# Frank Solutions for Chapter 9 Indices Class 9 Mathematics ICSE

### Exercise 9.1

1. Evaluate the following:

(i) 60

(ii) (1/2)-3

(iii) 232

(v) (0.008)(2/3)

(vi) (0.00243)(-3/5)

(viii) (i) 60 = 1

(ii) (1/2)-3 = (2)3

We get,

(1/2)-3 = 8

(iii) = 26

We get,

= 64

(v) (0.008)(2/3)

= (0.23)(2/3)

= (0.2)3× 2/3

= (0.2)2

We get,

(0.008)(2/3) = 0.04

(vi) (0.00243)(-3/5)

= 1/(0.00243)(3/5)

= 1/(0.35)(3/5)

= 1/(0.3)3

We get,

(0.00243)(-3/5) = (1/0.027)

(vii)

(viii)

2. Evaluate the following:

(a) 94 ÷ 27(-2/3)

(b) 7-4 × (343)(2/3) ÷ (49)(-1/2)

(c) (64/216)(2/3) × (16/36)(-3/2)

(a) 94 ÷ 27(-2/3) = {(3)2}4 ÷ {(3)3}(-2/3)

On further calculation, we get,

= (3)2 × 4 ÷ (3)3 × (- 2/3) [Using (am)n = amn]

= (3)8 ÷ (3)-2

= (3)8 – (-2) [Using am ÷ an = am – n]

= (3)8 + 2

= (3)10

This can be written as,

= (3)2 × 5

= {(3)2}5

= (9)5

We get,

= 59049

(b) 7-4 × (343)(2/3) ÷ (49)(- 1/2)

On calculating further, we get,

= 7-4 × (73)(2/3) ÷ (72)(-1/2)

= 7-4 × 73 × 2/3 ÷ 72×(-1/2)

= 7-4 × 72 ÷ 7-1

= 7-4 + 2 – (-1) [Using am×an = am + n and am ÷ an = am – n]

= 7-4 + 2 + 1

We get,

= 7-1

= (1/7) [Using a-m = (1/am)]

(c) (64/216)(2/3) × (16/36)(- 3/2)

On further calculation, we get,

= {(26)2/3/(63)2/3} × {(24)-3/2/(62)– 3/2}

= {(2)6 × 2/3/(6)3 × 2/3} × {(2)4 × (-32)/(6)2 × (- 3/2)}

[Using (am)n = amn]

= {(2)2 × 2/(6)2} × {(2)2 × (- 3)/(6)– 3}

= {(2)4/(6)2} × {(2)-6/(6)-3}

We get,

= {(2)4/(6)2} × {(6)3/(2)6}

[Using a-m = (1/am)

= {(2)4/(2)6} × {(6)3/(6)2}

= (2)4 – 6 × (6)3 – 2

[Using am ÷ an = am – n]

= (2)-2 × (6)1

= (1/22) × 6

= (1/4) × 6

We get,

= (3/2)

3. Write each of the following in the simplest form:

(a) (a3)× a4

(b) a2 × a3 ÷ a4

(c) a1/3 ÷ a-2/3

(d) a-3 × a2 × a0

(e) (b-2 – a-2) ÷ (b-1 – a-1)

(a) (a3)× a4 = (a)3×5 × a4

[Using (am)n = amn]

we get,

= (a)15 × a4

= a15 + 4

[Using am × an = am + n]

We get,

= a19

(b) a2 × a3 ÷ a4 = a2 + 3 – 4

[Using am × an = am + n and am ÷ an = am – n]

We get,

= a1

= a

(c) a1/3 ÷ a-2/3 = a(1/3) – (- 2/3)

[Using am ÷ an = am – n]

We get,

= a(1/3) + (2/3)

= a(1 + 2)/3

= a1

= 0

(d) a-3 × a2 × a0 = a-3 + 2 + 0

[Using am × an = am + n]

= a-1

We get,

= (1/a)

(e) (b-2 – a-2) ÷ (b-1 – a-1)

This can be written as,

= (1/b2 – 1/a2)/(1/b – 1/a)

= {(1/b)2 – (1/a)2}/(1/b – 1/a)

= {(1/b + 1/a) (1/b – 1/a)}/(1/b – 1/a)

We get,

= (1/b + 1/a)

4. Evaluate the following:

(i) (23 × 35 × 242)/(122 × 183 × 27)

(ii) (43 × 37 × 56)/(58 × 27 × 33)

(iii) (122 × 75-2 × 35 × 400)/(482 × 15-3 × 525)

(iv) (26 × 5-4 × 3-3 × 42)/(83 × 15-3 × 25-1)

(i) (23 × 35 × 242)/(122 × 183 × 27)

This can be written as,

= {23 × 35 × (23 × 3)2}/(22 × 3)2 × (2 × 32)3 × (33)

= (23 × 35 × 26 × 32)/(24 × 32 × 23 × 36 × 33)

= (29 × 37)/(27 × 311)

On further calculation, we get,

= (29 – 7/311 – 7)

= 22/34

We get,

= (4/81)

(ii) (43 × 37 × 56) / (58 × 27 × 33)

This can be written as,

= (22)3 × 37 – 3}/(58 – 6 × 27)

= (26 × 37 – 3)/(58 – 6 × 27)

On further calculation, we get,

= (26 × 34)/(5× 27)

= {34/(52 × 27 – 6)}

= {81/(52 × 21)}

We get,

= (81/50)

(iii) (122 × 75-2 × 35 × 400)/(482 × 15-3 × 525)

This can be written as,

= (122 × 35 × 400 × 153)/(482 × 525 × 752)

= {(22 × 3)2 × (7 × 5) × (24 × 52) × (3 × 5)3}/{(24 × 3)2 × (3 × 52 × 7) × (3 × 52)2}

On calculating further, we get,

= (24 + 4 × 32+ 3 × 51 + 2 + 3 × 7)/(28 × 32 + 1 + 2 × 54 + 2 × 7)

= (28 × 3× 56 × 7)/(28 × 35 × 56 × 7)

We get,

= 1

(iv) (26 × 5-4 × 3-3 × 42)/(83 × 15-3 × 25-1)

= (26 × 42 × 153 × 251)/(83 × 54 × 33)

This can be written as,

= {(26 × (22)2 × (3 × 5)3 × (52)1}/{(23)3 × 54 × 33}

= (26 + 4 × 33 × 53 + 2)/(29 × 33 × 54)

= (210 – 9 × 55 – 4)

= (2 × 5)

We get,

= 10

5. Simplify the following and express with positive index:

(a) 3p-2q3 ÷ 2p3q-2

(b) {(p-3)2/3}1/2

(a) 3p-2q3 ÷ 2p3q-2

This can be written as,

= (3p-2q3)/(2p3q-2)

= (3/2) {(p-2/p3) × (q3/q-2)}

= (3/2) {(p-2 ÷ p3) × (q3 ÷ q-2)}

= (3/2) {(p– 2 – 3) × (q3 – (- 2))} [Using am ÷ an = am – n]

= (3/2) {(p-5) × (q5)}

= (3/2) {(1/p5) × (q5)}

We get,

= (3q5/2p5)

(b) {(p-3)2/3}1/2

= p-3 × (2/3) × (1/2)

[Using (am)n = amn]

= p– 1

We get,

= (1/p)

6. Evaluate the following:

(i) {1 – (15/64)}– 1/2

(ii) (8/27)– 2/3 – (1/3)– 2 – 70

(iii) 95/2 – 3 × 50 – (1/81) – 1/2

(iv) (27)2/3 × (8)– 1/6 ÷ (18)– 1/2

(v) (16)3/4 + 2 (1/2)-1 × 30

(i) {1 – (15/64)}– 1/2

On taking LCM, we get,

= {(64 – 15)/64}– 1/2

= (49/64)– 1/2

= (64/49)1/2

We get,

= (8/7)

(ii) (8/27)– 2/3 – (1/3)–2 – 70

= (27/8)2/3 – (1/3)-2 – 70

= (27/8)2/3 – (3)– 1

On further calculation, we get,

= (3/2)3 × 2/3 – 9 – 1

= (3/2)2 – 10

= (9/4) – 10

On taking LCM, we get,

= {(9 – 40)/4}

= (-31/4)

