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Chapter | 1. Relations and Functions |

Exercise | 1.2 |

Class | Twelve |

Subject | Maths |

Category | NCERT Solutions for Class 12 |

NCERT Solutions for Class 12 Maths are available online for students to use and refer to while studying for their exams. The solutions provide step-by-step explanations for each problem in the textbook, making it easy for students to understand the concepts. With Class 12 Maths NCERT Solutions, students can be sure that they are getting the most accurate and up-to-date information available. The solutions are regularly updated by experts in the field, so students can be confident that they are using the most current information.

## NCERT Solutions for Class 12 Maths Exercise 1.2

### Relations and Functions Solutions

**Q 1) Show that the function f : R**_{*} → R_{* }defined by f(x) = 1/x is one-one and onto, where : R_{* }is the set of all non-zero real numbers. Is the result true, if the domain : R_{* }is replaced by N with co-domain being same as : R_{*}?

_{*}→ R

_{* }defined by f(x) = 1/x is one-one and onto, where : R

_{* }is the set of all non-zero real numbers. Is the result true, if the domain : R

_{* }is replaced by N with co-domain being same as : R

_{*}?

**Answer)**

Hence, function *g* is one-one but not onto.

**Q 2) Check the injectivity and surjectivity of the following functions:**

**(i) f: N → N given by f(x) = x ^{2}**

**(ii) f: Z → Z given by f(x) = x**

^{2 }**(iii) f: R → R given by f(x) = x**

^{2}**(iv) f: N → N given by f(x) = x**

^{3}**(v) f: Z → Z given by f(x) = x**

^{3}**Answer)**

**(i)** f: N → N is given by,

f(x) = x^{2}

It is seen that for x, y ∈N, f(x) = f(y)

⇒ x^{2} = y^{2} ⇒ x = y.

∴f is injective.

Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x^{2} = 2.

∴ f is not surjective.

Hence, function f is injective but not surjective.

**(ii) **f: Z → Z is given by,

f(x) = x^{2}

It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.

∴ f is not injective.

Now,−2 ∈ Z. But, there does not exist any element x ∈Z such that f(x) = x^{2} = −2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

**(iii)** f: R → R is given by,

f(x) = x^{2}

It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.

∴ f is not injective.

Now,−2 ∈ R. But, there does not exist any element x ∈ R such that f(x) = x^{2} = −2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

**(iv)** f: N → N given by,

f(x) = x^{3}

It is seen that for x, y ∈N, f(x) = f(y)

⇒ x^{3} = y^{3} ⇒ x = y.

∴f is injective.

Now, 2 ∈ N. But, there does not exist any element x in domain N such that f(x) = x^{3 }= 2.

∴ f is not surjective

Hence, function f is injective but not surjective.

**(v)** f: Z → Z is given by,

f(x) = x^{3}

It is seen that for x, y ∈ Z, f(x) = f(y)

⇒ x^{3} = y^{3} ⇒ x = y.

∴ f is injective.

Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x^{3} = 2.

∴ f is not surjective.

Hence, function f is injective but not surjective.

**Q 3) Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.**

**Answer)**

f: R → R is given by,

f(x) = [x]

It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1.

∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9.

∴ f is not one-one.

Now, consider 0.7 ∈ R.

It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ∈ R such that f(x) = 0.7.

∴ f is not onto.

Hence, the greatest integer function is neither one-one nor onto.

**Q 4) Show that the Modulus Function f : R → R, given by f(x) = |x| is neither one-one nor onto, where is |x| if x is positive or 0 and |-x| is -x if x is negative.**

**Answer)**

**Q 5) Show that the Signum Function ***f*: R → R , given by

*f*: R → R , given by

** is neither one-one nor onto.**

**Answer)**

*f*: R → R

It is seen that f(1) = f(2) = 1, but 1 ≠ 2.

∴ f is not one-one.

Now, as f(x) ) takes only 3 values (1, 0, or −1) for the element −2 in co-domain R , there does not exist any x in domain R such that

f(x) = −2.

∴ f is not onto.

Hence, the signum function is neither one-one nor onto.

**Q 6) Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. **

**Show that f is one-one.**

**Answer)**

It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.

f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.

∴ f (1) = 4, f (2) = 5, f (3) = 6

It is seen that the images of distinct elements of A under f are distinct.

Hence, function f is one-one.

