211. If   x +	1	= √3 , then the value of x3 +	1	is equal to
     	x		x3

1. 1. 1

   
   
   

   2. 3√3

   
   
   

   3. 0

   
   
   

   4. 3

   
   
   

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   x +	1	= √3
   	x

   
   ⇒ 
   

   	x +	1		3	= (√3)3 = 3√3
   		x

   
   

   ⇒  x3 +	1	+ 3		x +	1		= 3√3
   	x3				x

   
   

   ⇒  x3 +	1	+ 3√3 = 3√3
   	x3

   
   

   ⇒  x3 +	1	= 3√3 – 3√3 = 0
   	x3

   Correct Option: C

   x +	1	= √3
   	x

   
   ⇒ 
   

   	x +	1		3	= (√3)3 = 3√3
   		x

   
   

   ⇒  x3 +	1	+ 3		x +	1		= 3√3
   	x3				x

   
   

   ⇒  x3 +	1	+ 3√3 = 3√3
   	x3

   
   

   ⇒  x3 +	1	= 3√3 – 3√3 = 0
   	x3

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212. If   pq (p + q) = 1, then the value of	1	− p3 − q3 is equal to
     	p3q3

1. 1. 1

   
   
   

   2. 2

   
   
   

   3. 3

   
   
   

   4. 4

   
   
   

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1. View Hint View Answer Discuss in Forum

   pq (p + q) = 1
   

   ⇒  p + q =	1
   	pq

   
   On cubing both sides,
   

   (p + q)3 =	1
   	p3q3

   
   

   ⇒ p3 + q3 + 3pq (p + q) =	1
   	p3q3

   
   

   ⇒	1	− p3 − q3
   	p3q3

   
   = 3 pq (p + q) = 3 × 1 = 3

   Correct Option: C

   pq (p + q) = 1
   

   ⇒  p + q =	1
   	pq

   
   On cubing both sides,
   

   (p + q)3 =	1
   	p3q3

   
   

   ⇒ p3 + q3 + 3pq (p + q) =	1
   	p3q3

   
   

   ⇒	1	− p3 − q3
   	p3q3

   
   = 3 pq (p + q) = 3 × 1 = 3

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213. = x2 + y2 +	1	+	1	= 4, then the value of x2 + y2 is
     	x2		y2

1. 1. 2

   
   
   

   2. 4

   
   
   

   3. 8

   
   
   

   4. 16

   
   
   

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1. View Hint View Answer Discuss in Forum

   x2 + y2 +	1	+	1	− 4 = 0
   	x2		y2

   
   

   ⇒  x2 +	1	− 2 + y2 +	1	− 2 = 0
   	x2		y2

   
   

   ⇒

   x −

   1

   2

   +

   y −

   1

   2

   = 0

   x

   y

   
   

   ⇒  x −	1	= 0
   	x

   
   ⇒  x² − 1 = 0 ⇒ x = 1
   Similarly,
   y = 1
   ∴  x² + y² = 1 + 1 = 2

   Correct Option: A

   x2 + y2 +	1	+	1	− 4 = 0
   	x2		y2

   
   

   ⇒  x2 +	1	− 2 + y2 +	1	− 2 = 0
   	x2		y2

   
   

   ⇒

   x −

   1

   2

   +

   y −

   1

   2

   = 0

   x

   y

   
   

   ⇒  x −	1	= 0
   	x

   
   ⇒  x² − 1 = 0 ⇒ x = 1
   Similarly,
   y = 1
   ∴  x² + y² = 1 + 1 = 2

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214. If   n = 7 + 4 √3, then the value of		√n +	1		is :
     			√n

1. 1. 2√3

   
   
   

   2. 4

   
   
   

   3. –4

   
   
   

   4. – 2√3

   
   
   

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1. View Hint View Answer Discuss in Forum

   n = 7 + 4√3 = 7 + 2 × 2 × √3
   = 4 + 3 + 2 × 2 × √3
   = (2 + √3)²
   ∴ √n = 2 + √3
   

   ∴	1	=	1	= 2
   	√n		2 + √3

   
   

   =	1	×	2 - √3	= 2 - √3
   	2 + √3		2 - √3

   
   

   ∴ √n +	1	= 2 + √3 + 2 - √3 = 4
   	√n

   Correct Option: B

   n = 7 + 4√3 = 7 + 2 × 2 × √3
   = 4 + 3 + 2 × 2 × √3
   = (2 + √3)²
   ∴ √n = 2 + √3
   

   ∴	1	=	1	= 2
   	√n		2 + √3

   
   

   =	1	×	2 - √3	= 2 - √3
   	2 + √3		2 - √3

   
   

   ∴ √n +	1	= 2 + √3 + 2 - √3 = 4
   	√n

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215. If   a + b + c = 0, then the value of	a2 + b2 + c2	is
     	a2 − bc

1. 1. 0

   
   
   

   2. 1

   
   
   

   3. 2

   
   
   

   4. 3

   
   
   

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1. View Hint View Answer Discuss in Forum

   a + b + c = 0
   ⇒  b + c = –a
   On squaring both sides,
   ⇒  (b + c)² = a²
   ⇒  b² + c² + 2bc = a²
   ⇒  a² + b² + c² + 2bc = 2a²
   ⇒  a² + b² + c² = 2a² − 2bc = 2(a² − bc)
   

   ∴	a2 + b2 + c2	=	2(a2 − bc)	= 2
   	a2 − bc		a2 − bc

   Correct Option: C

   a + b + c = 0
   ⇒  b + c = –a
   On squaring both sides,
   ⇒  (b + c)² = a²
   ⇒  b² + c² + 2bc = a²
   ⇒  a² + b² + c² + 2bc = 2a²
   ⇒  a² + b² + c² = 2a² − 2bc = 2(a² − bc)
   

   ∴	a2 + b2 + c2	=	2(a2 − bc)	= 2
   	a2 − bc		a2 − bc

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