(iii) 95/2 – 3 × 50 – (1/81) – 1/2

On further calculation, we get,

= 32 × 5/2 – 3 × 1 – (81)1/2

= 35 – 3 – 92 × 1 / 2

= 243 – 3 – 9

We get,

= 231

(iv) (27)2/3 × (8)– 1/6 ÷ (18)– 1/2

This can be written as,

= 33 × 2/3 × {1/(23 × 1/6)} ÷ (1/18)1/2

= (32)/ 21/2) × (2 × 32)1/2

= (32/21/2) × 21/2 × 3

We get,

= 32 + 1

= 33

= 27

(v) (16)3/4 + 2 (1/2)-1 × 30

= 24× 3/4 + 2 × 2 × 1

On further calculation, we get,

= 23 + 4

= 8 + 4

We get,

= 12

= (1/22)1/2 + (0.1)-1 – 32

= (1/2) + (0.1)-1 – 32

= (1/2) + (1/0.1) – 9

= (1/2) + (10/1) – 9

= (1/2) + 1

On taking LCM, we get,

= {(1 + 2)/2}

= (3/2)

7. Simplify the following:

(a) (27x9)2/3

(b) (8x6y3)2/3

(c) (64a12/27b6)– 2/3

(d) (36m-4/49n-2)– 3/2

(e) (a1/3 + a–1/3) (a2/3 – 1 + a– 2/3)

(g) {(am){m – (1/m)}}(1/m + 1)

(h) xm + 2n. x3m – 8n ÷ x5m – 60

(i) (81)3/4 – (1/32)– 2/5 + 81/3. (1/2)-1. 20

(a) (27x9)2/3

This can be written as,

= (33x9)2/3

= (33)2/3(x9)2/3 [Using (a×b)n = an × bn]

On calculating further,

We get,

= (3)3× 2/3(x)9× 2/3 [Using (am)n = amn]

= (3)2x3×2

We get,

= 9x6

(b) (8x6y3)2/3

This can be written as,

= (23x6y3)2/3

= (23)2/3 (x6)2/3(y3)2/3 [Using (a × b)n = an × bn]

= (2)3× 2/3(x)6× 2/3(y)3× 2/3 [Using (am)n = amn]

= (2)2(x)2×2(y)2

We get,

= 4x4y2

(c) (64a12/27b6)– 2 / 3

This can be written as,

= {(26a12)/(33b6)}–2/3

= {26×(-2/3)a12× (-2/3)}/{33×(-2/3)b6×(-2/3)}

[Using (a×b)n = an×bn and (a/b)n = (an/bn)]

On further calculation, we get,

= (2–4a–8)/(3–2b–4)

= (32b4) / (24a8) [Using a-n = (1/an)]

We get,

= (9b4/16a8)

(d) (36m-4/49n-2)–3/2

This can be written as,

= {(62m-4)/(72n– 2)}–3/2

= {62× (-3/2) m-4×(-3/2)}/{72×(-3/2) n–2×(-3/2)}

[Using (a×b)n = an × bn and (a/b)n = (an/bn)]

On further calculation, we get,

= (6-3 m6)/(7– 3n3)

= (73m6)/(63n3)

[Using a–1 = (1/an)]

We get,

= (343m6)/(216n3)

(e) (a1/3 + a–1/3) (a2/3 – 1 + a–2/3)

= a1/3(a2/3 – 1 + a–2/3) + a– 1/3(a2/3 – 1 + a–2/ 3)

On simplification, we get,

= (a1/3×a2/3 – a1/3×1 + a1/3×a–2/3) + (a–1/3×a2/3 – a–1/3×1 + a–1/3×a–2/3)

= {a(1/3)+(2/3) – a1/3×1 + a(1/3)+(-2/3)} + {a(-1/3)+(2/3) – a–1/3×1 + a(-1/3)+ (-2/3)}

[Using am × an = am+n]

= {a1 – a1/3 + a–1/3} + {a1/3 – a–1/3 + a–1}

= {a – a1/3 + a–1/3 + a1/3 – a–1/3 + (1/a)}

We get,

= {a + (1/a)}

(f)

This can be written as,

= (x4y2)1/3 ÷ (x5y–5)1/6

On calculating further, we get,

= {x4×(1/3) y2×(1/3)} ÷ {x5×(1/6) y–5×(1/6)}

[Using (am)n = amn]

= {x(4/3) y(2/3)} ÷ {x(5/6) y(-5/6)}

= {x(4/3)y(2/3)}/{x(5/6) y(-5/6)}

= x(4/3)–(5/6) y(2/3)–(-5/6) [Using am ÷ an = am – n]

= x(1/2) y(3/2)

