286. If		4x +	1		= 5, x ≠ 0, then the value of	5x	is :
     			x			4x2 + 10x + 1

1. 1. 1

      2

   
   
   

   2. 1

      3

   
   
   

   3. 2

      3

   
   
   

   4. 3

   
   
   

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1. View Hint View Answer Discuss in Forum

   4x +	1	= 5
   	x

   
   

   Expression =	5x
   	4x2 + 1 + 10x

   
   

   =	5x
   	x		4x +	1	+ 10
   				x

   
   

   =	5	=	5	=	1
   	5 + 10		15		3

   Correct Option: B

   4x +	1	= 5
   	x

   
   

   Expression =	5x
   	4x2 + 1 + 10x

   
   

   =	5x
   	x		4x +	1	+ 10
   				x

   
   

   =	5	=	5	=	1
   	5 + 10		15		3

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287. If a² = b + c, b² = c + a, c² = a + b, then the value of 3
     

     	1	+	1	+	1		is :
     	a + 1		b + 1		c + 1

1. 1. 1

   
   
   

   2. 1/3

   
   
   

   3. 3

   
   
   

   4. 4

   
   
   

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1. View Hint View Answer Discuss in Forum

   a² = b + c
   ⇒  a² + a = a + b + c
   ⇒  a (a + 1) = a + b + c
   

   ⇒	1	=	a
   	a + 1		a + b + c

   
   Again,
   b² = c + a
   ⇒  b² + b = a + b + c
   ⇒  b (b + 1) = a + b + c
   

   ⇒	1	=	b
   	b + 1		a + b + c

   
   c² = a + b
   ⇒  c² + c = a + b + c
   ⇒  c (c + 1) = a + b + c
   

   ⇒	1	=	c
   	c + 1		a + b + c

   
   

   ∴  3		1	+	1	+	1
   		a + 1		b + 1		c + 1

   
   

   = 3		a	+	b	+	c
   		a + b + c		a + b + c		a + b + c

   
   

   = 3		a + b + c		= 3
   		a + b + c

   Correct Option: C

   a² = b + c
   ⇒  a² + a = a + b + c
   ⇒  a (a + 1) = a + b + c
   

   ⇒	1	=	a
   	a + 1		a + b + c

   
   Again,
   b² = c + a
   ⇒  b² + b = a + b + c
   ⇒  b (b + 1) = a + b + c
   

   ⇒	1	=	b
   	b + 1		a + b + c

   
   c² = a + b
   ⇒  c² + c = a + b + c
   ⇒  c (c + 1) = a + b + c
   

   ⇒	1	=	c
   	c + 1		a + b + c

   
   

   ∴  3		1	+	1	+	1
   		a + 1		b + 1		c + 1

   
   

   = 3		a	+	b	+	c
   		a + b + c		a + b + c		a + b + c

   
   

   = 3		a + b + c		= 3
   		a + b + c

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288. If   (a – 2) +	1	= – 1, then the value of (a + 2)2 +	1	is :
     	(a + 2)		(a + 2)2

1. 1. 7

   
   
   

   2. 11

   
   
   

   3. 23

   
   
   

   4. 27

   
   
   

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1. View Hint View Answer Discuss in Forum

   a – 2 +	1	= – 1
   	a + 2

   
   

   ⇒  (a – 2 + 4) +	1	= 4 – 1
   	a + 2

   
   

   ⇒  (a + 2) +	1	= 3
   	(a + 2)

   
   On squaring both sides,
   

   (a + 2)2 +	1	+ 2 × (a + 2) ×	1	= 9
   	(a + 2)2		(a + 2)

   
   

   ⇒ (a + 2)2 +	1
   	(a + 2)2

   
   = 9 – 2 = 7

   Correct Option: A

   a – 2 +	1	= – 1
   	a + 2

   
   

   ⇒  (a – 2 + 4) +	1	= 4 – 1
   	a + 2

   
   

   ⇒  (a + 2) +	1	= 3
   	(a + 2)

   
   On squaring both sides,
   

   (a + 2)2 +	1	+ 2 × (a + 2) ×	1	= 9
   	(a + 2)2		(a + 2)

   
   

   ⇒ (a + 2)2 +	1
   	(a + 2)2

   
   = 9 – 2 = 7

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289. If

     a +

     1

     = 1 and

     b +

     1

     = 1, then the value of

     c +

     1

     is :

     b

     c

     a

1. 1. 0

   
   
   

   2. 1

   
   
   

   3. –1

   
   
   

   4. 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   a +	1	= 1
   	b

   
   

   ⇒  a = 1 –	1	=	b − 1
   	b		b

   
   

   ⇒	1	=	b	..... (i)
   	a		b − 1

   
   

   Again,  b +	1	= 1
   	c

   
   

   ⇒	1	= 1 – b
   	c

   
   

   ⇒  c =	1	...(ii)
   	1 – b

   
   

   ∴  c +	1	=	1	+	b
   	a		1 – b		b – 1

   
   

   =	1	–	b
   	1 – b		1 – b

   
   

   =	1 – b	= 1
   	1 – b

   Correct Option: B

   a +	1	= 1
   	b

   
   

   ⇒  a = 1 –	1	=	b − 1
   	b		b

   
   

   ⇒	1	=	b	..... (i)
   	a		b − 1

   
   

   Again,  b +	1	= 1
   	c

   
   

   ⇒	1	= 1 – b
   	c

   
   

   ⇒  c =	1	...(ii)
   	1 – b

   
   

   ∴  c +	1	=	1	+	b
   	a		1 – b		b – 1

   
   

   =	1	–	b
   	1 – b		1 – b

   
   

   =	1 – b	= 1
   	1 – b

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290. If x = at² and y = 2at then

1. 1. x² = 4ay

   
   
   

   2. y² = 4ax

   
   
   

   3. x² + y² = a²

   
   
   

   4. x² − y² = a²

   
   
   

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1. View Hint View Answer Discuss in Forum

   x = at²
   y = 2at
   ⇒  y² = 4a²t²
   = 4a.at² = 4ax

   Correct Option: B

   x = at²
   y = 2at
   ⇒  y² = 4a²t²
   = 4a.at² = 4ax

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