486. If	4x	+ 2P = 12   for what value of P, x = 6 ?
     	3

1. 1. 6

   
   
   

   2. 4

   
   
   

   3. 2

   
   
   

   4. 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   When x = 6,
   

   4 × 6	+ 2P = 12
   3

   
   ⇒  8 + 2P = 12
   ⇒  2P = 12 – 8 = 4
   ⇒  P = 2

   Correct Option: C

   When x = 6,
   

   4 × 6	+ 2P = 12
   3

   
   ⇒  8 + 2P = 12
   ⇒  2P = 12 – 8 = 4
   ⇒  P = 2

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487. Let a = √6 − √5 , b = √5 − 2, c = 2 – √3
     Then point out the correct alternative among the four alternatives given below.

1. 1. b < a < c

   
   
   

   2. a < c < b

   
   
   

   3. b < c < a

   
   
   

   4. a < b < c

   
   
   

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1. View Hint View Answer Discuss in Forum

   1	=	1
   a		√6 – √5

   
   

   √6 + √5	= √6 + √5
   6 − 5

   
   Similarly,
   

   1	= √5 + 2;	1	= 2 + √3
   b		c

   
   

   ∴	1	>	1	>	1	⇒ a < b < c
   	a		b		c

   Correct Option: D

   1	=	1
   a		√6 – √5

   
   

   √6 + √5	= √6 + √5
   6 − 5

   
   Similarly,
   

   1	= √5 + 2;	1	= 2 + √3
   b		c

   
   

   ∴	1	>	1	>	1	⇒ a < b < c
   	a		b		c

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488. If   x = 5 − √21, then the value of	√x	is
     	√32 − x − √21

1. 1. 1	(√3 − √7)
      √2

   
   
   

   2. 1	(√7 − √3)
      √2

   
   
   

   3. 1	(√7 + √3)
      √2

   
   
   

   4. 1	(7 − √3)
      √2

   
   
   

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1. View Hint View Answer Discuss in Forum

   √x = √5 − √21
   

   ⇒  √x =	√10 − 2√21
   	√2

   
   

   =	√7 + 3 − 2 × √7× √3
   	√2

   
   

   =	√7 − √3
   	√2

   
   √32 − 2x = √32 − 2 ( 5 − √21)
   = √32 − 10 + 2√21)
   = √22 + 2√21)
   = √21 + 1 + 2 × √21 × 1
   = √21 + 1
   

   =	√7 − √3
   	√2(√21 + 1 − √21)

   
   

   =	1	= √7 − √3
   	√2

   Correct Option: B

   √x = √5 − √21
   

   ⇒  √x =	√10 − 2√21
   	√2

   
   

   =	√7 + 3 − 2 × √7× √3
   	√2

   
   

   =	√7 − √3
   	√2

   
   √32 − 2x = √32 − 2 ( 5 − √21)
   = √32 − 10 + 2√21)
   = √22 + 2√21)
   = √21 + 1 + 2 × √21 × 1
   = √21 + 1
   

   =	√7 − √3
   	√2(√21 + 1 − √21)

   
   

   =	1	= √7 − √3
   	√2

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489. If 6x – 5y = 13, 7x + 2y = 23 then 11x + 18y =

1. 1. –15

   
   
   

   2. 51

   
   
   

   3. 33

   
   
   

   4. 15

   
   
   

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1. View Hint View Answer Discuss in Forum

   6x – 5y = 13           ...(i)
   7x + 2y = 23           ...(ii)
   By equation (i) × 2 + (ii) × 5,
   

   12x – 10y = 26

   35x + 10y = 115

   _______________

   47x           = 141

   
   ⇒  x = 3
   From equation (i),
   6 × 3 – 5y = 13
   ⇒  18 – 5y = 13
   ⇒  5y = 5
   ⇒  y = 1
   ∴  11x + 18y = 11 × 3 + 18 × 1
   = 33 + 18 = 51

   Correct Option: B

   6x – 5y = 13           ...(i)
   7x + 2y = 23           ...(ii)
   By equation (i) × 2 + (ii) × 5,
   

   12x – 10y = 26

   35x + 10y = 115

   _______________

   47x           = 141

   
   ⇒  x = 3
   From equation (i),
   6 × 3 – 5y = 13
   ⇒  18 – 5y = 13
   ⇒  5y = 5
   ⇒  y = 1
   ∴  11x + 18y = 11 × 3 + 18 × 1
   = 33 + 18 = 51

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490. The value of (x^{b + c})^{b − c} (x^{c + a})^{c − a} (x^{a + b})^{a − b} (x ≠ 0) is

1. 1. 1

   
   
   

   2. 2

   
   
   

   3. –1

   
   
   

   4. 0

   
   
   

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1. View Hint View Answer Discuss in Forum

   (x^{b + c})^{b − c}. (x^{c + a})^{c − a}.(x^{a + b})^{a − b}
   = x^{b2− c2}.x^{c2− a2}.x^{a2− b2}
   = x^{b2 − c2 + c2− a2 + a2 − b2} = x⁰ = 1

   Correct Option: A

   (x^{b + c})^{b − c}. (x^{c + a})^{c − a}.(x^{a + b})^{a − b}
   = x^{b2− c2}.x^{c2− a2}.x^{a2− b2}
   = x^{b2 − c2 + c2− a2 + a2 − b2} = x⁰ = 1

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