117035
If \(f(x)=\frac{2 x+3}{3 x-2}, x \neq \frac{2}{3}\), then the function fof is
1 a constant function
2 an identity function
3 an even function
4 an exponential function
Explanation:
B Given, \(f(x)=\frac{2 x+3}{3 x-2}\) \(\text { fof }=f(f(x))\) \(\text { fof }(x)=\frac{2 \cdot\left(\frac{2 x+3}{3 x-2}\right)+3}{3 \cdot\left(\frac{2 x+3}{3 x-2}\right)-2}=\frac{\frac{4 x+6}{3 x-2}+3}{\frac{6 x+9}{3 x-2}-2}\) \(\text { fof }(x)=\frac{4 x+6+9 x-6}{6 x+9-6 x+4}=\frac{13 x}{13}=x\)Therefore, we can say that the composite function fof for the given function is an identity function.
MHT-CET 2020
Sets, Relation and Function
117044
The set \(A\) has 4 elements and the set \(B\) has 5 elements then the number of injective mappings that can be defined from \(A\) to \(B\) is
1 144
2 72
3 60
4 120
Explanation:
D Given, The set A has 4 elements and the set \(B\) has 5 elements. We know that, if set \(A\) has \(m\) elements and set \(B\) has \(n\) elements then the number of injective functions or one to one function is \(\frac{n !}{(n-m) !}\). Then, the number of injective mappings that can be defined from \(\mathrm{A}\) to \(\mathrm{B}\) is - \(=\frac{5 !}{(5-4) !}=\frac{5 !}{1 !}\) \(=5 !=5 \times 4 \times 3 \times 2 \times 1=120\)
Karnataka CET 2016
Sets, Relation and Function
117056
Let a function \(f: \mathbb{N} \rightarrow \mathbb{N}\) be defined by \(f(n)=\left[\begin{array}{cc}2 n, n=2,4,6,8, \ldots . . \\ n-1, n=3,7,11,15, \ldots . . \\ \frac{n+1}{2}, n=1,5,9,13, \ldots . .\end{array}\right.\) then, \(f\) is
1 one-one but not onto
2 onto but not one-one
3 neither one-one nor onto
4 one-one and onto
Explanation:
D Given, \(f(n)=\left\{\begin{array}{cc}2 n, & n=2,4,6,8 \ldots \ldots \\ n-1, & n=3,7,11,15 \ldots \ldots \\ \frac{n+1}{2}, & n=1,5,9,13 \ldots .\end{array}\right.\) If \(\mathrm{n}=2,4,6,8\), then \(2 \mathrm{n}\) in multiple of 4 . If \(n=3,7,11,15\) then (n-1) is not multiple of 4 . If \(\mathrm{n}=1,5,9,13\), then \(\left(\frac{\mathrm{n}+1}{2}\right)\) is the odd number. Hence, Every numbers give exactly one value. So, \(\mathrm{f}\) is one - one and onto.
Shift-I
Sets, Relation and Function
117059
The number bijective functions \(f:\{1,3,5,7\) \(\ldots . .99\} \rightarrow\{2,4,6,8, \ldots .100\}\), such that \(f(3) \geq\) \(f(9) \geq f(15) \geq f(21) \geq \ldots \geq f(99)\) is
1 \({ }^{50} \mathrm{P}_{17}\)
2 \({ }^{50} \mathrm{P}_{33}\)
3 \(33 ! \times 17\) !
4 \(\frac{50 !}{2}\)
Explanation:
B One to one functions define that each element of one set, say set (A) is mapped with a unique element of another set (B), solution of question, As function is one-one and onto, out of 50 elements of domain set 17 elements are following restriction. \(f(3) \geq f(a) \geq f(15) \ldots \ldots \geq f(99)\) So number of ways \(={ }^{50} \mathrm{C}_{17} 33\) ! \(={ }^{50} \mathrm{P}_{33}\)
117035
If \(f(x)=\frac{2 x+3}{3 x-2}, x \neq \frac{2}{3}\), then the function fof is
1 a constant function
2 an identity function
3 an even function
4 an exponential function
Explanation:
B Given, \(f(x)=\frac{2 x+3}{3 x-2}\) \(\text { fof }=f(f(x))\) \(\text { fof }(x)=\frac{2 \cdot\left(\frac{2 x+3}{3 x-2}\right)+3}{3 \cdot\left(\frac{2 x+3}{3 x-2}\right)-2}=\frac{\frac{4 x+6}{3 x-2}+3}{\frac{6 x+9}{3 x-2}-2}\) \(\text { fof }(x)=\frac{4 x+6+9 x-6}{6 x+9-6 x+4}=\frac{13 x}{13}=x\)Therefore, we can say that the composite function fof for the given function is an identity function.
MHT-CET 2020
Sets, Relation and Function
117044
The set \(A\) has 4 elements and the set \(B\) has 5 elements then the number of injective mappings that can be defined from \(A\) to \(B\) is
1 144
2 72
3 60
4 120
Explanation:
D Given, The set A has 4 elements and the set \(B\) has 5 elements. We know that, if set \(A\) has \(m\) elements and set \(B\) has \(n\) elements then the number of injective functions or one to one function is \(\frac{n !}{(n-m) !}\). Then, the number of injective mappings that can be defined from \(\mathrm{A}\) to \(\mathrm{B}\) is - \(=\frac{5 !}{(5-4) !}=\frac{5 !}{1 !}\) \(=5 !=5 \times 4 \times 3 \times 2 \times 1=120\)
Karnataka CET 2016
Sets, Relation and Function
117056
Let a function \(f: \mathbb{N} \rightarrow \mathbb{N}\) be defined by \(f(n)=\left[\begin{array}{cc}2 n, n=2,4,6,8, \ldots . . \\ n-1, n=3,7,11,15, \ldots . . \\ \frac{n+1}{2}, n=1,5,9,13, \ldots . .\end{array}\right.\) then, \(f\) is
1 one-one but not onto
2 onto but not one-one
3 neither one-one nor onto
4 one-one and onto
Explanation:
D Given, \(f(n)=\left\{\begin{array}{cc}2 n, & n=2,4,6,8 \ldots \ldots \\ n-1, & n=3,7,11,15 \ldots \ldots \\ \frac{n+1}{2}, & n=1,5,9,13 \ldots .\end{array}\right.\) If \(\mathrm{n}=2,4,6,8\), then \(2 \mathrm{n}\) in multiple of 4 . If \(n=3,7,11,15\) then (n-1) is not multiple of 4 . If \(\mathrm{n}=1,5,9,13\), then \(\left(\frac{\mathrm{n}+1}{2}\right)\) is the odd number. Hence, Every numbers give exactly one value. So, \(\mathrm{f}\) is one - one and onto.
Shift-I
Sets, Relation and Function
117059
The number bijective functions \(f:\{1,3,5,7\) \(\ldots . .99\} \rightarrow\{2,4,6,8, \ldots .100\}\), such that \(f(3) \geq\) \(f(9) \geq f(15) \geq f(21) \geq \ldots \geq f(99)\) is
1 \({ }^{50} \mathrm{P}_{17}\)
2 \({ }^{50} \mathrm{P}_{33}\)
3 \(33 ! \times 17\) !
