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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 1.3*. This

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*NCERT Solutions*## NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.3

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.3: Ques No 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.

**NCERT Solutions:**

Given that f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3} where f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}.

We need to find gof(x) = g[f(x)].

Put x = 1, we get gof(1) = g[f(1)] = g[2] = 3

Put x = 3, we get gof(3) = g[f(3)] = g[5] = 1

Put x = 4, we get gof(4) = g[f(4)] = g[1] = 3

Thus, gof = {(1, 3), (3, 4), (4, 3)}

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.3: Ques No 2.**

Let f, g and h be functions from R to R. Show that (f + g) oh = foh + goh and (f . g) oh = (foh) . (goh)

**NCERT Solutions:**

Proof of (f + g) oh = foh + goh

LHS

= [(f + g)oh](x)

= (f + g)[h(x)]

= f [h(x)] + g[h(x)]

= (foh)(x) + (goh)(x)

= {(foh) + (goh)}(x)

= RHS

Thus, we get {(f + g)oh}(x) = {(foh)(x) + (goh)}(x) for all x ∈R.

Proof of (f.g) oh = (foh) . (goh)

LHS

= [(f.g)oh](x)

= (f.g)[h(x)]

= f[h(x)] . g[h(x)]

= (foh)(x) . (goh)(x)

= {(foh).(goh)}(x)

= RHS

Thus, we get [(f.g)oh](x) = {(foh).(goh)}(x) for all x ∈R.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.3: Ques No 3.**

Find gof and fog, if

(i) f(x) = | x | and g(x) = | 5x – 2 |

**NCERT Solutions:**

Given that ?(x) = |x| and g(x) = |5x – 2|

Thus, we get gof(x) = g(f(x)) = |5(f(x)) – 2| = | 5| x | – 2|

And, fog(x) = f(g(x)) = | g(x) | = || 5x – 2|| = | 5x – 2|

(ii) f(x) = 8x^{3} and g(x) = x^{1/3}

**NCERT Solutions:**

Given that f(x) = 8x^{3} and g(x) = x^{1/3}

Thus, we get

gof(x) = g(f(x)) = g(x3) = (8x^{3})^{1/3 }= 2x

And, fog(x) = f(g(x)) = f(x^{1/3}) = 8(x^{1/3})3 = 8x

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.3: Ques No 4.**

If f(x) = (4x + 3)/(6x – 4), x ≠ 2/3, show that fof(x) = x for all x ≠ 2/3. What is the inverse of f ?

**NCERT Solutions:**

Given that f(x) = (4x + 3)/(6x – 4), x ≠ 2/3. Thus, we get

Thus, fof(x) = x for all x ≠ 2/3.

We get that fof(x) = I_{x}

Hence, the given function f is invertible and the inverse of f is f itself.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.3: Ques No 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)}

**NCERT Solutions:**

Given that f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}

Since f(1) = f(2) = f(3) = f(4) = 10, the given function is not one-one.

Hence, the given function has not inverse.

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

**NCERT Solutions:**

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

Since g(5) = g(7) = 4, the given function is not one-one.

Hence, the given function has not inverse.

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

**NCERT Solutions:**

Given that h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}

Since h = {(2, 7), (3, 9), (4, 11), (5, 13)}, the given function is one-one.

Since every element y of the set {7, 9, 11, 13} there exists an element x in the set {2, 3, 4, 5} such that h(x) = y, the given function is on-to function.

Thus, the given function is one-one on-to function. It has inverse function.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.3: Ques No 6.**

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.

**NCERT Solutions:**

Given that f : [–1, 1] → R, given by f (x) = x/(x + 2).

For one-one, f(x) = f(y)

⇒ x/(x + 2) = y/(y + 2)

⇒ x(y + 2) = y(x + 2)

⇒ xy + 2x = xy + 2y

⇒ 2x = 2y

⇒ x = y

Thus, the given function is one-one function.

Since f : [–1, 1] → Range f, the given function is on-to function.

Therefore, the given function f : [–1, 1] → Range is one-one on-to function and hence its inverse exists.

Let g: Range f → [−1, 1] be the inverse of f. Let y be an arbitrary element of range f. Since f: [−1, 1] → Range f is onto, we have y = f(x) for some x ∈ [−1, 1]

⇒ y = x/(x + 2)

⇒ xy + 2y = x

⇒ x(1 – y) = 2y

⇒ x = 2y/(1 – y) , y ≠ 1

Thus, f^{–1}(x) = 2x/(1 – x), x ≠ 1

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.3: Ques No 7.**

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

**NCERT Solutions:**

Given that f : R → R given by f(x) = 4x + 3

For one-one, f(x) = f(y)

⇒ 4x + 3 = 4y + 3

⇒ 4x = 4y

⇒ x = y

Thus, f is one-one function.

For on-to, let y = 4x + 3 ⇒ x = (y – 3)/4. Put x = (y – 3)/4 in f(x) = 4x + 3.

We get f(x) = 4(y – 3)/4 + 3 = y

Thus, the given function is on-to function.

Therefore, the given function has its inverse.

