Hello Students! In this post, you will find the complete **NCERT Solutions for Maths Class 12 Exercise 1.2**.

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**NCERT Solutions for Maths Class 12 Exercise 1.1**

**NCERT Solutions for Maths Class 12 Ex 1.2**

**Maths Class 12 Ex 1.2 ****Question 1.**

**Question 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?**

**Solution:**

**(a)** **The function f : R → R defined as f(x) = 1/x**** **

(i) We have f(x_{1}) = 1/x_{1} and f(x_{2}) = 1/x_{2}

If f(x_{1}) = f(x_{2})

1/x_{1 }= 1/x_{2}

⇒ x_{1} = x_{2 }

It means for each x ∈ R in domain, there is a unique
image in codomain R.

Therefore, f is one-one.

(ii) For every y ∈ R in codomain, there exists an
element x ∈ R in domain such that f(x) = y.

Therefore, f is onto.

**(b)** If
the domain R is replaced by N and codomain remains the same, then the function
is f : N **→** R

Let n_{1}, n_{2} ∈
N in domain.

If f(n_{1}) = f(n_{2})

⇒ 1/n_{1} = 1/n_{2}

⇒ n_{1} = n_{2}

⇒ f is one-one.

But for every y ∈ R in codomain, it may not have an
element x ∈ N in domain such that y is an
image of x. For example, 3/2, 4/5, -1 ∈
R in codomain, but these are not an image of any element of domain belonging to
N.

Therefore, f is not onto.

**Maths Class 12 Ex 1.2 ****Question 2.**

**Question 2.**

**Check the injectivity and surjectivity of the following functions:**

**(i) f : N → N given by f(x) = x²**

**(ii) f : Z → Z given by f(x) = x²**

**(iii) f : R → R given by f(x) = x²**

**(iv) f : N → N given by f(x) = x³**

**(v) f : Z → Z given by f(x) = x³**

**Solution:**

**(i) **The function f : N **→** N given by f(x) = x²

(a) Let x_{1, }x_{2 }∈
N, then f(x_{1}) = x_{1}² and f(x_{2}) = x_{2}²

If f(x_{1}) = f(x_{2})

⇒ x_{1}^{2} =
x_{2}^{2}

⇒ x_{1} = x_{2}

Therefore, f is one-one, i.e., it is injective.

(b) We have 2, 3 ∈ N in codomain, which are not an
image of any element x ∈ N in domain. It means there is no
natural number in domain whose square is 2 or 3.

Therefore, f is not onto, i.e., it is not surjective.

Hence, the function *f *is injective but not surjective.

**(ii)
**The function f : Z **→** Z
given by f(x) = x²

(a) Let 1, -1 ∈ Z in domain, then f(1) = f(-1) =
1

Here, f(1) = f(-1) = 1 but -1 and 1 are not equal. It means the
elements 1 and -1 have the same image 1.

Therefore, f is not one-one, i.e., it is not injective.

(b) We have -2, -3 ∈ Z in codomain, which are not an
image of any element x ∈ Z in domain. It means there is no
integer in domain whose square is -2 or -3.

Therefore, f is not onto, i.e., it is not surjective.

Hence, the function *f *is neither injective nor surjective.

**(iii)
**The function f : R **→ **R
given by f(x) = x²

(a) Let 1, -1 ∈ R in domain, then f(1) = f(-1) =
1

Here, f(1) = f(-1) = 1 but -1 and 1 are not equal. It means the
elements 1 and -1 have the same image 1.

Therefore, f is not one-one, i.e., it is not injective.

(b) We have -2, -3 ∈ R in codomain, which are not an
image of any element x ∈ R in domain. It means there is no
integer in domain whose square is -2 or -3.

Therefore, f is not onto, i.e., it is not surjective.

Hence, the function *f *is neither injective nor surjective.