4 \(\frac{50 !}{2}\)
Explanation:
B One to one functions define that each element of one set, say set (A) is mapped with a unique element of another set (B), solution of question, As function is one-one and onto, out of 50 elements of domain set 17 elements are following restriction. \(f(3) \geq f(a) \geq f(15) \ldots \ldots \geq f(99)\) So number of ways \(={ }^{50} \mathrm{C}_{17} 33\) ! \(={ }^{50} \mathrm{P}_{33}\)
117035
If \(f(x)=\frac{2 x+3}{3 x-2}, x \neq \frac{2}{3}\), then the function fof is
1 a constant function
2 an identity function
3 an even function
4 an exponential function
Explanation:
B Given, \(f(x)=\frac{2 x+3}{3 x-2}\) \(\text { fof }=f(f(x))\) \(\text { fof }(x)=\frac{2 \cdot\left(\frac{2 x+3}{3 x-2}\right)+3}{3 \cdot\left(\frac{2 x+3}{3 x-2}\right)-2}=\frac{\frac{4 x+6}{3 x-2}+3}{\frac{6 x+9}{3 x-2}-2}\) \(\text { fof }(x)=\frac{4 x+6+9 x-6}{6 x+9-6 x+4}=\frac{13 x}{13}=x\)Therefore, we can say that the composite function fof for the given function is an identity function.
MHT-CET 2020
Sets, Relation and Function
117044
The set \(A\) has 4 elements and the set \(B\) has 5 elements then the number of injective mappings that can be defined from \(A\) to \(B\) is
1 144
2 72
3 60
4 120
Explanation:
D Given, The set A has 4 elements and the set \(B\) has 5 elements. We know that, if set \(A\) has \(m\) elements and set \(B\) has \(n\) elements then the number of injective functions or one to one function is \(\frac{n !}{(n-m) !}\). Then, the number of injective mappings that can be defined from \(\mathrm{A}\) to \(\mathrm{B}\) is - \(=\frac{5 !}{(5-4) !}=\frac{5 !}{1 !}\) \(=5 !=5 \times 4 \times 3 \times 2 \times 1=120\)
Karnataka CET 2016
Sets, Relation and Function
117056
Let a function \(f: \mathbb{N} \rightarrow \mathbb{N}\) be defined by \(f(n)=\left[\begin{array}{cc}2 n, n=2,4,6,8, \ldots . . \\ n-1, n=3,7,11,15, \ldots . . \\ \frac{n+1}{2}, n=1,5,9,13, \ldots . .\end{array}\right.\) then, \(f\) is
1 one-one but not onto
2 onto but not one-one
3 neither one-one nor onto
4 one-one and onto
Explanation:
D Given, \(f(n)=\left\{\begin{array}{cc}2 n, & n=2,4,6,8 \ldots \ldots \\ n-1, & n=3,7,11,15 \ldots \ldots \\ \frac{n+1}{2}, & n=1,5,9,13 \ldots .\end{array}\right.\) If \(\mathrm{n}=2,4,6,8\), then \(2 \mathrm{n}\) in multiple of 4 . If \(n=3,7,11,15\) then (n-1) is not multiple of 4 . If \(\mathrm{n}=1,5,9,13\), then \(\left(\frac{\mathrm{n}+1}{2}\right)\) is the odd number. Hence, Every numbers give exactly one value. So, \(\mathrm{f}\) is one - one and onto.
Shift-I
Sets, Relation and Function
117059
The number bijective functions \(f:\{1,3,5,7\) \(\ldots . .99\} \rightarrow\{2,4,6,8, \ldots .100\}\), such that \(f(3) \geq\) \(f(9) \geq f(15) \geq f(21) \geq \ldots \geq f(99)\) is
1 \({ }^{50} \mathrm{P}_{17}\)
2 \({ }^{50} \mathrm{P}_{33}\)
3 \(33 ! \times 17\) !
