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## Karnataka 2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.3

### 2nd PUC Maths Relations and Functions NCERT Text Book Questions and Answers Ex 1.3

Question 1.
Let f: {1, 3, 4} →  {1,2, 5} and g : {1, 2, 5} {1,3} be given by f= {(1,2), (3,5), (4,1)} and
g = {(1, 3), (2, 3), (5,1)}. Write down gof.
gof (1) = g (f (1)) = g (2) = 3 gof (3)
= g {f (3)} = g (5) = 1
= gof (4) = g (f (4)) = g (1) = 3
∴ gof = {(1,3), (3,1), (4,3)}

Question 2.
Let f, g and h be functions from R to R. Show that
(f + g)oh =foh + goh
(f.g)oh = (foh) . (goh)
∀ x ∈ R
{(f + g)oh} (x) = (f + g) h (x)
= f (h (x)) + g (h (x)) = foh (x) + goh (x)
= {foh + goh} (x)
∴ (f + g) oh = foh + goh
((fg )oh) (x) = (f.g) (h (x))
= fh (x). gh (x)
= {foh (x)} . {goh (x)}
{foh. goh} (x)
(fg) oh = (foh) (goh) Question 3.
Find gof and fog, if
(i) f(x) = |x| and g(x) = |5x – 2| (ii) f (x) = 8x3 and g(x) = $$x^{\frac{1}{3}}$$ Question 4.
If $$f(x)=\frac{(4 x+3)}{(6 x-4)}, x \neq \frac{2}{3}$$,show that fof(x)= x,for all
$$x \neq \frac{2}{3}$$,what is the inverse of f? Question 5.
State with reason whether following functions have inverse
(i) f : {1,2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}

(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with
g = {(5, 4), (6, 3), (7, 4), (8, 2)}

(iii) h : {2, 3, 4, 5} → {7, 9,11,13} with
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
(i) f (1) = f (2) = f (3) = f (4) = 10
⇒ f is not one-one
∴ f does not have an inverse

(ii) g(5) = 4  g(6) = 3 g(7) = 4 g(8) = 2 g(5) = g(7) = 4
∴ g is not one-one, hence g does not have inverse

(iii) h(2) = 7, h(3) = 9, h(4) = 11, h(5)=13,
∴  h is one-one

Also range of h = {7,9,11,13} = co-domain
⇒ h is onto
Hence h has an inverse
h-1 = {(7,2), (9,3), (11,4), (13, 5)}. Question 6.
Show that f: [-1,1] → R, given by f(x) =$$\frac{x}{(x+2)}$$ is one-one. Find the inverse of the function f : [-1,1] → Range f.
(Hint: For $$y \in \text { Range } f, y=f(x)=\frac{x}{x+2}$$ for some x in [-1,1]
$$x=\frac{2 y}{(1-y)}$$ Question 7.
Consider f : R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.  Question 8.
Consider f : R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with the
inverse f-1 of/given by $$f^{-1}(y)=\sqrt{y-4}$$ . where R+ is the set of ail non-negative real numbers. Question 9.
Consider f : R+ → [- 5, ∞) given by f (x) = 9x3 + 6x – 5. Show that f is invertible with
$$f^{-1}(y)=\left(\frac{(\sqrt{y+6}) \cdot 1}{3}\right)$$ Question 10.
Let f : X → Y be an invertible function. Show that  f has unique inverse.  (Hint: suppose g1, and g2 are two inverses of f. Then for all y ∈ Y, fog1(y) as 1Y(y) = fog2(y). Use one-one ness of A
Let g1 and g2 are two inverse of f, then for all y ∈ y
fog1 (y) = f [g (y)] = y = Iy (y)
fog2 (y) = f [g2 (y)] = y = Iy (y)
(fog1) (y) = (fog2) (y) ∀ y ∈ y
f(g1,(y)) = f (g2 (y)) ∀ y ∈ y
∴ g, (y) = g2 (y) (∴ f is one-one)
g1 =g2
⇒ Inverse of f is unique.

Question 11.
Consider f : {1, 2, 3} → {a, b, c} given by f(1) = a,f (2) = b and f(3) = Find f-1 and show that
(f -1)-1 = f
Given that
f = {(1. a), (2, b), (3, c)}
f is one-one and onto
∴ f is invertible f-1 (a) = 1, f (b) = 2, f-1 (c) = 3,
∴ f1 (a, b, c} ⇒ {1,2, 3} is also one and onto
∵ Inverse of f is (f)-1 exist
(f1)-1 = {(1. a), (2, b), (3, c)}=f hence (f-1)-1 = f. Question 12.
Let f : X → Y be an invertible function. Show that the inverse of f-1 is f, i.e., that (f -1)-1 = f
Let f : X → Y is invertible
⇒ f is one and onto and
f-1 : Y ⇒ X is defined as f-1(y) = x
y : f (x) ∀ x ∈ X and y ∈ Y
let y1,y∈ y
f-1 (y1) = f-2(y2)
fof1 (y1) = fof1 (y2)
Iy (y1) = Iy(y2)
⇒ y1 = y2 ∴ f1 is one-one
∀ x ∈ X, ∋ y ∈ Y such that
f1 (y) = x, hence f1 is onto
hence invertible.
let g = (f1)-1
gof-1 = Iy and f-1og = lx
∀ x ∈ X, Ix (x) = x
fof-1(x) = f-1 [g(x)] =  x
fof-1 [g (x)] = f (x)
(fof-1) (g (x)) = f (x)
g (x) = f (x)
g = f
(f-1)-1 = f.

Question 13.
If f: R → R be given by $$f(x)=\left(3-x^{3}\right)^{\frac{1}{3}}$$,then fof (x) is
(A)$$x^{\frac{1}{3}}$$
(B) x3
(C) x
(D) (3 – x3)  Let  $$f: \mathbf{R}-\left\{-\frac{4}{3}\right\} \rightarrow \mathbf{R}$$,be a function as $$f(x)=+\frac{4 x}{3 x+4}$$.The inverse off is the mag g : Range
$$f \rightarrow \mathbf{R}-\left\{-\frac{4}{3}\right\}$$ given by
(A) $$g(y)=\frac{3 y}{3-4 y}$$
(B) $$g(y)=\frac{4 y}{4-3 y}$$
(C) $$g(y)=\frac{4 y}{3-4 y}$$
(D) $$g(y)=\frac{3 y}{4-3 y}$$ 