mathrm{C}_1 \times(1)^4+(-1)^2 \times{ }^2 \mathrm{C}_2 \times(2)^4\) \(=-2+16=14\)
[BITSAT-2010]
Sets, Relation and Function
117126
If \(A=\{x \mid x \in N, x \leq 5\}, B=\left\{x \mid x \in Z, x^2-5 x+6\right.\) \(=0\}\), then the number of onto functions from \(A\) to \(B\) is
1 2
2 30
3 23
4 32
Explanation:
Exp: (B) : Given, \(A=\{x \mid x \in N, x \leq 5\}\) \(B=\left\{x \mid x \in Z, x^2-5 x+6=0\right\}\) Then, \(\mathrm{A}=\{1,2,3,4,5\}\) \(B=x^2-5 x+6\) And, \(B=\{(x-2)(x-3)\}=0 \Rightarrow x=\{2,3\}\) \(B=\{2,3\}\) So, the number of onto functions from \(A\) to \(B\) is - \(=2^5-2\) \(=32-2\) \(=30\)
