For x = 9 and y = 4, the value of x2 + 2xy + y2 - 3 is :
172
100
166
169
Answer
Given,
⇒ x2 + 2xy + y2 - 3
⇒ (x + y)2 - 3
Substituting values we get :
⇒ (9 + 4)2 - 3
⇒ 132 - 3
⇒ 169 - 3
⇒ 166.
Hence, Option 3 is the correct option.
For x = 5 and y = 3, the value of x2 + y2 - 2xy + 7 is :
11
169
71
1
Answer
Given,
⇒ x2 + y2 - 2xy + 7
⇒ (x - y)2 + 7
Substituting values we get :
⇒ (5 - 3)2 + 7
⇒ 22 + 7
⇒ 4 + 7
⇒ 11.
Hence, Option 1 is the correct option.
If x + 1 x = 2 x + \dfrac{1}{x} = 2 x + x 1 = 2 x 2 + 1 x 2 + 5 x^2 + \dfrac{1}{x^2} + 5 x 2 + x 2 1 + 5
-1
2
9
7
Answer
( x + 1 x ) 2 = x 2 + 1 x 2 + ( 2 × x × 1 x ) ⇒ ( x + 1 x ) 2 = x 2 + 1 x 2 + 2 ⇒ x 2 + 1 x 2 = ( x + 1 x ) 2 − 2 \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + \Big(2 \times x \times \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)^2 - 2 \\[1em] ( x + x 1 ) 2 = x 2 + x 2 1 + ( 2 × x × x 1 ) ⇒ ( x + x 1 ) 2 = x 2 + x 2 1 + 2 ⇒ x 2 + x 2 1 = ( x + x 1 ) 2 − 2
Given,
x + 1 x = 2 x + \dfrac{1}{x} = 2 x + x 1 = 2
∴ x 2 + 1 x 2 = 2 2 − 2 ⇒ x 2 + 1 x 2 = 4 − 2 ⇒ x 2 + 1 x 2 = 2 \therefore x^2 + \dfrac{1}{x^2} = 2^2 - 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 4 - 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 2 ∴ x 2 + x 2 1 = 2 2 − 2 ⇒ x 2 + x 2 1 = 4 − 2 ⇒ x 2 + x 2 1 = 2
Adding 5 on both sides,
x 2 + 1 x 2 + 5 = 2 + 5 ⇒ x 2 + 1 x 2 + 5 = 7 x^2 + \dfrac{1}{x^2} + 5 = 2 + 5 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 5 = 7 x 2 + x 2 1 + 5 = 2 + 5 ⇒ x 2 + x 2 1 + 5 = 7
Hence, Option 4 is the correct option.
If x − 1 x = 8 x - \dfrac{1}{x} = 8 x − x 1 = 8 x 2 + 1 x 2 − 8 x^2 + \dfrac{1}{x^2} - 8 x 2 + x 2 1 − 8
56
58
70
-4
Answer
( x − 1 x ) 2 = x 2 + 1 x 2 − ( 2 × x × 1 x ) ⇒ ( x − 1 x ) 2 = x 2 + 1 x 2 − 2 ⇒ x 2 + 1 x 2 = ( x − 1 x ) 2 + 2 \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - \Big(2 \times x \times \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = \Big(x - \dfrac{1}{x}\Big)^2 + 2 \\[1em] ( x − x 1 ) 2 = x 2 + x 2 1 − ( 2 × x × x 1 ) ⇒ ( x − x 1 ) 2 = x 2 + x 2 1 − 2 ⇒ x 2 + x 2 1 = ( x − x 1 ) 2 + 2
Subtracting 8 on both sides,
⇒ x 2 + 1 x 2 − 8 = ( x − 1 x ) 2 + 2 − 8 ⇒ x 2 + 1 x 2 − 8 = 8 2 − 6 ⇒ x 2 + 1 x 2 − 8 = 64 − 6 ⇒ x 2 + 1 x 2 − 8 = 58. \Rightarrow x^2 + \dfrac{1}{x^2} - 8 = \Big(x - \dfrac{1}{x}\Big)^2 + 2 - 8 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} - 8 = 8^2 - 6 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} - 8 = 64 - 6 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} - 8 = 58. ⇒ x 2 + x 2 1 − 8 = ( x − x 1 ) 2 + 2 − 8 ⇒ x 2 + x 2 1 − 8 = 8 2 − 6 ⇒ x 2 + x 2 1 − 8 = 64 − 6 ⇒ x 2 + x 2 1 − 8 = 58.
Hence, Option 2 is the correct option.
If x 2 + 1 x 2 = 9 x^2 + \dfrac{1}{x^2} = 9 x 2 + x 2 1 = 9 x 4 + 1 x 4 + 5 x^4 + \dfrac{1}{x^4} + 5 x 4 + x 4 1 + 5
78
86
84
81
Answer
( x 2 + 1 x 2 ) 2 = x 4 + 1 x 4 + ( 2 × x 2 × 1 x 2 ) ⇒ ( x 2 + 1 x 2 ) 2 = x 4 + 1 x 4 + 2 ⇒ x 4 + 1 x 4 = ( x 2 + 1 x 2 ) 2 − 2 \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = x^4 + \dfrac{1}{x^4} + \Big(2 \times x^2 \times \dfrac{1}{x^2}\Big) \\[1em] \Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = x^4 + \dfrac{1}{x^4} + 2 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = \Big(x^2 + \dfrac{1}{x^2}\Big)^2 - 2 \\[1em] ( x 2 + x 2 1 ) 2 = x 4 + x 4 1 + ( 2 × x 2 × x 2 1 ) ⇒ ( x 2 + x 2 1 ) 2 = x 4 + x 4 1 + 2 ⇒ x 4 + x 4 1 = ( x 2 + x 2 1 ) 2 − 2
Adding 5 on both sides,
x 4 + 1 x 4 + 5 = ( x 2 + 1 x 2 ) 2 − 2 + 5 ⇒ x 4 + 1 x 4 + 5 = 9 2 + 3 ⇒ x 4 + 1 x 4 + 5 = 84. x^4 + \dfrac{1}{x^4} + 5 = \Big(x^2 + \dfrac{1}{x^2}\Big)^2 - 2 + 5\\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} + 5 = 9^2 + 3 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} + 5 = 84. x 4 + x 4 1 + 5 = ( x 2 + x 2 1 ) 2 − 2 + 5 ⇒ x 4 + x 4 1 + 5 = 9 2 + 3 ⇒ x 4 + x 4 1 + 5 = 84.
Hence, Option 3 is the correct option.
If x2 - 3x + 1 = 0, the value of
x 2 + 1 x 2 + 1 x^2 + \dfrac{1}{x^2} + 1 x 2 + x 2 1 + 1
8
10
5
10 9 \dfrac{10}{9} 9 10
Answer
Given,
⇒ x2 - 3x + 1 = 0
⇒ x2 + 1 = 3x
Dividing above equation by x, we get :
⇒ x 2 + 1 x = 3 x x ⇒ x + 1 x = 3 \Rightarrow \dfrac{x^2 + 1}{x} = \dfrac{3x}{x} \\[1em] \Rightarrow x + \dfrac{1}{x} = 3 ⇒ x x 2 + 1 = x 3 x ⇒ x + x 1 = 3
Squaring both sides we get :
⇒ ( x + 1 x ) 2 = 3 2 ⇒ x 2 + 1 x 2 + 2 × x × 1 x = 9 ⇒ x 2 + 1 x 2 + 2 = 9 ⇒ x 2 + 1 x 2 + 1 + 1 = 9 ⇒ x 2 + 1 x 2 + 1 = 9 − 1 ⇒ x 2 + 1 x 2 + 1 = 8. \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 3^2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 2 \times x \times \dfrac{1}{x} = 9 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 2 = 9 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 1 + 1 = 9 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 1 = 9 - 1 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 1 = 8. ⇒ ( x + x 1 ) 2 = 3 2 ⇒ x 2 + x 2 1 + 2 × x × x 1 = 9 ⇒ x 2 + x 2 1 + 2 = 9 ⇒ x 2 + x 2 1 + 1 + 1 = 9 ⇒ x 2 + x 2 1 + 1 = 9 − 1 ⇒ x 2 + x 2 1 + 1 = 8.
Hence, Option 1 is the correct option.
Evaluate ( 7 8 x + 4 5 y ) 2 \Big(\dfrac{7}{8}x + \dfrac{4}{5}y\Big)^2 ( 8 7 x + 5 4 y ) 2
Answer
Solving,
⇒ ( 7 8 x + 4 5 y ) 2 ⇒ ( 7 8 x ) 2 + ( 4 5 y ) 2 + 2 × 7 8 x × 4 5 y ⇒ 49 64 x 2 + 16 25 y 2 + 7 5 x y . \Rightarrow \Big(\dfrac{7}{8}x + \dfrac{4}{5}y\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{7}{8}x\Big)^2 + \Big(\dfrac{4}{5}y\Big)^2 + 2 \times \dfrac{7}{8}x \times \dfrac{4}{5}y \\[1em] \Rightarrow \dfrac{49}{64}x^2 + \dfrac{16}{25}y^2 + \dfrac{7}{5}xy. ⇒ ( 8 7 x + 5 4 y ) 2 ⇒ ( 8 7 x ) 2 + ( 5 4 y ) 2 + 2 × 8 7 x × 5 4 y ⇒ 64 49 x 2 + 25 16 y 2 + 5 7 x y .
Hence, ( 7 8 x + 4 5 y ) 2 = 49 64 x 2 + 16 25 y 2 + 7 5 x y . \Big(\dfrac{7}{8}x + \dfrac{4}{5}y\Big)^2 = \dfrac{49}{64}x^2 + \dfrac{16}{25}y^2 + \dfrac{7}{5}xy. ( 8 7 x + 5 4 y ) 2 = 64 49 x 2 + 25 16 y 2 + 5 7 x y .
Evaluate ( 2 x 7 − 7 y 4 ) 2 \Big(\dfrac{2x}{7} - \dfrac{7y}{4}\Big)^2 ( 7 2 x − 4 7 y ) 2
Answer
Solving,
⇒ ( 2 x 7 − 7 y 4 ) 2 ⇒ ( 2 x 7 ) 2 + ( 7 y 4 ) 2 − 2 × 2 x 7 × 7 y 4 ⇒ 4 x 2 49 + 49 y 2 16 − x y . \Rightarrow \Big(\dfrac{2x}{7} - \dfrac{7y}{4}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{2x}{7}\Big)^2 + \Big(\dfrac{7y}{4}\Big)^2 - 2 \times \dfrac{2x}{7} \times \dfrac{7y}{4} \\[1em] \Rightarrow \dfrac{4x^2}{49} + \dfrac{49y^2}{16} - xy. ⇒ ( 7 2 x − 4 7 y ) 2 ⇒ ( 7 2 x ) 2 + ( 4 7 y ) 2 − 2 × 7 2 x × 4 7 y ⇒ 49 4 x 2 + 16 49 y 2 − x y .
Hence, ( 2 x 7 − 7 y 4 ) 2 = 4 x 2 49 + 49 y 2 16 − x y . \Big(\dfrac{2x}{7} - \dfrac{7y}{4}\Big)^2 = \dfrac{4x^2}{49} + \dfrac{49y^2}{16} - xy. ( 7 2 x − 4 7 y ) 2 = 49 4 x 2 + 16 49 y 2 − x y .
Evaluate ( a 2 b + 2 b a ) 2 − ( a 2 b − 2 b a ) 2 − 4 \Big(\dfrac{a}{2b} + \dfrac{2b}{a}\Big)^2 - \Big(\dfrac{a}{2b} - \dfrac{2b}{a}\Big)^2 - 4 ( 2 b a + a 2 b ) 2 − ( 2 b a − a 2 b ) 2 − 4
Answer
Solving,
⇒ ( a 2 b + 2 b a ) 2 − ( a 2 b − 2 b a ) 2 − 4 ⇒ ( a 2 b ) 2 + ( 2 b a ) 2 + 2 × a 2 b × 2 b a − [ ( a 2 b ) 2 + ( 2 b a ) 2 − 2 × a 2 b × 2 b a ] − 4 ⇒ a 2 4 b 2 + 4 b 2 a 2 + 2 − [ a 2 4 b 2 + 4 b 2 a 2 − 2 ] − 4 ⇒ a 2 4 b 2 + 4 b 2 a 2 + 2 − a 2 4 b 2 − 4 b 2 a 2 + 2 − 4 ⇒ 4 − 4 ⇒ 0. \Rightarrow \Big(\dfrac{a}{2b} + \dfrac{2b}{a}\Big)^2 - \Big(\dfrac{a}{2b} - \dfrac{2b}{a}\Big)^2 - 4 \\[1em] \Rightarrow \Big(\dfrac{a}{2b}\Big)^2 + \Big(\dfrac{2b}{a}\Big)^2 + 2 \times \dfrac{a}{2b} \times \dfrac{2b}{a} - \Big[\Big(\dfrac{a}{2b}\Big)^2 + \Big(\dfrac{2b}{a}\Big)^2 - 2 \times \dfrac{a}{2b} \times \dfrac{2b}{a}\Big] - 4 \\[1em] \Rightarrow \dfrac{a^2}{4b^2} + \dfrac{4b^2}{a^2} + 2 - \Big[\dfrac{a^2}{4b^2} + \dfrac{4b^2}{a^2} - 2\Big] - 4 \\[1em] \Rightarrow \dfrac{a^2}{4b^2} + \dfrac{4b^2}{a^2} + 2 - \dfrac{a^2}{4b^2} - \dfrac{4b^2}{a^2} + 2 - 4 \\[1em] \Rightarrow 4 - 4 \\[1em] \Rightarrow 0. ⇒ ( 2 b a + a 2 b ) 2 − ( 2 b a − a 2 b ) 2 − 4 ⇒ ( 2 b a ) 2 + ( a 2 b ) 2 + 2 × 2 b a × a 2 b − [ ( 2 b a ) 2 + ( a 2 b ) 2 − 2 × 2 b a × a 2 b ] − 4 ⇒ 4 b 2 a 2 + a 2 4 b 2 + 2 − [ 4 b 2 a 2 + a 2 4 b 2 − 2 ] − 4 ⇒ 4 b 2 a 2 + a 2 4 b 2 + 2 − 4 b 2 a 2 − a 2 4 b 2 + 2 − 4 ⇒ 4 − 4 ⇒ 0.
Hence, ( a 2 b + 2 b a ) 2 − ( a 2 b − 2 b a ) 2 − 4 \Big(\dfrac{a}{2b} + \dfrac{2b}{a}\Big)^2 - \Big(\dfrac{a}{2b} - \dfrac{2b}{a}\Big)^2 - 4 ( 2 b a + a 2 b ) 2 − ( 2 b a − a 2 b ) 2 − 4
Evaluate (4a + 3b)2 - (4a - 3b)2 + 48ab
Answer
Solving,
⇒ (4a + 3b)2 - (4a - 3b)2 + 48ab
⇒ (4a)2 + (3b)2 + 2 × 4a × 3b - [(4a)2 + (3b)2 - 2 × 4a × 3b] + 48ab
⇒ 16a2 + 9b2 + 24ab - [16a2 + 9b2 - 24ab] + 48ab
⇒ 16a2 + 9b2 + 24ab - 16a2 - 9b2 + 24ab + 48ab
⇒ 96ab.
Hence, (4a + 3b)2 - (4a - 3b)2 + 48ab = 96ab.
If x + y = 7 2 \dfrac{7}{2} 2 7 5 2 \dfrac{5}{2} 2 5
(i) x - y
(ii) x2 - y2
Answer
(i) By formula,
(x - y)2 = (x + y)2 - 4xy
⇒ ( x − y ) 2 = ( 7 2 ) 2 − 4 × 5 2 ⇒ ( x − y ) 2 = 49 4 − 10 ⇒ ( x − y ) 2 = 49 − 40 4 ⇒ ( x − y ) 2 = 9 4 ⇒ ( x − y ) = 9 4 ⇒ x − y = ± 3 2 . \Rightarrow (x - y)^2 = \Big(\dfrac{7}{2}\Big)^2 - 4 \times \dfrac{5}{2} \\[1em] \Rightarrow (x - y)^2 = \dfrac{49}{4} - 10 \\[1em] \Rightarrow (x - y)^2 = \dfrac{49 - 40}{4} \\[1em] \Rightarrow (x - y)^2 = \dfrac{9}{4} \\[1em] \Rightarrow (x - y) = \sqrt{\dfrac{9}{4}} \\[1em] \Rightarrow x - y = \pm\dfrac{3}{2}. ⇒ ( x − y ) 2 = ( 2 7 ) 2 − 4 × 2 5 ⇒ ( x − y ) 2 = 4 49 − 10 ⇒ ( x − y ) 2 = 4 49 − 40 ⇒ ( x − y ) 2 = 4 9 ⇒ ( x − y ) = 4 9 ⇒ x − y = ± 2 3 .
Hence, x - y = ± 3 2 \pm \dfrac{3}{2} ± 2 3
(ii) Solving,
⇒ x 2 − y 2 ⇒ ( x − y ) ( x + y ) ⇒ ± 3 2 × 7 2 ⇒ ± 21 4 . \Rightarrow x^2 - y^2 \\[1em] \Rightarrow (x - y)(x + y) \\[1em] \Rightarrow \pm \dfrac{3}{2} \times \dfrac{7}{2} \\[1em] \Rightarrow \pm \dfrac{21}{4}. ⇒ x 2 − y 2 ⇒ ( x − y ) ( x + y ) ⇒ ± 2 3 × 2 7 ⇒ ± 4 21 .
