Tags Ncert Solutions Maths class 12th

Ncert Solutions Maths class 12th

Get insights from 2.5k questions on Ncert Solutions Maths class 12th, answered by students, alumni, and experts. You may also ask and answer any question you like about Ncert Solutions Maths class 12th

Follow Ask Question

2.5k

Questions

Discussions

Active Users

Followers

New answer posted

10 months ago

Ncert Solutions Maths class 12th Relations and Functions

Exercise-1.4

43. Determine whether or not each of the definition of given below gives a binary operation. In the event that * is not a binary operation, give justification for this.

(i) On Z⁺ , define * by a * b = a − b

(ii) On Z⁺, define * by a * b = ab

(iii) On R, define * by a * b = ab²

(iv) On Z⁺, define * by a * b = |a − b|

(v) On Z⁺, define * by a * b = a

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

(i) On Z⁺, * is defined by a * b = a − b.

It is not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 − 2= −1 ∉ Z⁺.

(ii) On Z⁺, * is defined by a * b = ab.

It is seen that for each a, b ∈ Z⁺, there is a unique element ab in Z⁺.

This means that * carries each pair (a, b) to a unique element a * b = ab in Z⁺.

Therefore, * is a binary operation.

(iii) On R, * is defined by a * b = ab².

It is seen that for each a, b ∈ R, there is a unique element ab² in R.

This means that * carries each pair (a, b) to a unique element a * b = ab² in R.

Therefore, * is a binary operation.

(iv) On Z⁺, * is defined by a * b = |a −

...more

0 0

New answer posted

10 months ago

Ncert Solutions Maths class 12th Relations and Functions

42. Let $f : R - {- \frac{4}{3}} \to R$ be a function defined as $f (x) = \frac{4 x}{3 x + 4}$ . The inverse of f is map g Range $f \to R - {- \frac{4}{3}}$

$g (y) = \frac{3 y}{3 - 4 y}$

$g (y) = \frac{4 y}{4 - 3 y}$

$g (y) = \frac{4 y}{3 - 4 y}$

$g (y) = \frac{3 y}{4 - 3 y}$

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

It is given that $f : R - {- \frac{4}{3}} \to R$ is defined as $f (x) = \frac{4 x}{3 x - 4}$

Let y be an arbitrary element of Range f.

Then, there exists $x \in R - {- \frac{4}{3}}$ such that $y = f (x)$

$\Rightarrow y = \frac{4 x}{3 x + 4}$

$\Rightarrow 3 x y + 4 y = 4 x$ let us define g: Range $f \to R - {- \frac{4}{3}} a s, g (y) = \frac{4 y}{4 - 3 y}$

$\begin{array}{l} \Rightarrow x (4 - 3 y) = 4 y \\ N o w, (g o f) (x) = g (f (x)) = g (\frac{4 x}{3 x + 4}) \\ \Rightarrow x = \frac{4 y}{4 - 3 y} \\ = \frac{4 (4 \frac{x}{3 x + 4})}{4 - 3 (\frac{4 x}{3 x + 4})} = \frac{16 x}{12 x + 16 - 12 x} = \frac{16 x}{16} = x \\ A n d, (f o g) (y) = f (g (y)) = f (\frac{4 y}{4 - 3 y}) \\ = \frac{4 (\frac{4 y}{4 - 3 y})}{3 (\frac{4 y}{4 - 3 y}) + 4} = \frac{16 y}{12 y + 16 - 12 y} = \frac{16 y}{16} = y \\ ∴ g o f = I_{R - (- \frac{4}{3})} a n d, f o g = I_" R a n g e, f " \end{array}$

Thus, g is the inverse of f i.e., $f^{- 1} = g .$

Hence, the inverse of f is the map g: Range $f \to R - {- \frac{4}{3}}$ which is given by

$g (y) = \frac{4 y}{4 - 3 y}$

The correct answer is B.

0 0

New answer posted

10 months ago

Ncert Solutions Maths class 12th Relations and Functions

41. If $f : R \to R$ be given by $f (x) = {(3 - x^{3})}^{\frac{1}{3}}$ , then f of(x) is

(A) 1/x³

(B) x³

(C) x

(D) (3 − x³)

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

$f : R \to R$ is given as $f (x) = {(3 - x^{\land} 3)}^{\land} (1 / 3)$

$\begin{array}{l} f (x) = {(3 - x^{3})}^{\frac{1}{3}} \\ ∴ f o f (x) = f (f (x)) = f ({(3 - x^{3})}^{\frac{1}{3}}) = {[3 - {({(3 - x^{3})}^{\frac{1}{3}})}^{3}]}^{\frac{1}{3}} \\ = {[3 - (3 - x^{3})]}^{\frac{1}{3}} = {(x^{3})}^{\frac{1}{3}} = x \\ ∴ f o f (x) = x \end{array}$

The correct answer is C.

