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

1. 1. 9

   
   
   

   2. –9

   
   
   

   3. 1

   
   
   

   4. 2

   
   
   

---

1. View Hint View Answer Discuss in Forum

   x2 + 3x + 1	=	1
   x2 − 3x + 1		2

   
   ⇒  2x² + 6x + 2 = x² – 3x + 1
   ⇒  2x² – x² + 2 – 1 = – 6x – 3x
   ⇒  x² + 1 = – 9x
   

   ⇒	x2 + 1	= –9
   	x

   
   

   ⇒  x +	1	= –9
   	x

   Correct Option: B

   x2 + 3x + 1	=	1
   x2 − 3x + 1		2

   
   ⇒  2x² + 6x + 2 = x² – 3x + 1
   ⇒  2x² – x² + 2 – 1 = – 6x – 3x
   ⇒  x² + 1 = – 9x
   

   ⇒	x2 + 1	= –9
   	x

   
   

   ⇒  x +	1	= –9
   	x

---

322. If (a + b)² = 100 and (a – b) = 4, then ab equals to :

1. 1. 116

   
   
   

   2. 84

   
   
   

   3. 21

   
   
   

   4. 53

   
   
   

---

1. View Hint View Answer Discuss in Forum

   4ab = (a + b)² – (a – b)²
   ⇒  4ab = 100 – (4)² = 100 – 16
   ⇒  4ab = 84
   

   ⇒  ab =	84	= 21
   	4

   Correct Option: C

   4ab = (a + b)² – (a – b)²
   ⇒  4ab = 100 – (4)² = 100 – 16
   ⇒  4ab = 84
   

   ⇒  ab =	84	= 21
   	4

---

---

323. If 3a2 = b2 ≠ 0, then the value of	(a + b)3 - (a - b)3	is
     	(a + b)2 - (a - b)2

     
     

1. 1. 	3b
      	2

      
      

   
   
   

   2. b
      

   
   
   

   3. 	b
      	2

      
      

   
   
   

   4. 	2b
      	3

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Expression =	(a + b)3 - (a - b)3
   	(a + b)2 + (a - b)2

   
   

   Expression =	(a3 + b3 + 3a2b + 3ab2) - (a3 - b3 - 3a2b + 3ab2)
   	(a2 + b2 + 2ab) + (a2 + b2 - 2ab)

   
   

   Expression =	(a3 + b3 + 3a2b + 3ab2 - a3 + b3 + 3a2b - 3ab2)
   	(a2 + b2 + a2 + b2)

   
   

   Expression =	(6a2b + 2b3)
   	2(a2 + b2)

   
   

   Expression =	2b(3a2 + b2)
   	2(a2 + b2)

   
   

   	Expression =	b(b2 + b2)
   			b2	+ b2
   			3

   
   

   	Expression =	b × 2b2
   			4b2
   			3

   
   

   Expression =	b(3 × 2)	=	3b
   	4		2

   Correct Option: A

   Expression =	(a + b)3 - (a - b)3
   	(a + b)2 + (a - b)2

   
   

   Expression =	(a3 + b3 + 3a2b + 3ab2) - (a3 - b3 - 3a2b + 3ab2)
   	(a2 + b2 + 2ab) + (a2 + b2 - 2ab)

   
   

   Expression =	(a3 + b3 + 3a2b + 3ab2 - a3 + b3 + 3a2b - 3ab2)
   	(a2 + b2 + a2 + b2)

   
   

   Expression =	(6a2b + 2b3)
   	2(a2 + b2)

   
   

   Expression =	2b(3a2 + b2)
   	2(a2 + b2)

   
   

   	Expression =	b(b2 + b2)
   			b2	+ b2
   			3

   
   

   	Expression =	b × 2b2
   			4b2
   			3

   
   

   Expression =	b(3 × 2)	=	3b
   	4		2

---

324. 138. If a² + b² + c² – ab – bc – ca = 0, then
     

1. 1. a = b = c
      

   
   
   

   2. a ≠ b = c
      

   
   
   

   3. a = b ≠ c
      

   
   
   

   4. a ≠ b ≠ c

   
   
   

---

1. View Hint View Answer Discuss in Forum

   a² + b² + c² – ab – bc – ca = 0
   ⇒ 2a² + 2b² + 2c² – 2ab – 2bc – 2ca = 0
   ⇒ a² + b² - 2ab + b² + c² – 2bc + c² + a² – 2ca = 0
   ⇒ (a - b)² + (b - c)² + (c - a)² = 0
   It is possible only when,
   a – b = 0 ⇒ a = b
   b – c = 0 ⇒ b = c
   c – a = 0 ⇒ c = a
   ∴ a = b = c

   Correct Option: A

   a² + b² + c² – ab – bc – ca = 0
   ⇒ 2a² + 2b² + 2c² – 2ab – 2bc – 2ca = 0
   ⇒ a² + b² - 2ab + b² + c² – 2bc + c² + a² – 2ca = 0
   ⇒ (a - b)² + (b - c)² + (c - a)² = 0
   It is possible only when,
   a – b = 0 ⇒ a = b
   b – c = 0 ⇒ b = c
   c – a = 0 ⇒ c = a
   ∴ a = b = c

---

---

325. If x +	1	= 0 , then the value of x5 +	1		is
     	x		x5

     
     

1. 1. 2
      

   
   
   

   2. – 1
      

   
   
   

   3. 1
      

   
   
   

   4. 0

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Using Rule 1 and 8,
   

   x +	1	= 0
   	x

   
   On squaring both sides,
   

   ⇒		x +	1		2	= 0
   			x

   
   

   ⇒ x2 +	1	+ 2 = 0
   	x2

   
   

   ⇒ x2 +	1	= – 2 ..... (i)(not admissible)
   	x2

   
   

   On cubing , x +	1	= 0
   	x

   
   

   ⇒ x3 +	1	+ 3 × 0 = 0
   	x3

   
   

   ⇒ x3 +	1	= 0
   	x3

   
   

   ∴		x2 +	1			x3 +	1		= 2 × 2 = 4
   			x2				x3

   
   

   ⇒ x5 +	1	+ x +	1	= 0
   	x5		x

   
   

   ⇒ x5 +	1	= 0
   	x5

   
   

   Correct Option: D

   Using Rule 1 and 8,
   

   x +	1	= 0
   	x

   
   On squaring both sides,
   

   ⇒		x +	1		2	= 0
   			x

   
   

   ⇒ x2 +	1	+ 2 = 0
   	x2

   
   

   ⇒ x2 +	1	= – 2 ..... (i)(not admissible)
   	x2

   
   

   On cubing , x +	1	= 0
   	x

   
   

   ⇒ x3 +	1	+ 3 × 0 = 0
   	x3

   
   

   ⇒ x3 +	1	= 0
   	x3

   
   

   ∴		x2 +	1			x3 +	1		= 2 × 2 = 4
   			x2				x3

   
   

   ⇒ x5 +	1	+ x +	1	= 0
   	x5		x

   
   

   ⇒ x5 +	1	= 0
   	x5

   
   

---