Hence, x2 - y2 = ± 21 4 \pm \dfrac{21}{4} ± 4 21
If a - b = 0.9 and ab = 0.36; find :
(i) a + b
(ii) a2 - b2
Answer
(i) By formula,
⇒ (a + b)2 = (a - b)2 + 4ab
⇒ (a + b)2 = (0.9)2 + 4 × 0.36
⇒ (a + b)2 = 0.81 + 1.44
⇒ (a + b)2 = 2.25
⇒ a + b = 2.25 = ± 1.5 \sqrt{2.25} = \pm1.5 2.25 = ± 1.5
Hence, a + b = ± 1.5 \pm 1.5 ± 1.5
(ii) Solving,
⇒ a2 - b2
⇒ (a - b)(a + b)
⇒ 0.9 × ± 1.5 \pm 1.5 ± 1.5
⇒ ± 1.35 \pm 1.35 ± 1.35
Hence, a2 - b2 = ± 1.35 \pm 1.35 ± 1.35
If a - b = 4 and a + b = 6; find :
(i) a2 + b2
(ii) ab
Answer
(i) We know that,
(a - b)2 = a2 + b2 - 2ab .............(1)
(a + b)2 = a2 + b2 + 2ab ..............(2)
Adding equation (1) and (2), we get :
(a - b)2 + (a + b)2 = 2(a2 + b2 )
(a2 + b2 ) = ( a − b ) 2 + ( a + b ) 2 2 \dfrac{(a - b)^2 + (a + b)^2}{2} 2 ( a − b ) 2 + ( a + b ) 2
Substituting values we get :
⇒ a 2 + b 2 = 4 2 + 6 2 2 ⇒ a 2 + b 2 = 16 + 36 2 ⇒ a 2 + b 2 = 52 2 ⇒ a 2 + b 2 = 26. \Rightarrow a^2 + b^2 = \dfrac{4^2 + 6^2}{2} \\[1em] \Rightarrow a^2 + b^2 = \dfrac{16 + 36}{2} \\[1em] \Rightarrow a^2 + b^2 = \dfrac{52}{2} \\[1em] \Rightarrow a^2 + b^2 = 26. ⇒ a 2 + b 2 = 2 4 2 + 6 2 ⇒ a 2 + b 2 = 2 16 + 36 ⇒ a 2 + b 2 = 2 52 ⇒ a 2 + b 2 = 26.
Hence, a2 + b2 = 26.
(ii) We know that,
(a - b)2 = a2 + b2 - 2ab .............(1)
(a + b)2 = a2 + b2 + 2ab ..............(2)
Subtracting equation (1) from (2), we get :
⇒ (a + b)2 - (a - b)2 = a2 + b2 + 2ab - (a2 + b2 - 2ab)
⇒ (a + b)2 - (a - b)2 = 4ab
⇒ ab = ( a + b ) 2 − ( a − b ) 2 4 \dfrac{(a + b)^2 - (a - b)^2}{4} 4 ( a + b ) 2 − ( a − b ) 2
Substituting values we get :
⇒ a b = 6 2 − 4 2 4 = 36 − 16 4 = 20 4 = 5. \Rightarrow ab = \dfrac{6^2 - 4^2}{4} \\[1em] = \dfrac{36 - 16}{4} \\[1em] =\dfrac{20}{4} \\[1em] = 5. ⇒ ab = 4 6 2 − 4 2 = 4 36 − 16 = 4 20 = 5.
Hence, ab = 5.
If a + 1 a = 6 a + \dfrac{1}{a} = 6 a + a 1 = 6
(i) a − 1 a a - \dfrac{1}{a} a − a 1
(ii) a 2 − 1 a 2 a^2 - \dfrac{1}{a^2} a 2 − a 2 1
Answer
(i) By formula,
⇒ ( a − 1 a ) 2 = a 2 + 1 a 2 − ( 2 × a × 1 a ) ⇒ ( a − 1 a ) 2 = a 2 + 1 a 2 − 2 ⇒ ( a − 1 a ) 2 = a 2 + 1 a 2 + 2 − 4 ⇒ ( a − 1 a ) 2 = ( a + 1 a ) 2 − 4 ⇒ ( a − 1 a ) 2 = 6 2 − 4 ⇒ ( a − 1 a ) 2 = 36 − 4 ⇒ ( a − 1 a ) 2 = 32 ⇒ a − 1 a = 32 = ± 4 2 . \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} - \Big(2 \times a \times \dfrac{1}{a}\Big) \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} - 2 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} + 2 - 4 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = \Big(a + \dfrac{1}{a}\Big)^2 - 4 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 6^2 - 4 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 36 - 4 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 32 \\[1em] \Rightarrow a - \dfrac{1}{a} = \sqrt{32} = \pm 4\sqrt{2}. ⇒ ( a − a 1 ) 2 = a 2 + a 2 1 − ( 2 × a × a 1 ) ⇒ ( a − a 1 ) 2 = a 2 + a 2 1 − 2 ⇒ ( a − a 1 ) 2 = a 2 + a 2 1 + 2 − 4 ⇒ ( a − a 1 ) 2 = ( a + a 1 ) 2 − 4 ⇒ ( a − a 1 ) 2 = 6 2 − 4 ⇒ ( a − a 1 ) 2 = 36 − 4 ⇒ ( a − a 1 ) 2 = 32 ⇒ a − a 1 = 32 = ± 4 2 .
Hence, a − 1 a = ± 4 2 a - \dfrac{1}{a} = \pm 4\sqrt{2} a − a 1 = ± 4 2
(ii) Solving,
⇒ a 2 − 1 a 2 ⇒ ( a − 1 a ) ( a + 1 a ) ⇒ ± 4 2 × 6 ⇒ ± 24 2 . \Rightarrow a^2 - \dfrac{1}{a^2} \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)\Big(a + \dfrac{1}{a}\Big) \\[1em] \Rightarrow \pm 4\sqrt{2} \times 6 \\[1em] \Rightarrow \pm 24\sqrt{2}. ⇒ a 2 − a 2 1 ⇒ ( a − a 1 ) ( a + a 1 ) ⇒ ± 4 2 × 6 ⇒ ± 24 2 .
Hence, a 2 − 1 a 2 = ± 24 2 a^2 - \dfrac{1}{a^2} = \pm 24\sqrt{2} a 2 − a 2 1 = ± 24 2
If a − 1 a = 8 a - \dfrac{1}{a} = 8 a − a 1 = 8
(i) a + 1 a a + \dfrac{1}{a} a + a 1
(ii) a 2 − 1 a 2 a^2 - \dfrac{1}{a^2} a 2 − a 2 1
Answer
(i) By formula,
⇒ ( a + 1 a ) 2 − ( a − 1 a ) 2 \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 ⇒ ( a + a 1 ) 2 − ( a − a 1 ) 2
Substituting values we get :
⇒ ( a + 1 a ) 2 − 8 2 = 4 ⇒ ( a + 1 a ) 2 − 64 = 4 ⇒ ( a + 1 a ) 2 = 68 ⇒ a + 1 a = 68 = ± 2 17 . \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - 8^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - 64 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 68 \\[1em] \Rightarrow a + \dfrac{1}{a} = \sqrt{68} = \pm 2\sqrt{17}. ⇒ ( a + a 1 ) 2 − 8 2 = 4 ⇒ ( a + a 1 ) 2 − 64 = 4 ⇒ ( a + a 1 ) 2 = 68 ⇒ a + a 1 = 68 = ± 2 17 .
Hence, a + 1 a = ± 2 17 . a + \dfrac{1}{a} = \pm 2\sqrt{17}. a + a 1 = ± 2 17 .
(ii) By formula,
⇒ a 2 − 1 a 2 = ( a + 1 a ) ( a − 1 a ) \Rightarrow a^2 - \dfrac{1}{a^2} = \Big(a + \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a}\Big) ⇒ a 2 − a 2 1 = ( a + a 1 ) ( a − a 1 )
Substituting values we get :
⇒ a 2 − 1 a 2 = ± 2 17 × 8 = ± 16 17 . \Rightarrow a^2 - \dfrac{1}{a^2} = \pm 2\sqrt{17} \times 8 \\[1em] = \pm 16\sqrt{17}. ⇒ a 2 − a 2 1 = ± 2 17 × 8 = ± 16 17 .
Hence, a 2 − 1 a 2 = ± 16 17 a^2 - \dfrac{1}{a^2} = \pm 16\sqrt{17} a 2 − a 2 1 = ± 16 17
If a2 - 3a + 1 = 0 and a ≠ 0; find :
(i) a + 1 a a + \dfrac{1}{a} a + a 1
(ii) a 2 + 1 a 2 a^2 + \dfrac{1}{a^2} a 2 + a 2 1
Answer
(i) Given,
⇒ a2 - 3a + 1 = 0
⇒ a2 + 1 = 3a
Dividing above equation by a, we get :
⇒ a 2 + 1 a = 3 a a \Rightarrow \dfrac{a^2 + 1}{a} = \dfrac{3a}{a} ⇒ a a 2 + 1 = a 3 a
⇒ a + 1 a = 3 \Rightarrow a + \dfrac{1}{a} = 3 ⇒ a + a 1 = 3
Hence, a + 1 a = 3 a + \dfrac{1}{a} = 3 a + a 1 = 3
(ii) By formula,
⇒ ( a + 1 a ) 2 = a 2 + 1 a 2 + 2 ⇒ 3 2 = a 2 + 1 a 2 + 2 ⇒ a 2 + 1 a 2 = 9 − 2 ⇒ a 2 + 1 a 2 = 7. \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} + 2 \\[1em] \Rightarrow 3^2 = a^2 + \dfrac{1}{a^2} + 2 \\[1em] \Rightarrow a^2 + \dfrac{1}{a^2} = 9 - 2 \\[1em] \Rightarrow a^2 + \dfrac{1}{a^2} = 7. ⇒ ( a + a 1 ) 2 = a 2 + a 2 1 + 2 ⇒ 3 2 = a 2 + a 2 1 + 2 ⇒ a 2 + a 2 1 = 9 − 2 ⇒ a 2 + a 2 1 = 7.
Hence, a 2 + 1 a 2 = 7 a^2 + \dfrac{1}{a^2} = 7 a 2 + a 2 1 = 7
If a2 - 5a - 1 = 0 and a ≠ 0, find :
(i) a − 1 a a - \dfrac{1}{a} a − a 1
(ii) a + 1 a a + \dfrac{1}{a} a + a 1
(iii) a 2 − 1 a 2 a^2 - \dfrac{1}{a^2} a 2 − a 2 1
Answer
(i) Given,
⇒ a2 - 5a - 1 = 0
⇒ a2 - 1 = 5a
Dividing above equation by a, we get :
⇒ a 2 − 1 a = 5 a a \Rightarrow \dfrac{a^2 - 1}{a} = \dfrac{5a}{a} ⇒ a a 2 − 1 = a 5 a
⇒ a − 1 a = 5 \Rightarrow a - \dfrac{1}{a} = 5 ⇒ a − a 1 = 5
Hence, a − 1 a = 5 a - \dfrac{1}{a} = 5 a − a 1 = 5
(ii) By formula,
⇒ ( a + 1 a ) 2 − ( a − 1 a ) 2 = 4 ⇒ ( a + 1 a ) 2 − 5 2 = 4 ⇒ ( a + 1 a ) 2 − 25 = 4 ⇒ ( a + 1 a ) 2 = 29 ⇒ a + 1 a = ± 29 . \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - 5^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - 25 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 29 \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm \sqrt{29}. ⇒ ( a + a 1 ) 2 − ( a − a 1 ) 2 = 4 ⇒ ( a + a 1 ) 2 − 5 2 = 4 ⇒ ( a + a 1 ) 2 − 25 = 4 ⇒ ( a + a 1 ) 2 = 29 ⇒ a + a 1 = ± 29 .
Hence, a + 1 a = ± 29 a + \dfrac{1}{a} = \pm \sqrt{29} a + a 1 = ± 29
(iii) By formula,
⇒ ( a 2 − 1 a 2 ) = ( a + 1 a ) ( a − 1 a ) = ± 29 × 5 = ± 5 29 . \Rightarrow \Big(a^2 - \dfrac{1}{a^2}\Big) = \Big(a + \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a}\Big) \\[1em] = \pm \sqrt{29} \times 5 \\[1em] = \pm 5\sqrt{29}. ⇒ ( a 2 − a 2 1 ) = ( a + a 1 ) ( a − a 1 ) = ± 29 × 5 = ± 5 29 .
Hence, a 2 − 1 a 2 = ± 5 29 a^2 - \dfrac{1}{a^2} = \pm 5\sqrt{29} a 2 − a 2 1 = ± 5 29
If 3x + 4y = 16 and xy = 4; find the value of 9x2 + 16y2 .
Answer
Solving,
⇒ (3x + 4y)2 = (3x)2 + (4y)2 + 2 × 3x × 4y
⇒ (3x + 4y)2 = 9x2 + 16y2 + 24xy
Substituting values we get :
⇒ 162 = 9x2 + 16y2 + 24 × 4
⇒ 256 = 9x2 + 16y2 + 96
⇒ 9x2 + 16y2 = 256 - 96 = 160.
Hence, 9x2 + 16y2 = 160.
The difference between two positive numbers is 5 and the sum of their squares is 73. Find the product of these numbers.
Answer
Let two positive numbers be x and y.
Given,
Difference between two positive numbers is 5.
∴ x - y = 5
⇒ x = y + 5 ...........(1)
Given,
Sum of squares of numbers is 73.
∴ x2 + y2 = 73
Substituting value of x from equation (1) in above equation, we get :
⇒ (y + 5)2 + y2 = 73
⇒ y2 + 52 + 2 × y × 5 + y2 = 73
⇒ 2y2 + 25 + 10y = 73
⇒ 2y2 + 10y + 25 - 73 = 0
⇒ 2y2 + 10y - 48 = 0
⇒ 2(y2 + 5y - 24) = 0
⇒ y2 + 5y - 24 = 0
⇒ y2 + 8y - 3y - 24 = 0
⇒ y(y + 8) - 3(y + 8) = 0
⇒ (y - 3)(y + 8) = 0
⇒ y - 3 = 0 or y + 8 = 0
⇒ y = 3 or y = -8.
If y = 3, x = y + 5 = 3 + 5 = 8, xy = 24,
If y = -8, x = y + 5 = -8 + 5 = -3, xy = 24.
Hence, product of numbers = 24.
If x2 - 2x + 1 = 0; the value of x 4 + 1 x 4 x^4 + \dfrac{1}{x^4} x 4 + x 4 1
9
7
3
2
Answer
Given,
⇒ x2 - 2x + 1 = 0
⇒ x2 + 1 = 2x
Dividing above equation by x, we get :
⇒ x 2 + 1 x = 2 x x \dfrac{x^2 + 1}{x} = \dfrac{2x}{x} x x 2 + 1 = x 2 x
⇒ x + 1 x x + \dfrac{1}{x} x + x 1
Squaring both sides we get :
⇒ ( x + 1 x ) 2 = 2 2 ⇒ x 2 + 1 x 2 + 2 × x × 1 x = 4 ⇒ x 2 + 1 x 2 + 2 = 4 ⇒ x 2 + 1 x 2 = 2. \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 2^2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 2 \times x \times \dfrac{1}{x} = 4 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 2 = 4 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 2. ⇒ ( x + x 1 ) 2 = 2 2 ⇒ x 2 + x 2 1 + 2 × x × x 1 = 4 ⇒ x 2 + x 2 1 + 2 = 4 ⇒ x 2 + x 2 1 = 2.
Squaring both sides we get :
⇒ ( x 2 + 1 x 2 ) 2 = 2 2 ⇒ ( x 2 ) 2 + ( 1 x 2 ) 2 + 2 × x 2 × 1 x 2 = 4 ⇒ x 4 + 1 x 4 + 2 = 4 ⇒ x 4 + 1 x 4 = 4 − 2 ⇒ x 4 + 1 x 4 = 2. \Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = 2^2 \\[1em] \Rightarrow (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{x^2} = 4 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} + 2 = 4 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 4 - 2 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 2. ⇒ ( x 2 + x 2 1 ) 2 = 2 2 ⇒ ( x 2 ) 2 + ( x 2 1 ) 2 + 2 × x 2 × x 2 1 = 4 ⇒ x 4 + x 4 1 + 2 = 4 ⇒ x 4 + x 4 1 = 4 − 2 ⇒ x 4 + x 4 1 = 2.
Hence, Option 4 is the correct option.
If a = 5, b = -2 and c = -3, a3 + b3 + c3 is equal to :
-90
90
60
-60
Answer
By property,
If a + b + c = 0, then
a3 + b3 + c3 = 3abc
Since, 5 + (-2) + (-3) = 5 - 2 - 3 = 0.
∴ a3 + b3 + c3 = 3 × 5 × -2 × -3 = 90.
Hence, Option 2 is the correct option.
If x 2 + 1 x 2 = 2 x^2 + \dfrac{1}{x^2} = 2 x 2 + x 2 1 = 2 x 2 − 1 x 2 x^2 - \dfrac{1}{x^2} x 2 − x 2 1
0
2
± 2 \pm 2 ± 2
8
Answer
By formula,
⇒ ( x 2 + 1 x 2 ) 2 − ( x 2 − 1 x 2 ) 2 \Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 - \Big(x^2 - \dfrac{1}{x^2}\Big)^2 ⇒ ( x 2 + x 2 1 ) 2 − ( x 2 − x 2 1 ) 2
Substituting values we get :
⇒ 2 2 − ( x 2 − 1 x 2 ) 2 = 4 ⇒ 4 − ( x 2 − 1 x 2 ) 2 = 4 ⇒ ( x 2 − 1 x 2 ) 2 = 4 − 4 ⇒ ( x 2 − 1 x 2 ) 2 = 0 ⇒ x 2 − 1 x 2 = 0. \Rightarrow 2^2 - \Big(x^2 - \dfrac{1}{x^2}\Big)^2 = 4 \\[1em] \Rightarrow 4 - \Big(x^2 - \dfrac{1}{x^2}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big)^2 = 4 - 4 \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big)^2 = 0 \\[1em] \Rightarrow x^2 - \dfrac{1}{x^2} = 0. ⇒ 2 2 − ( x 2 − x 2 1 ) 2 = 4 ⇒ 4 − ( x 2 − x 2 1 ) 2 = 4 ⇒ ( x 2 − x 2 1 ) 2 = 4 − 4 ⇒ ( x 2 − x 2 1 ) 2 = 0 ⇒ x 2 − x 2 1 = 0.
Hence, Option 1 is the correct option.
If x + 3y + 2z = 0, the value of x3 + 27y3 + 8z3 is :
9xyz
6xyz
18xyz
xyz
Answer
By property,
If a + b + c = 0, then
a3 + b3 + c3 = 3abc ....(1)
Comparing equation x + 3y + 2z = 0, with a + b + c = 0, we get :
a = x, b = 3y and c = 2z.
Substituting values in equation (1), we get :
x3 + (3y)3 + (2z)3 = 3 × x × 3y × 2z = 18xyz.
Hence, Option 3 is the correct option.
Find the cube of 3a - 2b.
Answer
Solving,
(3a - 2b)3 = (3a)3 - (2b)3 - 3 × 3a × 2b(3a - 2b)
= 27a3 - 8b3 - 18ab(3a - 2b)
= 27a3 - 8b3 - 54a2 b + 36ab2 .
Hence, cube of 3a - 2b = 27a3 - 8b3 - 54a2 b + 36ab2 .
Find the cube of 5a + 3b.
Answer
Solving,
(5a + 3b)3 = (5a)3 + (3b)3 + 3 × 5a × 3b(5a + 3b)
= 125a3 + 27b3 + 45ab(5a + 3b)
= 125a3 + 27b3 + 225a2 b + 135ab2 .
