116. If   (2 + √3) = b (2 – √3)b = 1, then the value of	1	+	1	is
     	a		b

1. 1. 1

   
   
   

   2. 2

   
   
   

   3. 2√3

   
   
   

   4. 4

   
   
   

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   (2 + √3)a = (2 – √3)b = 1
   

   ⇒  a =	1
   	2 + √3

   
   

   ∴	1	= 2 + √3
   	a

   
   Similarly,
   

   b =	1
   	2 − √3

   
   

   1	= 2 − √3
   b

   
   

   ∴	1	+	1	= 2 + √3 + 2 − √3 = 4
   	a		b

   Correct Option: D

   (2 + √3)a = (2 – √3)b = 1
   

   ⇒  a =	1
   	2 + √3

   
   

   ∴	1	= 2 + √3
   	a

   
   Similarly,
   

   b =	1
   	2 − √3

   
   

   1	= 2 − √3
   b

   
   

   ∴	1	+	1	= 2 + √3 + 2 − √3 = 4
   	a		b

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117. If a + b + c = 3, a2 + b2 + c2 = 6 and	1	+	1	+	1	= 1, where a, b, c are all non-zero, then ‘abc’ is equal to
     	a		b		c

1. 1. 2

      3

   
   
   

   2. 3

      2

   
   
   

   3. 1

      2

   
   
   

   4. 1

      3

   
   
   

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   a + b + c = 3; a² + b² + c² = 6
   ∴  (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
   ⇒  3² = 6 + 2 (ab + bc + ca)
   ⇒  9 – 6 = 2(ab + bc + ca)
   

   ⇒ ab + bc + ca =	3
   	2

   
   

   ∴	1	+	1	+	1	= 1
   	a		b		c

   
   

   ⇒	bc + ac + ab	= 1
   	abc

   
   

   ⇒  abc = ab + bc + ca =	3
   	2

   Correct Option: B

   a + b + c = 3; a² + b² + c² = 6
   ∴  (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
   ⇒  3² = 6 + 2 (ab + bc + ca)
   ⇒  9 – 6 = 2(ab + bc + ca)
   

   ⇒ ab + bc + ca =	3
   	2

   
   

   ∴	1	+	1	+	1	= 1
   	a		b		c

   
   

   ⇒	bc + ac + ab	= 1
   	abc

   
   

   ⇒  abc = ab + bc + ca =	3
   	2

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118. If a, b, c are positive and a + b + c = 1, then the least value of	1	+	1	+	1	is
     	a		b		c

1. 1. 9

   
   
   

   2. 5

   
   
   

   3. 3

   
   
   

   4. 1

   
   
   

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   The value of	1	+	1	+	1	will be minimum, if values of a, b and c be maximum.
   	a		b		c

   
   a + b + c = 1
   ∴ Values of a, b and c will be maximum if
   a = b = c
   

   ∴ a = b = c =	1
   	3

   
   

   ∴	1	+	1	+	1	= 3 + 3 + 3 = 9
   	a		b		c

   Correct Option: A

   The value of	1	+	1	+	1	will be minimum, if values of a, b and c be maximum.
   	a		b		c

   
   a + b + c = 1
   ∴ Values of a, b and c will be maximum if
   a = b = c
   

   ∴ a = b = c =	1
   	3

   
   

   ∴	1	+	1	+	1	= 3 + 3 + 3 = 9
   	a		b		c

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119. The reciprocal of x +	1	is
     	x

1. 1. x

      x2 + 1

   
   
   

   2. x

      x + 1

   
   
   

   3. x −	1
      	x

   
   
   

   4. 1	+ x
      x

   
   
   

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   x +	1	=	x² + 1
   	x		x

   
   

   ∴ Its reciprocal =	x² + 1
   	10

   Correct Option: B

   x +	1	=	x² + 1
   	x		x

   
   

   ∴ Its reciprocal =	x² + 1
   	10

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120. If	5x	=	1	, then the value of		x +	1		is
     	2x2 5x + 1		3				2x

1. 1. 15

   
   
   

   2. 10

   
   
   

   3. 20

   
   
   

   4. 5

   
   
   

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1. View Hint View Answer Discuss in Forum

   	5x	=	1
   	2x² + 5x + 1		3

   
   Dividing Numerator and Denominator by
   

   5				=	1
   2x + 5 +	1				3
   	x

   
   On dividing N^r and D^r by 2,
   

   	5				=
   	3							1
   x +	5	+	1					3
   	2		2x

   
   

   ⇒		x +	1		+	5	=	15
   			2x			2		2

   
   

   ⇒ x +	1	=	15	-	5	=	10	= 5
   	2x		2		2		2

   Correct Option: D

   	5x	=	1
   	2x² + 5x + 1		3

   
   Dividing Numerator and Denominator by
   

   5				=	1
   2x + 5 +	1				3
   	x

   
   On dividing N^r and D^r by 2,
   

   	5				=
   	3							1
   x +	5	+	1					3
   	2		2x

   
   

   ⇒		x +	1		+	5	=	15
   			2x			2		2

   
   

   ⇒ x +	1	=	15	-	5	=	10	= 5
   	2x		2		2		2

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