Here you will get NCERT Solutions for Class 12 Maths Chapter 1 – Exercise 1.2 in the text as well as pdf format. As we all know that NCERT books are very helpful in Class 12 CBSE Board. Here we have given question from Chapter 1 of Class 12th Maths subject along with their respective answers.

NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 Overview | |
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Book | National Council of Educational Research and Training (NCERT) |

Class | 12th Class |

Subject | NCERT Solution Class 12th Maths |

Chapter | Chapter 1 – Relations and Functions |

## Chapter 1: Relations and Functions – Exercise 1.2

**Q-1: Prove that the function f: R∗→R∗ which is defined by f(a) = 1a which is one- one and onto, where R∗ is the set of all of the non- zero real numbers. Check whether the result is true or not, if the domain, say, R∗ is replaced by M having co- domain as same as R∗?**

**Solutions**

**Q-2: Check the following functions for their injectivity and surjectivity:**

**(i) f: N →N which is given by f(a) = a ^{2}**

**(ii) f: Z →Z which is given by f(a) = a ^{2}**

**(iii) f: R → R which is given by f(a) = a ^{2}**

**(iv) f: N →N which is given by f(a) = a ^{3}**

**(v) f: Z → Z which is given by f(a) = a ^{3}**

**Solutions**

**Q-3: Show that the GIF (Greatest Integer Function) f: R →R which is given by f(a) = [a], which is neither one- one nor onto, where [a] notifies the greatest integer which must be less than or equal to a.**

**Solutions**

**Q-4: Prove that the Modulus Function f: R → R which is given by f(a) = [a], which is neither one- one nor it is onto, where |a| is a, if a will be positive or 0 and |a| is –a, if a will be negative.**

**Solutions**

**Q-5: Prove that the Signum function f: R → R which is given by f(a) =**

**1, if a > 0**

**0, if a = 0**

**-1, if x < 0 is neither onto nor one- one.**

**Solutions**

**Q-6: Let us take P as {2, 3, 4}, Q as {5, 6, 7, 8} and let f = {(2, 5), (3, 6), (4, 7)} which are the function from P and Q. Prove that the function f is one- one.**

**Solution**

**Q-7. Check whether the function in each of the following conditions is one- one, onto or bijective. Justify your answer in each cases.**

**(i) f : R → R which is defines by f(a) = 4 – 5a**

**(ii) f : R → R which is defined by f(a) = 2 + a ^{2}**

**Solution**

**Q-8: Let us consider P and Q be the two sets. Prove that f : P × Q → Q × P so that (x, y) = (y, x) is bijective function.**

**Solution**

**Q-9: Let f : N → N which will be defined by f(p) =**

**p+1/2, if p is odd**

**p2, if p is even, for all p ɛ N.**

**Check and justify your answer that whether the function f is bijective or not?**

**Solution**

**Q-10: Let P = A – {3} and Q = A – {1}. Let us consider the function f : P → Q which defined by f(a) =(a–3/a–4). Check and Justify your answer, whether the function f is onto and one- one?**

**Solution**

**Q-11: Let us assume, f : R → R will be defined as g(a) = a ^{5}. Select the correct answer from the following:**

**(a) f is many- one onto (b) f is one- one onto**

**(c) f is neither one- one nor onto (d) f is one- one but, not onto**

**Solution**

**Q-12: Let us assume, f : R → R will be defined as g(a) = 4a . Select the correct answer from the following:**

**(a) f is many- one onto (b) f is one- one onto**

**(c) f is neither one- one nor onto (d) f is one- one but, not onto**

**Solution**

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