Hence, cube of 5a + 3b = 125a3 + 27b3 + 225a2 b + 135ab2 .
Find the cube of 2a + 1 2 a \dfrac{1}{2a} 2 a 1
Answer
Solving,
( 2 a + 1 2 a ) 3 = ( 2 a ) 3 + ( 1 2 a ) 3 + 3 × 2 a × 1 2 a × ( 2 a + 1 2 a ) ⇒ 8 a 3 + 1 8 a 3 + 3 ( 2 a + 1 2 a ) ⇒ 8 a 3 + 1 8 a 3 + 6 a + 3 2 a . \Big(2a + \dfrac{1}{2a}\Big)^3 = (2a)^3 + \Big(\dfrac{1}{2a}\Big)^3 + 3 \times 2a \times \dfrac{1}{2a} \times \Big(2a + \dfrac{1}{2a}\Big) \\[1em] \Rightarrow 8a^3 + \dfrac{1}{8a^3} + 3\Big(2a + \dfrac{1}{2a}\Big) \\[1em] \Rightarrow 8a^3 + \dfrac{1}{8a^3} + 6a + \dfrac{3}{2a}. ( 2 a + 2 a 1 ) 3 = ( 2 a ) 3 + ( 2 a 1 ) 3 + 3 × 2 a × 2 a 1 × ( 2 a + 2 a 1 ) ⇒ 8 a 3 + 8 a 3 1 + 3 ( 2 a + 2 a 1 ) ⇒ 8 a 3 + 8 a 3 1 + 6 a + 2 a 3 .
Hence, ( 2 a + 1 2 a ) 3 = 8 a 3 + 1 8 a 3 + 6 a + 3 2 a \Big(2a + \dfrac{1}{2a}\Big)^3 = 8a^3 + \dfrac{1}{8a^3} + 6a + \dfrac{3}{2a} ( 2 a + 2 a 1 ) 3 = 8 a 3 + 8 a 3 1 + 6 a + 2 a 3
Find the cube of 3 a − 1 a 3a - \dfrac{1}{a} 3 a − a 1
Answer
Solving,
⇒ ( 3 a − 1 a ) 3 = ( 3 a ) 3 − ( 1 a ) 3 − 3 × 3 a × 1 a × ( 3 a − 1 a ) = 27 a 3 − 1 a 3 − 9 ( 3 a − 1 a ) = 27 a 3 − 1 a 3 − 27 a + 9 a . \Rightarrow \Big(3a - \dfrac{1}{a}\Big)^3 = (3a)^3 - \Big(\dfrac{1}{a}\Big)^3 - 3 \times 3a \times \dfrac{1}{a} \times \Big(3a - \dfrac{1}{a}\Big) \\[1em] = 27a^3 - \dfrac{1}{a^3} - 9\Big(3a - \dfrac{1}{a}\Big) \\[1em] = 27a^3 - \dfrac{1}{a^3} - 27a + \dfrac{9}{a}. ⇒ ( 3 a − a 1 ) 3 = ( 3 a ) 3 − ( a 1 ) 3 − 3 × 3 a × a 1 × ( 3 a − a 1 ) = 27 a 3 − a 3 1 − 9 ( 3 a − a 1 ) = 27 a 3 − a 3 1 − 27 a + a 9 .
Hence, ( 3 a − 1 a ) 3 = 27 a 3 − 1 a 3 − 27 a + 9 a . \Big(3a - \dfrac{1}{a}\Big)^3 = 27a^3 - \dfrac{1}{a^3} - 27a + \dfrac{9}{a}. ( 3 a − a 1 ) 3 = 27 a 3 − a 3 1 − 27 a + a 9 .
If a 2 + 1 a 2 = 47 a^2 + \dfrac{1}{a^2} = 47 a 2 + a 2 1 = 47
(i) a + 1 a a + \dfrac{1}{a} a + a 1
(ii) a 3 + 1 a 3 a^3 + \dfrac{1}{a^3} a 3 + a 3 1
Answer
(i) By formula,
⇒ ( a + 1 a ) 2 = a 2 + 1 a 2 + 2 ⇒ ( a + 1 a ) 2 = 47 + 2 ⇒ ( a + 1 a ) 2 = 49 ⇒ a + 1 a = 49 ⇒ a + 1 a = ± 7. \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} + 2 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 47 + 2 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 49 \\[1em] \Rightarrow a + \dfrac{1}{a} = \sqrt{49} \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm 7. ⇒ ( a + a 1 ) 2 = a 2 + a 2 1 + 2 ⇒ ( a + a 1 ) 2 = 47 + 2 ⇒ ( a + a 1 ) 2 = 49 ⇒ a + a 1 = 49 ⇒ a + a 1 = ± 7.
Hence, a + 1 a = ± 7. a + \dfrac{1}{a} = \pm 7. a + a 1 = ± 7.
(ii) By formula,
( a + 1 a ) 3 = a 3 + 1 a 3 + 3 ( a + 1 a ) \Big(a + \dfrac{1}{a}\Big)^3 = a^3 + \dfrac{1}{a^3} + 3\Big(a + \dfrac{1}{a}\Big) ( a + a 1 ) 3 = a 3 + a 3 1 + 3 ( a + a 1 )
Substituting a + 1 a = 7 a + \dfrac{1}{a} = 7 a + a 1 = 7
⇒ 7 3 = a 3 + 1 a 3 + 3 × 7 ⇒ 343 = a 3 + 1 a 3 + 21 ⇒ a 3 + 1 a 3 = 343 − 21 ⇒ a 3 + 1 a 3 = 322. \Rightarrow 7^3 = a^3 + \dfrac{1}{a^3} + 3 \times 7 \\[1em] \Rightarrow 343 = a^3 + \dfrac{1}{a^3} + 21 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = 343 - 21 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = 322. ⇒ 7 3 = a 3 + a 3 1 + 3 × 7 ⇒ 343 = a 3 + a 3 1 + 21 ⇒ a 3 + a 3 1 = 343 − 21 ⇒ a 3 + a 3 1 = 322.
Substituting a + 1 a = − 7 a + \dfrac{1}{a} = -7 a + a 1 = − 7
⇒ ( − 7 ) 3 = a 3 + 1 a 3 + 3 × − 7 ⇒ − 343 = a 3 + 1 a 3 − 21 ⇒ a 3 + 1 a 3 = − 343 + 21 ⇒ a 3 + 1 a 3 = − 322. \Rightarrow (-7)^3 = a^3 + \dfrac{1}{a^3} + 3 \times -7 \\[1em] \Rightarrow -343 = a^3 + \dfrac{1}{a^3} - 21 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = -343 + 21 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = -322. ⇒ ( − 7 ) 3 = a 3 + a 3 1 + 3 × − 7 ⇒ − 343 = a 3 + a 3 1 − 21 ⇒ a 3 + a 3 1 = − 343 + 21 ⇒ a 3 + a 3 1 = − 322.
Hence, a 3 + 1 a 3 = ± 322. a^3 + \dfrac{1}{a^3} = \pm 322. a 3 + a 3 1 = ± 322.
If a 2 + 1 a 2 = 18 a^2 + \dfrac{1}{a^2} = 18 a 2 + a 2 1 = 18
(i) a − 1 a a - \dfrac{1}{a} a − a 1
(ii) a 3 − 1 a 3 a^3 - \dfrac{1}{a^3} a 3 − a 3 1
Answer
(i) By formula,
⇒ ( a − 1 a ) 2 = a 2 + 1 a 2 − 2 ⇒ ( a − 1 a ) 2 = 18 − 2 ⇒ ( a − 1 a ) 2 = 16 ⇒ a − 1 a = 16 ⇒ a − 1 a = ± 4. \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} - 2 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 18 - 2 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 16 \\[1em] \Rightarrow a - \dfrac{1}{a} = \sqrt{16} \\[1em] \Rightarrow a - \dfrac{1}{a} = \pm 4. ⇒ ( a − a 1 ) 2 = a 2 + a 2 1 − 2 ⇒ ( a − a 1 ) 2 = 18 − 2 ⇒ ( a − a 1 ) 2 = 16 ⇒ a − a 1 = 16 ⇒ a − a 1 = ± 4.
Hence, a − 1 a = ± 4 a - \dfrac{1}{a} = \pm 4 a − a 1 = ± 4
(ii) By formula,
( a − 1 a ) 3 = a 3 − 1 a 3 − 3 ( a − 1 a ) \Big(a - \dfrac{1}{a}\Big)^3 = a^3 - \dfrac{1}{a^3} -3\Big(a - \dfrac{1}{a}\Big) ( a − a 1 ) 3 = a 3 − a 3 1 − 3 ( a − a 1 )
Substituting a − 1 a = 4 a - \dfrac{1}{a} = 4 a − a 1 = 4
⇒ 4 3 = a 3 − 1 a 3 − 3 × 4 ⇒ 64 = a 3 − 1 a 3 − 12 ⇒ 64 + 12 = a 3 − 1 a 3 ⇒ a 3 − 1 a 3 = 76. \Rightarrow 4^3 = a^3 - \dfrac{1}{a^3} - 3 \times 4 \\[1em] \Rightarrow 64 = a^3 - \dfrac{1}{a^3} - 12 \\[1em] \Rightarrow 64 + 12 = a^3 - \dfrac{1}{a^3} \\[1em] \Rightarrow a^3 - \dfrac{1}{a^3} = 76. ⇒ 4 3 = a 3 − a 3 1 − 3 × 4 ⇒ 64 = a 3 − a 3 1 − 12 ⇒ 64 + 12 = a 3 − a 3 1 ⇒ a 3 − a 3 1 = 76.
Substituting a − 1 a = − 4 a - \dfrac{1}{a} = -4 a − a 1 = − 4
⇒ ( − 4 ) 3 = a 3 − 1 a 3 − 3 × − 4 ⇒ − 64 = a 3 − 1 a 3 + 12 ⇒ − 64 − 12 = a 3 − 1 a 3 ⇒ a 3 − 1 a 3 = − 76. \Rightarrow (-4)^3 = a^3 - \dfrac{1}{a^3} - 3 \times -4 \\[1em] \Rightarrow -64 = a^3 - \dfrac{1}{a^3} + 12 \\[1em] \Rightarrow -64 - 12 = a^3 - \dfrac{1}{a^3} \\[1em] \Rightarrow a^3 - \dfrac{1}{a^3} = -76. ⇒ ( − 4 ) 3 = a 3 − a 3 1 − 3 × − 4 ⇒ − 64 = a 3 − a 3 1 + 12 ⇒ − 64 − 12 = a 3 − a 3 1 ⇒ a 3 − a 3 1 = − 76.
Hence, a 3 − 1 a 3 = ± 76 a^3 - \dfrac{1}{a^3} = \pm 76 a 3 − a 3 1 = ± 76
If a + 1 a = p a + \dfrac{1}{a} = p a + a 1 = p
a 3 + 1 a 3 = p ( p 2 − 3 ) a^3 + \dfrac{1}{a^3} = p(p^2 - 3) a 3 + a 3 1 = p ( p 2 − 3 )
Answer
By formula,
a 3 + 1 a 3 = ( a + 1 a ) 3 − 3 ( a + 1 a ) = p 3 − 3 × p = p 3 − 3 p = p ( p 2 − 3 ) a^3 + \dfrac{1}{a^3} = \Big(a + \dfrac{1}{a}\Big)^3 - 3\Big(a + \dfrac{1}{a}\Big) \\[1em] = p^3 - 3 \times p \\[1em] = p^3 - 3p \\[1em] = p(p^2 - 3) a 3 + a 3 1 = ( a + a 1 ) 3 − 3 ( a + a 1 ) = p 3 − 3 × p = p 3 − 3 p = p ( p 2 − 3 )
Hence, proved that a 3 + 1 a 3 = p ( p 2 − 3 ) a^3 + \dfrac{1}{a^3} = p(p^2 - 3) a 3 + a 3 1 = p ( p 2 − 3 )
If a + 2b = 5; then show that :
a3 + 8b3 + 30ab = 125.
Answer
Given,
⇒ a + 2b = 5
Cubing both sides we get :
⇒ (a + 2b)3 = 53
⇒ a3 + (2b)3 + 3 × a × 2b (a + 2b) = 125
⇒ a3 + 8b3 + 6ab(a + 2b) = 125
⇒ a3 + 8b3 + 6ab × 5 = 125
⇒ a3 + 8b3 + 30ab = 125.
Hence, proved that a3 + 8b3 + 30ab = 125.
If ( a + 1 a ) 2 = 3 \Big(a + \dfrac{1}{a}\Big)^2 = 3 ( a + a 1 ) 2 = 3
a 3 + 1 a 3 = 0. a^3 + \dfrac{1}{a^3} = 0. a 3 + a 3 1 = 0.
Answer
Given,
⇒ ( a + 1 a ) 2 = 3 ⇒ a + 1 a = ± 3 \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 3 \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm\sqrt{3} ⇒ ( a + a 1 ) 2 = 3 ⇒ a + a 1 = ± 3
By formula,
⇒ a 3 + 1 a 3 = ( a + 1 a ) 3 − 3 ( a + 1 a ) \Rightarrow a^3 + \dfrac{1}{a^3} = \Big(a + \dfrac{1}{a}\Big)^3 - 3\Big(a + \dfrac{1}{a}\Big) ⇒ a 3 + a 3 1 = ( a + a 1 ) 3 − 3 ( a + a 1 )
Substituting a + 1 a = − 3 a + \dfrac{1}{a} = -\sqrt{3} a + a 1 = − 3
⇒ a 3 + 1 a 3 = ( − 3 ) 3 − 3 × − 3 = − 3 3 + 3 3 = 0. \Rightarrow a^3 + \dfrac{1}{a^3} = (-\sqrt{3})^3 - 3 \times -\sqrt{3} \\[1em] = -3\sqrt{3} + 3\sqrt{3} \\[1em] = 0. ⇒ a 3 + a 3 1 = ( − 3 ) 3 − 3 × − 3 = − 3 3 + 3 3 = 0.
Substituting a + 1 a = 3 a + \dfrac{1}{a} = \sqrt{3} a + a 1 = 3
⇒ a 3 + 1 a 3 = ( 3 ) 3 − 3 × 3 = 3 3 − 3 3 = 0. \Rightarrow a^3 + \dfrac{1}{a^3} = (\sqrt{3})^3 - 3 \times \sqrt{3} \\[1em] = 3\sqrt{3} - 3\sqrt{3} \\[1em] = 0. ⇒ a 3 + a 3 1 = ( 3 ) 3 − 3 × 3 = 3 3 − 3 3 = 0.
Hence, proved that a 3 + 1 a 3 = 0. a^3 + \dfrac{1}{a^3} = 0. a 3 + a 3 1 = 0.
If a + 2b + c = 0; then show that :
a3 + 8b3 + c3 = 6abc.
Answer
Given,
a + 2b + c = 0 ........(1)
By property,
If x + y + z = 0, then
⇒ x3 + y3 + z3 = 3xyz .........(2)
Comparing,
Eq. 1 with x + y + z = 0, we get :
x = a, y = 2b and z = c.
Then,
Substituting value of x, y and z in Eq. 2, we get :
⇒ a3 + (2b)3 + c3 = 3 × a × 2b × c
⇒ a3 + 8b3 + c3 = 6abc.
Hence, proved that a3 + 8b3 + c3 = 6abc.
Use property to evaluate:
(i) 93 - 53 - 43
(ii) 383 + (-26)3 + (-12)3
Answer
(i) By property,
If a + b + c = 0, then
⇒ a3 + b3 + c3 = 3abc
Given,
⇒ 93 - 53 - 43
⇒ 93 + (-5)3 + (-4)3
Let a = 9, b = -5 and c = -4.
∴ a + b + c = 9 + (-5) + (-4) = 9 - 5 - 4 = 0
∴ 93 + (-5)3 + (-4)3 = 3 × 9 × (-5) × (-4) = 540.
Hence, 93 - 53 - 43 = 540.
(ii) By property,
If a + b + c = 0, then
⇒ a3 + b3 + c3 = 3abc
Given,
⇒ 383 + (-26)3 + (-12)3
Let a = 38, b = -26 and c = -12.
∴ a + b + c = 38 + (-26) + (-12) = 38 - 26 - 12 = 0
∴ 383 + (-26)3 + (-12)3 = 3 × 38 × (-26) × (-12) = 35,568.
Hence, 383 + (-26)3 + (-12)3 = 35568.
If a≠ 0 and a − 1 a = 3 a - \dfrac{1}{a} = 3 a − a 1 = 3
(i) a 2 + 1 a 2 a^2 + \dfrac{1}{a^2} a 2 + a 2 1
(ii) a 3 − 1 a 3 a^3 - \dfrac{1}{a^3} a 3 − a 3 1
Answer
(i) By formula,
⇒ a 2 + 1 a 2 = ( a − 1 a ) 2 + 2 = 3 2 + 2 = 9 + 2 = 11. \Rightarrow a^2 + \dfrac{1}{a^2} = \Big(a - \dfrac{1}{a}\Big)^2 + 2 \\[1em] = 3^2 + 2 \\[1em] = 9 + 2 = 11. ⇒ a 2 + a 2 1 = ( a − a 1 ) 2 + 2 = 3 2 + 2 = 9 + 2 = 11.
Hence, a 2 + 1 a 2 = 11 a^2 + \dfrac{1}{a^2} = 11 a 2 + a 2 1 = 11
(ii) By formula,
⇒ ( a − 1 a ) 3 = a 3 − 1 a 3 − 3 ( a − 1 a ) ⇒ 3 3 = a 3 − 1 a 3 − 3 × 3 ⇒ 27 = a 3 − 1 a 3 − 9 ⇒ a 3 − 1 a 3 = 36. \Rightarrow \Big(a - \dfrac{1}{a}\Big)^3 = a^3 - \dfrac{1}{a^3} - 3\Big(a -\dfrac{1}{a}\Big) \\[1em] \Rightarrow 3^3 = a^3 - \dfrac{1}{a^3} - 3 \times 3 \\[1em] \Rightarrow 27 = a^3 - \dfrac{1}{a^3} - 9 \\[1em] \Rightarrow a^3 - \dfrac{1}{a^3} = 36. ⇒ ( a − a 1 ) 3 = a 3 − a 3 1 − 3 ( a − a 1 ) ⇒ 3 3 = a 3 − a 3 1 − 3 × 3 ⇒ 27 = a 3 − a 3 1 − 9 ⇒ a 3 − a 3 1 = 36.
Hence, a 3 − 1 a 3 = 36 a^3 - \dfrac{1}{a^3} = 36 a 3 − a 3 1 = 36
If a ≠ 0 and a − 1 a = 4 a - \dfrac{1}{a} = 4 a − a 1 = 4
(i) a 2 + 1 a 2 a^2 + \dfrac{1}{a^2} a 2 + a 2 1
(ii) a 4 + 1 a 4 a^4 + \dfrac{1}{a^4} a 4 + a 4 1
(iii) a 3 − 1 a 3 a^3 - \dfrac{1}{a^3} a 3 − a 3 1
Answer
(i) By formula,
⇒ a 2 + 1 a 2 = ( a − 1 a ) 2 + 2 \Rightarrow a^2 + \dfrac{1}{a^2} = \Big(a - \dfrac{1}{a}\Big)^2 + 2 ⇒ a 2 + a 2 1 = ( a − a 1 ) 2 + 2
Substituting values we get :
⇒ a 2 + 1 a 2 = 4 2 + 2 = 16 + 2 = 18. \Rightarrow a^2 + \dfrac{1}{a^2} = 4^2 + 2 \\[1em] = 16 + 2 \\[1em] = 18. ⇒ a 2 + a 2 1 = 4 2 + 2 = 16 + 2 = 18.
Hence, a 2 + 1 a 2 = 18 a^2 + \dfrac{1}{a^2} = 18 a 2 + a 2 1 = 18
(ii) By formula,
⇒ a 4 + 1 a 4 = ( a 2 + 1 a 2 ) 2 − 2 \Rightarrow a^4 + \dfrac{1}{a^4} = \Big(a^2 + \dfrac{1}{a^2}\Big)^2 - 2 ⇒ a 4 + a 4 1 = ( a 2 + a 2 1 ) 2 − 2
Substituting values we get :
⇒ a 4 + 1 a 4 = 18 2 − 2 = 324 − 2 = 322. \Rightarrow a^4 + \dfrac{1}{a^4} = 18^2 - 2 \\[1em] = 324 - 2 \\[1em] = 322. ⇒ a 4 + a 4 1 = 1 8 2 − 2 = 324 − 2 = 322.
Hence, a 4 + 1 a 4 = 322 a^4 + \dfrac{1}{a^4} = 322 a 4 + a 4 1 = 322
(iii) By formula,
⇒ a 3 − 1 a 3 = ( a − 1 a ) 3 + 3 ( a − 1 a ) \Rightarrow a^3 - \dfrac{1}{a^3} = \Big(a - \dfrac{1}{a}\Big)^3 + 3\Big(a - \dfrac{1}{a}\Big) ⇒ a 3 − a 3 1 = ( a − a 1 ) 3 + 3 ( a − a 1 )
Substituting values we get :
⇒ a 3 − 1 a 3 = 4 3 + 3 × 4 = 64 + 12 = 76. \Rightarrow a^3 - \dfrac{1}{a^3} = 4^3 + 3 \times 4 \\[1em] = 64 + 12 \\[1em] = 76. ⇒ a 3 − a 3 1 = 4 3 + 3 × 4 = 64 + 12 = 76.
Hence, a 3 − 1 a 3 = 76 a^3 - \dfrac{1}{a^3} = 76 a 3 − a 3 1 = 76
If x ≠ 0 and x + 1 x = 2 x + \dfrac{1}{x} = 2 x + x 1 = 2
x 2 + 1 x 2 = x 3 + 1 x 3 = x 4 + 1 x 4 x^2 + \dfrac{1}{x^2} = x^3 + \dfrac{1}{x^3} = x^4 + \dfrac{1}{x^4} x 2 + x 2 1 = x 3 + x 3 1 = x 4 + x 4 1
Answer
By formula,
⇒ x 2 + 1 x 2 = ( x + 1 x ) 2 − 2 \Rightarrow x^2 + \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)^2 - 2 ⇒ x 2 + x 2 1 = ( x + x 1 ) 2 − 2
Substituting values we get :
⇒ x 2 + 1 x 2 = 2 2 − 2 = 4 − 2 = 2. \Rightarrow x^2 + \dfrac{1}{x^2} = 2^2 - 2 \\[1em] = 4 - 2 \\[1em] = 2. ⇒ x 2 + x 2 1 = 2 2 − 2 = 4 − 2 = 2.
By formula,
⇒ x 3 + 1 x 3 = ( x + 1 x ) 3 − 3 ( x + 1 x ) \Rightarrow x^3 + \dfrac{1}{x^3} = \Big(x + \dfrac{1}{x}\Big)^3 - 3\Big(x + \dfrac{1}{x}\Big) ⇒ x 3 + x 3 1 = ( x + x 1 ) 3 − 3 ( x + x 1 )
Substituting values we get :
⇒ x 3 + 1 x 3 = 2 3 − 3 × 2 = 8 − 6 = 2. \Rightarrow x^3 + \dfrac{1}{x^3} = 2^3 - 3 \times 2 \\[1em] = 8 - 6 \\[1em] = 2. ⇒ x 3 + x 3 1 = 2 3 − 3 × 2 = 8 − 6 = 2.
By formula,
⇒ x 4 + 1 x 4 = ( x 2 + 1 x 2 ) 2 − 2 \Rightarrow x^4 + \dfrac{1}{x^4} = \Big(x^2 + \dfrac{1}{x^2}\Big)^2 - 2 ⇒ x 4 + x 4 1 = ( x 2 + x 2 1 ) 2 − 2
Substituting values we get :
⇒ x 4 + 1 x 4 = 2 2 − 2 = 4 − 2 = 2. \Rightarrow x^4 + \dfrac{1}{x^4} = 2^2 - 2 \\[1em] = 4 - 2 \\[1em] = 2. ⇒ x 4 + x 4 1 = 2 2 − 2 = 4 − 2 = 2.
Hence, proved that x 2 + 1 x 2 = x 3 + 1 x 3 = x 4 + 1 x 4 x^2 + \dfrac{1}{x^2} = x^3 + \dfrac{1}{x^3} = x^4 + \dfrac{1}{x^4} x 2 + x 2 1 = x 3 + x 3 1 = x 4 + x 4 1
If 2x - 3y = 10 and xy = 16, find the value of 8x3 - 27y3 .
Answer
Given,
2x - 3y = 10
Cubing both sides we get :
⇒ (2x - 3y)3 = 103
⇒ (2x)3 - (3y)3 - 3 × 2x × 3y(2x - 3y) = 1000
⇒ 8x3 - 27y3 - 18xy(2x - 3y) = 1000
⇒ 8x3 - 27y3 - 18 × 16 × 10 = 1000
⇒ 8x3 - 27y3 - 2880 = 1000
⇒ 8x3 - 27y3 = 3880.
Hence, 8x3 - 27y3 = 3880.
Expand (3x + 5y + 2z)(3x - 5y + 2z)
Answer
Given,
⇒ (3x + 5y + 2z)(3x - 5y + 2z)
⇒ 9x2 - 15xy + 6xz + 15xy - 25y2 + 10yz + 6zx - 10yz + 4z2
⇒ 9x2 - 25y2 + 4z2 + 12xz.
Hence, (3x + 5y + 2z)(3x - 5y + 2z) = 9x2 - 25y2 + 4z2 + 12xz.
Expand (3x - 5y - 2z)(3x - 5y + 2z)
Answer
Given,
⇒ ( 3 x − 5 y − 2 z ) ( 3 x − 5 y + 2 z ) ⇒ 9 x 2 − 15 x y + 6 z x − 15 x y + 25 y 2 − 10 y z − 6 z x + 10 y z − 4 z 2 ⇒ 9 x 2 − 15 x y + 6 z x − 15 x y + 25 y 2 − 10 y z − 6 z x + 10 y z − 4 z 2 ⇒ 9 x 2 − 30 x y + 25 y 2 − 4 z 2 . ⇒ (3x - 5y - 2z)(3x - 5y + 2z) \\[1em] ⇒ 9x^2 - 15xy + 6zx - 15xy + 25y^2 - 10yz - 6zx + 10yz - 4z^2 \\[1em] ⇒ 9x^2 - 15xy + \cancel{6zx} - 15xy + 25y^2 - \cancel{10yz} - \cancel{6zx} + \cancel{10yz} - 4z^2 \\[1em] ⇒ 9x^2 - 30xy + 25y^2 - 4z^2 . ⇒ ( 3 x − 5 y − 2 z ) ( 3 x − 5 y + 2 z ) ⇒ 9 x 2 − 15 x y + 6 z x − 15 x y + 25 y 2 − 10 yz − 6 z x + 10 yz − 4 z 2 ⇒ 9 x 2 − 15 x y + 6 z x − 15 x y + 25 y 2 − 10 yz − 6 z x + 10 yz − 4 z 2 ⇒ 9 x 2 − 30 x y + 25 y 2 − 4 z 2 .
Hence, (3x - 5y - 2z)(3x - 5y + 2z) = 9x2 - 30xy + 25y2 - 4z2 .
The sum of two numbers is 9 and their product is 20. Find the sum of their :
(i) squares
(ii) cubes.
Answer
Let two numbers be a and b.
Given,
Sum of numbers = 9 and Product = 20.
∴ a + b = 9 and ab = 20.
(i) Sum of squares = a2 + b2
By formula,
⇒ (a + b)2 = a2 + b2 + 2ab
Substituting values we get :
⇒ 92 = a2 + b2 + 2 × 20
⇒ 81 = a2 + b2 + 40
⇒ a2 + b2 = 81 - 40 = 41.
Hence, sum of squares = 41.
(ii) Sum of cubes = a3 + b3
By formula,
⇒ (a + b)3 = a3 + b3 + 3ab(a + b)
⇒ 93 = a3 + b3 + 3 × 20 × 9
⇒ 729 = a3 + b3 + 540
⇒ a3 + b3 = 729 - 540
⇒ a3 + b3 = 189.
Hence, a3 + b3 = 189.
Two positive numbers x and y are such that x > y. If the difference of these numbers is 5 and their product is 24, find :
(i) sum of these numbers.
(ii) difference of their cubes.
(iii) sum of their cubes.
Answer
Given,
Difference of numbers = 5 and product = 24.
Let the two numbers be x and y.
∴ x - y = 5 and xy = 24.
(i) By formula,
(x + y)2 = (x - y)2 + 4xy
Substituting values we get :
⇒ (x + y)2 = 52 + 4 × 24
⇒ (x + y)2 = 25 + 96
⇒ (x + y)2 = 121
⇒ (x + y) = 121 = ± 11 \sqrt{121} = \pm 11 121 = ± 11
Since, numbers are positive so sum cannot be negative.
Hence, sum of these numbers = 11.
(ii) By formula,
⇒ x3 - y3 = (x - y)3 + 3xy(x - y)
⇒ x3 - y3 = 53 + 3 × 24 × 5
⇒ x3 - y3 = 125 + 360
⇒ x3 - y3 = 485.
Hence, difference of cubes of numbers = 485.
(iii) By formula,
⇒ x3 + y3 = (x + y)3 - 3xy(x + y)
Substituting x + y = 11, we get :
⇒ x3 + y3 = 113 - 3 × 24 × 11
⇒ x3 + y3 = 1331 - 792
⇒ x3 + y3 = 539.
Hence, sum of cubes of numbers = 539.
If 4x2 + y2 = a and xy = b, find the value of 2x + y.
Answer
⇒ (2x + y)2 = (2x)2 + y2 + 2 × 2x × y
⇒ (2x + y)2 = 4x2 + y2 + 4xy
Substituting values we get,
⇒ (2x + y)2 = a + 4b
⇒ (2x + y) = ± a + 4 b \pm \sqrt{a + 4b} ± a + 4 b
Hence, (2x + y) = ± a + 4 b \pm \sqrt{a + 4b} ± a + 4 b
The expansion of (x - 3y)(x + 5y) is :
x2 + 2xy + 15y2
x2 + 2xy - 15y2
x2 - 2xy + 15y2
x2 - 2xy - 15y2
Answer
Expansion of (x - a)(x + b) = x2 - (a - b)x - ab
∴ Expansion of (x - 3y)(x + 5y) = x2 - (3y - 5y)x - 3y × 5y
= x2 - (-2y)x - 15y2
= x2 + 2xy - 15y2 .
Hence, Option 2 is the correct option.
x2 - (a + b)x + ab is the expansion of :
(x - b)(x - a)
(x - b)(x + a)
(x + b)(x - a)
(x + a)(x + b)
Answer
Given,
⇒ x2 - (a + b)x + ab
⇒ x2 - ax - bx + ab
⇒ x(x - a) - b(x - a)
⇒ (x - a)(x - b).
Hence, Option 1 is the correct option.
If a + b - c = 4 and a2 + b2 + c2 = 14, the value of ab - bc - ca is :
2
1
0.5
-0.5
Answer
By formula,
⇒ (a + b - c)2 = a2 + b2 + c2 + 2(ab - bc - ca)
Substituting values we get :
⇒ 42 = 14 + 2(ab - bc - ca)
⇒ 16 = 14 + 2(ab - bc - ca)
⇒ 2(ab - bc - ca) = 16 - 14
⇒ 2(ab - bc - ca) = 2
⇒ ab - bc - ca = 1
Hence, Option 2 is the correct option.
( a − 1 2 a ) 2 \Big(a - \dfrac{1}{2a}\Big)^2 ( a − 2 a 1 ) 2
( a 2 + 1 4 a 2 − 2 ) \Big(a^2 + \dfrac{1}{4a^2} - 2\Big) ( a 2 + 4 a 2 1 − 2 )
( a 2 + 1 4 a 2 + 2 ) \Big(a^2 + \dfrac{1}{4a^2} + 2\Big) ( a 2 + 4 a 2 1 + 2 )
( a 2 + 1 4 a 2 − 1 ) \Big(a^2 + \dfrac{1}{4a^2} - 1\Big) ( a 2 + 4 a 2 1 − 1 )
( a 2 − 1 4 a 2 − 2 ) \Big(a^2 - \dfrac{1}{4a^2} - 2\Big) ( a 2 − 4 a 2 1 − 2 )
Answer
Given,
⇒ ( a − 1 2 a ) 2 \Big(a - \dfrac{1}{2a}\Big)^2 ( a − 2 a 1 ) 2
Expanding,
⇒ ( a − 1 2 a ) ( a − 1 2 a ) ⇒ a 2 − a × 1 2 a − 1 2 a × a − 1 2 a × − 1 2 a ⇒ a 2 − 1 2 − 1 2 + 1 4 a 2 ⇒ a 2 + 1 4 a 2 − 1. \Rightarrow \Big(a - \dfrac{1}{2a}\Big)\Big(a - \dfrac{1}{2a}\Big) \\[1em] \Rightarrow a^2 - a \times \dfrac{1}{2a} - \dfrac{1}{2a} \times a - \dfrac{1}{2a} \times -\dfrac{1}{2a} \\[1em] \Rightarrow a^2 - \dfrac{1}{2} - \dfrac{1}{2} + \dfrac{1}{4a^2} \\[1em] \Rightarrow a^2 + \dfrac{1}{4a^2} - 1. ⇒ ( a − 2 a 1 ) ( a − 2 a 1 ) ⇒ a 2 − a × 2 a 1 − 2 a 1 × a − 2 a 1 × − 2 a 1 ⇒ a 2 − 2 1 − 2 1 + 4 a 2 1 ⇒ a 2 + 4 a 2 1 − 1.
Hence, Option 3 is the correct option.
Expand (x + 8)(x + 10)
Answer
Given,
⇒ (x + 8)(x + 10)
Expanding,
⇒ x2 + 10x + 8x + 80
⇒ x2 + 18x + 80.
Hence, expansion of (x + 8)(x + 10) = x2 + 18x + 80.
Expand (x + 8)(x - 10)
Answer
Given,
⇒ (x + 8)(x - 10)
Expanding,
⇒ x2 - 10x + 8x - 80
⇒ x2 - 2x - 80.
Hence, expansion of (x + 8)(x - 10) = x2 - 2x - 80.
Expand (x - 8)(x + 10)
Answer
Given,
⇒ (x - 8)(x + 10)
Expanding,
⇒ x2 + 10x - 8x - 80
⇒ x2 + 2x - 80.
Hence, expansion of (x - 8)(x + 10) = x2 + 2x - 80.
Expand (x - 8)(x - 10)
Answer
Given,
⇒ (x - 8)(x - 10)
Expanding,
⇒ x2 - 10x - 8x + 80
⇒ x2 - 18x + 80.
Hence, expansion of (x - 8)(x - 10) = x2 - 18x + 80.
Expand ( 2 x − 1 x ) ( 3 x + 2 x ) \Big(2x - \dfrac{1}{x}\Big)\Big(3x + \dfrac{2}{x}\Big) ( 2 x − x 1 ) ( 3 x + x 2 )
Answer
Given,
( 2 x − 1 x ) ( 3 x + 2 x ) \Big(2x - \dfrac{1}{x}\Big)\Big(3x + \dfrac{2}{x}\Big) ( 2 x − x 1 ) ( 3 x + x 2 )
Expanding,
⇒ 2 x × 3 x + 2 x × 2 x − 1 x × 3 x − 1 x × 2 x ⇒ 6 x 2 + 4 − 3 − 2 x 2 ⇒ 6 x 2 + 1 − 2 x 2 . \Rightarrow 2x \times 3x + 2x \times \dfrac{2}{x} - \dfrac{1}{x} \times 3x - \dfrac{1}{x} \times \dfrac{2}{x} \\[1em] \Rightarrow 6x^2 + 4 - 3 - \dfrac{2}{x^2} \\[1em] \Rightarrow 6x^2 + 1 - \dfrac{2}{x^2}. ⇒ 2 x × 3 x + 2 x × x 2 − x 1 × 3 x − x 1 × x 2 ⇒ 6 x 2 + 4 − 3 − x 2 2 ⇒ 6 x 2 + 1 − x 2 2 .
Hence, ( 2 x − 1 x ) ( 3 x + 2 x ) = 6 x 2 + 1 − 2 x 2 . \Big(2x - \dfrac{1}{x}\Big)\Big(3x + \dfrac{2}{x}\Big) = 6x^2 + 1 - \dfrac{2}{x^2}. ( 2 x − x 1 ) ( 3 x + x 2 ) = 6 x 2 + 1 − x 2 2 .
Expand ( 3 a + 2 b ) ( 2 a − 3 b ) \Big(3a + \dfrac{2}{b}\Big)\Big(2a - \dfrac{3}{b}\Big) ( 3 a + b 2 ) ( 2 a − b 3 )
Answer
Given,
( 3 a + 2 b ) ( 2 a − 3 b ) \Big(3a + \dfrac{2}{b}\Big)\Big(2a - \dfrac{3}{b}\Big) ( 3 a + b 2 ) ( 2 a − b 3 )
Expanding,
⇒ 3 a × 2 a + 3 a × − 3 b + 2 b × 2 a + 2 b × − 3 b ⇒ 6 a 2 − 9 a b + 4 a b − 6 b 2 ⇒ 6 a 2 − 5 a b − 6 b 2 . \Rightarrow 3a \times 2a + 3a \times -\dfrac{3}{b} + \dfrac{2}{b} \times 2a + \dfrac{2}{b} \times -\dfrac{3}{b} \\[1em] \Rightarrow 6a^2 - \dfrac{9a}{b} + \dfrac{4a}{b} - \dfrac{6}{b^2} \\[1em] \Rightarrow 6a^2 - \dfrac{5a}{b} - \dfrac{6}{b^2}. ⇒ 3 a × 2 a + 3 a × − b 3 + b 2 × 2 a + b 2 × − b 3 ⇒ 6 a 2 − b 9 a + b 4 a − b 2 6 ⇒ 6 a 2 − b 5 a − b 2 6 .
Hence, ( 3 a + 2 b ) ( 2 a − 3 b ) = 6 a 2 − 5 a b − 6 b 2 . \Big(3a + \dfrac{2}{b}\Big)\Big(2a - \dfrac{3}{b}\Big) = 6a^2 - \dfrac{5a}{b} - \dfrac{6}{b^2}. ( 3 a + b 2 ) ( 2 a − b 3 ) = 6 a 2 − b 5 a − b 2 6 .
Expand (x + y - z)2
Answer
Given,
⇒ (x + y - z)2
Expanding,
⇒ x2 + y2 + z2 + 2xy - 2yz - 2zx
⇒ x2 + y2 + z2 + 2(xy - yz - zx).
Hence, (x + y - z)2 = x2 + y2 + z2 + 2(xy - yz - zx).
Expand (x - 2y + 2)2
Answer
Given,
⇒ (x - 2y + 2)2
Expanding,
⇒ (x - 2y + 2)(x - 2y + 2)
⇒ x2 - 2xy + 2x - 2xy + 4y2 - 4y + 2x - 4y + 4
⇒ x2 + 4y2 + 4 - 4xy - 8y + 4x.
Hence, (x - 2y + 2)2 = x2 + 4y2 + 4 - 4xy - 8y + 4x.
Expand (5a - 3b + c)2
Answer
Given,
⇒ (5a - 3b + c)2
Expanding,
⇒ (5a - 3b + c)(5a - 3b + c)
⇒ (5a)2 - 15ab + 5ac - 15ab + 9b2 - 3bc + 5ac - 3bc + c2
⇒ 25a2 + 9b2 + c2 - 30ab - 6bc + 10ac.
Hence, (5a - 3b + c)2 = 25a2 + 9b2 + c2 - 30ab - 6bc + 10ac.
Expand (5x - 3y - 2)2
Answer
Given,
⇒ (5x - 3y - 2)2
Expanding,
⇒ (5x - 3y - 2)(5x - 3y - 2)
⇒ 25x2 - 15xy - 10x - 15xy + 9y2 + 6y - 10x + 6y + 4
⇒ 25x2 + 9y2 - 30xy - 20x + 12y + 4.
Hence, (5x - 3y - 2)2 = 25x2 + 9y2 - 30xy - 20x + 12y + 4.
Expand ( x − 1 x + 5 ) 2 \Big(x - \dfrac{1}{x} + 5\Big)^2 ( x − x 1 + 5 ) 2
Answer
Given,
⇒ ( x − 1 x + 5 ) 2 \Big(x - \dfrac{1}{x} + 5\Big)^2 ( x − x 1 + 5 ) 2
Expanding,
⇒ ( x − 1 x + 5 ) ( x − 1 x + 5 ) ⇒ x 2 − 1 + 5 x − 1 + 1 x 2 − 5 x + 5 x − 5 x + 25 ⇒ x 2 + 1 x 2 − 10 x + 23 + 10 x . \Rightarrow \Big(x - \dfrac{1}{x} + 5\Big)\Big(x - \dfrac{1}{x} + 5\Big) \\[1em] \Rightarrow x^2 - 1 + 5x - 1 + \dfrac{1}{x^2} - \dfrac{5}{x} + 5x - \dfrac{5}{x} + 25 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} - \dfrac{10}{x} + 23 + 10x. ⇒ ( x − x 1 + 5 ) ( x − x 1 + 5 ) ⇒ x 2 − 1 + 5 x − 1 + x 2 1 − x 5 + 5 x − x 5 + 25 ⇒ x 2 + x 2 1 − x 10 + 23 + 10 x .
Hence, ( x − 1 x + 5 ) 2 = x 2 + 1 x 2 − 10 x + 23 + 10 x . \Big(x - \dfrac{1}{x} + 5\Big)^2 = x^2 + \dfrac{1}{x^2} - \dfrac{10}{x} + 23 + 10x. ( x − x 1 + 5 ) 2 = x 2 + x 2 1 − x 10 + 23 + 10 x .
If a + b + c = 12 and a2 + b2 + c2 = 50; find ab + bc + ca.
Answer
By formula,
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca).
Substituting values we get :
⇒ 122 = 50 + 2(ab + bc + ca)
⇒ 144 = 50 + 2(ab + bc + ca)
⇒ 2(ab + bc + ca) = 144 - 50
⇒ 2(ab + bc + ca) = 94
⇒ ab + bc + ca = 47.
Hence, ab + bc + ca = 47.
If a2 + b2 + c2 = 35 and ab + bc + ca = 23; find a + b + c.
Answer
By formula,
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca).
Substituting values we get :
⇒ (a + b + c)2 = 35 + 2 × 23
⇒ (a + b + c)2 = 35 + 46
⇒ (a + b + c)2 = 81
⇒ (a + b + c) = 81 = ± 9 \sqrt{81} = \pm 9 81 = ± 9
Hence, (a + b + c) = ± 9 \pm 9 ± 9
If a + b + c = p and ab + bc + ca = q; find a2 + b2 + c2 .
Answer
By formula,
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca).
Substituting values we get :
⇒ p2 = a2 + b2 + c2 + 2q
⇒ a2 + b2 + c2 = p2 - 2q.
Hence, a2 + b2 + c2 = p2 - 2q.
If a2 + b2 + c2 = 50 and ab + bc + ca = 47, find a + b + c.
Answer
By formula,
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca).
Substituting values we get :
⇒ (a + b + c)2 = 50 + 2 × 47
⇒ (a + b + c)2 = 50 + 94
⇒ (a + b + c)2 = 144
⇒ (a + b + c) = 144 = ± 12 \sqrt{144} = \pm 12 144 = ± 12
Hence, (a + b + c) = ± 12 \pm 12 ± 12
If x + y - z = 4 and x2 + y2 + z2 = 30, then find the value of xy - yz - zx.
Answer
By formula,
(x + y - z)2 = x2 + y2 + z2 + 2(xy - yz - zx)
Substituting values we get :
⇒ 42 = 30 + 2(xy - yz - zx)
⇒ 16 = 30 + 2(xy - yz - zx)
⇒ 2(xy - yz - zx) = 16 - 30
⇒ 2(xy - yz - zx) = -14
⇒ xy - yz - zx = -7.
Hence, xy - yz - zx = -7.
The longest road that can be placed in a rectangular box = 20 cm and the sum of its length breadth and height is 30 cm. Find the total surface area of the box.
Answer
Given, the longest road that can be placed in a rectangular box, d = 20 cm
The longest rod which can be kept inside a rectangular box will be equal to the diagonal of the box.
By formula,
⇒ Diagonal2 = l2 + b2 + h2
⇒ 202 = l2 + b2 + h2
⇒ 400 = l2 + b2 + h2 ..................(1)
Given,
The sum of its length breadth and height equals to 30 cm.
⇒ l + b + h = 30 cm.
Squaring both sides, we get :
⇒ (l + b + h)2 = 302
⇒ l2 + b2 + h2 + 2(lb + bh + hl) = 900
From equation (1), we get
⇒ 400 + 2(lb + bh + hl) = 900
⇒ 2(lb + bh + hl) = 900 - 400
⇒ 2(lb + bh + hl) = 500
Hence, the total surface area of the box = 500 cm2 .
If a + 1 a = 2.5 and a − 1 a = 1.5 a + \dfrac{1}{a} = 2.5 \text{ and } a - \dfrac{1}{a} = 1.5 a + a 1 = 2.5 and a − a 1 = 1.5 ( a + 1 a ) 2 − ( a − 1 a ) 2 \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 ( a + a 1 ) 2 − ( a − a 1 ) 2
4
2
1
8.5
Answer
Solving,
⇒ ( a + 1 a ) 2 − ( a − 1 a ) 2 ⇒ ( 2.5 ) 2 − ( 1.5 ) 2 ⇒ 6.25 − 2.25 ⇒ 4. \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 \\[1em] \Rightarrow (2.5)^2 - (1.5)^2 \\[1em] \Rightarrow 6.25 - 2.25 \\[1em] \Rightarrow 4. ⇒ ( a + a 1 ) 2 − ( a − a 1 ) 2 ⇒ ( 2.5 ) 2 − ( 1.5 ) 2 ⇒ 6.25 − 2.25 ⇒ 4.
Hence, Option 1 is the correct option.
In (3x + 2)(x - 4), the coefficient of x is :
-14
8
-10
3
Answer
Given,
⇒ (3x + 2)(x - 4)
⇒ 3x2 - 12x + 2x - 8
⇒ 3x2 - 10x - 8.
∴ Coefficient of x is -10.
Hence, Option 3 is the correct option.
(x + y - z)(x - y + z) is equal to :
x2 - y2 - z2 - 2yz
x2 - y2 - z2 + 2yz
x2 - y2 + z2 - 2yz
x2 - y2 + z2 + 2yz
Answer
Given,
⇒ (x + y - z)(x - y + z)
Expanding,
⇒ x2 - xy + xz + xy - y2 + yz - zx + yz - z2
⇒ x2 - y2 - z2 + 2yz.
Hence, Option 2 is the correct option.
If (3x - 4y)2 = 9x2 + axy + 16y2 ; the value of a is :
-24
24
-12
12
Answer
Solving,
⇒ (3x - 4y)2 = 9x2 + axy + 16y2
⇒ 9x2 + 16y2 - 24xy = 9x2 + axy + 16y2
From above equation,
axy = -24xy
a = − 24 x y x y -\dfrac{24xy}{xy} − x y 24 x y
Hence, Option 1 is the correct option.
If x + 2y + 3z = 0 and x3 + 4y3 + 9z3 = 18xyz; evaluate :
( x + 2 y ) 2 x y + ( 2 y + 3 z ) 2 y z + ( 3 z + x ) 2 z x \dfrac{(x + 2y)^2}{xy} + \dfrac{(2y + 3z)^2}{yz} + \dfrac{(3z + x)^2}{zx} x y ( x + 2 y ) 2 + yz ( 2 y + 3 z ) 2 + z x ( 3 z + x ) 2
Answer
Given,
x + 2y + 3z = 0
∴ x + 2y = -3z, 2y + 3z = -x and 3z + x = -2y
Solving,
⇒ ( x + 2 y ) 2 x y + ( 2 y + 3 z ) 2 y z + ( 3 z + x ) 2 z x ⇒ ( − 3 z ) 2 x y + ( − x ) 2 y z + ( − 2 y ) 2 z x ⇒ 9 z 2 x y + x 2 y z + 4 y 2 z x ⇒ 9 z 3 + x 3 + 4 y 3 x y z ⇒ x 3 + 4 y 3 + 9 z 3 x y z ⇒ 18 x y z x y z ⇒ 18. \Rightarrow \dfrac{(x + 2y)^2}{xy} + \dfrac{(2y + 3z)^2}{yz} + \dfrac{(3z + x)^2}{zx} \\[1em] \Rightarrow \dfrac{(-3z)^2}{xy} + \dfrac{(-x)^2}{yz} + \dfrac{(-2y)^2}{zx}\\[1em] \Rightarrow \dfrac{9z^2}{xy} + \dfrac{x^2}{yz} + \dfrac{4y^2}{zx} \\[1em] \Rightarrow \dfrac{9z^3 + x^3 + 4y^3}{xyz} \\[1em] \Rightarrow \dfrac{x^3 + 4y^3 + 9z^3}{xyz} \\[1em] \Rightarrow \dfrac{18xyz}{xyz} \\[1em] \Rightarrow 18. ⇒ x y ( x + 2 y ) 2 + yz ( 2 y + 3 z ) 2 + z x ( 3 z + x ) 2 ⇒ x y ( − 3 z ) 2 + yz ( − x ) 2 + z x ( − 2 y ) 2 ⇒ x y 9 z 2 + yz x 2 + z x 4 y 2 ⇒ x yz 9 z 3 + x 3 + 4 y 3 ⇒ x yz x 3 + 4 y 3 + 9 z 3 ⇒ x yz 18 x yz ⇒ 18.
Hence, ( x + 2 y ) 2 x y + ( 2 y + 3 z ) 2 y z + ( 3 z + x ) 2 z x \dfrac{(x + 2y)^2}{xy} + \dfrac{(2y + 3z)^2}{yz} + \dfrac{(3z + x)^2}{zx} x y ( x + 2 y ) 2 + yz ( 2 y + 3 z ) 2 + z x ( 3 z + x ) 2
If a + 1 a = m a + \dfrac{1}{a} = m a + a 1 = m
(i) a − 1 a a - \dfrac{1}{a} a − a 1
(ii) a 2 − 1 a 2 a^2 - \dfrac{1}{a^2} a 2 − a 2 1
Answer
(i) By formula,
⇒ ( a + 1 a ) 2 − ( a − 1 a ) 2 = 4 \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = 4 ⇒ ( a + a 1 ) 2 − ( a − a 1 ) 2 = 4
Substituting values we get :
⇒ m 2 − ( a − 1 a ) 2 = 4 ⇒ ( a − 1 a ) 2 = m 2 − 4 ⇒ a − 1 a = ± m 2 − 4 . \Rightarrow m^2 - \Big(a - \dfrac{1}{a}\Big)^2 = 4 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = m^2 - 4 \\[1em] \Rightarrow a - \dfrac{1}{a} = \pm \sqrt{m^2 - 4}. ⇒ m 2 − ( a − a 1 ) 2 = 4 ⇒ ( a − a 1 ) 2 = m 2 − 4 ⇒ a − a 1 = ± m 2 − 4 .
Hence, a − 1 a = ± m 2 − 4 . a - \dfrac{1}{a} = \pm \sqrt{m^2 - 4}. a − a 1 = ± m 2 − 4 .
(ii) By formula,
⇒ a 2 − 1 a 2 = ( a − 1 a ) ( a + 1 a ) \Rightarrow a^2 - \dfrac{1}{a^2} = \Big(a - \dfrac{1}{a}\Big)\Big(a + \dfrac{1}{a}\Big) ⇒ a 2 − a 2 1 = ( a − a 1 ) ( a + a 1 )
Substituting values we get :
⇒ ⇒ a 2 − 1 a 2 = ± m 2 − 4 × m = ± m m 2 − 4 . \Rightarrow \Rightarrow a^2 - \dfrac{1}{a^2} = \pm \sqrt{m^2 - 4} \times m \\[1em] = \pm m\sqrt{m^2 - 4}. ⇒⇒ a 2 − a 2 1 = ± m 2 − 4 × m = ± m m 2 − 4 .
Hence, a 2 − 1 a 2 = ± m m 2 − 4 . a^2 - \dfrac{1}{a^2} = \pm m\sqrt{m^2 - 4}. a 2 − a 2 1 = ± m m 2 − 4 .
In the expansion of (2x2 - 8)(x - 4)2 ; find the value of :
(i) coefficient of x3
(ii) coefficient of x2
(iii) constant term.
Answer
Given,
⇒ (2x2 - 8)(x - 4)2
Expanding,
⇒ (2x2 - 8)(x2 + 16 - 8x)
⇒ 2x4 + 32x2 - 16x3 - 8x2 - 128 + 64x
⇒ 2x4 - 16x3 + 24x2 + 64x - 128.
(i) Hence, coefficient of x3 is -16.
(ii) Hence, coefficient of x2 is 24.
(iii) Hence, constant term = -128.
If x > 0 and x 2 + 1 9 x 2 = 25 36 x^2 + \dfrac{1}{9x^2} = \dfrac{25}{36} x 2 + 9 x 2 1 = 36 25 x 3 + 1 27 x 3 x^3 + \dfrac{1}{27x^3} x 3 + 27 x 3 1
Answer
Given,
⇒ x 2 + 1 9 x 2 = 25 36 ⇒ ( x + 1 3 x ) 2 − 2 3 = 25 36 ⇒ ( x + 1 3 x ) 2 = 25 36 + 2 3 ⇒ ( x + 1 3 x ) 2 = 25 + 24 36 ⇒ ( x + 1 3 x ) 2 = 49 36 ⇒ x + 1 3 x = 49 36 ⇒ x + 1 3 x = ± 7 6 . \Rightarrow x^2 + \dfrac{1}{9x^2} = \dfrac{25}{36} \\[1em] \Rightarrow \Big(x + \dfrac{1}{3x}\Big)^2 - \dfrac{2}{3} = \dfrac{25}{36} \\[1em] \Rightarrow \Big(x + \dfrac{1}{3x}\Big)^2 = \dfrac{25}{36} + \dfrac{2}{3} \\[1em] \Rightarrow \Big(x + \dfrac{1}{3x}\Big)^2 = \dfrac{25 + 24}{36} \\[1em] \Rightarrow \Big(x + \dfrac{1}{3x}\Big)^2 = \dfrac{49}{36}\\[1em] \Rightarrow x + \dfrac{1}{3x} = \sqrt{\dfrac{49}{36}} \\[1em] \Rightarrow x + \dfrac{1}{3x} = \pm \dfrac{7}{6}. ⇒ x 2 + 9 x 2 1 = 36 25 ⇒ ( x + 3 x 1 ) 2 − 3 2 = 36 25 ⇒ ( x + 3 x 1 ) 2 = 36 25 + 3 2 ⇒ ( x + 3 x 1 ) 2 = 36 25 + 24 ⇒ ( x + 3 x 1 ) 2 = 36 49 ⇒ x + 3 x 1 = 36 49 ⇒ x + 3 x 1 = ± 6 7 .
Since, x is > 0,
∴ x + 1 3 x = 7 6 x + \dfrac{1}{3x} = \dfrac{7}{6} x + 3 x 1 = 6 7
By formula,
⇒ ( x 3 + 1 27 x 3 ) = ( x + 1 3 x ) 3 − ( x + 1 3 x ) \Rightarrow \Big(x^3 + \dfrac{1}{27x^3}\Big) = \Big(x + \dfrac{1}{3x}\Big)^3 - \Big(x + \dfrac{1}{3x}\Big) ⇒ ( x 3 + 27 x 3 1 ) = ( x + 3 x 1 ) 3 − ( x + 3 x 1 )
Substituting x + 1 3 x = 7 6 x + \dfrac{1}{3x} = \dfrac{7}{6} x + 3 x 1 = 6 7
⇒ ( x 3 + 1 27 x 3 ) = ( 7 6 ) 3 − ( 7 6 ) = 343 216 − 7 6 = 343 − 252 216 = 91 216 . \Rightarrow \Big(x^3 + \dfrac{1}{27x^3}\Big) = \Big(\dfrac{7}{6}\Big)^3 - \Big(\dfrac{7}{6}\Big) \\[1em] = \dfrac{343}{216} - \dfrac{7}{6} \\[1em] = \dfrac{343 - 252}{216} \\[1em] = \dfrac{91}{216}. ⇒ ( x 3 + 27 x 3 1 ) = ( 6 7 ) 3 − ( 6 7 ) = 216 343 − 6 7 = 216 343 − 252 = 216 91 .
Hence, x 3 + 1 27 x 3 = 91 216 x^3 + \dfrac{1}{27x^3} = \dfrac{91}{216} x 3 + 27 x 3 1 = 216 91
If 2(x2 + 1) = 5x, find :
(i) x − 1 x x - \dfrac{1}{x} x − x 1
(ii) x 3 − 1 x 3 x^3 - \dfrac{1}{x^3} x 3 − x 3 1
Answer
(i) Given,
⇒ 2 ( x 2 + 1 ) = 5 x ⇒ x 2 + 1 x = 5 2 ⇒ x + 1 x = 5 2 . \Rightarrow 2(x^2 + 1) = 5x \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = \dfrac{5}{2} \\[1em] \Rightarrow x + \dfrac{1}{x} = \dfrac{5}{2}. ⇒ 2 ( x 2 + 1 ) = 5 x ⇒ x x 2 + 1 = 2 5 ⇒ x + x 1 = 2 5 .
By formula,
⇒ ( x + 1 x ) 2 − ( x − 1 x ) 2 = 4 ⇒ ( 5 2 ) 2 − ( x − 1 x ) 2 = 4 ⇒ ( x − 1 x ) 2 = 25 4 − 4 ⇒ ( x − 1 x ) 2 = 25 − 16 4 ⇒ ( x − 1 x ) 2 = 9 4 ⇒ ( x − 1 x ) = 9 4 ⇒ ( x − 1 x ) = ± 3 2 . \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(\dfrac{5}{2}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{25}{4} - 4\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{25 - 16}{4} \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{9}{4} \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \sqrt{\dfrac{9}{4}} \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \pm \dfrac{3}{2}. ⇒ ( x + x 1 ) 2 − ( x − x 1 ) 2 = 4 ⇒ ( 2 5 ) 2 − ( x − x 1 ) 2 = 4 ⇒ ( x − x 1 ) 2 = 4 25 − 4 ⇒ ( x − x 1 ) 2 = 4 25 − 16 ⇒ ( x − x 1 ) 2 = 4 9 ⇒ ( x − x 1 ) = 4 9 ⇒ ( x − x 1 ) = ± 2 3 .
Hence, ( x − 1 x ) = ± 3 2 . \Big(x - \dfrac{1}{x}\Big) = \pm \dfrac{3}{2}. ( x − x 1 ) = ± 2 3 .
(ii) By formula,
⇒ ( x 3 − 1 x 3 ) = ( x − 1 x ) 3 + 3 ( x − 1 x ) \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = \Big(x - \dfrac{1}{x}\Big)^3 + 3\Big(x - \dfrac{1}{x}\Big) \\[1em] ⇒ ( x 3 − x 3 1 ) = ( x − x 1 ) 3 + 3 ( x − x 1 )
Substituting ( x − 1 x ) = 3 2 \Big(x - \dfrac{1}{x}\Big) = \dfrac{3}{2} ( x − x 1 ) = 2 3
⇒ ( x 3 − 1 x 3 ) = ( 3 2 ) 3 + 3 × 3 2 = 27 8 + 9 2 = 27 + 36 8 = 63 8 . \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = \Big(\dfrac{3}{2}\Big)^3 + 3 \times \dfrac{3}{2} \\[1em] = \dfrac{27}{8} + \dfrac{9}{2} \\[1em] = \dfrac{27 + 36}{8} \\[1em] = \dfrac{63}{8}. ⇒ ( x 3 − x 3 1 ) = ( 2 3 ) 3 + 3 × 2 3 = 8 27 + 2 9 = 8 27 + 36 = 8 63 .
Substituting ( x − 1 x ) = − 3 2 \Big(x - \dfrac{1}{x}\Big) = -\dfrac{3}{2} ( x − x 1 ) = − 2 3
⇒ ( x 3 − 1 x 3 ) = ( − 3 2 ) 3 + 3 × − 3 2 = − 27 8 − 9 2 = − 27 − 36 8 = − 63 8 . \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = \Big(-\dfrac{3}{2}\Big)^3 + 3 \times -\dfrac{3}{2} \\[1em] = -\dfrac{27}{8} - \dfrac{9}{2} \\[1em] = \dfrac{-27 - 36}{8} \\[1em] = \dfrac{-63}{8}. ⇒ ( x 3 − x 3 1 ) = ( − 2 3 ) 3 + 3 × − 2 3 = − 8 27 − 2 9 = 8 − 27 − 36 = 8 − 63 .
Hence, ( x 3 − 1 x 3 ) = ± 63 8 \Big(x^3 - \dfrac{1}{x^3}\Big) = \pm \dfrac{63}{8} ( x 3 − x 3 1 ) = ± 8 63
If a2 + b2 = 34 and ab = 12; find :
(i) 3(a + b)2 + 5(a - b)2
(ii) 7(a - b)2 - 2(a + b)2
Answer
(i) Expanding,
⇒ 3(a + b)2 + 5(a - b)2
⇒ 3(a2 + b2 + 2ab) + 5(a2 + b2 - 2ab)
⇒ 3(34 + 2 × 12) + 5(34 - 2 × 12)
⇒ 3(34 + 24) + 5(34 - 24)
⇒ 3 × 58 + 5 × 10
⇒ 174 + 50
⇒ 224.
Hence, 3(a + b)2 + 5(a - b)2 = 224.
(ii) Expanding,
⇒ 7(a - b)2 - 2(a + b)2
⇒ 7(a2 + b2 - 2ab) - 2(a2 + b2 + 2ab)
⇒ 7(34 - 2 × 12) - 2(34 + 2 × 12)
⇒ 7(34 - 24) - 2(34 + 24)
⇒ 7 × 10 - 2 × 58
⇒ 70 - 116
⇒ -46.
Hence, 7(a - b)2 - 2(a + b)2 = -46.
If 3x - 4 x \dfrac{4}{x} x 4 27 x 3 − 64 x 3 27x^3 - \dfrac{64}{x^3} 27 x 3 − x 3 64
Answer
Given,
⇒ 3 x − 4 x = 4 \Rightarrow 3x - \dfrac{4}{x} = 4 ⇒ 3 x − x 4 = 4
Cubing both sides we get :
⇒ ( 3 x − 4 x ) 3 = 4 3 ⇒ ( 3 x ) 3 − ( 4 x ) 3 − 3 × 3 x × 4 x × ( 3 x − 4 x ) = 64 ⇒ ( 3 x ) 3 − ( 4 x ) 3 − 36 × 4 = 64 ⇒ 27 x 3 − 64 x 3 − 144 = 64 ⇒ 27 x 3 − 64 x 3 = 64 + 144 ⇒ 27 x 3 − 64 x 3 = 208. \Rightarrow \Big(3x - \dfrac{4}{x}\Big)^3 = 4^3 \\[1em] \Rightarrow (3x)^3 - \Big(\dfrac{4}{x}\Big)^3 - 3 \times 3x \times \dfrac{4}{x} \times \Big(3x - \dfrac{4}{x}\Big) = 64 \\[1em] \Rightarrow (3x)^3 - \Big(\dfrac{4}{x}\Big)^3 - 36 \times 4 = 64 \\[1em] \Rightarrow 27x^3 - \dfrac{64}{x^3} - 144 = 64 \\[1em] \Rightarrow 27x^3 - \dfrac{64}{x^3} = 64 + 144 \\[1em] \Rightarrow 27x^3 - \dfrac{64}{x^3} = 208. ⇒ ( 3 x − x 4 ) 3 = 4 3 ⇒ ( 3 x ) 3 − ( x 4 ) 3 − 3 × 3 x × x 4 × ( 3 x − x 4 ) = 64 ⇒ ( 3 x ) 3 − ( x 4 ) 3 − 36 × 4 = 64 ⇒ 27 x 3 − x 3 64 − 144 = 64 ⇒ 27 x 3 − x 3 64 = 64 + 144 ⇒ 27 x 3 − x 3 64 = 208.
Hence, 27 x 3 − 64 x 3 = 208. 27x^3 - \dfrac{64}{x^3} = 208. 27 x 3 − x 3 64 = 208.
If x 2 + 1 x 2 x^2 + \dfrac{1}{x^2} x 2 + x 2 1
7x3 + 8x - 7 x 3 − 8 x \dfrac{7}{x^3} - \dfrac{8}{x} x 3 7 − x 8
Answer
By formula,
⇒ ( x − 1 x ) 2 = x 2 + 1 x 2 − 2 ⇒ ( x − 1 x ) 2 = 7 − 2 ⇒ ( x − 1 x ) 2 = 5 ⇒ x − 1 x = ± 5 . \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 7 - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 5 \\[1em] \Rightarrow x - \dfrac{1}{x} = \pm \sqrt{5}. ⇒ ( x − x 1 ) 2 = x 2 + x 2 1 − 2 ⇒ ( x − x 1 ) 2 = 7 − 2 ⇒ ( x − x 1 ) 2 = 5 ⇒ x − x 1 = ± 5 .
By formula,
⇒ ( x − 1 x ) 3 = x 3 − 1 x 3 − 3 ( x − 1 x ) \Rightarrow \Big(x - \dfrac{1}{x}\Big)^3 = x^3 - \dfrac{1}{x^3} - 3\Big(x - \dfrac{1}{x}\Big) ⇒ ( x − x 1 ) 3 = x 3 − x 3 1 − 3 ( x − x 1 )
Substituting x − 1 x = 5 x - \dfrac{1}{x} = \sqrt{5} x − x 1 = 5
⇒ ( 5 ) 3 = x 3 − 1 x 3 − 3 × 5 ⇒ 5 5 = x 3 − 1 x 3 − 3 5 ⇒ x 3 − 1 x 3 = 8 5 . \Rightarrow (\sqrt{5})^3 = x^3 - \dfrac{1}{x^3} - 3 \times \sqrt{5} \\[1em] \Rightarrow 5\sqrt{5} = x^3 - \dfrac{1}{x^3} - 3\sqrt{5} \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = 8\sqrt{5}. ⇒ ( 5 ) 3 = x 3 − x 3 1 − 3 × 5 ⇒ 5 5 = x 3 − x 3 1 − 3 5 ⇒ x 3 − x 3 1 = 8 5 .
Substituting x − 1 x = − 5 x - \dfrac{1}{x} = -\sqrt{5} x − x 1 = − 5
⇒ ( − 5 ) 3 = x 3 − 1 x 3 − 3 × − 5 ⇒ − 5 5 = x 3 − 1 x 3 + 3 5 ⇒ x 3 − 1 x 3 = − 8 5 . \Rightarrow (-\sqrt{5})^3 = x^3 - \dfrac{1}{x^3} - 3 \times -\sqrt{5} \\[1em] \Rightarrow -5\sqrt{5} = x^3 - \dfrac{1}{x^3} + 3\sqrt{5} \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = -8\sqrt{5}. ⇒ ( − 5 ) 3 = x 3 − x 3 1 − 3 × − 5 ⇒ − 5 5 = x 3 − x 3 1 + 3 5 ⇒ x 3 − x 3 1 = − 8 5 .
Solving given equation,
⇒ 7 x 3 + 8 x − 7 x 3 − 8 x ⇒ 7 x 3 − 7 x 3 + 8 x − 8 x ⇒ 7 ( x 3 − 1 x 3 ) + 8 ( x − 1 x ) \Rightarrow 7x^3 + 8x - \dfrac{7}{x^3} - \dfrac{8}{x} \\[1em] \Rightarrow 7x^3 - \dfrac{7}{x^3} + 8x - \dfrac{8}{x} \\[1em] \Rightarrow 7\Big(x^3 - \dfrac{1}{x^3}\Big) + 8\Big(x - \dfrac{1}{x}\Big) ⇒ 7 x 3 + 8 x − x 3 7 − x 8 ⇒ 7 x 3 − x 3 7 + 8 x − x 8 ⇒ 7 ( x 3 − x 3 1 ) + 8 ( x − x 1 )
Substituting x 3 − 1 x 3 = 8 5 and x − 1 x = 5 x^3 - \dfrac{1}{x^3} = 8\sqrt{5}\text{ and } x - \dfrac{1}{x} = \sqrt{5} x 3 − x 3 1 = 8 5 and x − x 1 = 5
⇒ 7 × 8 5 + 8 × 5 ⇒ 56 5 + 8 5 ⇒ 64 5 . \Rightarrow 7 \times 8\sqrt{5} + 8 \times \sqrt{5} \\[1em] \Rightarrow 56\sqrt{5} + 8\sqrt{5} \\[1em] \Rightarrow 64\sqrt{5}. ⇒ 7 × 8 5 + 8 × 5 ⇒ 56 5 + 8 5 ⇒ 64 5 .
Substituting x 3 − 1 x 3 = − 8 5 and x − 1 x = − 5 x^3 - \dfrac{1}{x^3} = -8\sqrt{5}\text{ and } x - \dfrac{1}{x} = -\sqrt{5} x 3 − x 3 1 = − 8 5 and x − x 1 = − 5
⇒ 7 × − 8 5 + 8 × − 5 ⇒ − 56 5 − 8 5 ⇒ − 64 5 . \Rightarrow 7 \times -8\sqrt{5} + 8 \times -\sqrt{5} \\[1em] \Rightarrow -56\sqrt{5} - 8\sqrt{5} \\[1em] \Rightarrow -64\sqrt{5}. ⇒ 7 × − 8 5 + 8 × − 5 ⇒ − 56 5 − 8 5 ⇒ − 64 5 .
Hence, 7x3 + 8x - 7 x 3 − 8 x = ± 64 5 \dfrac{7}{x^3} - \dfrac{8}{x} = \pm 64\sqrt{5} x 3 7 − x 8 = ± 64 5
If x = 1 x − 5 x = \dfrac{1}{x - 5} x = x − 5 1 x 2 − 1 x 2 x^2 - \dfrac{1}{x^2} x 2 − x 2 1
Answer
Given,
⇒ x = 1 x − 5 ⇒ x ( x − 5 ) = 1 ⇒ x 2 − 5 x = 1 ⇒ x 2 − 1 = 5 x ⇒ x 2 − 1 x = 5 x x ⇒ x − 1 x = 5. \Rightarrow x = \dfrac{1}{x - 5} \\[1em] \Rightarrow x(x - 5) = 1 \\[1em] \Rightarrow x^2 - 5x = 1 \\[1em] \Rightarrow x^2 - 1 = 5x \\[1em] \Rightarrow \dfrac{x^2 - 1}{x} = \dfrac{5x}{x} \\[1em] \Rightarrow x - \dfrac{1}{x} = 5. ⇒ x = x − 5 1 ⇒ x ( x − 5 ) = 1 ⇒ x 2 − 5 x = 1 ⇒ x 2 − 1 = 5 x ⇒ x x 2 − 1 = x 5 x ⇒ x − x 1 = 5.
By formula,
⇒ ( x + 1 x ) 2 − ( x − 1 x ) 2 = 4 ⇒ ( x + 1 x ) 2 − 5 2 = 4 ⇒ ( x + 1 x ) 2 − 25 = 4 ⇒ ( x + 1 x ) 2 = 25 + 4 ⇒ ( x + 1 x ) 2 = 29 ⇒ x + 1 x = ± 29 . \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - 5^2 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - 25 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 25 + 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 29 \\[1em] \Rightarrow x + \dfrac{1}{x} = \pm \sqrt{29}. ⇒ ( x + x 1 ) 2 − ( x − x 1 ) 2 = 4 ⇒ ( x + x 1 ) 2 − 5 2 = 4 ⇒ ( x + x 1 ) 2 − 25 = 4 ⇒ ( x + x 1 ) 2 = 25 + 4 ⇒ ( x + x 1 ) 2 = 29 ⇒ x + x 1 = ± 29 .
By formula,
⇒ x 2 − 1 x 2 = ( x − 1 x ) ( x + 1 x ) = 5 × ± 29 = ± 5 29 . \Rightarrow x^2 - \dfrac{1}{x^2} = \Big(x - \dfrac{1}{x}\Big)\Big(x + \dfrac{1}{x}\Big)\\[1em] = 5 \times \pm \sqrt{29} \\[1em] = \pm 5\sqrt{29}. ⇒ x 2 − x 2 1 = ( x − x 1 ) ( x + x 1 ) = 5 × ± 29 = ± 5 29 .
Hence, x 2 − 1 x 2 = ± 5 29 x^2 - \dfrac{1}{x^2} = \pm 5\sqrt{29} x 2 − x 2 1 = ± 5 29
If x = 1 5 − x \dfrac{1}{5 - x} 5 − x 1 x 3 + 1 x 3 x^3 + \dfrac{1}{x^3} x 3 + x 3 1
Answer
Given,
⇒ x = 1 5 − x ⇒ x ( 5 − x ) = 1 ⇒ 5 x − x 2 = 1 ⇒ x 2 + 1 = 5 x ⇒ x 2 + 1 x = 5 x x ⇒ x + 1 x = 5. \Rightarrow x = \dfrac{1}{5 - x} \\[1em] \Rightarrow x(5 - x) = 1 \\[1em] \Rightarrow 5x - x^2 = 1 \\[1em] \Rightarrow x^2 + 1 = 5x \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = \dfrac{5x}{x} \\[1em] \Rightarrow x + \dfrac{1}{x} = 5. ⇒ x = 5 − x 1 ⇒ x ( 5 − x ) = 1 ⇒ 5 x − x 2 = 1 ⇒ x 2 + 1 = 5 x ⇒ x x 2 + 1 = x 5 x ⇒ x + x 1 = 5.
