216. If   a + b + c = 0, then the value of	1	+	1	+	1	is
     	(a + b)(b + c)		(a + c)(b + a)		(c + a)(c + b)

1. 1. 1

   
   
   

   2. 0

   
   
   

   3. –1

   
   
   

   4. –2

   
   
   

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   1	+	1	+	1
   (a + b)(b + c)		(a + c)(b + a)		(c + a)(c + b)

   
   

   =	c + a + b + c + a + b
   	(a + b)(b + c)(c + a)

   
   

   =	2(a + b + c )
   	(a + b)(b + c)(c + a)

   
   = 0 [because;  a + b + c = 0]

   Correct Option: B

   1	+	1	+	1
   (a + b)(b + c)		(a + c)(b + a)		(c + a)(c + b)

   
   

   =	c + a + b + c + a + b
   	(a + b)(b + c)(c + a)

   
   

   =	2(a + b + c )
   	(a + b)(b + c)(c + a)

   
   = 0 [because;  a + b + c = 0]

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217. If x + y + z = 0, then	x2	+	y2	+	z2	= ?
     	yz		zx		xy

1. 1. (xyz)²

   
   
   

   2. x² + y² + z²

   
   
   

   3. 9

   
   
   

   4. 3

   
   
   

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   x2	+	y2	+	z2
   yz		zx		xy

   
   

   =	x3 + y3 + z3	=	3xyz	= 3
   	xyz		xyz

   Correct Option: D

   x2	+	y2	+	z2
   yz		zx		xy

   
   

   =	x3 + y3 + z3	=	3xyz	= 3
   	xyz		xyz

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218. If   x – y =	x + y	=	xy	, the numerical value of xy is
     	7		4

1. 1. 4

      3

   
   
   

   2. 3

      4

   
   
   

   3. 1

      4

   
   
   

   4. 1

      3

   
   
   

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   x – y =	x + y	=	xy	= k
   	7		4

   
   ⇒  x – y = k
   x + y = 7k
   ∴  (x + y)² – (x – y)² = 49k² – k²
   ⇒  4xy = 48k²
   ⇒  16k = 48k²
   

   ⇒  k =	1
   	3

   
   

   ∴  xy = 4k = 4 ×	1	=	4
   	3		3

   Correct Option: A

   x – y =	x + y	=	xy	= k
   	7		4

   
   ⇒  x – y = k
   x + y = 7k
   ∴  (x + y)² – (x – y)² = 49k² – k²
   ⇒  4xy = 48k²
   ⇒  16k = 48k²
   

   ⇒  k =	1
   	3

   
   

   ∴  xy = 4k = 4 ×	1	=	4
   	3		3

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219. If   a2 + b2 + 2b + 4a + 5 = 0, then the value of	a − b	is
     	a + b

1. 1. 3

   
   
   

   2. –3

   
   
   

   3. 1

      3

   
   
   

   4. −	1
      	3

   
   
   

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   a² + b² + 2b + 4a + 5 = 0
   ⇒  a² + 4a + b² + 2b + 5 = 0
   ⇒  a² + 4a + 4 + b² + 2b + 1 = 0
   ⇒  (a + 2)² + (b + 1)² = 0
   It is possible only when
   a + 2 = 0 ⇒ a = –2
   and, b + 1 = 0 ⇒ b = – 1
   

   ∴	a − b	=	−2 + 1
   	a + b		−2 − 1

   
   

   =	− 1	=	1
   	−3		3

   Correct Option: C

   a² + b² + 2b + 4a + 5 = 0
   ⇒  a² + 4a + b² + 2b + 5 = 0
   ⇒  a² + 4a + 4 + b² + 2b + 1 = 0
   ⇒  (a + 2)² + (b + 1)² = 0
   It is possible only when
   a + 2 = 0 ⇒ a = –2
   and, b + 1 = 0 ⇒ b = – 1
   

   ∴	a − b	=	−2 + 1
   	a + b		−2 − 1

   
   

   =	− 1	=	1
   	−3		3

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220. If   m +	1	= 4, find the value of (m − 2)2 +	1	.
     	m − 2		(m − 2)2

1. 1. –2

   
   
   

   2. 0

   
   
   

   3. 2

   
   
   

   4. 4

   
   
   

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   m +	1	= 4
   	m − 2

   
   

   ⇒  m +	1	– 2 = 4 – 2
   	m − 2

   
   

   ⇒  (m − 2) +	1	= 4 – 2 = 2
   	(m − 2)

   
   On squaring both sides,
   

   (m − 2)2 +	1	+ 2(m − 2)		1		= 4
   	(m − 2)2			m − 2

   
   

   ⇒  (m − 2)2 +	1	= 4 − 2 = 2
   	(m − 2)2

   
   Second Method :
   

   m +	1	= 4
   	m − 2

   
   

   m − 2 +	1	= 4 − 2
   	m − 2

   
   

   m − 2 +	1	= 2
   	m − 2

   
   

   (m − 2)2 +	1	= 2
   	(m − 2)2

   Correct Option: C

   m +	1	= 4
   	m − 2

   
   

   ⇒  m +	1	– 2 = 4 – 2
   	m − 2

   
   

   ⇒  (m − 2) +	1	= 4 – 2 = 2
   	(m − 2)

   
   On squaring both sides,
   

   (m − 2)2 +	1	+ 2(m − 2)		1		= 4
   	(m − 2)2			m − 2

   
   

   ⇒  (m − 2)2 +	1	= 4 − 2 = 2
   	(m − 2)2

   
   Second Method :
   

   m +	1	= 4
   	m − 2

   
   

   m − 2 +	1	= 4 − 2
   	m − 2

   
   

   m − 2 +	1	= 2
   	m − 2

   
   

   (m − 2)2 +	1	= 2
   	(m − 2)2

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