0 0

New answer posted

10 months ago

Ncert Solutions Maths class 12th Relations and Functions

40. Let f: X → Y be an invertible function. Show that the inverse of f−1 is f, i.e., (f⁻¹)⁻¹ = f.

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

Let f: X → Y be an invertible function.

Then, there exists a function g: Y → X such that gof = IX and fog = IY.

Here, f⁻¹ = g.

Now, gof = IX and fog = IY

⇒ f⁻¹ of = IX and fof⁻¹ = IY

Hence, f⁻¹ : Y → X is invertible and f is the inverse of f⁻¹

i.e., (f−1)⁻¹ = f.

0 0

New answer posted

10 months ago

Ncert Solutions Maths class 12th Relations and Functions

39. 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⁾⁻¹ = f.

0 Follower 9 Views

Vishal Baghel

Contributor-Level 10

$f (1) = a, f (2) b, a n d, f (3) = c$

If we define $g : {a, b, c} \to {1, 2, 3} a s, g (a) = 1, g (b) = 2, g (c) = 3,$ then we have:

$\begin{array}{l} (f o g) (a) = f (g (a)) = f (1) = a \\ (f o g) (b) = f (g (b)) = f (2) = b \\ (f o g) (c) = f (g (c)) = f (3) = c \\ A n d \\ (g o f) (1) = g (f (1)) = f (a) = 1 \\ (g o f) (2) = g (f (2)) = f (b) = 2 \\ (g o f) (3) = g (f (3)) = f (c) = 3 \\ ∴ g o f = I_{X} a n d, f o g = I_{Y} \\ W h e r e, X = {1, 2, 3}, a n d, Y = {a, b, c} . \end{array}$

Thus, the inverse of f exists and $f^{- 1} = g .$

$∴ f^{- 1} : {a, b, c} \to {1, 2, 3}$ is given by,

$\begin{array}{l} f^{- 1} (a) = 1, f^{- 1} (b) = 2, f^{- 1} (c) = 3 \end{array}$

Let us now find the inverse of $f^{- 1}$ i.e., find the inverse of g.

If we define $h : {1, 2, 3} \to {a, b, c} a s$

$h (1) = a, h (2) = b, h (3) = c$ , then we have

$\begin{array}{l} (g o h) (1) = g (h (1)) = g (a) = 1 \\ (g o h) (2) = g (h (2)) = g (b) = 2 \\ (g o h) (3) = g (h (3)) = g (c) = 3 \\ A n d \\ (h o g) (a) = h (g (a)) = h (1) = a \\ (h o g) (b) = h (g (b)) = h (2) = b \\ (h o g) (c) = h (g (c)) = h (3) = c \\ ∴ g o h = I_{X} a n d, h o g = I_{Y} \\ W h e r e, X = {1, 2, 3}, a n d, Y = {a, b, c} . \end{array}$

Thus, the inverse of g exists and $g^{- 1} = h \Rightarrow {(f^{- 1})}^{- 1} = h .$

It can be noted that h=f.

Hence, ${(f^{- 1})}^{- 1} = f .$

0 0

New answer posted

10 months ago

Ncert Solutions Maths class 12th Relations and Functions

38. 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).

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

Let $f : X \to Y$ be an invertible function.

Also, suppose f has two inverses (say g₁ and g₂ ).

Then, for all y ∈ Y, we have:

$\begin{array}{l} f o g_{1} (y) = I_{y} (y) = f o g_{2} (y) \\ \Rightarrow f (g_{1} (y)) = f (g_{2} (y)) \end{array}$ [f is invertible => f is one-one]

$\Rightarrow g_{1} = g_{2}$ [g is one-one]

Hence, f has a unique inverse.

0 0

New answer posted

10 months ago

Ncert Solutions Maths class 12th Relations and Functions

36. 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.

0 Follower 28 Views

Vishal Baghel

Contributor-Level 10

$f : R_{+} \to [4, \infty)$ is given as $f (x) = x^{2} + 4$ .

One-one:

$\begin{array}{l} L e t, f (x) = f (y) . \\ \Rightarrow x^{2} + 4 = y^{2} + 4 \\ \Rightarrow x^{2} = y^{2} \\ \Rightarrow x = y [a s, x = y \in R] \end{array}$

$∴ f$ is a one-one function.