= x(1/2) (y3)(1/2)

[Using (am)n = amn]

y3

We get,

(g) {(am){m–(1/m)}}(1/m +1)

= (a)m×{m–(1/m)}×{1/(m+1)}

[Using (am)= amn]

Now,

Consider,

m×{m–(1/m)} × {1/(m + 1)}

= (m2 – 1) × {1/(m + 1)}

= m2 × {1/(m + 1)} – 1×{1/(m + 1)}

= {m2/(m + 1)} – {1/(m + 1)}

= {(m2 – 1)}/{(m + 1)}

= {(m – 1)(m + 1)}/(m + 1)

We get,

= (m – 1)

Therefore, (a) m × {m – (1 / m)} × {1 / (m + 1)} = am – 1

(h) xm+2n.x3m–8n ÷ x5m–60

= xm+2n+3m–8n–5m–(-60)

[Using am×an = am+n and am ÷ an = am – n]

= xm+2n+3m–8n–5m+60

We get,

= x–m–6n+60

(i) (81)3/4 – (1/32)–2/5 + 81/3.(1/2)-1.20

This can be written as,

= (34)(3/4) – (1/25)(-2/5) + (23)(1/3).(1/2)–1×1

[Using a0 = 1]

= 34×(3/4) – {1/2 5×(-2/5)} + 23×(1/3).(2)1

= 33 – {1/2–2} + 21.(2)1

= 33 – 22 + 2(1+1)

[Using am × an = am + n and a– n = (1/an)]

= 33 – 22 + 22

We get,

= 33

= 27

(j)

8. Simplify the following:

(i) (5x ×7 – 5x)/(5x+2 – 5x+1)

(ii) (3x+1 + 3x)/(3x+3 – 3x+1)

(iii) (2m×3 – 2m)/(2m+4 – 2m+1)

(iv) (5n+2 – 6.5n+1)/(13.5n – 2.5n+1)

(i) (5x×7 – 5x)/(5x+2 – 5x+1)

On taking common terms, we get,

= {5x(7–1)}/{5x+1(5–1)}

= (5x–x–1×6)/4 [Using am ÷ an = am – n]

= (5–1×6)/4

= 6/(5×4)

We get,

= (3/10)

(ii) (3x+1 + 3x)/(3x+3 – 3x+1)

On taking common terms, we get,

= {3x(3+1)}/{3x(33–3)}

= {4/(27–3)}

= (4/24)

We get,

= (1/6)

(iii) (2m×3 – 2m)/(2m+4 – 2m+1)

On taking common terms, we get,

= {2m(3–1)}/{2m(24– 2)}

= {2/(16–2)}

= (2/14)

We get,

= (1/7)

(iv) (5n+2 – 6.5n+1)/(13.5n – 2.5n+1)

On taking common terms, we get,

= {5n(52 – 6×5)}/{5n(13 – 2×5)}

= (25 – 30)/(13 – 10)

We get,

= (- 5/3)

9. Solve for x:

(a) 22x + 1 = 8

(b) 3 × 7x = 7 × 3x

(c) 2x + 3 + 2x + 1 = 320

(d) 9 × 3x = (27)2x – 5

(e) 22x + 3 – 9 × 2+ 1 = 0

(f) 1 = px

(g) p3 × p– 2 = px

(h) p– 5 = (1/px + 1)

(i) 22x + 2x + 2 – 4 × 23 = 0

(j) 9 x 81x = 1/27x – 3

(k) 22x – 1 – 9 × 2x – 2 + 1 = 0

(l) 5xz:5x= 25:1

(a) 22x + 1 = 8

This can be written as,

22x + 1 = 23

⇒ 2x + 1 = 3

⇒ 2x = 3 – 1

⇒ 2x = 2

We get,

x = 1

(b) 3 × 7x = 7 × 3x

⇒ (7x/7) = (3x/3)

⇒ 7x–1 = 3x – 1 [Using am ÷ an = am – n]

⇒ 7x–1= 3x–1 × 1

⇒ 7x–1 = 3x–1 × 7[Using a0 = 1]

⇒ x – 1 = 0

We get,

x = 1

(c) 2x+3 + 2x+1 = 320

This can be written as,

2x+3 + 2x+1 = 26×5

⇒ 2x. 23 + 2x.21 = 26 ×5

On taking common terms, we get,

2x (23 + 21) = 26 × 5

⇒ 2x (8 + 2) = 26 × 5

⇒ 2x (10) = 26 × 5

⇒ 2x (10/5) = 26

⇒ 2x. 2 = 26

⇒ (2x.2)/26 = 1

⇒ 2x+1–6 = 1× 20

⇒ 2x–5 = 1× 20

⇒ x – 5 = 0

We get,

x = 5

(d) 9×3x = (27)2x–5

This can be written as,

32 × 3x = (33)2x–5

⇒ 32 × 3x = 33×(2x –5)

On further calculation, we get,

32 + x = 36x – 15

⇒ 1 = (36x – 15)/(32 + x)

⇒ 1 = 36x – 15 – 2 – x

⇒ 30 = 35x – 17

⇒ 5x – 17 = 0

⇒ 5x = 17

We get,

x = (17/5)

(e) 22x+3 – 9×2+ 1 = 0

This can be written as,

22x.23 – 9×2x + 1 = 0

Put 2x = t

Then,

22x = t2

So,

22x.23 – 9×2x + 1 = 0 becomes,

⇒ 8t2 – 9t + 1 = 0

⇒ 8t2 – 8t – 1t + 1 = 0

On taking common terms, we get,

8t (t – 1) – 1 (t – 1) = 0

⇒ (t – 1) = 0 or (8t – 1) = 0

⇒ t = 1 or t = (1/8)

⇒ 2x = 1 or 2x = (1/23)

⇒ 2x = 20 or 2x = 2-3

Hence,

⇒ x = 0 or x = – 3

(f) 1 = px

p0 = px [Using a0 = 1]

Therefore,

x = 0

(g) p3 × p– 2 = px

⇒ p3 + (- 2) = px [Using am × an = am+n]

⇒ p3 – 2 = px

⇒ p1 = px

Hence,

x = 1

(h) p– 5 = (1/px+1)

⇒ p-5×px+1 = 1

⇒ p–5+x+1 = 1 [Using am × an = am + n]

⇒ px – 4 = p0

⇒ x – 4 = 0

We get,

x = 4

(i) 22x + 2x+2 – 4×23 = 0

This can be written as,

22x + 2x+2 – 22 ×23 = 0

⇒ 22x + 2x.22 – 22+3 = 0 [Using am × an = am + n]

⇒ 22x + 2x.22 – 25 = 0

⇒ 22x + 2x.4 – 32 = 0

Put 2x = t

Then, 22x = t2

22x + 2x.4 – 32 = 0 becomes,

⇒ t2 + 4t – 32 = 0

⇒ t2 + 8t – 4t – 32 = 0

On taking common terms, we get,

t (t + 8) – 4 (t + 8) = 0

⇒ t + 8 = 0 or t – 4 = 0

⇒ t = – 8 or t = 4

⇒ 2x = – 8 or 2x = 4

⇒ 2x = – 23 or 2x = 22

Now,

Consider second equation,

2x = 22

We get,

x = 2

(j) 9×81x = 1/27x – 3

This can be written as,

32×34x = 1/33 (x – 3)

⇒ 32×34x = 1/33x – 9

Using (am)n = amn

32 ×34x ×33x–9 = 1

⇒ 32+4x+3x–9 = 1×30

On further calculation, we get,

2 + 4x + 3x – 9 = 0

⇒ 7x – 7 = 0

⇒ 7x = 7

⇒ x = 1

(k) 22x–1 – 9×2x–2 + 1 = 0

This can be written as,

22x.2-1 – 9×2x. 2–2 +1 = 0

Let 2= t,

So, 22x = t2

Then,

22x.2–1 – 9×2x.2–2 + 1 = 0 becomes,

⇒ (t2/2) – 9 × (t/22) + 1 = 0

⇒ (t2/2) – (9t/4) + 1 = 0

Taking LCM, we get,

2t2 – 9t + 4 = 0

⇒ 2t2 – 8t – t + 4 = 0

⇒ 2t (t – 4) – 1 (t – 4) = 0

⇒ t – 4 = 0 or 2t – 1 = 0

⇒ t = 4 or t = (1/2)

Hence,

2x = 4 or 2x = (1/2)

⇒ 2x = 22 or 2x = 2–1

Therefore,

x = 2 or x = –1

Therefore,

x = 2 or x = – 1

(n)

