2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.2

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Karnataka 2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.2

2nd PUC Maths Relations and Functions NCERT Text Book Questions and Answers Ex 1.2

Question 1.
Show that the function f : R₀ → R₀ defined by \(f(x)=\frac{1}{x}\)  is one-one and onto, where R₀ is the set of all non-zero real numbers. Is the result true, if the domain R₀ is replaced by N with co-domain being same as R₀?
Answer:

f : N → R_0,  then the function will remains one – one however mot onto.
Here the range of \(\mathrm{f}=\left\{1, \frac{1}{2}, \frac{1}{3}, \ldots \ldots .\right\} \neq \mathrm{R}_{\mathrm{e}}\)

Question 2.
Check the injectivity and surjectivity of the following functions:
(i) f : N → N given by f (x) = x²
Answer:

hence f is injective, for some elements like, 2, 3 etc has no preimage in N such that
f (x) = 2 hence not surjective.

(ii) f : Z → Z given by f(x) = x²
Answer:
f is not one-one or injective as f (1) = f (-1) ⇒ 1 = -1
f is not on to or surjective as \(f(x)=2 \Rightarrow x=\sqrt{2} \notin Z\)

(iii) f : R → R given by f(x) = x²
Answer:
f is not injective
as f (x₁) = f (x₂) ⇒ x₁= ± x₂
Range off does not contain any negative real. Hence not surjective.

(iv) f : N → N given by f(x) = x²
Answer:
Let x₁, x₂ ∈ N
f(x₁) = f(x₂) ⇒ \(x_{1}^{3}=x_{2}^{3}\)
hence one-one
Range of f= {1,8,27,…… } ≠ N
f is not surjective.

(v) f : Z → Z given by f (x) = x³
Answer:
Let x₁, x₂ ∈ Z
\(\begin{aligned} f\left(x_{1}\right)=f\left(x_{2}\right) & \Rightarrow x_{1}^{3}=x_{2}^{3} \\& \Rightarrow x_{1}=x_{2} \end{aligned}\)
f is one-one or injective
Range of f = {0, ±1, ±8, ±27,…………….. } ≠Z
f is not surjective

(vi) f : R → R given by f (x) = x³
Answer:

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.
Answer:
∀ x ∈ [0, 1)
f (x) = [x] = 0
f is not one-one, or injective
ex: \(0, \frac{1}{2} \in R \Rightarrow f(0)=[0]=0\)
\(f\left(\frac{1}{2}\right)=\left[\frac{1}{2}\right]=0\)
Range of f is the set of integers ≠ R hence f is not onto.

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.
Answer:
Let x₁, x₂ ∈ R f (x.) = f (x₁)
\(\begin{aligned}f\left(x_{1}\right)=f\left(x_{2}\right)&\Rightarrow\left|x_{1}\right|=\left|x_{2}\right| \\& \Rightarrow x_{1}=\pm x_{2} \end{aligned}\)
\(|-2|=|+2|=2\)hence f is not one-one.
Range of functions is only non negative real numbers
Range of f = {0, ∞) ≠ R
∴ f is not onto.

Question 5.
Show that the Signum Function f : R → R, given by
\(f(x)=\left\{\begin{array}{ll}{1} & {\text { if } x>0} \\{0} & {\text { if } x=0} \\{-1} & {\text { if } x<0}\end{array}\right.\)
is neither one-one nor onto.
Answer:
f (x) = 1 ∀ x ∈ [0, ∞ )
hence f is not one-one Range of f = {-1,0, 1} hence f is not onto.

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.
Answer:
x₁, x₂ ∈ A, f (x₁) = f (x₂)
⇒ x₁ = x₂
∴ hence one-one
f (1) = 4, f (2) = 5, f (3) ≠ R, hence f is one-one.

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
Answer:
Let x₁ , x₂ ∈ R
f (x₁) = f (x₂) ⇒ 3- 4x₁, = 3- 4x₂
⇒ x₁ = x₂ hence one-one function

hence f is not onto

(ii) f : R→ R defined by f(x) = 1 + x²
Answer:
Let x₁, x₂ ∈ R
f (x₁) = f (x₂)
⇒ 1 + x₁² = 1 + x₂²
⇒ x₁ = ±x₂
∴  f is not one-one
Also range of f contains only real numbers which is greater than or equal to 1≠ R
∴ f is not onto hence f is not bijective.

Question 8.
Let A and B be sets. Show that
f : A x B → B x A such that
f (a, b) = (b, a) is bijective function.
Answer:
Let (p, q), (r, s) g A x B f (p,q) = f (r, s)
⇒ (q, p) = (s, r) ⇒ q = s and p = r
⇒ (p, q) = (r, s), hence one-one
∀ (b, a) ∈ B x A, 3 (a, b) ∈ A x B
such that f (a, b) = f (b, a), hence on to hence f is bijective function

Question 9.
Let f : N → N be defined by
\(\mathbf{f}(\mathbf{n})=\left\{\begin{array}{c} {\frac{\mathbf{n}+\mathbf{1}}{2}, \text { if } \mathbf{n} \text { is odd }} \\{\mathbf{All}  \mathbf{n} \in \mathbf{N}} \\{\frac{\mathbf{n}}{2}, \text { if } \mathbf{n} \text { is even }}\end{array}\right.\)
State whether the function / is bijective. Justify your answer.
Answer:
\(f(2 n-1)=\frac{2 n-1+1}{2}=n\)
\(f(2 n)=\frac{2 n}{2}=n\)
f (2n – 1) = f (2n) = n, but 2n -1 ≠ 2n
en, f (3) = 2, f (4) = 2, but 3 ≠ 4
hence f is not one-one
Here range of f = N
as for any N ∈ N, ∋ 2n ∈ N such that
f(2n) = n
hence f is on to. But it is not bijective

Question 10.
Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by \(f(x)=\left(\frac{x-2}{x-3}\right)\). is f one-one onto? justify your answer.
Answer:

Question 11.
Let f : R → R be defined as f(x) = x⁴. 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.
Answer:
f (x,) = f (x₂) ⇒ x₁⁴ = x₂⁴
= x₁⁴ = x₂⁴ = 0
= (x₁ – x₂) (x₁ + x₂) (x₁² = x₂²)
x₁ = x₂ or x₁ = – x₂
∴ f is not one-one
f (x) = x⁴ ≥ o ∀ x ∈ R
Range of f = [0, ∞) ∈ R
hence not onto
‍∴ correct Answer is “D”

Question 12.
Let f : R → R be defined as f (x) = x⁵. 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 either one-one nor onto.
Answer: