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# NCERT Solutions for Class 12 Maths Relations and Functions

Hi Students, Welcome to **Amans Maths Blogs (AMB)**. In this post, you will get the * NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.2*. This

**can be downloaded in PDF file. The downloading link is given at last.**

*NCERT Solutions*## NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.2

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.2: Ques No 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_{∗}.

**NCERT Solutions:**

Given that the function f : R_{∗} → R_{∗} defined by f(x) = 1/x.

For One-One,

Since f(x) = f(y) ⇒ 1/x = 1/y ⇒ x = y, thus the given function is one-one function.

For On To,

If 1/x = y, then x = 1/y. Put x = 1/y in f(x) = 1/x, then f(y) = 1/(1/y) = y. Thus, the given function is on to function.

Hence, the given function is one-one on to function.

Now, the function f : N → R_{∗} defined by g(x) = 1/x.

For One-One,

Since g(x) = g(x_{2}) ⇒ 1/x_{1} = 1/x_{2} ⇒ x_{1} = x_{2}, thus the given function is one-one function.

For On To,

Since x = 1.2 ∈ R, the no value of x in N such that g(x) = 1/1.2. Thus, the given function is not on to function.

Hence, the given function is one-one but not on to function.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.2: Ques No 2.**

Check the injectivity and surjectivity of the following functions:

(i) f : N → N given by f(x) = x^{2}

**NCERT Solutions:**

Since f: N → N is given by f(x) = x^{2}. Then, it is seen that for x, y ∈ N, f(x) = f(y) ⇒ x^{2} = y^{2} ⇒ x = y. Thus, the given function is injective.

Now, let 2 ∈ N. But, there does not exist any x in N such that f(x) = x^{2} = 2. Thus, the given function is not surjective.

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

(ii) f : Z → Z given by f(x) = x^{2}

**NCERT Solutions:**

Given that f: Z → Z is given by f(x) = x^{2}.

Since f(−1) = f(1) = 1, but −1 ≠ 1, then the given function f is not injective.

Now, let −2 ∈ Z. Since there does not exist any element x ∈ Z such that

f(x) = −2 or x^{2} = −2. Thus, the given function is f is not surjective.

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

(iii) f : R → R given by f(x) = x^{2}

**NCERT Solutions:**

Given that f : R → R given by f(x) = x^{2}

Since f(−1) = f(1) = 1, but −1 ≠ 1, then the given function f is not injective.

Now, let −2 ∈ R. Since there does not exist any element x ∈ R such that

f(x) = −2 or x^{2} = −2. Thus, the given function is f is not surjective.

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

(iv) f : N → N given by f(x) = x^{3}

**NCERT Solutions:**

Given that f : N → N given by f(x) = x^{3}

Since f(x) = f(y) ⇒ x^{3} = y^{3} ⇒ x = y, then the given function f is injective.

Now, let 2 ∈ N. Since there does not exist any element x ∈ R such that

f(x) = 2 or x^{3} = 2. Thus, the given function is f is not surjective.

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

(v) f : Z → Z given by f(x) = x^{3}

**NCERT Solutions:**

Given that f : Z → Z given by f(x) = x^{3}

Since f(x) = f(y) ⇒ x^{3} = y^{3} ⇒ x = y, then the given function f is injective.

Now, let 2 ∈ Z. Since there does not exist any element x ∈ Z such that

f(x) = 2 or x^{3} = 2. Thus, the given function is f is not surjective.

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

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.2: Ques No 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.

**NCERT Solutions:**

Given that f : R → R, given by f(x) = [x]

Since f(1.3) = [1.3] = 1 and f(1.8) = [1.8] = 1, then f(1.3) = f(1.8) = 1 but 1.3 ≠ 1.8. Thus, the given function is not one-one function.

Now, let 0.8 ∈ R. Then, there is no value of x in R such that f(x) = [x] = 0.8. Thus, the given function is not on to function.

Therefore, the greatest integer function is neither one-one nor on to function.