117058
Let \(A=\left(x_1, x_2, x_3, \ldots . ., x_7\right), B=\left(y_1, y_2, y_3\right)\). The total number of functions \(f: \mathrm{A} \rightarrow \mathrm{B}\) that are onto and there are exactly three elements \(x\) in A such that \(f(x)=y_2\) is equal to
1 490
2 510
3 630
4 None of these
Explanation:
A Given, \(\mathrm{A}=\left\{\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4, \mathrm{x}_5, \mathrm{x}_6, \mathrm{x}_7\right\}\) And, \(\mathrm{B}=\left\{\mathrm{y}_1, \mathrm{y}_2, \mathrm{y}_3\right\}\) There are exactly 3 element in A such. \(\mathrm{f}(\mathrm{x})=\mathrm{y}_2\) No. of ways to select 3 elements \(={ }^7 \mathrm{C}_3\) Now the each remaining element in \(\mathrm{A}\) can be \(\left(\mathrm{y}_1\right.\) or \(\left.\mathrm{y}_3\right)\) \(\Rightarrow 2 \times 2 \times 2 \times 2 \rightarrow\) (left 4 element each have two choice) \(\Rightarrow 16\) But \(\rightarrow 16-2=14\) ways Total No. of ways \(={ }^7 \mathrm{C}_3 \times 14\) \(=\frac{7 !}{3 ! \times 4 !} \times 14\) \(=490 \text { ways }\)
AMU-2019
Sets, Relation and Function
117060
Which of the following function is injective map?
: For option (a) \(f(x)=x^2+2, x \in(-\infty, \infty)\) \(\mathrm{f}(1)=\mathrm{f}(-1)\) but \(1 \neq-1\) (Non injective) For option (b) \(f(x)=|x+2|, x \in[-2, \infty)]\) Let, \(f(x)=f(y), x, y \in[-2, \infty)\) \(|x+2|=|y+2|\) \(x+2=y+2\) \(\mathrm{x}=\mathrm{y}\) (injective function) For option (c) \(f(x)=(x-4)(x-5) x \in(-\infty, \infty)\) \(f(4)=f(5)\) but \(4 \neq 5\) (Non injective) For option (d) \(f(x)=\frac{4 x^2+3 x-5}{4+3 x-5 x^2} x \in(-\infty, \infty)\) \(\mathrm{f}(1)=\mathrm{f}(-1)\) but \(1 \neq-1\) (Non injective) So, option (b) is an injectivefunction.
AMU-2021
Sets, Relation and Function
117062
The set of zeros of the function \(f(x)=0\) is nonempty, when \(f(x)\) equals
1 \(e^{-\mathrm{x}}+\mathrm{x}\)
2 \(|\mathrm{x}|+(\mathrm{x}-2)^2\)
3 \(x-\ln x\)
4 \(x+e^x\)
Explanation:
B For option (b) \(f(x)=|x|+(x-2)^2\) \(\text { Put } f(x)=0 \text { for } x>0\) \(x+x^2+4-4 x=0\) \(x^2-3 x+4=0\) \(x=\frac{3 \pm \sqrt{9-4 \times 4}}{2}=\frac{3 \pm \sqrt{-7}}{2}\) No real zeros. For \(\mathrm{x}\lt 0\), then \(\mathrm{f}(\mathrm{x})=-\mathrm{x}+(\mathrm{x}-2)^2\) \(\text { Put } f(x)=0\) \(-x+x^2+4-4 x=0\) \(x^2-4 x-x+4=0\) \(x(x-4)-1(x-4)=0\) \((x-1)(x-4)=0\) \(x=1,4\) Therefore, set of zero in non empty when \(\mathrm{f}(\mathrm{x})=|\mathrm{x}|+(\mathrm{x}-2)^2\)
117058
Let \(A=\left(x_1, x_2, x_3, \ldots . ., x_7\right), B=\left(y_1, y_2, y_3\right)\). The total number of functions \(f: \mathrm{A} \rightarrow \mathrm{B}\) that are onto and there are exactly three elements \(x\) in A such that \(f(x)=y_2\) is equal to
1 490
2 510
3 630
4 None of these
Explanation:
A Given, \(\mathrm{A}=\left\{\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4, \mathrm{x}_5, \mathrm{x}_6, \mathrm{x}_7\right\}\) And, \(\mathrm{B}=\left\{\mathrm{y}_1, \mathrm{y}_2, \mathrm{y}_3\right\}\) There are exactly 3 element in A such. \(\mathrm{f}(\mathrm{x})=\mathrm{y}_2\) No. of ways to select 3 elements \(={ }^7 \mathrm{C}_3\) Now the each remaining element in \(\mathrm{A}\) can be \(\left(\mathrm{y}_1\right.\) or \(\left.\mathrm{y}_3\right)\) \(\Rightarrow 2 \times 2 \times 2 \times 2 \rightarrow\) (left 4 element each have two choice) \(\Rightarrow 16\) But \(\rightarrow 16-2=14\) ways Total No. of ways \(={ }^7 \mathrm{C}_3 \times 14\) \(=\frac{7 !}{3 ! \times 4 !} \times 14\) \(=490 \text { ways }\)
AMU-2019
Sets, Relation and Function
117060
Which of the following function is injective map?
: For option (a) \(f(x)=x^2+2, x \in(-\infty, \infty)\) \(\mathrm{f}(1)=\mathrm{f}(-1)\) but \(1 \neq-1\) (Non injective) For option (b) \(f(x)=|x+2|, x \in[-2, \infty)]\) Let, \(f(x)=f(y), x, y \in[-2, \infty)\) \(|x+2|=|y+2|\) \(x+2=y+2\) \(\mathrm{x}=\mathrm{y}\) (injective function) For option (c) \(f(x)=(x-4)(x-5) x \in(-\infty, \infty)\) \(f(4)=f(5)\) but \(4 \neq 5\) (Non injective) For option (d) \(f(x)=\frac{4 x^2+3 x-5}{4+3 x-5 x^2} x \in(-\infty, \infty)\) \(\mathrm{f}(1)=\mathrm{f}(-1)\) but \(1 \neq-1\) (Non injective) So, option (b) is an injectivefunction.
AMU-2021
Sets, Relation and Function
117062
The set of zeros of the function \(f(x)=0\) is nonempty, when \(f(x)\) equals
1 \(e^{-\mathrm{x}}+\mathrm{x}\)
2 \(|\mathrm{x}|+(\mathrm{x}-2)^2\)
3 \(x-\ln x\)
4 \(x+e^x\)
Explanation:
B For option (b) \(f(x)=|x|+(x-2)^2\) \(\text { Put } f(x)=0 \text { for } x>0\) \(x+x^2+4-4 x=0\) \(x^2-3 x+4=0\) \(x=\frac{3 \pm \sqrt{9-4 \times 4}}{2}=\frac{3 \pm \sqrt{-7}}{2}\) No real zeros. For \(\mathrm{x}\lt 0\), then \(\mathrm{f}(\mathrm{x})=-\mathrm{x}+(\mathrm{x}-2)^2\) \(\text { Put } f(x)=0\) \(-x+x^2+4-4 x=0\) \(x^2-4 x-x+4=0\) \(x(x-4)-1(x-4)=0\) \((x-1)(x-4)=0\) \(x=1,4\) Therefore, set of zero in non empty when \(\mathrm{f}(\mathrm{x})=|\mathrm{x}|+(\mathrm{x}-2)^2\)
117058
Let \(A=\left(x_1, x_2, x_3, \ldots . ., x_7\right), B=\left(y_1, y_2, y_3\right)\). The total number of functions \(f: \mathrm{A} \rightarrow \mathrm{B}\) that are onto and there are exactly three elements \(x\) in A such that \(f(x)=y_2\) is equal to
1 490
2 510
3 630
4 None of these
Explanation:
A Given, \(\mathrm{A}=\left\{\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4, \mathrm{x}_5, \mathrm{x}_6, \mathrm{x}_7\right\}\) And, \(\mathrm{B}=\left\{\mathrm{y}_1, \mathrm{y}_2, \mathrm{y}_3\right\}\) There are exactly 3 element in A such. \(\mathrm{f}(\mathrm{x})=\mathrm{y}_2\) No. of ways to select 3 elements \(={ }^7 \mathrm{C}_3\) Now the each remaining element in \(\mathrm{A}\) can be \(\left(\mathrm{y}_1\right.\) or \(\left.\mathrm{y}_3\right)\) \(\Rightarrow 2 \times 2 \times 2 \times 2 \rightarrow\) (left 4 element each have two choice) \(\Rightarrow 16\) But \(\rightarrow 16-2=14\) ways Total No. of ways \(={ }^7 \mathrm{C}_3 \times 14\) \(=\frac{7 !}{3 ! \times 4 !} \times 14\) \(=490 \text { ways }\)
AMU-2019
Sets, Relation and Function
117060
Which of the following function is injective map?
: For option (a) \(f(x)=x^2+2, x \in(-\infty, \infty)\) \(\mathrm{f}(1)=\mathrm{f}(-1)\) but \(1 \neq-1\) (Non injective) For option (b) \(f(x)=|x+2|, x \in[-2, \infty)]\) Let, \(f(x)=f(y), x, y \in[-2, \infty)\) \(|x+2|=|y+2|\) \(x+2=y+2\) \(\mathrm{x}=\mathrm{y}\) (injective function) For option (c) \(f(x)=(x-4)(x-5) x \in(-\infty, \infty)\) \(f(4)=f(5)\) but \(4 \neq 5\) (Non injective) For option (d) \(f(x)=\frac{4 x^2+3 x-5}{4+3 x-5 x^2} x \in(-\infty, \infty)\) \(\mathrm{f}(1)=\mathrm{f}(-1)\) but \(1 \neq-1\) (Non injective) So, option (b) is an injectivefunction.
AMU-2021
Sets, Relation and Function
117062
The set of zeros of the function \(f(x)=0\) is nonempty, when \(f(x)\) equals
1 \(e^{-\mathrm{x}}+\mathrm{x}\)
2 \(|\mathrm{x}|+(\mathrm{x}-2)^2\)
3 \(x-\ln x\)
4 \(x+e^x\)
Explanation:
B For option (b) \(f(x)=|x|+(x-2)^2\) \(\text { Put } f(x)=0 \text { for } x>0\) \(x+x^2+4-4 x=0\) \(x^2-3 x+4=0\) \(x=\frac{3 \pm \sqrt{9-4 \times 4}}{2}=\frac{3 \pm \sqrt{-7}}{2}\) No real zeros. For \(\mathrm{x}\lt 0\), then \(\mathrm{f}(\mathrm{x})=-\mathrm{x}+(\mathrm{x}-2)^2\) \(\text { Put } f(x)=0\) \(-x+x^2+4-4 x=0\) \(x^2-4 x-x+4=0\) \(x(x-4)-1(x-4)=0\) \((x-1)(x-4)=0\) \(x=1,4\) Therefore, set of zero in non empty when \(\mathrm{f}(\mathrm{x})=|\mathrm{x}|+(\mathrm{x}-2)^2\)
117058
Let \(A=\left(x_1, x_2, x_3, \ldots . ., x_7\right), B=\left(y_1, y_2, y_3\right)\). The total number of functions \(f: \mathrm{A} \rightarrow \mathrm{B}\) that are onto and there are exactly three elements \(x\) in A such that \(f(x)=y_2\) is equal to
1 490
2 510
3 630
4 None of these
Explanation:
A Given, \(\mathrm{A}=\left\{\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4, \mathrm{x}_5, \mathrm{x}_6, \mathrm{x}_7\right\}\) And, \(\mathrm{B}=\left\{\mathrm{y}_1, \mathrm{y}_2, \mathrm{y}_3\right\}\) There are exactly 3 element in A such. \(\mathrm{f}(\mathrm{x})=\mathrm{y}_2\) No. of ways to select 3 elements \(={ }^7 \mathrm{C}_3\) Now the each remaining element in \(\mathrm{A}\) can be \(\left(\mathrm{y}_1\right.\) or \(\left.\mathrm{y}_3\right)\) \(\Rightarrow 2 \times 2 \times 2 \times 2 \rightarrow\) (left 4 element each have two choice) \(\Rightarrow 16\) But \(\rightarrow 16-2=14\) ways Total No. of ways \(={ }^7 \mathrm{C}_3 \times 14\) \(=\frac{7 !}{3 ! \times 4 !} \times 14\) \(=490 \text { ways }\)
AMU-2019
Sets, Relation and Function
117060
Which of the following function is injective map?
: For option (a) \(f(x)=x^2+2, x \in(-\infty, \infty)\) \(\mathrm{f}(1)=\mathrm{f}(-1)\) but \(1 \neq-1\) (Non injective) For option (b) \(f(x)=|x+2|, x \in[-2, \infty)]\) Let, \(f(x)=f(y), x, y \in[-2, \infty)\) \(|x+2|=|y+2|\) \(x+2=y+2\) \(\mathrm{x}=\mathrm{y}\) (injective function) For option (c) \(f(x)=(x-4)(x-5) x \in(-\infty, \infty)\) \(f(4)=f(5)\) but \(4 \neq 5\) (Non injective) For option (d) \(f(x)=\frac{4 x^2+3 x-5}{4+3 x-5 x^2} x \in(-\infty, \infty)\) \(\mathrm{f}(1)=\mathrm{f}(-1)\) but \(1 \neq-1\) (Non injective) So, option (b) is an injectivefunction.
AMU-2021
Sets, Relation and Function
117062
The set of zeros of the function \(f(x)=0\) is nonempty, when \(f(x)\) equals
1 \(e^{-\mathrm{x}}+\mathrm{x}\)
2 \(|\mathrm{x}|+(\mathrm{x}-2)^2\)
3 \(x-\ln x\)
4 \(x+e^x\)
Explanation:
B For option (b) \(f(x)=|x|+(x-2)^2\) \(\text { Put } f(x)=0 \text { for } x>0\) \(x+x^2+4-4 x=0\) \(x^2-3 x+4=0\) \(x=\frac{3 \pm \sqrt{9-4 \times 4}}{2}=\frac{3 \pm \sqrt{-7}}{2}\) No real zeros. For \(\mathrm{x}\lt 0\), then \(\mathrm{f}(\mathrm{x})=-\mathrm{x}+(\mathrm{x}-2)^2\) \(\text { Put } f(x)=0\) \(-x+x^2+4-4 x=0\) \(x^2-4 x-x+4=0\) \(x(x-4)-1(x-4)=0\) \((x-1)(x-4)=0\) \(x=1,4\) Therefore, set of zero in non empty when \(\mathrm{f}(\mathrm{x})=|\mathrm{x}|+(\mathrm{x}-2)^2\)