4 \(\frac{50 !}{2}\)
Explanation:
B One to one functions define that each element of one set, say set (A) is mapped with a unique element of another set (B), solution of question, As function is one-one and onto, out of 50 elements of domain set 17 elements are following restriction. \(f(3) \geq f(a) \geq f(15) \ldots \ldots \geq f(99)\) So number of ways \(={ }^{50} \mathrm{C}_{17} 33\) ! \(={ }^{50} \mathrm{P}_{33}\)
117035
If \(f(x)=\frac{2 x+3}{3 x-2}, x \neq \frac{2}{3}\), then the function fof is
1 a constant function
2 an identity function
3 an even function
4 an exponential function
Explanation:
B Given, \(f(x)=\frac{2 x+3}{3 x-2}\) \(\text { fof }=f(f(x))\) \(\text { fof }(x)=\frac{2 \cdot\left(\frac{2 x+3}{3 x-2}\right)+3}{3 \cdot\left(\frac{2 x+3}{3 x-2}\right)-2}=\frac{\frac{4 x+6}{3 x-2}+3}{\frac{6 x+9}{3 x-2}-2}\) \(\text { fof }(x)=\frac{4 x+6+9 x-6}{6 x+9-6 x+4}=\frac{13 x}{13}=x\)Therefore, we can say that the composite function fof for the given function is an identity function.
MHT-CET 2020
Sets, Relation and Function
117044
The set \(A\) has 4 elements and the set \(B\) has 5 elements then the number of injective mappings that can be defined from \(A\) to \(B\) is
1 144
2 72
3 60
4 120
Explanation:
D Given, The set A has 4 elements and the set \(B\) has 5 elements. We know that, if set \(A\) has \(m\) elements and set \(B\) has \(n\) elements then the number of injective functions or one to one function is \(\frac{n !}{(n-m) !}\). Then, the number of injective mappings that can be defined from \(\mathrm{A}\) to \(\mathrm{B}\) is - \(=\frac{5 !}{(5-4) !}=\frac{5 !}{1 !}\) \(=5 !=5 \times 4 \times 3 \times 2 \times 1=120\)
Karnataka CET 2016
Sets, Relation and Function
117056
Let a function \(f: \mathbb{N} \rightarrow \mathbb{N}\) be defined by \(f(n)=\left[\begin{array}{cc}2 n, n=2,4,6,8, \ldots . . \\ n-1, n=3,7,11,15, \ldots . . \\ \frac{n+1}{2}, n=1,5,9,13, \ldots . .\end{array}\right.\) then, \(f\) is
1 one-one but not onto
2 onto but not one-one
3 neither one-one nor onto
4 one-one and onto
Explanation:
D Given, \(f(n)=\left\{\begin{array}{cc}2 n, & n=2,4,6,8 \ldots \ldots \\ n-1, & n=3,7,11,15 \ldots \ldots \\ \frac{n+1}{2}, & n=1,5,9,13 \ldots .\end{array}\right.\) If \(\mathrm{n}=2,4,6,8\), then \(2 \mathrm{n}\) in multiple of 4 . If \(n=3,7,11,15\) then (n-1) is not multiple of 4 . If \(\mathrm{n}=1,5,9,13\), then \(\left(\frac{\mathrm{n}+1}{2}\right)\) is the odd number. Hence, Every numbers give exactly one value. So, \(\mathrm{f}\) is one - one and onto.
Shift-I
Sets, Relation and Function
117059
The number bijective functions \(f:\{1,3,5,7\) \(\ldots . .99\} \rightarrow\{2,4,6,8, \ldots .100\}\), such that \(f(3) \geq\) \(f(9) \geq f(15) \geq f(21) \geq \ldots \geq f(99)\) is
1 \({ }^{50} \mathrm{P}_{17}\)
2 \({ }^{50} \mathrm{P}_{33}\)
3 \(33 ! \times 17\) !
4 \(\frac{50 !}{2}\)
Explanation:
B One to one functions define that each element of one set, say set (A) is mapped with a unique element of another set (B), solution of question, As function is one-one and onto, out of 50 elements of domain set 17 elements are following restriction. \(f(3) \geq f(a) \geq f(15) \ldots \ldots \geq f(99)\) So number of ways \(={ }^{50} \mathrm{C}_{17} 33\) ! \(={ }^{50} \mathrm{P}_{33}\)
