With all our research we gather information on NCERT Solutions for Class 12 Maths Chapter 1 – Exercise 1.3. This solution will help CBSE Board candidate to solve Class 12th Maths subject of Exercise 1.3. Here we have given question from Relations and Functions Chapter of Class 12th Maths subject along with their respective answers.

NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.3 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 1.3

**Q-1: Let us consider, g : {3, 5, 6} → {2, 4, 7} and f : {2, 4, 7} → {3, 5} will begiven by g = {(3, 4), (5, 7), (6, 3)} and f = {(3, 5), (4, 5), (7, 3)}. Find fog.**

**Solution**

**Q-2: Let us consider**

**f, g and h be the functions from R to R. Prove that:**

**(g + f)oh = goh + foh**

**(g.f)oh = (goh).(foh)**

**Solution**

**Q-3: Find fog and gof, if**

**(i) g(a) = |a| and f(a) = |5a – 2|**

**(ii) g(a) = 8a ^{3} and f(a) = x^{1/3} **

**Solution**

**Q-4: If g(a) = (4a+3) / (6a–4), a ≠ 2/3. Prove that gog(a) = a, for every a ≠ 2/3. What will be the inverse of the function g?**

**Solution**

**Q-5: Explain with suitable reason that, whether the following functions has any inverse**

**(i) g : {2, 3, 4, 5} → {11} with**

**g = {(2, 11), (3, 11), (4, 11), (5, 11)}**

**(ii) f : {6, 7, 8, 9} → {2, 3, 4, 5} with**

**f = {(6, 5), (7, 4), (8, 5), (9, 3)}**

**(iii) h : {3, 4, 5, 6} → {8, 10, 12, 14} with**

**h = {(3, 8), (4, 10), (5, 12), (6, 14)}**

**Solution**

**Q-6: Prove that:**

**g : [-2, 2] → R,**

**which is given by g(a) = a/(a+2) will be one- one. What will be the inverse of the function g : [-2, 2] → Range g.**

**Solution**

**Q-7: Let us consider g : R → R which is given by g(a) = 4a + 3. Prove that the function g is invertible. Also find the inverse of g.**

**Solution**

**Q-8: Let us consider a function, g : R _{+} → [4, **

**) which is given by g(a) = a**

^{2}+ 4. Prove that the function g is invertible with the inverse g^{-1}of the given function g by g^{-1 }(b) = √b-4, where R_{+}will be the set of all the non- negative real number.**Solution**

**Q-9: Let us consider a function, g: R _{+ }→ [-5, ∞) which is given by g(a) = 9a^{2} + 6a – 5. Prove that, the function g is invertible with g^{-1}(b) = (√b+6)–1/3).**

**Solution**

**Q-10: Let us consider a function g : P → Q be an invertible function. Prove that the function g have a unique inverse**

**Solution**

**Q-11. Let us consider the function g: {2, 3, 4} → {x, y, z}, generally given by g(2) = x, g(3) = y, g(4) = z. Find g ^{-1} and prove that (g^{-1})^{-1} = g.**

**Solution**

**Q-12: Let us consider g : P → Q be an invertible function. Prove that the inverse of g ^{-1} is g, i.e., (g^{-1})^{-1} = g.**

**Solution**

**Q-13: If g : R → R is given by g(a) = (3–a ^{3})^{1/3}, then gog(a) is**

**(a) 1 / a ^{3}**

**(b) a ^{3}**

**(c) a**

**(d) (3 – a ^{3)}**

**Solution**

**Q-14: Let us consider a function g: R – {−4/3} → R as g(a) = 4a/3a+4. The inverse of the function g is map f: Range g → R – {−4/3} is given by**

**(a) f(b) = 3b/3–4b (b) f(b) = 4b/4–3b**

**(c) f(b) = 4b/3–4b (d) f(b) = 3b/4–3b**

**Solution**

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