0 0

New answer posted

10 months ago

Ncert Solutions Maths class 12th Relations and Functions

35. Consider f : R→ R given by f(x) = 4x + 3 Show that f is invertible. Find the inverse of f.

0 Follower 16 Views

Vishal Baghel

Contributor-Level 10

$f : R \to R$ is given by,

$\begin{array}{l} f (x) = 4 x + 3 \\ O n e - o n e : \\ L e t, f (x) = f (y) . \\ \Rightarrow 4 x + 3 = 4 y + 3 \\ \Rightarrow 4 x = 4 y \\ \Rightarrow x = y \end{array}$

$∴$ f is a one-one function.

Onto:

$\begin{array}{l} F o r, y \in R, l e t, y = 4 x + 3 . \\ \Rightarrow x = \frac{y - 3}{4} \in R \end{array}$

Therefore, for any $y \in R$ , there exists $x = \frac{y - 3}{4} \in R$ such that

$f (x) = f (\frac{y - 3}{4}) = 4 (\frac{y - 3}{4}) + 3 = y$

$∴$ f is onto.

Thus, f is one-one and onto and therefore, $f^{- 1}$ exists.

Let us define $g : R \to R$ by $g (x) = \frac{y - 3}{4}$

$\begin{array}{l} N o w, (g o f) (x) = g (f (x)) = g (4 x + 3) = \frac{(4 x + 3) - 3}{4} = x \\ (f o g) (y) = f (g (y)) = f (\frac{y - 3}{4}) = 4 (\frac{y - 3}{4}) + 3 = y - 3 + 3 = y \\ ∴ g o f = f o g = I_{R} \end{array}$

Hence, f is invertible and the inverse of f is given by

$f^{- 1} = g (y) = \frac{y - 3}{4}$

0 0

New answer posted

10 months ago

Ncert Solutions Maths class 12th Relations and Functions

34. Show that f: [-1, 1] → R given by f(x) = x/x+2 is one-one. Find the inverse of the function f: [-1, 1] → Range f

0 Follower 42 Views

Vishal Baghel

Contributor-Level 10

$f : [- 1, 1] \to R$ is given as $f (x) = \frac{x}{x + 2}$

$\begin{array}{l} L e t, f (x) = f (y) . \\ \Rightarrow \frac{x}{x + 2} = \frac{y}{y + 2} \\ \Rightarrow x y + 2 x = x y + 2 y \\ \Rightarrow 2 x = 2 y \\ \Rightarrow x = y \end{array}$

$∴$ f is a one-one function.

It is clear that $f : [- 1, 1] \to$ Range f is onto.

$∴$ $f : [- 1, 1] \to$ Range f is one-one onto and therefore, the inverse of the function:

$f : [- 1, 1] \to$ Range f exists.

Let g: Range $f \to [- 1, 1]$ be the inverse of f.

Let y be an arbitrary element of range f.

Since $f : [- 1, 1] \to$ Range f is onto, we have:

$\begin{array}{l} \Rightarrow y = \frac{x}{x + 2} \\ \Rightarrow x y + 2 y = x \\ \Rightarrow x (1 - y) = 2 y \\ \Rightarrow x = \frac{2 y}{1 - y}, y \neq 1 \\ g (y) = \frac{2 y}{1 - y}, y \neq 1 \\ N o w, (g o f) (x) = g (f (x)) = g (\frac{x}{x + 2}) = \frac{2 (\frac{x}{x + 2})}{1 - \frac{x}{x + 2}} = \frac{2 x}{x + 2 - x} = 2 \frac{x}{2} = x \\ (f o g) (y) = f (g (y)) = f (\frac{2 y}{1 - y}) = \frac{\frac{2 y}{(1 - y)}}{(\frac{2 y}{1 - y}) + 2} = \frac{2 y}{2 y + 2 - 2 y} = \frac{2 y}{2} = y \\ ∴ g o f = I_{- 1, 1}, a n d, f o g - I_{R a n g e, f} \\ ∴ f^{- 1} = g \\ ∴ f^{- 1} (y) = \frac{2 y}{1 - y}, y \neq 1 \end{array}$

0 0

New answer posted

10 months ago

Ncert Solutions Maths class 12th Relations and Functions

33. 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)}

0 Follower 13 Views

Vishal Baghel

Contributor-Level 10

(i) f: {1, 2, 3, 4} → {10} defined as:

f = { (1, 10), (2, 10), (3, 10), (4, 10)}

From the given definition of f, we can see that f is a many one function as: f (1) = f (2) = f (3) = f (4) = 10

∴f is not one-one.

Hence, function f does not have an inverse.

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

g = { (5, 4), (6, 3), (7, 4), (8, 2)}

From the given definition of g, it is seen that g is a many one function as: g (5) = g (7) = 4.

∴g is not one-one,

Hence, function g does not have an inverse.

(iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} defined as:

h = { (2, 7), (3, 9), (4, 11), (5, 13)}

It is seen that

...more

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
687k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Ncert Solutions Maths class 12th

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience