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

## 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(x2) ⇒ 1/x1 = 1/x2 ⇒ x1 = x2, 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) = x2

NCERT Solutions:

Since f: N → N is given by f(x) = x2. Then, it is seen that for x, y ∈ N, f(x) = f(y) ⇒ x2 = y2 ⇒ 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) = x2 = 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) = x2

NCERT Solutions:

Given that f: Z → Z is given by f(x) = x2.

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 x2 = −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) = x2

NCERT Solutions:

Given that f : R → R given by f(x) = x2

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 x2 = −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) = x3

NCERT Solutions:

Given that f : N → N given by f(x) = x3

Since f(x) = f(y) ⇒ x3 = y3 ⇒ 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 x3 = 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) = x3

NCERT Solutions:

Given that f : Z → Z given by f(x) = x3

Since f(x) = f(y) ⇒ x3 = y3 ⇒ 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 x3 = 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.

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