221. If   x =	4ab	(a ≠ b) , the value of	x + 2a	+	x + 2b	is;
     	a + b		x − 2a		x − 2b

1. 1. a

   
   
   

   2. b

   
   
   

   3. 2ab

   
   
   

   4. 2

   
   
   

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   x =	4ab	⇒	x	=	2b
   	a + b		2a		a + b

   
   By componendo and dividendo,
   

   x + 2a	=	2b + a + b	=	3b + a
   x − 2a		2b − a − b		b − a

   
   

   Similarly,	x	=	2a
   	2b		a + b

   
   

   ⇒	x + 2a	=	2b + a + b	=	3a + b
   	x − 2a		2b − a − b		a − b

   
   

   ∴	x + 2a	+	x + 2b
   	x − 2a		x − 2b

   
   

   =	3b + a	+	3a + b
   	b − a		a − b

   
   

   =	3b + a − 3a − b	=	2b − 2a
   	b − a		b − a

   =	2(b − a)	= 2
   	b − a

   Correct Option: D

   x =	4ab	⇒	x	=	2b
   	a + b		2a		a + b

   
   By componendo and dividendo,
   

   x + 2a	=	2b + a + b	=	3b + a
   x − 2a		2b − a − b		b − a

   
   

   Similarly,	x	=	2a
   	2b		a + b

   
   

   ⇒	x + 2a	=	2b + a + b	=	3a + b
   	x − 2a		2b − a − b		a − b

   
   

   ∴	x + 2a	+	x + 2b
   	x − 2a		x − 2b

   
   

   =	3b + a	+	3a + b
   	b − a		a − b

   
   

   =	3b + a − 3a − b	=	2b − 2a
   	b − a		b − a

   =	2(b − a)	= 2
   	b − a

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222. If   x +	9	= 6, then the value of		x2 +	9		is :
     	x				x2

1. 1. 8

   
   
   

   2. 9

   
   
   

   3. 10

   
   
   

   4. 12

   
   
   

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   x +	9	= 6
   	x

   
   ⇒  x² – 6x + 9 = 0
   ⇒  (x – 3)² = 0 ⇒ x = 3
   

   ∴		x2 +	9		=		9 +	9		= 10
   			x2					9

   Correct Option: C

   x +	9	= 6
   	x

   
   ⇒  x² – 6x + 9 = 0
   ⇒  (x – 3)² = 0 ⇒ x = 3
   

   ∴		x2 +	9		=		9 +	9		= 10
   			x2					9

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223. If   x = √3 + √2, then the value of		x2 −	1		is :
     			x2

1. 1. 4

   
   
   

   2. 6

   
   
   

   3. 9

   
   
   

   4. 10

   
   
   

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   x = √3 + √2
   

   1	=	1
   x		√3 + √2

   
   

   =	1	×	3 − 2√2	= √3 − √2
   	√3 + √2		√3 − √2

   
   

   ∴  x +	1	= 2√3
   	x

   
   

   ∴		x2 +	1		=		x +	1		2	− 2
   			x2					x

   
   = (2√3)² − 2
   = 12 – 2 = 10

   Correct Option: D

   x = √3 + √2
   

   1	=	1
   x		√3 + √2

   
   

   =	1	×	3 − 2√2	= √3 − √2
   	√3 + √2		√3 − √2

   
   

   ∴  x +	1	= 2√3
   	x

   
   

   ∴		x2 +	1		=		x +	1		2	− 2
   			x2					x

   
   = (2√3)² − 2
   = 12 – 2 = 10

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224. If   x = 3 + 2 √2, then the value of		√x −	1		is :
     			√x

1. 1. 1

   
   
   

   2. 2

   
   
   

   3. 2√2

   
   
   

   4. 3√3

   
   
   

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   x = 3 + 2√2
   

   ∴	1	=	1
   	x		3 + 2√2

   
   

   =	1	×	3 − 2√2
   	3 + 2√2		3 − 2√2

   
   

   =	3 − 2√2	= 3 − 2√2
   	9 − 8

   
   

   ∴		√x +	1		2	= x +	1	− 2
   			√x				x

   
   = 3 + 2√2 + 3 − 2√2
   = 4
   

   ∴	√x −	1	= 2
   		√x

   Correct Option: B

   x = 3 + 2√2
   

   ∴	1	=	1
   	x		3 + 2√2

   
   

   =	1	×	3 − 2√2
   	3 + 2√2		3 − 2√2

   
   

   =	3 − 2√2	= 3 − 2√2
   	9 − 8

   
   

   ∴		√x +	1		2	= x +	1	− 2
   			√x				x

   
   = 3 + 2√2 + 3 − 2√2
   = 4
   

   ∴	√x −	1	= 2
   		√x

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225. If a² + b² + c² + 3 = 2 (a + b + c) then the value of (a + b + c) is

1. 1. 2

   
   
   

   2. 3

   
   
   

   3. 4

   
   
   

   4. 5

   
   
   

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   a² + b² + c² + 3
   = 2a + 2b + 2c
   ⇒  a² – 2a + 1 + b² – 2b + 1 + c² – 2c + 1 = 0
   ⇒  (a – 1)² +(b –1)² +(c – 1)² = 0
   ⇒  a – 1 = 0 ⇒ a = 1;
     b – 1 = 0 ⇒ b = 1
   and, c – 1 = 0 ⇒ c = 1
   ∴  a + b + c = 3

   Correct Option: B

   a² + b² + c² + 3
   = 2a + 2b + 2c
   ⇒  a² – 2a + 1 + b² – 2b + 1 + c² – 2c + 1 = 0
   ⇒  (a – 1)² +(b –1)² +(c – 1)² = 0
   ⇒  a – 1 = 0 ⇒ a = 1;
     b – 1 = 0 ⇒ b = 1
   and, c – 1 = 0 ⇒ c = 1
   ∴  a + b + c = 3

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