**(iv)
**The function f : N **→** N
given by f(x) = x^{3}

(a) Let x_{1, }x_{2 }∈
N, then f(x_{1}) = x_{1}^{3} and f(x_{2}) = x_{2}^{3}

If f(x_{1}) = f(x_{2})

⇒ x_{1}^{3} =
x_{2}^{3}

⇒ x_{1} = x_{2}

Therefore, f is one-one, i.e., it is injective.

(b) We have 2, 3 ∈ N in codomain, which are not an
image of any element x ∈ N in domain. It means there is no
natural number in domain whose cube is 2 or 3.

Therefore, f is not onto, i.e., it is not surjective.

Hence, the function *f *is injective but not surjective.

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

(a) Let x_{1, }x_{2 }∈ Z,
then f(x_{1}) = x_{1}^{3} and f(x_{2}) = x_{2}^{3}

If f(x_{1}) = f(x_{2})

⇒ x_{1}^{3} =
x_{2}^{3}

⇒ x_{1} = x_{2}

Therefore, f is one-one, i.e., it is injective.

(b) We have 2, 3 ∈ Z in codomain, which are not an
image of any element x ∈ Z in domain. It means there is no
natural number in domain whose cube is 2 or 3.

Therefore, f is not onto, i.e., it is not surjective.

Hence, the function *f *is injective but not surjective.

**Maths Class 12 Ex 1.2 ****Question 3.**

**Question 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.**

**Solution:**

The function f : R **→** R
given by f(x) = [x]

(a) Let 1.3, 1.4 ∈ R in domain, then f(1.3) = 1 and
f(1.4) = 1

Here, f(1.3) = f(1.4) = 1 but 1.3 ≠ 1.4

Therefore, f is not one-one.

(b) We have 1.1, 1.2 ∈ R in codomain, which are not an
image of any element x ∈ R in domain because the ranges
are only integers here.

Therefore, f is not onto.

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

**Maths Class 12 Ex 1.2 ****Question 4.**

**Question 4.**

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

**Solution:**

The function f : R **→ **R
given by f(x) = |x|

(a) Let 2, -2 ∈ R in domain, then f(2) = f(-2) = 2

Here, f(2) = f(-2) = 2 but -2 and 2 are not equal. It means the
elements 2 and -2 have the same image 2. Therefore, f is not one-one.

(b) We have -2, -3 ∈ R in codomain, which are not an
image of any element x ∈ R in domain. Therefore, f is not
onto.

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

**Maths Class 12 Ex 1.2 ****Question 5.**

**Question 5.**

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

**f(x) = 1, if x > 0**

**f(x) = 0, if x = 0**

**f(x) = –1, if x < 0**

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

**Solution:**

The function f : R **→ **R
given by

f(x) = 1, if x > 0

f(x) = 0, if x = 0

f(x) = -1, if x < 0

(a) Let 1, 2 ∈ R in domain, then f(1) = f(2) = 1

Here, f(1) = f(2) = 1 but 1 ≠ 2

Similarly, for -1, -2 ∈ R in domain, f(-1) = f(-2) = –1, but
-1 ≠ -2

Therefore, f is not one-one.

(b) Except –1, 0, 1, there is no element y ∈ R
in codomain which is an image of an element x ∈ R
in domain such that f(x) = y.

Therefore, f is not onto.

Hence, f is neither one-one nor onto.

**Maths Class 12 Ex 1.2 ****Question 6.**

**Question 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.**

**Solution:**

We have A = {1, 2, 3}, B = {4, 5, 6, 7} and the function f = {(1, 4), (2, 5),
(3, 6)}

From the given function, we can see that 4 is the image of 1, 5 is
the image of 2 and 6 is the image of 3. Every element of A has a unique image
in B.

Therefore, f is one-one.

**Maths Class 12 Ex 1.2 ****Question 7.**

**Question 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 + x²**

**Solution:**

**(i)** The function f : R **→** R defined by **f(x) = **3 – 4x.

Let x_{1},_{ }x_{2} ∈ R
in domain, then f(x_{1}) = 3 – 4x_{1} and f(x_{2}) = 3
– 4x_{2}

(a) If f(x_{1}) = f(x_{2}), then 3 – 4x_{1} = 3 –
4x_{2}

⇒ x_{1} = x_{2}.

