546. If x = 6 +	1	, then the value of x4 +	1	is
     	x		x4

     
     

1. 1. 1448
      

   
   
   

   2. 1442
      

   
   
   

   3. 1444
      

   
   
   

   4. 1446

   
   
   

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1. View Hint View Answer Discuss in Forum

   x = 6 +	1
   	x

   
   

   ⇒ x -	1	= 6
   	x

   
   On squaring both sides,
   

   ⇒ x2 +	1	- 2 = 36
   	x2

   
   

   ⇒ x2 +	1	= 36 + 2 = 38
   	x2

   
   On squaring again,
   

   ⇒ x4 +	1	+ 2 = 1444
   	x4

   
   

   ⇒ x4 +	1	= 1444 - 2 = 1442
   	x4

   
   

   Correct Option: B

   x = 6 +	1
   	x

   
   

   ⇒ x -	1	= 6
   	x

   
   On squaring both sides,
   

   ⇒ x2 +	1	- 2 = 36
   	x2

   
   

   ⇒ x2 +	1	= 36 + 2 = 38
   	x2

   
   On squaring again,
   

   ⇒ x4 +	1	+ 2 = 1444
   	x4

   
   

   ⇒ x4 +	1	= 1444 - 2 = 1442
   	x4

   
   

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547. If x +	1	= 5, then x6 +	1	is
     	x		x6

     
     

1. 1. 12098
      

   
   
   

   2. 12048
      

   
   
   

   3. 14062
      

   
   
   

   4. 12092

   
   
   

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1. View Hint View Answer Discuss in Forum

   x +	1	= 5
   	x

   
   On cubing both sides,
   

   ⇒		x +	1		3	= (5)3
   			x

   
   

   ⇒ x3 +	1	+ 3x.	1		x +	1		= 125
   	x3		x			x

   
   

   ⇒ x3 +	1	+ 3 × 5 = 125
   	x3

   
   

   ⇒ x3 +	1	= 125 - 15 = 110
   	x3

   
   On squaring both sides,
   

   x6 +	1	+ 2.x3.	1	= 12100
   	x6		x3

   
   

   ⇒ x6 +	1	= 12100 – 2 = 12098
   	x6

   Correct Option: A

   x +	1	= 5
   	x

   
   On cubing both sides,
   

   ⇒		x +	1		3	= (5)3
   			x

   
   

   ⇒ x3 +	1	+ 3x.	1		x +	1		= 125
   	x3		x			x

   
   

   ⇒ x3 +	1	+ 3 × 5 = 125
   	x3

   
   

   ⇒ x3 +	1	= 125 - 15 = 110
   	x3

   
   On squaring both sides,
   

   x6 +	1	+ 2.x3.	1	= 12100
   	x6		x3

   
   

   ⇒ x6 +	1	= 12100 – 2 = 12098
   	x6

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548. If x2 – 3x + 1 = 0, then the value of	x6 + x4 + x2 + 1	will be
     	x3

     
     

1. 1. 18
      

   
   
   

   2. 15
      

   
   
   

   3. 21
      

   
   
   

   4. 30

   
   
   

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1. View Hint View Answer Discuss in Forum

   x² – 3 x + 1 = 0
   ⇒ x² + 1 = 3 x
   

   ⇒	x2 + 1	= 3
   	x

   
   

   ⇒ x +	1	= 3 ........... (i)
   	x

   
   

   ∴ Expression =	x6 + x4 + x2 + 1
   	x3

   
   

   Expression =	x6	+	x4	+	x2	+	1
   	x3		x3		x3		x3

   
   

   Expression = x3 + x +	1	+	1
   	x		x3

   
   

   Expression =		x3 +	1		+		x +	1		= 2 × 2 = 4
   			x3					x

   
   

   Expression =

   x +

   1

   3

   - 3 . x .

   1

   x +

   1

   +

   x +

   1

   x

   x

   x

   x

   
   Expression = 3³ – 3 × 3 + 3 = 27 – 9 + 3 = 21
   

   Correct Option: C

   x² – 3 x + 1 = 0
   ⇒ x² + 1 = 3 x
   

   ⇒	x2 + 1	= 3
   	x

   
   

   ⇒ x +	1	= 3 ........... (i)
   	x

   
   

   ∴ Expression =	x6 + x4 + x2 + 1
   	x3

   
   

   Expression =	x6	+	x4	+	x2	+	1
   	x3		x3		x3		x3

   
   

   Expression = x3 + x +	1	+	1
   	x		x3

   
   

   Expression =		x3 +	1		+		x +	1		= 2 × 2 = 4
   			x3					x

   
   

   Expression =

   x +

   1

   3

   - 3 . x .

   1

   x +

   1

   +

   x +

   1

   x

   x

   x

   x

   
   Expression = 3³ – 3 × 3 + 3 = 27 – 9 + 3 = 21
   

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549. If x4 +	1	= 119, and x > 1, then find the positive value of x3 -	1	.
     	x4		x3

     
     

1. 1. 25
      

   
   
   

   2. 27
      

   
   
   

   3. 36
      

   
   
   

   4. 49

   
   
   

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1. View Hint View Answer Discuss in Forum

   x4 +	1	= 119
   	x4

   
   

   ⇒		x2 +	1		2	- 2 = 119
   			x2

   
   

   ⇒		x2 +	1		2	= 119 + 2 = 121
   			x2

   
   

   ⇒		x2 +	1		2	= (11)2
   			x2

   
   

   ⇒ x2 +	1	= 11
   	x2

   
   

   ⇒		x -	1		2	+ 2 = 11
   			x

   
   

   ⇒		x -	1		2	= 11 - 2 = 9 = 32
   			x

   
   

   ⇒ x -	1	= 3
   	x

   
   On cubing both sides,
   

   ⇒		x -	1		3	= 27
   			x

   
   

   ⇒ x3 -	1	- 3x.	1		x -	1		= 27
   	x3		x			x

   
   

   ⇒ x3 -	1	- 3 × 3 = 27
   	x3

   
   

   ⇒ x3 -	1	= 27 + 9 = 36
   	x3

   Correct Option: C

   x4 +	1	= 119
   	x4

   
   

   ⇒		x2 +	1		2	- 2 = 119
   			x2

   
   

   ⇒		x2 +	1		2	= 119 + 2 = 121
   			x2

   
   

   ⇒		x2 +	1		2	= (11)2
   			x2

   
   

   ⇒ x2 +	1	= 11
   	x2

   
   

   ⇒		x -	1		2	+ 2 = 11
   			x

   
   

   ⇒		x -	1		2	= 11 - 2 = 9 = 32
   			x

   
   

   ⇒ x -	1	= 3
   	x

   
   On cubing both sides,
   

   ⇒		x -	1		3	= 27
   			x

   
   

   ⇒ x3 -	1	- 3x.	1		x -	1		= 27
   	x3		x			x

   
   

   ⇒ x3 -	1	- 3 × 3 = 27
   	x3

   
   

   ⇒ x3 -	1	= 27 + 9 = 36
   	x3

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550. If

     p

     +

     q

     +

     r

     = 1 and

     a

     +

     b

     +

     c

     = 0

     a

     b

     c

     p

     q

     r

     
     

     where p, q, r and a, b, c are non-zero, then the value of	p2	+	q2	+	r2	is
     	a2		b2		c2

     
     

1. 1. –1
      

   
   
   

   2. 0
      

   
   
   

   3. 1
      

   
   
   

   4. 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let	p	= x ,	q	= y ,	r	= z
   	a		b		c

   
   ∴ x + y + z = 1
   

   and	1	+	1	+	1	= 0
   	x		y		z

   
   

   ⇒	yz + xz + xy	= 0
   	xyz

   
   ⇒ yz + xz + xy = 0
   ∴ x + y + z = 1
   On squaring both sides ,
   x² + y² + z² + 2xy + 2yz + 2zx = 1
   ⇒ x² + y² + z² + 0 = 1
   ⇒ x² + y² + z² = 1

   Correct Option: C

   Let	p	= x ,	q	= y ,	r	= z
   	a		b		c

   
   ∴ x + y + z = 1
   

   and	1	+	1	+	1	= 0
   	x		y		z

   
   

   ⇒	yz + xz + xy	= 0
   	xyz

   
   ⇒ yz + xz + xy = 0
   ∴ x + y + z = 1
   On squaring both sides ,
   x² + y² + z² + 2xy + 2yz + 2zx = 1
   ⇒ x² + y² + z² + 0 = 1
   ⇒ x² + y² + z² = 1

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