Since y = 4x + 3 ⇒ x = (y – 3)/4, then the inverse of the given function is

f^{–1}(x) = (x – 3)/4.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.3: Ques No 8.**

Consider f : R^{+} → [4, ∞) given by f(x) = x^{2} + 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.

**NCERT Solutions:**

Given that f : R^{+} → [4, ∞) given by f(x) = x^{2} + 4.

For one-one, f(x) = f(y)

⇒ x^{2} + 4 = y^{2} + 4

⇒ x^{2} = y^{2}

⇒ x = y [x and y in R^{+}]

Thus, f is one-one function.

For on-to,

For y ∈ [4, ∞), let y = x^{2} + 4

⇒ ?^{2} = ? − 4

⇒ x = √(y – 4) >= 0

Thus, the given function is on-to function.

Therefore, the given function is one-one on-to function and hence its inverse exists.

And, the inverse of the given function is f^{-1}(y) = √(y – 4).

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.3: Ques No 9.**

Consider f : R^{+} → [– 5, ∞) given by f (x) = 9x^{2} + 6x – 5. Show that f is invertible with

.

**NCERT Solutions:**

Given that f : R^{+} → [–5, ∞) given by f(x) = 9x^{2} + 6x – 5.

For one-one, f(x) = f(y)

⇒ 9x^{2} + 6x – 5 = 9y^{2} + 6y – 5

⇒ 9(x^{2} – y^{2}) + 6(x – y) = 0

⇒ (x – y) [9(x – y) + 6] = 0

⇒ (x – y) = 0

⇒ x = y

Thus, f is one-one function.

For on-to function,

Let y = 9x^{2} + 6x – 5

⇒ y = 9x^{2} + 6x + 1 – 6

⇒ y = (3x + 1)^{2} – 6

⇒ (3x + 1)^{2} = y + 6

⇒ (3x + 1) = √(y + 6)

⇒ 3x = √(y + 6) – 1

⇒ x = (√(y + 6) – 1) / 3

Now, put x = (√(y + 6) – 1) / 3 in f(x) = 9x^{2} + 6x – 5, we get

Since f(x) = y, then the given function is on-to function.

Therefore, the given function is one-one and on-to function and hence its inverse function exists.

And, its inverse function is

.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.3: Ques No 10.**

Let f : X → Y be an invertible function. Show that f has unique inverse.

**NCERT Solutions:**

Given that f: X → Y be an invertible function.

Let f has two inverses g_{1} and g_{2}.

Then, for all y ∈Y, we have

fog_{1}(y) = I_{Y}(y) = fog_{2}(y)

⇒ f(g_{1} (y)) = f(g_{2} (y))

⇒ g_{1} (y) = g_{2} (y) [as f is invertible ⇒ f is one – one]

⇒ ?1 = ?2 [as ? is one – one]

Hence, the given function f has a unique inverse.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.3: Ques No 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})^{–1} = f.

**NCERT Solutions:**

Given that f : {1, 2, 3} → {a, b, c} given by f(1) = a, f(2) = b and f(3) = c.

Now,

(fog)(a) = f(g(a)) = f(1) = a

(fog)(b) = f(g(b)) = f(2) = b

(fog)(c) = f(g(c)) = f(3) = c

And,

(gof)(1) = g(f(1)) = f(a) = 1

(gof)(2) = g(f(2)) = f(b) = 2

(gof)(3) = g(f(3)) = f(c) = 3

Since gof = I_{x} and fog = I_{y}, where x = {1, 2, 3} and y= {a, b, c}.

Thus, the inverse of the given function f exists and f^{−1} = g.

Therefore, f^{−1} :{ a, b, c} → {1, 2, 3} is given by f^{−1} (a) = 1, f^{−1} (b) = 2, f^{–1} (c) = 3.

Now, let h: {1, 2, 3} → {a, b, c} as h (1) = a, h (2) = b, h (3) = c. Then,

(goh)(1) = g(h(1)) = g(a) = 1

(goh)(2) = g(h(2)) = g(b) = 2

(goh)(3) = g(h(3)) = g(c) = 3

And,

(hog)(a) = h(g(a)) = h(1) = a

(hog)(b) = h(g(b)) = h(2) = b

(hog)(c) = h(g(c)) = h(3) = c

Thus, goh = I_{x} and hog = I_{y}, where x = {1, 2, 3} and y = {a, b, c}.

Thus, the inverse of g exists and g^{−1} = h ⇒ (f^{−1} )^{−1} = h.

Since we have h = f. Hence, (f^{−1})^{−1} = f.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.3: Ques No 12.**

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

**NCERT Solutions:**

Given that f: X → Y be an invertible function.

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

Since f−1 = g, then gof = I_{x} and fog = I_{y}

⇒ f^{−1}of = I_{x} and fof^{−1} = I_{y}

Hence, f^{−1}: Y → X is invertible and f is the inverse of f^{−1} i.e., (f^{−1})^{−1} = f.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.3: Ques No 13.**

If f: R → R be given by f(x) = (3 – x^{3})^{1/3} then fof(x) is

**NCERT Solutions:**

**NCERT Solution for Class 12 Maths Relations and Functions ****Exercise**** 1.3: Ques No 14.**

**NCERT Solutions:**