117079
The mapping \(f: N \rightarrow N\) given by \(f(n)=n^3+3, n\) \(\in N\) where \(N\) is the set of natural number, is
1 One to one and onto
2 One to one but not onto
3 Onto but not one to one
4 Neither one to one nor onto
Explanation:
B Given, \(\mathrm{f}(\mathrm{n})=\mathrm{x}^3+3\) \(\text { If } \mathrm{n}=1 \text {, value of } \mathrm{f}(\mathrm{n})=4\) \(\text { If } \mathrm{n}=2 \text {, value of } \mathrm{f}(\mathrm{n})=11\) \(\text { If } \mathrm{n}=3 \text {, value of } \mathrm{f}(\mathrm{n})=30\) They \(f(n)\) is one - one but clearly range is not equal to co-domain so, \(\mathrm{f}(\mathrm{n})\) is not onto.
J and K CET-2017
Sets, Relation and Function
117080
Which of the following functions is neither even nor odd ?
1 \(f(x)=5 x+\sin (4 x)\)
2 \(f(x)=4 x^3+7 \tan x\)
3 \(f(x)=7 x^4+8 x^2-6 x\)
4 \(f(x)=5 x^2+\cos (6 x)\)
Explanation:
C By option (a) \(f(-x)=-5 x+\sin (-4 x)\) \(=-(5 x+\sin 4 x)=-f(x)\) So, \(\quad \mathrm{f}(\mathrm{x})\) is odd (b) \(\mathrm{f}(-\mathrm{x})=-4 \mathrm{x}^3+7 \tan (-\mathrm{x})\) \(=-\left(4 \mathrm{x}^3+7 \tan \mathrm{x}\right)=-\mathrm{f}(\mathrm{x})\) So, \(\quad \mathrm{f}(\mathrm{x})\) is odd. (c) \(\quad \mathrm{f}(-\mathrm{x})=7 \mathrm{x}^4+8 \mathrm{x}^2+6 \mathrm{x} \neq \mathrm{f}(\mathrm{x})\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is neither even nor odd (d) \(\quad \mathrm{f}(-\mathrm{x})=5 \mathrm{x}^2+\cos (-6 \mathrm{x})=5 \mathrm{x}^2+\cos 6 \mathrm{x}=\mathrm{f}(\mathrm{x})\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is even function
J and K CET-2018
Sets, Relation and Function
117081
The function \(f: R \rightarrow R\) given by \(f(x)=x^2, x \in R\), is
1 one to one but not onto
2 not one to one but onto
3 both one to one and onto
4 neither one to one nor onto
Explanation:
D One-one function or iniective function \(\Rightarrow\) A function \(\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}\) is said to be a one-one function, if different elements in a have different images or associated with different element in B i.e if. \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) \Rightarrow \mathrm{x}_1=\mathrm{x}_2 \forall \mathrm{x}_1 / \mathrm{x}_2 \in \mathbf{A} .\) Onto function/ suriective function \(\Rightarrow\) Any function \(\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}\) is said to be onto if every element in \(B\) has atleast one pre- image in \(A\). Explanation- \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) \(f(x)=x^2\) \(\text { For } \mathrm{n}=-2, \mathrm{f}(\mathrm{x})=4\) \(\mathrm{n}=+2, \mathrm{f}(\mathrm{x})=4\) So, \(f(x)\) is a non - injective (many one) function for \(f(x)\) \(=-5\) no \((\mathrm{x})\) exist is real set. So, \(f(x)\) is a non- surfective function or onto function.
J and K CET-2016
Sets, Relation and Function
117082
The function \(f(x)=\frac{1}{2-\cos 3 x}, x \in\left[0, \frac{\pi}{3}\right]\) is
1 one one, but not onto
2 onto, but not one one
3 one to one as well as onto
4 neither one to one nor onto
Explanation:
C Given, \(f(x)=\frac{1}{2-\cos 3 x}, x \in\left[0, \frac{\pi}{3}\right]\) For one - one let \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) \(\frac{1}{2-\cos 3 x_1}=\frac{1}{2-\cos 3 x_2}\) \(2-\cos 3 x_1=2-\cos 3 x_2\) \(\cos 3 x_1=\cos 3 x_2\) \(x_1=x_2\) \(f\) is one-one for onto Let \(y=f(x), y \in\) codomain. \(y=\frac{1}{2-\cos 3 x}\) \(y(2-\cos 3 x)=1\) \(2-\cos 3 x=\frac{1}{y}\) \(\cos 3 x=2-\frac{1}{y}\) \(x=\frac{1}{3} \cos ^{-1}\left(2-\frac{1}{y}\right)\) Here, for all \(\mathrm{y} \in\). co domain there exist \(\mathrm{x} \in\) codomain, so \(\mathrm{f}(\mathrm{x})\) is onto. Here for all \(\mathrm{y} \in\) co domain there exist \(\mathrm{x} \in\) domain. So, \(\mathrm{f}(\mathrm{x})\) is onto.
117079
The mapping \(f: N \rightarrow N\) given by \(f(n)=n^3+3, n\) \(\in N\) where \(N\) is the set of natural number, is
1 One to one and onto
2 One to one but not onto
3 Onto but not one to one
4 Neither one to one nor onto
Explanation:
B Given, \(\mathrm{f}(\mathrm{n})=\mathrm{x}^3+3\) \(\text { If } \mathrm{n}=1 \text {, value of } \mathrm{f}(\mathrm{n})=4\) \(\text { If } \mathrm{n}=2 \text {, value of } \mathrm{f}(\mathrm{n})=11\) \(\text { If } \mathrm{n}=3 \text {, value of } \mathrm{f}(\mathrm{n})=30\) They \(f(n)\) is one - one but clearly range is not equal to co-domain so, \(\mathrm{f}(\mathrm{n})\) is not onto.
J and K CET-2017
Sets, Relation and Function
117080
Which of the following functions is neither even nor odd ?
1 \(f(x)=5 x+\sin (4 x)\)
2 \(f(x)=4 x^3+7 \tan x\)
3 \(f(x)=7 x^4+8 x^2-6 x\)
4 \(f(x)=5 x^2+\cos (6 x)\)
Explanation:
C By option (a) \(f(-x)=-5 x+\sin (-4 x)\) \(=-(5 x+\sin 4 x)=-f(x)\) So, \(\quad \mathrm{f}(\mathrm{x})\) is odd (b) \(\mathrm{f}(-\mathrm{x})=-4 \mathrm{x}^3+7 \tan (-\mathrm{x})\) \(=-\left(4 \mathrm{x}^3+7 \tan \mathrm{x}\right)=-\mathrm{f}(\mathrm{x})\) So, \(\quad \mathrm{f}(\mathrm{x})\) is odd. (c) \(\quad \mathrm{f}(-\mathrm{x})=7 \mathrm{x}^4+8 \mathrm{x}^2+6 \mathrm{x} \neq \mathrm{f}(\mathrm{x})\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is neither even nor odd (d) \(\quad \mathrm{f}(-\mathrm{x})=5 \mathrm{x}^2+\cos (-6 \mathrm{x})=5 \mathrm{x}^2+\cos 6 \mathrm{x}=\mathrm{f}(\mathrm{x})\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is even function
J and K CET-2018
Sets, Relation and Function
117081
The function \(f: R \rightarrow R\) given by \(f(x)=x^2, x \in R\), is
1 one to one but not onto
2 not one to one but onto
3 both one to one and onto
4 neither one to one nor onto
Explanation:
D One-one function or iniective function \(\Rightarrow\) A function \(\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}\) is said to be a one-one function, if different elements in a have different images or associated with different element in B i.e if. \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) \Rightarrow \mathrm{x}_1=\mathrm{x}_2 \forall \mathrm{x}_1 / \mathrm{x}_2 \in \mathbf{A} .\) Onto function/ suriective function \(\Rightarrow\) Any function \(\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}\) is said to be onto if every element in \(B\) has atleast one pre- image in \(A\). Explanation- \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) \(f(x)=x^2\) \(\text { For } \mathrm{n}=-2, \mathrm{f}(\mathrm{x})=4\) \(\mathrm{n}=+2, \mathrm{f}(\mathrm{x})=4\) So, \(f(x)\) is a non - injective (many one) function for \(f(x)\) \(=-5\) no \((\mathrm{x})\) exist is real set. So, \(f(x)\) is a non- surfective function or onto function.