Therefore, f is one-one.

(b) Let f(x) = 3 – 4x = y

Then x = (3 – y)/4

It means for every value of y ∈ R
in codomain, there exists an element y ∈ R
in domain such that f(x) = y. Therefore, f is onto.

Hence, the function f is bijective.

**(ii)** The
function f : R **→ **R defined
by f(x) = 1 + x²

(a) Let 1, -1 ∈ R in domain, then f(1) = 1 + 1 =
2 and f(-1) = 1 + 1 = 2

Here, f(1) = f(-1) = 2, i.e., 1 and -1 have the same image 2. But 1 ≠ -1.

Therefore, f is not one-one.

(b) We have -2, -3 ∈ R in codomain, which are not an
image of any element x ∈ R in domain. Therefore, f is not
onto.

Hence, the function f is not bijective.

**Maths Class 12 Ex 1.2 ****Question 8.**

**Question 8.**

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

**Solution:**

We have the function f : (A **×** B) **→** B **×** A such that f(a, b) = (b, a)

(a) Since f(a_{1}, b_{1}) = (b_{1}, a_{1}) and
f(a_{2}, b_{2}) = (b_{2}, a_{2})

If f(a_{1}, b_{1}) = f(a_{2}, b_{2})

⇒ (b_{1}, a_{1}) =
(b_{2}, a_{2})

⇒ b_{1} = b_{2} and
a_{1} = a_{2}

Therefore, f is one-one.

(b) For every element (q, p) ∈ **B × A** in codomain, there
exists an element (p, q) ∈ **A × B** in domain such that **f(p, q) = (q, p). Therefore,** f is onto.

Thus, f is a bijective function.

**Maths Class 12 Ex 1.2 ****Question 9.**

**Question 9.**

**Let f : N → N be defined by**

**for all n****∈**

**N.**

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

**Solution:**

The function f : N **→** N,
defined by

The elements 1, 2 **∈**** N** in domain of
f have the same image 1 in its codomain

(b) Every element in codomain N is an image of an element in domain N. For example, 1 in codomain is the image of 1 and 2 in domain.

Therefore, f is onto.

Hence, f is not bijective.

**Maths Class 12 Ex 1.2 ****Question 10.**

**Question 10.**

**Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = (x - 2)/(x - 3)**

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

**Solution:**

The function f : A **→** B where
A = R – {3}, B = R – {1} is defined by

**Maths Class 12 Ex 1.2 ****Question 11.**

**Question 11.**

**Let 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**

**Solution:**

**(D)** For 1, -1 ∈ **R in domain,** f(1)
= (1)^{4} = 1 and f(-1) = (-1)^{4} = 1

We have f(1) = f(-1) = 1 but 1 ≠ -1. It means for 1 and -1, we have the same
image 1.

Therefore, f is not one-one.

For -2 ∈ **R** in the codomain of f, there is no element in domain
whose image is -2.

Therefore, f is not onto. Thus, f is neither one-one nor onto.

Hence, option (D) is correct.

**Maths Class 12 Ex 1.2 ****Question 12.**

**Question 12.**

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

**Solution:**

**(A)** The function f : R **→** R is defined by f(x) = 3x

(a) Let x_{1}, x_{2 }∈ **R in domain, then** f(x_{1})
= 3x_{1} and f(x_{2}) = 3x_{2}

If f(x_{1}) = f(x_{2})

⇒ 3x_{1} = 3x_{2}

⇒ x_{1} = x_{2}

⇒ Therefore, f is one-one.

(b) For every element y ∈ **R** in codomain, there exists an element x ∈ **R in domain such that y = 3x or x
= y/3**.

Therefore, f is onto. Thus, f is one-one and onto.

Hence, option (A) is correct.

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