NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3.
- Relations and Functions Class 12 Ex 1.1
- Relations and Functions Class 12 Ex 1.2
- Relations and Functions Class 12 Ex 1.4
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 12 |
| Subject | Maths |
| Chapter | Chapter 1 |
| Chapter Name | Relations and Functions |
| Exercise | Ex 1.3 |
| Number of Questions Solved | 14 |
| Category | NCERT Solutions |
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3
Ex 1.3 Class 12 Maths 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 g of.
Solution:
f= {(1,2),(3,5),(4,1)}
g= {(1,3),(2,3),(5,1)}
f(1) = 2, g(2) = 3 => gof(1) = 3
f(3) = 5, g(5)= 1 =>gof(3) = 1
f(4) = 1, g(1) = 3 => gof(4) = 3
=> gof= {(1,3), (3,1), (4,3)}
Ex 1.3 Class 12 Maths 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)
Solution:
f + R –> R, g: R –> R, h: R –> R
(i) (f+g)oh(x)=(f+g)[h(x)]
= f[h(x)]+g[h(x)]
={foh} (x)+ {goh} (x)
=>(f + g) oh = foh + goh
(ii) (f • g) oh (x) = (f • g) [h (x)]
= f[h (x)] • g [h (x)]
= {foh} (x) • {goh} (x)
=> (f • g) oh = (foh) • (goh)
Ex 1.3 Class 12 Maths Question 3.
Find gof and fog, if
(i) f (x) = |x| and g (x) = |5x – 2|
(ii) f (x) = 8x³ and g (x) = \({ x }^{ 1/3 }\).
Solution:
(i) f(x) = |x|, g(x) = |5x – 2|
(a) gof(x) = g(f(x)) = g|x|= |5| x | – 2|
(b) fog(x) = f(g (x)) = f(|5x – 2|) = ||5 x – 2|| = |5x-2|
(ii) f(x) = 8x³ and g(x) = \({ x }^{ 1/3 }\)
(a) gof(x) = g(f(x)) = g(8x³) = \({ { (8x }^{ 3 }) }^{ 1/3 }\) = 2x
(b) fog (x) = f(g (x))=f(\({ x }^{ 1/3 }\)) = 8.(\({ x }^{ 1/3 }\))³ = 8x
Ex 1.3 Class 12 Maths Question 4.
If \(f(x)=\frac { 4x+3 }{ 6x-4 } x\neq \frac { 2 }{ 3 } \), show that fof (x) = x, for all \(x\neq \frac { 2 }{ 3 } \). What is the inverse of f?
Solution:
\(f(x)=\frac { 4x+3 }{ 6x-4 } x\neq \frac { 2 }{ 3 } \)
(a) fof (x) = f(f(x)) = \(f\frac { 4x+3 }{ 6x-4 } \)
Ex 1.3 Class 12 Maths 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: {1,2,3,4,5}–>{7,9,11,13} with h = {(2,7), (3,9), (4,11), (5,13)}
Solution:
f: {1,2,3,4} –> {10} with f = {(1,10), (2,10), (3,10), (4,10)}
(i) f is not one-one since 1,2,3,4 have the same image 4.
=> f has no inverse.
(ii) g: {5,6,7,8} –> {1,2,3,4} with g = {(5,4), (6,3) , (7,4), (8,2)}
Here also 5 and 7 have the same image
∴ g is not one-one. Therefore g is not invertible.
(iii) f has an inverse
Ex 1.3 Class 12 Maths 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 ∈ Range f, y = f (x) = \(\frac { x }{ (x+2) } \) for some x in [- 1,1], i.e., x = \(\frac { 2y }{ (1-y) } \)
Solution:
Ex 1.3 Class 12 Maths Question 7.
Consider f: R –> R given by f (x) = 4x + 3. Show that f is invertible. Find the inverse of f.
Solution:
f: R—>R given by f(x) = 4x + 3
f (x1) = 4x1 + 3, f (x2) = 4x2 + 3
If f(x1) = f(x2), then 4x1 + 3 = 4x2 + 3
or 4x1 = 4x2 or x1 = x2
f is one-one
Also let y = 4x + 3, or 4x = y – 3
∴ \(x=\frac { y-3 }{ 4 } \)
For each value of y ∈ R and belonging to co-domain of y has a pre-image in its domain.
∴ f is onto i.e. f is one-one and onto
f is invertible and f-1 (y) = g (y) = \(\frac { y-3 }{ 4 } \)
Ex 1.3 Class 12 Maths Question 8.