J and K CET-2016
Sets, Relation and Function
117082
The function \(f(x)=\frac{1}{2-\cos 3 x}, x \in\left[0, \frac{\pi}{3}\right]\) is
1 one one, but not onto
2 onto, but not one one
3 one to one as well as onto
4 neither one to one nor onto
Explanation:
C Given, \(f(x)=\frac{1}{2-\cos 3 x}, x \in\left[0, \frac{\pi}{3}\right]\) For one - one let \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) \(\frac{1}{2-\cos 3 x_1}=\frac{1}{2-\cos 3 x_2}\) \(2-\cos 3 x_1=2-\cos 3 x_2\) \(\cos 3 x_1=\cos 3 x_2\) \(x_1=x_2\) \(f\) is one-one for onto Let \(y=f(x), y \in\) codomain. \(y=\frac{1}{2-\cos 3 x}\) \(y(2-\cos 3 x)=1\) \(2-\cos 3 x=\frac{1}{y}\) \(\cos 3 x=2-\frac{1}{y}\) \(x=\frac{1}{3} \cos ^{-1}\left(2-\frac{1}{y}\right)\) Here, for all \(\mathrm{y} \in\). co domain there exist \(\mathrm{x} \in\) codomain, so \(\mathrm{f}(\mathrm{x})\) is onto. Here for all \(\mathrm{y} \in\) co domain there exist \(\mathrm{x} \in\) domain. So, \(\mathrm{f}(\mathrm{x})\) is onto.
117079
The mapping \(f: N \rightarrow N\) given by \(f(n)=n^3+3, n\) \(\in N\) where \(N\) is the set of natural number, is
1 One to one and onto
2 One to one but not onto
3 Onto but not one to one
4 Neither one to one nor onto
Explanation:
B Given, \(\mathrm{f}(\mathrm{n})=\mathrm{x}^3+3\) \(\text { If } \mathrm{n}=1 \text {, value of } \mathrm{f}(\mathrm{n})=4\) \(\text { If } \mathrm{n}=2 \text {, value of } \mathrm{f}(\mathrm{n})=11\) \(\text { If } \mathrm{n}=3 \text {, value of } \mathrm{f}(\mathrm{n})=30\) They \(f(n)\) is one - one but clearly range is not equal to co-domain so, \(\mathrm{f}(\mathrm{n})\) is not onto.
J and K CET-2017
Sets, Relation and Function
117080
Which of the following functions is neither even nor odd ?
1 \(f(x)=5 x+\sin (4 x)\)
2 \(f(x)=4 x^3+7 \tan x\)
3 \(f(x)=7 x^4+8 x^2-6 x\)
4 \(f(x)=5 x^2+\cos (6 x)\)
Explanation:
C By option (a) \(f(-x)=-5 x+\sin (-4 x)\) \(=-(5 x+\sin 4 x)=-f(x)\) So, \(\quad \mathrm{f}(\mathrm{x})\) is odd (b) \(\mathrm{f}(-\mathrm{x})=-4 \mathrm{x}^3+7 \tan (-\mathrm{x})\) \(=-\left(4 \mathrm{x}^3+7 \tan \mathrm{x}\right)=-\mathrm{f}(\mathrm{x})\) So, \(\quad \mathrm{f}(\mathrm{x})\) is odd. (c) \(\quad \mathrm{f}(-\mathrm{x})=7 \mathrm{x}^4+8 \mathrm{x}^2+6 \mathrm{x} \neq \mathrm{f}(\mathrm{x})\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is neither even nor odd (d) \(\quad \mathrm{f}(-\mathrm{x})=5 \mathrm{x}^2+\cos (-6 \mathrm{x})=5 \mathrm{x}^2+\cos 6 \mathrm{x}=\mathrm{f}(\mathrm{x})\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is even function
J and K CET-2018
Sets, Relation and Function
117081
The function \(f: R \rightarrow R\) given by \(f(x)=x^2, x \in R\), is
1 one to one but not onto
2 not one to one but onto
3 both one to one and onto
4 neither one to one nor onto
Explanation:
D One-one function or iniective function \(\Rightarrow\) A function \(\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}\) is said to be a one-one function, if different elements in a have different images or associated with different element in B i.e if. \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) \Rightarrow \mathrm{x}_1=\mathrm{x}_2 \forall \mathrm{x}_1 / \mathrm{x}_2 \in \mathbf{A} .\) Onto function/ suriective function \(\Rightarrow\) Any function \(\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}\) is said to be onto if every element in \(B\) has atleast one pre- image in \(A\). Explanation- \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) \(f(x)=x^2\) \(\text { For } \mathrm{n}=-2, \mathrm{f}(\mathrm{x})=4\) \(\mathrm{n}=+2, \mathrm{f}(\mathrm{x})=4\) So, \(f(x)\) is a non - injective (many one) function for \(f(x)\) \(=-5\) no \((\mathrm{x})\) exist is real set. So, \(f(x)\) is a non- surfective function or onto function.
J and K CET-2016
Sets, Relation and Function
117082
The function \(f(x)=\frac{1}{2-\cos 3 x}, x \in\left[0, \frac{\pi}{3}\right]\) is
1 one one, but not onto
2 onto, but not one one
3 one to one as well as onto
4 neither one to one nor onto
Explanation:
C Given, \(f(x)=\frac{1}{2-\cos 3 x}, x \in\left[0, \frac{\pi}{3}\right]\) For one - one let \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) \(\frac{1}{2-\cos 3 x_1}=\frac{1}{2-\cos 3 x_2}\) \(2-\cos 3 x_1=2-\cos 3 x_2\) \(\cos 3 x_1=\cos 3 x_2\) \(x_1=x_2\) \(f\) is one-one for onto Let \(y=f(x), y \in\) codomain. \(y=\frac{1}{2-\cos 3 x}\) \(y(2-\cos 3 x)=1\) \(2-\cos 3 x=\frac{1}{y}\) \(\cos 3 x=2-\frac{1}{y}\) \(x=\frac{1}{3} \cos ^{-1}\left(2-\frac{1}{y}\right)\) Here, for all \(\mathrm{y} \in\). co domain there exist \(\mathrm{x} \in\) codomain, so \(\mathrm{f}(\mathrm{x})\) is onto. Here for all \(\mathrm{y} \in\) co domain there exist \(\mathrm{x} \in\) domain. So, \(\mathrm{f}(\mathrm{x})\) is onto.
117079
The mapping \(f: N \rightarrow N\) given by \(f(n)=n^3+3, n\) \(\in N\) where \(N\) is the set of natural number, is
1 One to one and onto
2 One to one but not onto
3 Onto but not one to one
4 Neither one to one nor onto
Explanation:
B Given, \(\mathrm{f}(\mathrm{n})=\mathrm{x}^3+3\) \(\text { If } \mathrm{n}=1 \text {, value of } \mathrm{f}(\mathrm{n})=4\) \(\text { If } \mathrm{n}=2 \text {, value of } \mathrm{f}(\mathrm{n})=11\) \(\text { If } \mathrm{n}=3 \text {, value of } \mathrm{f}(\mathrm{n})=30\) They \(f(n)\) is one - one but clearly range is not equal to co-domain so, \(\mathrm{f}(\mathrm{n})\) is not onto.
J and K CET-2017
Sets, Relation and Function
117080
Which of the following functions is neither even nor odd ?