**Q 7) In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.**

**(i) f: R → R defined by f(x) = 3 − 4x****(ii) f: R → R defined by f(x) = 1 + x ^{2}**

**Answer)**

**(i)** *f*: **R **→ **R** is defined as *f*(*x*) = 3 − 4*x*.

∴ *f* is one-one.

For any real number (*y)* in **R**, there exists in **R** such that

∴*f *is onto.

Hence, *f *is bijective.

**(ii)** *f*: **R** →** R** is defined as

For instance,

∴ *f* is not one-one.

Consider an element −2 in co-domain **R**.

It is seen that is positive for all *x* ∈ **R**.

Thus, there does not exist any* x* in domain **R** such that *f* (*x*) = −2.

∴ *f* is not onto.

Hence, *f* is neither one-one nor onto.

**Q 8) Let A and B be sets. Show that f: A × B → B × A such that (a, b) = (b, a) is bijective function.**

**Answer)**

*f*: *A* × *B* → *B* × *A* is defined as *f*(*a*, *b*) = (*b*, *a*).

∴ *f* is one-one.

Now, let (*b*, *a*) ∈ *B* × *A* be any element.

Then, there exists (*a*, *b*) ∈*A* × *B* such that *f*(*a*, *b*) = (*b*, *a*). [By definition of *f*]

∴ *f* is onto.

Hence,* f* is bijective.

**Q 9) Let ***f*: N → N be defined by

*f*: N → N be defined by

**State whether the function f is bijective. Justify your answer.**

**Answer)** *f*: **N** → **N** is defined as

It can be observed that:

∴ *f* is not one-one.

Consider a natural number (*n)* in co-domain **N**.

Case **I: ***n* is odd

∴*n* = 2*r* + 1 for some *r* ∈ **N. **Then, there exists 4*r *+ 1∈**N** such that

Case **II: ***n* is even

∴*n* = 2*r* for some *r* ∈ **N. **Then,there exists 4*r* ∈**N** such that

∴ *f* is onto.

Hence, *f* is not a bijective function.

**Q 10) Let A = R − {3} and B = R − {1}. Consider the function ***f*: A → B defined by

*f*: A → B defined by

** Is f one-one and onto? Justify your answer.**

**Answer)**

A = **R** − {3}, B = **R** − {1}

*f*: A → B is defined as

∴ *f* is one-one.

Let *y* ∈B = **R** − {1}. Then, *y* ≠ 1.

The function *f *is onto if there exists *x* ∈A such that *f*(*x*) = *y.*

Now,

Thus, for any *y *∈ B, there exists such that

Hence, function* f* is one-one and onto.

**Q 11) Let ***f*: R → R be defined as* f*(*x*) = *x*^{4}. Choose the correct answer.

*f*: R → R be defined as

*f*(

*x*) =

*x*

^{4}. Choose the correct answer.

**(A) f is one-one onto (B) f is many-one onto**

(C) f is one-one but not onto (D) f is neither one-one nor onto

**Answer)** *f*: **R** → **R** is defined as *f*(*x*) = *x*^{4}

Let *x*, *y *∈ **R** such that *f*(*x*) = *f*(*y*).

∴ f (x_{1}) = f (x_{2}) does not imply that x_{1 }= x_{2}

For instance,

∴ *f* is not one-one.

Consider an element −-2 in co-domain **R**. It is clear that there does not exist any *x* in domain **R** such that* f*(*x*) = −-2.

∴ *f* is not onto.

Hence, function *f* is neither one-one nor onto.

The correct answer is D.

**Q 12) Let ***f*: R → R be defined as *f*(*x*) = 3*x*. Choose the correct answer.

*f*: R → R be defined as

*f*(

*x*) = 3

*x*. Choose the correct answer.

**(A) f is one-one onto (B) f is many-one onto**

(C) f is one-one but not onto (D) f is neither one-one nor onto

**Answer)**

*f*: **R** → **R** is defined as *f*(*x*) = 3*x*.

Let *x*, *y *∈ **R** such that *f*(*x*) = *f*(*y*).

⇒ 3*x* = 3*y*

⇒ *x* = *y*

∴*f *is one-one.

Also, for any real number (*y)* in co-domain **R**, there exists in **R** such that

∴*f *is onto.

Hence, function *f* is one-one and onto.

The correct answer is A.

That’s it. These were the solutions of NCERT Class 12 Maths Exercise 1.2. Our team hopes that you have found these solutions helpful for you. If you have any doubt related to this chapter then feel free to comment your doubts below. Our team will try their best to help you with your doubts.