By formula,
⇒ ( x + 1 x ) 3 = x 3 + 1 x 3 + 3 ( x + 1 x ) ⇒ 5 3 = x 3 + 1 x 3 + 3 × 5 ⇒ 125 = x 3 + 1 x 3 + 15 ⇒ x 3 + 1 x 3 = 125 − 15 = 110. \Rightarrow \Big(x + \dfrac{1}{x}\Big)^3 = x^3 + \dfrac{1}{x^3} + 3\Big(x + \dfrac{1}{x}\Big) \\[1em] \Rightarrow 5^3 = x^3 + \dfrac{1}{x^3} + 3 \times 5 \\[1em] \Rightarrow 125 = x^3 + \dfrac{1}{x^3} + 15 \\[1em] \Rightarrow x^3 + \dfrac{1}{x^3} = 125 - 15 = 110. ⇒ ( x + x 1 ) 3 = x 3 + x 3 1 + 3 ( x + x 1 ) ⇒ 5 3 = x 3 + x 3 1 + 3 × 5 ⇒ 125 = x 3 + x 3 1 + 15 ⇒ x 3 + x 3 1 = 125 − 15 = 110.
Hence, x 3 + 1 x 3 = 110. x^3 + \dfrac{1}{x^3} = 110. x 3 + x 3 1 = 110.
If 3a + 5b + 4c = 0, show that :
27a3 + 125b3 + 64c3 = 180abc.
Answer
By property,
If x + y + z = 0, then :
x3 + y3 + z3 = 3xyz ........(1)
Comparing,
3a + 5b + 4c = 0 with x + y + z = 0, we get :
x = 3a, y = 5b and z = 4c.
Substituting values in equation (1), we get :
⇒ (3a)3 + (5b)3 + (4c)3 = 3 × 3a × 5b × 4c
⇒ 27a3 + 125b3 + 64c3 = 180abc.
Hence, proved that 27a3 + 125b3 + 64c3 = 180abc.
The sum of two Whole numbers is 7 and the sum of their cubes is 133, find the sum of their squares.
Answer
Let two numbers be x and y.
Given,
The sum of two numbers is 7 and the sum of their cubes is 133.
x + y = 7 and x3 + y3 = 133
By formula,
⇒ x3 + y3 = (x + y)3 - 3xy(x + y)
⇒ 133 = 73 - 3xy × 7
⇒ 133 = 343 - 21xy
⇒ 21xy = 343 - 133
⇒ 21xy = 210
⇒ xy = 10.
By formula,
⇒ (x + y)2 = x2 + y2 + 2xy
⇒ 72 = x2 + y2 + 2 × 10
⇒ 49 = x2 + y2 + 20
⇒ x2 + y2 = 49 - 20
⇒ x2 + y2 = 29.
Hence, sum of squares of numbers = 29.
In each of the following, find the value of 'a' :
(i) 4x2 + ax + 9 = (2x + 3)2
(ii) 4x2 + ax + 9 = (2x - 3)2
(iii) 9x2 + (7a - 5)x + 25 = (3x + 5)2
Answer
(i) Given,
⇒ 4x2 + ax + 9 = (2x + 3)2
⇒ 4x2 + ax + 9 = (2x)2 + 32 + 2 × 2x × 3
⇒ 4x2 + ax + 9 = 4x2 + 9 + 12x
⇒ ax = 12x
⇒ a = 12.
Hence, a = 12.
(ii) Given,
⇒ 4x2 + ax + 9 = (2x - 3)2
⇒ 4x2 + ax + 9 = (2x)2 + 32 - 2 × 2x × 3
⇒ 4x2 + ax + 9 = 4x2 + 9 - 12x
⇒ ax = -12x
⇒ a = -12.
Hence, a = -12.
(iii) Given,
⇒ 9x2 + (7a - 5)x + 25 = (3x + 5)2
⇒ 9x2 + (7a - 5)x + 25 = (3x)2 + 52 + 2 × 3x × 5
⇒ 9x2 + (7a - 5)x + 25 = 9x2 + 30x + 25
⇒ 7a - 5 = 30
⇒ 7a = 35
⇒ a = 35 7 \dfrac{35}{7} 7 35
Hence, a = 5.
If x 2 + 1 x = 3 1 3 \dfrac{x^2 + 1}{x} = 3\dfrac{1}{3} x x 2 + 1 = 3 3 1
(i) x − 1 x x - \dfrac{1}{x} x − x 1
(ii) x 3 − 1 x 3 x^3 - \dfrac{1}{x^3} x 3 − x 3 1
Answer
(i) Given,
⇒ x 2 + 1 x = 3 1 3 ⇒ x + 1 x = 10 3 . \Rightarrow \dfrac{x^2 + 1}{x} = 3\dfrac{1}{3} \\[1em] \Rightarrow x + \dfrac{1}{x} = \dfrac{10}{3}. ⇒ x x 2 + 1 = 3 3 1 ⇒ x + x 1 = 3 10 .
By formula,
⇒ ( x + 1 x ) 2 − ( x − 1 x ) 2 = 4 ⇒ ( 10 3 ) 2 − ( x − 1 x ) 2 = 4 ⇒ 100 9 − ( x − 1 x ) 2 = 4 ⇒ ( x − 1 x ) 2 = 100 9 − 4 ⇒ ( x − 1 x ) 2 = 100 − 36 9 ⇒ ( x − 1 x ) 2 = 64 9 ⇒ x − 1 x = 64 9 ⇒ x − 1 x = 8 3 = 2 2 3 . \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(\dfrac{10}{3}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \dfrac{100}{9} - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{100}{9} - 4 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{100 - 36}{9} \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \dfrac{64}{9} \\[1em] \Rightarrow x - \dfrac{1}{x} = \sqrt{\dfrac{64}{9}} \\[1em] \Rightarrow x - \dfrac{1}{x} = \dfrac{8}{3} = 2\dfrac{2}{3}. ⇒ ( x + x 1 ) 2 − ( x − x 1 ) 2 = 4 ⇒ ( 3 10 ) 2 − ( x − x 1 ) 2 = 4 ⇒ 9 100 − ( x − x 1 ) 2 = 4 ⇒ ( x − x 1 ) 2 = 9 100 − 4 ⇒ ( x − x 1 ) 2 = 9 100 − 36 ⇒ ( x − x 1 ) 2 = 9 64 ⇒ x − x 1 = 9 64 ⇒ x − x 1 = 3 8 = 2 3 2 .
Hence, x − 1 x = 2 2 3 x - \dfrac{1}{x} = 2\dfrac{2}{3} x − x 1 = 2 3 2
(ii) By formula,
⇒ x 3 − 1 x 3 = ( x − 1 x ) 3 + 3 ( x − 1 x ) = ( 8 3 ) 3 + 3 × 8 3 = 512 27 + 24 3 = 512 + 216 27 = 728 27 = 26 26 27 . \Rightarrow x^3 - \dfrac{1}{x^3} = \Big(x - \dfrac{1}{x}\Big)^3 + 3\Big(x - \dfrac{1}{x}\Big) \\[1em] = \Big( \dfrac{8}{3}\Big)^3 + 3 \times \dfrac{8}{3} \\[1em] = \dfrac{512}{27} + \dfrac{24}{3} \\[1em] = \dfrac{512 + 216}{27} \\[1em] = \dfrac{728}{27} \\[1em] = 26\dfrac{26}{27}. ⇒ x 3 − x 3 1 = ( x − x 1 ) 3 + 3 ( x − x 1 ) = ( 3 8 ) 3 + 3 × 3 8 = 27 512 + 3 24 = 27 512 + 216 = 27 728 = 26 27 26 .
Hence, x 3 − 1 x 3 = 26 26 27 x^3 - \dfrac{1}{x^3} = 26\dfrac{26}{27} x 3 − x 3 1 = 26 27 26
The difference between two positive numbers is 4 and the difference between their cubes is 316. Find :
(i) their product
(ii) the sum of their squares.
Answer
Let two positive numbers be x and y with x > y.
Given,
The difference between two positive numbers is 4 and the difference between their cubes is 316.
x - y = 4 and x3 - y3 = 316.
(i) Given,
⇒ x - y = 4
Cubing both sides we get :
⇒ (x - y)3 = 43
⇒ x3 - y3 - 3xy(x - y) = 64
⇒ 316 - 3xy × 4 = 64
⇒ 316 - 12xy = 64
⇒ 12xy = 316 - 64
⇒ 12xy = 252
⇒ xy = 252 12 \dfrac{252}{12} 12 252
Hence, product of numbers = 21.
(ii) Given,
x - y = 4
Squaring both sides we get :
⇒ (x - y)2 = 42
⇒ x2 + y2 - 2xy = 16
⇒ x2 + y2 - 2 × 21 = 16
⇒ x2 + y2 - 42 = 16
⇒ x2 + y2 = 16 + 42
⇒ x2 + y2 = 58.
Hence, sum of squares = 58.
If x2 + 1 x 2 \dfrac{1}{\text{x}^2} x 2 1 1 x \dfrac{1}{\text{x}} x 1
1
-1
± 1 \pm 1 ± 1
0
Answer
Given, x2 + 1 x 2 \dfrac{1}{\text{x}^2} x 2 1
As we know
⇒ ( x − 1 x ) 2 = x 2 + 1 x 2 − 2 × x × 1 x ⇒ ( x − 1 x ) 2 = x 2 + 1 x 2 − 2 ⇒ ( x − 1 x ) 2 = 3 − 2 ⇒ ( x − 1 x ) 2 = 1 ⇒ x − 1 x = 1 ⇒ x − 1 x = ± 1 \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \times x \times \dfrac{1}{x}\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 3 - 2\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 1\\[1em] \Rightarrow x - \dfrac{1}{x} = \sqrt{1}\\[1em] \Rightarrow x - \dfrac{1}{x} = \pm 1 ⇒ ( x − x 1 ) 2 = x 2 + x 2 1 − 2 × x × x 1 ⇒ ( x − x 1 ) 2 = x 2 + x 2 1 − 2 ⇒ ( x − x 1 ) 2 = 3 − 2 ⇒ ( x − x 1 ) 2 = 1 ⇒ x − x 1 = 1 ⇒ x − x 1 = ± 1
Hence, option 3 is the correct option.
If a3 + b3 + c3 = 3abc, then a + b + c is equal to:
1
-1
± 1 \pm 1 ± 1
0
Answer
Given, a3 + b3 + c3 = 3abc
By formula,
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
⇒ 3abc - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
⇒ 0 = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
⇒ (a + b + c) = 0 or (a2 + b2 + c2 - ab - bc - ca) = 0
Hence, option 4 is the correct option.
196 × 196 × 196 + 204 × 204 × 204 ( 196 ) 2 + ( 204 ) 2 − 196 × 204 \dfrac{196 \times 196 \times 196 + 204 \times 204 \times 204}{(196)^2 + (204)^2 - 196 \times 204} ( 196 ) 2 + ( 204 ) 2 − 196 × 204 196 × 196 × 196 + 204 × 204 × 204
400
-8
8
none of these
Answer
Given, 196 × 196 × 196 + 204 × 204 × 204 ( 196 ) 2 + ( 204 ) 2 − 196 × 204 \dfrac{196 \times 196 \times 196 + 204 \times 204 \times 204}{(196)^2 + (204)^2 - 196 \times 204} ( 196 ) 2 + ( 204 ) 2 − 196 × 204 196 × 196 × 196 + 204 × 204 × 204
Using the formula, (a3 + b3 ) = (a + b)(a2 - ab + b2 )
⇒ 196 3 + 204 3 ( 196 ) 2 + ( 204 ) 2 − 196 × 204 = ( 196 + 204 ) ( ( 196 ) 2 + ( 204 ) 2 − 196 × 204 ) ( 196 ) 2 + ( 204 ) 2 − 196 × 204 = ( 196 + 204 ) = 400. \Rightarrow \dfrac{196^3 + 204^3}{(196)^2 + (204)^2 - 196 \times 204}\\[1em] = \dfrac{\Big(196 + 204\Big)\Big((196)^2 + (204)^2 - 196 \times 204\Big)}{(196)^2 + (204)^2 - 196 \times 204}\\[1em] = (196 + 204)\\[1em] = 400. ⇒ ( 196 ) 2 + ( 204 ) 2 − 196 × 204 19 6 3 + 20 4 3 = ( 196 ) 2 + ( 204 ) 2 − 196 × 204 ( 196 + 204 ) ( ( 196 ) 2 + ( 204 ) 2 − 196 × 204 ) = ( 196 + 204 ) = 400.
Hence, option 1 is the correct option.
If a = x + 1 x \dfrac{1}{\text{x}} x 1 1 x \dfrac{1}{\text{x}} x 1 2 - b2 is:
x2 + 1 x 2 \dfrac{1}{\text{x}^2} x 2 1
x2 - 1 x 2 \dfrac{1}{\text{x}^2} x 2 1
4
2 ( x 2 + 1 x 2 ) 2\Big(x^2 + \dfrac{1}{\text{x}^2}\Big) 2 ( x 2 + x 2 1 )
Answer
Given, a = x + 1 x \dfrac{1}{\text{x}} x 1 1 x \dfrac{1}{\text{x}} x 1
⇒ a 2 − b 2 = ( x + 1 x ) 2 − ( x − 1 x ) 2 = x 2 + ( 1 x ) 2 + 2 × x × 1 x − [ x 2 + ( 1 x ) 2 − 2 × x × 1 x ] = x 2 + 1 x 2 + 2 − [ x 2 + 1 x 2 − 2 ] = x 2 + 1 x 2 + 2 − x 2 − 1 x 2 + 2 = 4 \Rightarrow a^2 - b^2 = \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2\\[1em] = x^2 + \Big(\dfrac{1}{x}\Big)^2 + 2 \times x \times \dfrac{1}{x} - \Big[x^2 + \Big(\dfrac{1}{x}\Big)^2 - 2 \times x \times \dfrac{1}{x}\Big]\\[1em] = x^2 + \dfrac{1}{x^2} + 2 - \Big[x^2 + \dfrac{1}{x^2} - 2\Big]\\[1em] = x^2 + \dfrac{1}{x^2} + 2 - x^2 - \dfrac{1}{x^2} + 2\\[1em] = 4 ⇒ a 2 − b 2 = ( x + x 1 ) 2 − ( x − x 1 ) 2 = x 2 + ( x 1 ) 2 + 2 × x × x 1 − [ x 2 + ( x 1 ) 2 − 2 × x × x 1 ] = x 2 + x 2 1 + 2 − [ x 2 + x 2 1 − 2 ] = x 2 + x 2 1 + 2 − x 2 − x 2 1 + 2 = 4
Hence, option 3 is the correct option.
Statement 1: x > 0 and x 2 + 1 x 2 x^2 + \dfrac{1}{x^2} x 2 + x 2 1 x 2 − 1 x 2 x^2 - \dfrac{1}{x^2} x 2 − x 2 1
Statement 2: ( x + 1 x ) 2 = x 2 + 1 x 2 + 2 (x + \dfrac{1}{x})^2 = x^2 + \dfrac{1}{x^2} + 2 ( x + x 1 ) 2 = x 2 + x 2 1 + 2
( x − 1 x ) 2 = x 2 + 1 x 2 − 2 (x - \dfrac{1}{x})^2 = x^2 + \dfrac{1}{x^2} - 2 ( x − x 1 ) 2 = x 2 + x 2 1 − 2
and, ( x 2 − 1 x 2 ) = ( x + 1 x ) ( x − 1 x ) (x^2 - \dfrac{1}{x^2}) = (x + \dfrac{1}{x})(x - \dfrac{1}{x}) ( x 2 − x 2 1 ) = ( x + x 1 ) ( x − x 1 )
Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.
Answer
Given, x2 + 1 x 2 \dfrac{1}{\text{x}^2} x 2 1
As we know,
⇒ ( x − 1 x ) 2 = x 2 + 1 x 2 − 2 × x × 1 x ⇒ ( x − 1 x ) 2 = x 2 + 1 x 2 − 2 ⇒ ( x − 1 x ) 2 = 2 − 2 ⇒ ( x − 1 x ) 2 = 0 ⇒ x − 1 x = 0 ⇒ x − 1 x = 0 ⇒ ( x + 1 x ) 2 = x 2 + 1 x 2 + 2 × x × 1 x ⇒ ( x + 1 x ) 2 = x 2 + 1 x 2 + 2 ⇒ ( x + 1 x ) 2 = 2 + 2 ⇒ ( x + 1 x ) 2 = 4 ⇒ x + 1 x = 4 ⇒ x + 1 x = ± 2 \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \times x \times \dfrac{1}{x}\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 2 - 2\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 0\\[1em] \Rightarrow x - \dfrac{1}{x} = \sqrt{0}\\[1em] \Rightarrow x - \dfrac{1}{x} = 0\\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 \times x \times \dfrac{1}{x}\\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2\\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 2 + 2\\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 4\\[1em] \Rightarrow x + \dfrac{1}{x} = \sqrt{4}\\[1em] \Rightarrow x + \dfrac{1}{x} = \pm 2\\[1em] ⇒ ( x − x 1 ) 2 = x 2 + x 2 1 − 2 × x × x 1 ⇒ ( x − x 1 ) 2 = x 2 + x 2 1 − 2 ⇒ ( x − x 1 ) 2 = 2 − 2 ⇒ ( x − x 1 ) 2 = 0 ⇒ x − x 1 = 0 ⇒ x − x 1 = 0 ⇒ ( x + x 1 ) 2 = x 2 + x 2 1 + 2 × x × x 1 ⇒ ( x + x 1 ) 2 = x 2 + x 2 1 + 2 ⇒ ( x + x 1 ) 2 = 2 + 2 ⇒ ( x + x 1 ) 2 = 4 ⇒ x + x 1 = 4 ⇒ x + x 1 = ± 2
Now, we know that ( x 2 − 1 x 2 ) = ( x + 1 x ) ( x − 1 x ) (x^2 - \dfrac{1}{x^2}) = (x + \dfrac{1}{x})(x - \dfrac{1}{x}) ( x 2 − x 2 1 ) = ( x + x 1 ) ( x − x 1 )
= ± \pm ±
= 0.
∴ Both the statements are true.
Hence, option 1 is the correct option.
Assertion (A): x2 - 5x - 1 = 0
⇒ x - 1 x \dfrac{1}{x} x 1
Reason (R): x2 - 5x- 1 = 0
⇒ x - 1 x \dfrac{1}{x} x 1
But x - 1 x \dfrac{1}{x} x 1
A is true, but R is false.
A is false, but R is true.
Both A and R are true, and R is the correct reason for A.
Both A and R are true, and R is the incorrect reason for A.
Answer
Given, x2 - 5x - 1 = 0
⇒ x2 - 1 = 5x
⇒ x 2 − 1 x \dfrac{x^2 - 1}{x} x x 2 − 1
⇒ x 2 x − 1 x \dfrac{x^2}{x} - \dfrac{1}{x} x x 2 − x 1
⇒ x - 1 x \dfrac{1}{x} x 1
So, x - 1 x \dfrac{1}{x} x 1
∴ A is false, but R is true.