1 \(f(x)=5 x+\sin (4 x)\)
2 \(f(x)=4 x^3+7 \tan x\)
3 \(f(x)=7 x^4+8 x^2-6 x\)
4 \(f(x)=5 x^2+\cos (6 x)\)
Explanation:
C By option (a) \(f(-x)=-5 x+\sin (-4 x)\) \(=-(5 x+\sin 4 x)=-f(x)\) So, \(\quad \mathrm{f}(\mathrm{x})\) is odd (b) \(\mathrm{f}(-\mathrm{x})=-4 \mathrm{x}^3+7 \tan (-\mathrm{x})\) \(=-\left(4 \mathrm{x}^3+7 \tan \mathrm{x}\right)=-\mathrm{f}(\mathrm{x})\) So, \(\quad \mathrm{f}(\mathrm{x})\) is odd. (c) \(\quad \mathrm{f}(-\mathrm{x})=7 \mathrm{x}^4+8 \mathrm{x}^2+6 \mathrm{x} \neq \mathrm{f}(\mathrm{x})\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is neither even nor odd (d) \(\quad \mathrm{f}(-\mathrm{x})=5 \mathrm{x}^2+\cos (-6 \mathrm{x})=5 \mathrm{x}^2+\cos 6 \mathrm{x}=\mathrm{f}(\mathrm{x})\) \(\therefore \quad \mathrm{f}(\mathrm{x})\) is even function
J and K CET-2018
Sets, Relation and Function
117081
The function \(f: R \rightarrow R\) given by \(f(x)=x^2, x \in R\), is
1 one to one but not onto
2 not one to one but onto
3 both one to one and onto
4 neither one to one nor onto
Explanation:
D One-one function or iniective function \(\Rightarrow\) A function \(\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}\) is said to be a one-one function, if different elements in a have different images or associated with different element in B i.e if. \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) \Rightarrow \mathrm{x}_1=\mathrm{x}_2 \forall \mathrm{x}_1 / \mathrm{x}_2 \in \mathbf{A} .\) Onto function/ suriective function \(\Rightarrow\) Any function \(\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}\) is said to be onto if every element in \(B\) has atleast one pre- image in \(A\). Explanation- \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) \(f(x)=x^2\) \(\text { For } \mathrm{n}=-2, \mathrm{f}(\mathrm{x})=4\) \(\mathrm{n}=+2, \mathrm{f}(\mathrm{x})=4\) So, \(f(x)\) is a non - injective (many one) function for \(f(x)\) \(=-5\) no \((\mathrm{x})\) exist is real set. So, \(f(x)\) is a non- surfective function or onto function.
J and K CET-2016
Sets, Relation and Function
117082
The function \(f(x)=\frac{1}{2-\cos 3 x}, x \in\left[0, \frac{\pi}{3}\right]\) is
1 one one, but not onto
2 onto, but not one one
3 one to one as well as onto
4 neither one to one nor onto
Explanation:
C Given, \(f(x)=\frac{1}{2-\cos 3 x}, x \in\left[0, \frac{\pi}{3}\right]\) For one - one let \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) \(\frac{1}{2-\cos 3 x_1}=\frac{1}{2-\cos 3 x_2}\) \(2-\cos 3 x_1=2-\cos 3 x_2\) \(\cos 3 x_1=\cos 3 x_2\) \(x_1=x_2\) \(f\) is one-one for onto Let \(y=f(x), y \in\) codomain. \(y=\frac{1}{2-\cos 3 x}\) \(y(2-\cos 3 x)=1\) \(2-\cos 3 x=\frac{1}{y}\) \(\cos 3 x=2-\frac{1}{y}\) \(x=\frac{1}{3} \cos ^{-1}\left(2-\frac{1}{y}\right)\) Here, for all \(\mathrm{y} \in\). co domain there exist \(\mathrm{x} \in\) codomain, so \(\mathrm{f}(\mathrm{x})\) is onto. Here for all \(\mathrm{y} \in\) co domain there exist \(\mathrm{x} \in\) domain. So, \(\mathrm{f}(\mathrm{x})\) is onto.
