Class 12 Maths Chapter 1 Relations and Functions – Exercise 1.2 NCERT Solutions

Introduction

Exercise 1.2 focuses on types of functions — one‑one, onto, bijective, and composition of functions. Students also learn how to prove invertibility and find inverse functions. This exercise is crucial for building the foundation of calculus and higher algebra.

Formulas Used

1. Composition of Functions:

$$(f\circ g)(x)=f(g(x))$$

2. Inverse Function: If $f:A\to B$ is bijective, then

$$f^{-1}:B\to A$$

3. One‑One Function: If $f(a)=f(b)⟹a=b$.

4. Onto Function: Range = Codomain.

5. Bijective Function: Function that is both one‑one and onto.

Students Frequently Make Mistakes

• Confusing composition order: $(f\circ g)(x)\mathrm{\ne }(g\circ f)(x)$.

• Forgetting to check onto condition properly.

• Errors in proving invertibility.

• Misinterpreting domain restrictions.

• Mixing up range and codomain.

NCERT Questions with Step‑by‑Step Solutions (10)

Q1. If $f(x)=x^2,g(x)=x+1$, find $(f\circ g)(x)$.

$$(f\circ g)(x)=f(g(x))=f(x+1)=(x+1{)}^2$$

Q2. Find $(g\circ f)(x)$.

$$(g\circ f)(x)=g(f(x))=g(x^2)=x^2+1$$

Q3. If $f(x)=2x+3$, find inverse. Let $y=2x+3$.

$$x=\frac{y-3}{2}⟹f^{-1}(y)=\frac{y-3}{2}$$

Q4. Show that $f(x)=x^3$ is bijective.

• One‑one: $f(a)=f(b)⟹a^3=b^3⟹a=b$. ✔

• Onto: For any $y\in \mathbb{R}$, take $x=\sqrt[3]{y}$. ✔ Hence bijective.

Q5. Find inverse of $f(x)=\frac{x-1}{x+1}$. Let $y=\frac{x-1}{x+1}$.

$$y(x+1)=x-1⟹x(y-1)=-(y+1)⟹x=\frac{-(y+1)}{y-1}$$

So, $f^{-1}(y)=\frac{-(y+1)}{y-1}$.

Q6. If $f(x)=\sin x,g(x)=\cos x$, find $(f\circ g)(x)$.

$$(f\circ g)(x)=f(g(x))=\sin (\cos x)$$

Q7. Find $(g\circ f)(x)$.

$$(g\circ f)(x)=g(f(x))=g(\sin x)=\cos (\sin x)$$

Q8. Show that $f(x)=x^2$ is not one‑one on $\mathbb{R}$. Since $f(2)=f(-2)=4$. ✘ Not one‑one.

Q9. Restrict domain of $f(x)=x^2$ to $[0,\mathrm{\infty })$. Show invertibility. Now one‑one and onto $[0,\mathrm{\infty })$. Inverse:

$$f^{-1}(x)=\sqrt{x}$$

Q10. If $f(x)=\ln x,g(x)=e^x$, find $(f\circ g)(x)$ and $(g\circ f)(x)$.

$$(f\circ g)(x)=f(e^x)=\ln (e^x)=x$$

$$(g\circ f)(x)=g(\ln x)=e^{\ln x}=x$$

FAQs (10)

FAQ1. What is composition of functions? $(f\circ g)(x)=f(g(x))$.

FAQ2. Is composition commutative? No, generally $(f\circ g)(x)\mathrm{\ne }(g\circ f)(x)$.

FAQ3. What is inverse function? Function that reverses mapping of original function.

FAQ4. When is function invertible? If bijective.

FAQ5. What is one‑one function? Distinct inputs → distinct outputs.

FAQ6. What is onto function? Range = codomain.

FAQ7. What is bijective function? Function that is both one‑one and onto.

FAQ8. Why restrict domain for invertibility? To make function one‑one.

FAQ9. What is identity function in composition? If $(f\circ f^{-1})(x)=x$.

FAQ10. Why is Exercise 1.2 important? It builds foundation for inverse functions and calculus.

Conclusion

Exercise 1.2 has 10 solved questions and 10 FAQs that strengthen your understanding of composition, inverse, and types of functions. This builds the foundation for advanced calculus in Class 12 Maths.

visit:www.fuzymathacademy.com