Consider f: R+ –> [4, ∞] given by f (x) = x² + 4. Show that f is invertible with the inverse f-1 of f given by f-1 (y) = √y-4 , where R+ is the set of all non-negative real numbers.
Solution:
f(x1) = x12 + 4 and f(x2) = x22 + 4
f(x1) = f(x2) => x12 + 4 = x22 + 4
or x12 = x22 => x1 = x2 As x ∈ R
∴ x>0, x12 = x22 => x1 = x2 =>f is one-one
Let y = x² + 4 or x² = y – 4 or x = ±√y-4
x being > 0, -ve sign not to be taken
x = √y – 4
∴ f-1 (y) = g(y) = √y-4 ,y ≥ 4
For every y ≥ 4, g (y) has real positive value.
∴ The inverse of f is f-1 (y) = √y-4
Ex 1.3 Class 12 Maths Question 9.
Consider f: R+ –> [- 5, ∞) given by f (x) = 9x² + 6x – 5. Show that f is invertible with
\({ f }^{ -1 }(y)=\left( \frac { \left( \sqrt { y+6 } \right) -1 }{ 3 } \right) \)
Solution:
Let y be an arbitrary element in range of f.
Let y = 9x² + 6x – 5 = 9x² + 6x + 1 – 6
=> y = (3x + 1)² – 6
=> y + 6 = (3x + 1)²
=> 3x + 1 = √y + 6
Ex 1.3 Class 12 Maths 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)=Iy(y)=fog2(y).Use one-one ness of f.
Solution:
If f is invertible gof (x) = Ix and fog (y) = Iy
∴ f is one-one and onto.
Let there be two inverse g1 and g2
fog1 (y) = Iy, fog2 (y) = Iy
Iy being unique for a given function f
=> g1 (y) = g2 (y)
f is one-one and onto
f has a unique inverse.
Ex 1.3 Class 12 Maths Question 11.
Consider f: {1,2,3} –> {a, b, c} given by f (1) = a, f (2)=b and f (3)=c. Find f-1 and show that (f-1)f-1=f.
Solution:
f: {1,2, 3,} –> {a,b,c} so that f(1) = a, f(2) = b, f(3) = c
Now let X = {1,2,3}, Y = {a,b,c}
∴ f: X –> Y
∴ f-1: Y –> X such that f-1 (a)= 1, f-1(b) = 2; f-1(c) = 3
Inverse of this function may be written as
(f-1)-1 : X –> Y such that
(f-1)-1 (1) = a, (f-1)-1 (2) = b, (f-1)-1 (3) = c
We also have f: X –> Y such that
f(1) = a,f(2) = b,f(3) = c => (f-1)-1 = f
Ex 1.3 Class 12 Maths Question 12.
Let f: X –> Y be an invertible function. Show that the inverse of f-1 is f, i.e., (f-1)-1 = f.
Solution:
f: X —> Y is an invertible function
f is one-one and onto
=> g : Y –> X, where g is also one-one and onto such that
gof (x) = Ix and fog (y) = Iy => g = f-1
Now f-1 o (f-1)-1 = I
and fo[f-1o (f-1)-1] =fol
or (fof-1)-1 o (f-1)-1 = f
=> Io (f-1)-1 = f
=> (f-1)-1 = f
Ex 1.3 Class 12 Maths 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)
Solution:
f: R-> R defined by f(x) = \({ \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } } \)
fof (x) = f[f(x)] = \({f{ \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } }} \)
= \({ \left[ 3-{ \left\{ { \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } } \right\} }^{ 3 } \right] }^{ \frac { 1 }{ 3 } } \)
= \({ \left[ 3-{ \left\{ { \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } } \right\} } \right] }\)
= \({ \left( { x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } }\)
= x
Ex 1.3 Class 12 Maths Question 14.
Let f: \(R-\left\{ -\frac { 4 }{ 3 } \right\} \rightarrow R\) be a function defined as f (x) = \(\frac { 4x }{ 3x+4 } \) . The inverse of f is the map g: Range f–> \(R-\left\{ -\frac { 4 }{ 3 } \right\} \rightarrow R\) given by
(a) \(g(y)=\frac { 3y }{ 3-4y } \)
(b) \(g(y)=\frac { 4y }{ 4-3y } \)
(c) \(g(y)=\frac { 4y }{ 3-4y } \)
(d) \(g(y)=\frac { 3y }{ 4-3y } \)
Solution:
(b)
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