Hence, option 2 is the correct option.
Assertion (A): ( x + 1 x ) 2 − ( x − 1 x ) 2 = 4 \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 ( x + x 1 ) 2 − ( x − x 1 ) 2 = 4
Reason (R): (a + b)2 - (a - b)2 = 4ab
A is true, but R is false.
A is false, but R is true.
Both A and R are true, and R is the correct reason for A.
Both A and R are true, and R is the incorrect reason for A.
Answer
Given, (a + b)2 - (a - b)2
= (a2 + b2 + 2ab) - (a2 + b2 - 2ab)
= a2 + b2 + 2ab - a2 - b2 + 2ab
= 2ab + 2ab
= 4ab.
So, reason (R) is true.
Substituting the value of a = x and b = 1 x \dfrac{1}{x} x 1
⇒ (a + b)2 - (a - b)2 = 4ab
⇒ ( x + 1 x ) 2 − ( x − 1 x ) 2 = 4 × x × 1 x \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \times x \times \dfrac{1}{x} ⇒ ( x + x 1 ) 2 − ( x − x 1 ) 2 = 4 × x × x 1
= 4.
So, assertion (A) is true.
∴ Both A and R are true, and R is the correct reason for A.
Hence, option 3 is the correct option.
Assertion (A): If x > y, x + y = 6 and x - y = 2 then x2 + y2 = 40
Reason (R): (x + y)2 + (x - y)2 = 2(x2 + y2 )
A is true, but R is false.
A is false, but R is true.
Both A and R are true, and R is the correct reason for A.
Both A and R are true, and R is the incorrect reason for A.
Answer
By formula,
⇒ (x + y)2 = x2 + y2 + 2xy ....(1)
⇒ (x - y)2 = x2 + y2 - 2xy .....(2)
Adding equation (1) and (2), we get :
⇒ (x + y)2 + (x - y)2 = x2 + y2 + 2xy + x2 + y2 - 2xy
⇒ (x + y)2 + (x - y)2 = 2(x2 + y2 )
So, reason (R) is true.
Given,
x + y = 6 and x - y = 2
By formula,
⇒ 2(x2 + y2 ) = (x + y)2 + (x - y)2
⇒ 2(x2 + y2 ) = 62 + 22
⇒ 2(x2 + y2 ) = 36 + 4
⇒ 2(x2 + y2 ) = 40
⇒ x2 + y2 = 40 2 \dfrac{40}{2} 2 40
⇒ x2 + y2 = 20.
So, assertion (A) is false.
∴ A is false, but R is true.
Hence, option 2 is the correct option.
Assertion (A): 53 - 33 - 23 = 3 x 5 x -3 x -2
Reason (R): ∵ 5 - 3 - 2 = 0
⇒ 53 - 33 - 23 = 53 + (-3)3 + (-2)3
A is true, but R is false.
A is false, but R is true.
Both A and R are true, and R is the correct reason for A.
Both A and R are true, and R is the incorrect reason for A.
Answer
By property,
If a + b + c = 0, then
⇒ a3 + b3 + c3 = 3abc
Given,
⇒ 53 - 33 - 23
⇒ 53 + (-3)3 + (-2)3
Let a = 5, b = -3 and c = -2.
∴ a + b + c = 5 + (-3) + (-2) = 5 - 3 - 2 = 0
⇒ a3 + b3 + c3 = 3abc
⇒ 53 + (-3)3 + (-2)3 = 3 × 5 × (-3) × (-2)
∴ Both A and R are true, and R is the correct reason for A.
Hence, option 3 is the correct option.
Simplify (x + 6)(x + 4)(x - 2)
Answer
Given,
⇒ (x + 6)(x + 4)(x - 2)
⇒ (x + 6)(x2 - 2x + 4x - 8)
⇒ (x + 6)(x2 + 2x - 8)
⇒ x3 + 2x2 - 8x + 6x2 + 12x - 48
⇒ x3 + 8x2 + 4x - 48.
Hence, (x + 6)(x + 4)(x - 2) = x3 + 8x2 + 4x - 48.
Simplify (x - 6)(x - 4)(x + 2)
Answer
Given,
⇒ (x - 6)(x - 4)(x + 2)
⇒ (x - 6)(x2 + 2x - 4x - 8)
⇒ (x - 6)(x2 - 2x - 8)
⇒ x3 - 2x2 - 8x - 6x2 + 12x + 48
⇒ x3 - 8x2 + 4x + 48.
Hence, (x - 6)(x - 4)(x + 2) = x3 - 8x2 + 4x + 48.
Simplify (x - 6)(x - 4)(x - 2)
Answer
Given,
⇒ (x - 6)(x - 4)(x - 2)
⇒ (x - 6)(x2 - 2x - 4x + 8)
⇒ (x - 6)(x2 - 6x + 8)
⇒ x3 - 6x2 + 8x - 6x2 + 36x - 48
⇒ x3 - 12x2 + 44x - 48.
Hence, (x - 6)(x - 4)(x - 2) = x3 - 12x2 + 44x - 48.
Simplify (x + 6)(x - 4)(x - 2)
Answer
Given,
⇒ (x + 6)(x - 4)(x - 2)
⇒ (x + 6)(x2 - 2x - 4x + 8)
⇒ (x + 6)(x2 - 6x + 8)
⇒ x3 - 6x2 + 8x + 6x2 - 36x + 48
⇒ x3 - 28x + 48.
Hence, (x + 6)(x - 4)(x - 2) = x3 - 28x + 48.
Simplify using following identity;
(a ± \pm ± 2 ∓ \mp ∓ 2 ) = a3 ± \pm ± 3
(i) (2x + 3y)(4x2 - 6xy + 9y2 )
(ii) ( a 3 − 3 b ) ( a 2 9 + a b + 9 b 2 ) \Big(\dfrac{a}{3} - 3b\Big)\Big(\dfrac{a^2}{9} + ab + 9b^2\Big) ( 3 a − 3 b ) ( 9 a 2 + ab + 9 b 2 )
Answer
(i) Given,
⇒ (2x + 3y)(4x2 - 6xy + 9y2 )
⇒ (2x + 3y)[(2x)2 - 2x × 3y + (3y)2 ]
Comparing above equation with (a ± \pm ± 2 ∓ \mp ∓ ± \pm ± 2 ) = a3 ± \pm ± 3 , we get :
a = 2x and b = 3y
∴ (2x + 3y)[(2x)2 - 2x × 3y + (3y)2 ] = (2x)3 + (3y)3
= 8x3 + 27y3 .
Hence, (2x + 3y)(4x2 - 6xy + 9y2 ) = 8x3 + 27y3 .
(ii) Given,
⇒ ( a 3 − 3 b ) ( a 2 9 + a b + 9 b 2 ) ⇒ ( a 3 − 3 b ) [ ( a 3 ) 2 + a 3 × 3 b + ( 3 b ) 2 ] \Rightarrow \Big(\dfrac{a}{3} - 3b\Big)\Big(\dfrac{a^2}{9} + ab + 9b^2\Big) \\[1em] \Rightarrow \Big(\dfrac{a}{3} - 3b\Big)\Big[\Big(\dfrac{a}{3}\Big)^2 + \dfrac{a}{3} \times 3b + (3b)^2\Big]\\[1em] ⇒ ( 3 a − 3 b ) ( 9 a 2 + ab + 9 b 2 ) ⇒ ( 3 a − 3 b ) [ ( 3 a ) 2 + 3 a × 3 b + ( 3 b ) 2 ]
Using identity given in question :
∴ ( a 3 − 3 b ) [ ( a 3 ) 2 + a 3 × 3 b + ( 3 b ) 2 ] = ( a 3 ) 3 − ( 3 b ) 3 = a 3 27 − 27 b 3 . \therefore \Big(\dfrac{a}{3} - 3b\Big)\Big[\Big(\dfrac{a}{3}\Big)^2 + \dfrac{a}{3} \times 3b + (3b)^2\Big] = \Big(\dfrac{a}{3}\Big)^3 - (3b)^3\\[1em] = \dfrac{a^3}{27} - 27b^3. ∴ ( 3 a − 3 b ) [ ( 3 a ) 2 + 3 a × 3 b + ( 3 b ) 2 ] = ( 3 a ) 3 − ( 3 b ) 3 = 27 a 3 − 27 b 3 .
Hence, ( a 3 − 3 b ) ( a 2 9 + a b + 9 b 2 ) = a 3 27 − 27 b 3 \Big(\dfrac{a}{3} - 3b\Big)\Big(\dfrac{a^2}{9} + ab + 9b^2\Big) = \dfrac{a^3}{27} - 27b^3 ( 3 a − 3 b ) ( 9 a 2 + ab + 9 b 2 ) = 27 a 3 − 27 b 3
Using suitable identity, evaluate :
(i) (104)3
(ii) (97)3
Answer
(i) Given,
⇒ (104)3
⇒ (100 + 4)3
Using identity :
(a + b)3 = a3 + b3 + 3ab(a + b)
⇒ (100 + 4)3 = (100)3 + 43 + 3 × 100 × 4 × (100 + 4)
⇒ (100 + 4)3 = 1000000 + 64 + 124800
⇒ 1124864.
Hence, (104)3 = 1124864.
(ii) Given,
⇒ (97)3
⇒ (100 - 3)3
⇒ (100)3 - 33 - 3 × 100 × 3 × (100 - 3)
⇒ 1000000 - 27 - 900 × 97
⇒ 1000000 - 27 - 87300
⇒ 912673.
Hence, (97)3 = 912673.
Simplify :
( x 2 − y 2 ) 3 + ( y 2 − z 2 ) 3 + ( z 2 − x 2 ) 3 ( x − y ) 3 + ( y − z ) 3 + ( z − x ) 3 \dfrac{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}{(x - y)^3 + (y- z)^3 + (z - x)^3} ( x − y ) 3 + ( y − z ) 3 + ( z − x ) 3 ( x 2 − y 2 ) 3 + ( y 2 − z 2 ) 3 + ( z 2 − x 2 ) 3
Answer
If a + b + c = 0; we have :
a3 + b3 + c3 = 3abc.
Since, (x2 - y2 ) + (y2 - z2 ) + (z2 - x2 ) = 0.
∴ ( x 2 − y 2 ) 3 + ( y 2 − z 2 ) 3 + ( z 2 − x 2 ) 3 = 3 ( x 2 − y 2 ) ( y 2 − z 2 ) ( z 2 − x 2 ) (x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3 = 3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2) ( x 2 − y 2 ) 3 + ( y 2 − z 2 ) 3 + ( z 2 − x 2 ) 3 = 3 ( x 2 − y 2 ) ( y 2 − z 2 ) ( z 2 − x 2 )
Also,
(x - y) + (y - z) + (z - x) = 0
∴ ( x − y ) 3 + ( y − z ) 3 + ( z − x ) 3 = 3 ( x − y ) ( y − z ) ( z − x ) (x - y)^3 + (y- z)^3 + (z - x)^3 = 3(x - y)(y - z)(z - x) ( x − y ) 3 + ( y − z ) 3 + ( z − x ) 3 = 3 ( x − y ) ( y − z ) ( z − x )
Dividing equation (1) by (2), we get :
⇒ ( x 2 − y 2 ) 3 + ( y 2 − z 2 ) 3 + ( z 2 − x 2 ) 3 ( x − y ) 3 + ( y − z ) 3 + ( z − x ) 3 = 3 ( x 2 − y 2 ) ( y 2 − z 2 ) ( z 2 − x 2 ) 3 ( x − y ) ( y − z ) ( z − x ) = 3 ( x − y ) ( x + y ) ( y − z ) ( y + z ) ( z − x ) ( z + x ) 3 ( x − y ) ( y − z ) ( z − x ) = ( x + y ) ( y + z ) ( z + x ) . \Rightarrow \dfrac{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}{(x - y)^3 + (y- z)^3 + (z - x)^3} = \dfrac{3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2)}{3(x - y)(y - z)(z - x)} \\[1em] = \dfrac{3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x)}{3(x - y)(y - z)(z - x)} \\[1em] = (x + y)(y + z)(z + x). ⇒ ( x − y ) 3 + ( y − z ) 3 + ( z − x ) 3 ( x 2 − y 2 ) 3 + ( y 2 − z 2 ) 3 + ( z 2 − x 2 ) 3 = 3 ( x − y ) ( y − z ) ( z − x ) 3 ( x 2 − y 2 ) ( y 2 − z 2 ) ( z 2 − x 2 ) = 3 ( x − y ) ( y − z ) ( z − x ) 3 ( x − y ) ( x + y ) ( y − z ) ( y + z ) ( z − x ) ( z + x ) = ( x + y ) ( y + z ) ( z + x ) .
Hence, ( x 2 − y 2 ) 3 + ( y 2 − z 2 ) 3 + ( z 2 − x 2 ) 3 ( x − y ) 3 + ( y − z ) 3 + ( z − x ) 3 \dfrac{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}{(x - y)^3 + (y- z)^3 + (z - x)^3} ( x − y ) 3 + ( y − z ) 3 + ( z − x ) 3 ( x 2 − y 2 ) 3 + ( y 2 − z 2 ) 3 + ( z 2 − x 2 ) 3
Evaluate:
0.8 × 0.8 × 0.8 + 0.5 × 0.5 × 0.5 0.8 × 0.8 − 0.8 × 0.5 + 0.5 × 0.5 \dfrac{0.8 \times 0.8 \times 0.8 + 0.5 \times 0.5 \times 0.5}{0.8 \times 0.8 - 0.8 \times 0.5 + 0.5 \times 0.5} 0.8 × 0.8 − 0.8 × 0.5 + 0.5 × 0.5 0.8 × 0.8 × 0.8 + 0.5 × 0.5 × 0.5
Answer
Given,
⇒ 0.8 × 0.8 × 0.8 + 0.5 × 0.5 × 0.5 0.8 × 0.8 − 0.8 × 0.5 + 0.5 × 0.5 ⇒ ( 0.8 ) 3 + ( 0.5 ) 3 ( 0.8 ) 2 + ( 0.5 ) 2 − 0.8 × 0.5 \Rightarrow \dfrac{0.8 \times 0.8 \times 0.8 + 0.5 \times 0.5 \times 0.5}{0.8 \times 0.8 - 0.8 \times 0.5 + 0.5 \times 0.5} \\[1em] \Rightarrow \dfrac{(0.8)^3 + (0.5)^3}{(0.8)^2 + (0.5)^2 - 0.8 \times 0.5} ⇒ 0.8 × 0.8 − 0.8 × 0.5 + 0.5 × 0.5 0.8 × 0.8 × 0.8 + 0.5 × 0.5 × 0.5 ⇒ ( 0.8 ) 2 + ( 0.5 ) 2 − 0.8 × 0.5 ( 0.8 ) 3 + ( 0.5 ) 3
Using the formula, (a3 + b3 ) = (a + b)(a2 - ab + b2 )
= ( 0.8 + 0.5 ) [ ( 0.8 ) 2 + ( 0.5 ) 2 − 0.8 × 0.5 ] ( 0.8 ) 2 + ( 0.5 ) 2 − 0.8 × 0.5 = ( 0.8 + 0.5 ) = 1.3 = \dfrac{(0.8 + 0.5)[(0.8)^2 + (0.5)^2 - 0.8 \times 0.5]}{(0.8)^2 + (0.5)^2 - 0.8 \times 0.5}\\[1em] = (0.8 + 0.5)\\[1em] = 1.3 = ( 0.8 ) 2 + ( 0.5 ) 2 − 0.8 × 0.5 ( 0.8 + 0.5 ) [( 0.8 ) 2 + ( 0.5 ) 2 − 0.8 × 0.5 ] = ( 0.8 + 0.5 ) = 1.3
Hence, 0.8 × 0.8 × 0.8 + 0.5 × 0.5 × 0.5 0.8 × 0.8 − 0.8 × 0.5 + 0.5 × 0.5 \dfrac{0.8 \times 0.8 \times 0.8 + 0.5 \times 0.5 \times 0.5}{0.8 \times 0.8 - 0.8 \times 0.5 + 0.5 \times 0.5} 0.8 × 0.8 − 0.8 × 0.5 + 0.5 × 0.5 0.8 × 0.8 × 0.8 + 0.5 × 0.5 × 0.5
If a - 2b + 3c = 0; state the value of a3 - 8b3 + 27c3 .
Answer
By property,
If x + y + z = 0, then :
x3 + y3 + z3 = 3xyz.
Given,
⇒ a - 2b + 3c = 0
⇒ a + (-2b) + 3c = 0
∴ a3 + (-2b)3 + (3c)3 = 3 × a × (-2b) × 3c
⇒ a3 - 8b3 + 27c3 = -18abc.
Hence, a3 - 8b3 + 27c3 = -18abc.
If x + 5y = 10; find the value of x3 + 125y3 + 150xy - 1000.
Answer
Given,
⇒ x + 5y = 10
Cubing both sides we get :
⇒ (x + 5y)3 = 103
⇒ x3 + (5y)3 + 3 × x × 5y × (x + 5y) = 1000
⇒ x3 + 125y3 + 15xy × 10 = 1000
⇒ x3 + 125y3 + 150xy - 1000 = 0.
Hence, x3 + 125y3 + 150xy - 1000 = 0.
If a + b = 11 and a2 + b2 = 65; find a3 + b3 .
Answer
By formula,
⇒ (a + b)2 = a2 + b2 + 2ab
⇒ 112 = 65 + 2ab
⇒ 121 = 65 + 2ab
⇒ 2ab = 56
⇒ ab = 56 2 \dfrac{56}{2} 2 56
By formula,
⇒ (a + b)3 = a3 + b3 + 3ab(a + b)
⇒ 113 = a3 + b3 + 3 × 28 × 11
⇒ 1331 = a3 + b3 + 924
⇒ a3 + b3 = 1331 - 924 = 407.
Hence, a3 + b3 = 407.
If x, y and z are three different numbers, then prove that :
x2 + y2 + z2 - xy - yz - zx is always positive.
Answer
Given,
⇒ x2 + y2 + z2 - xy - yz - zx
Multiplying the above equation by 2,
⇒ 2(x2 + y2 + z2 - xy - yz - zx )
⇒ 2x2 + 2y2 + 2z2 - 2xy - 2yz - 2zx
⇒ x2 + x2 + y2 + y2 + z2 + z2 - 2xy - 2yz - 2zx
⇒ x2 + y2 - 2xy + y2 + z2 - 2yz + z2 + x2 - 2zx
⇒ (x - y)2 + (y - z)2 + (z - x)2
From above equation we can see that for distinct value of x, y and z given equation is always positive.