(o) 9x + 4 = 32 × (27)x + 1

This can be written as,

9x+4 = 32 ×(33)x+1

⇒ 32(x+4) = 32×33x+3

⇒ 32x+8 = 32+3x+3

Hence,

2x + 8 = 2 + 3x +3

⇒ 2x + 8 = 3x + 5

⇒ 3x – 2x = 8 – 5

We get,

x = 3

10. Find the value of k in each of the following:

(iv) (1/3)– 4 ÷ 9 (-1/3) = 3k

This can be written as,

(3–1)–4 ÷ (32)–1/3 = 3k

⇒ 34 ÷ 3(-2/3) = 3k

⇒ 34–(- 2/3) = 3k

⇒ 34+ 2/3 = 3k

⇒ 3(14/3) = 3k

We get,

k = (14/3)

11. If a = 2(1/3)– 2(- 1/3), prove that 2a3+ 6a = 3

Given

a = 2(1/3) – 2(- 1/3)

This can be written as,

a = 2(1/3) – {1/2(1/3)}

On taking cube on both sides, we get,

a3 = [2(1/3) – {1/2(1/3)}]3

On further calculation, we get,

a3 = 2 – (1/2) – 3 [2(1/3) – {1/2(1/3)}]

⇒ a3 = {(4 – 1)/2} – 3a

⇒ a3 = (3/2) – 3a

We get,

2a3 + 6a = 3

12. If x = 3(2/3)+ 3(1/3), prove that x3– 9x – 12 = 0

Given

x = 32/3 + 31/3

⇒ x3 = 32 + 3 + 3 × 32/3 × 31/3 (32/3 + 31/3)

⇒ x3 = 9 + 3 + 3 × 32/3 + 1/3 (x)

⇒ x3 = 12 + 9x

We get,

x3 – 9x – 12 = 0

We get,

a = kx, b = ky , c = kz

Also given that,

abc = 1

⇒ kx × ky × kz = 1

⇒ kx + y + z = k0

We get,

x + y + z = 0

14. If ax= by= cz and b2 = ac, prove that y = 2xz/(z + x).

Let ax = by = cz = k

So,

a = k1/x, b = k1/y, c = k1/z

Also given that,

b2 = ac

⇒ k2/y = k1/x × k1/z

⇒ k2/y = k1/x + 1/z

⇒ (2/y) = (1/x) + (1/z)

⇒ (2/y) = (z + x)/zx

We get,

y = {2zx/(z + x)}

15. Show that: {1/(1 + ap–q)} + {1/(1 + aq–p)} = 1

Consider LHS of the equation, i.e,

{1/(1 + ap–q)} + {1/(1 + aq–p)}

On taking LCM, we get,

= {(1 + aq–p) + (1 + ap–q)}/(1 + ap–q) (1 + aq–p)

= (2 + a–(p–q) + ap–q)/(1 + ap–q) (1 + a–(p–q))

= 2 + a–(p–q) + ap–q/(1 + a–(p–q) + ap–q + ap–q. a–(p–q))

= 2 + a–(p–q) + ap–q/(1 + a–(p–q) + ap–q + ap–q–p+q

= 2 + a–(p–q) + ap–q/(1 + a–(p–q) + ap–q + a0)

= 2 + a–(p–q) + ap–q/(1 + a–(p–q) + ap–q + 1)

= 2 + a–(p–q) + ap–q/(2 + a–(p–q) + ap–q)

We get,

= 1

= RHS

Therefore,

LHS = RHS

Hence, proved

16. Find the value of (8p)P if 9p+2 – 9p = 240.

17. If ax = by = cz and abc = 1, show that 1/x + 1y + 1/z = 0.

18. If x 1/3 + y 1/3 + z 1/3 = 0, prove that (x + y + z)3 = 27xyz

19. If 2250 = 2a .3b. 5c, find a, b and c. Hence, calculated the value of 3a × 2-b × 5-c

20. If 2400 = 2x × 3y × 5z, find the numerical value of x, y, z. Find the value of 2-x × 3y × 5z as fraction.

21. If 2x = 3y = 12z; show that 1/z = 1/y + 2/x.

22. (a) Find the value of ‘a’ and ‘b’ if:

(b) Find the value of ‘a’ and ‘b’ if:

(a)

(b)

23. (a) Prove the following:

(c) Prove the following:

(d) Prove the following: