116. In what ratio, the line joining (– 1, 1) and (5, 7) is divided by the line x + y = 4 ?

1. 1. 2 : 1

   
   
   

   2. 1 : 3

   
   
   

   3. 1 : 2

   
   
   

   4. 1 : 4

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let the ratio k : 1
   
   = Using section formula,
   

   x =	5 × k + 1 × -1
   	k + 1

   
   

   ⇒ x =	5k - 1
   	k + 1

   
   

   y =	7 × k + 1 × 1
   	k + 1

   
   

   ⇒ y =	7k + 1
   	k + 1

   
   Putting the value of x and y in the equation of line, we get
   

   5k - 1	+	7k + 1	= 4
   k + 1		k + 1

   
   12k = 4(k + 1)
   → 12k = 4k + 4
   8k = 4
   k = 1/2
   ∴ Ratio is 1 : 2.

   Correct Option: C

   Let the ratio k : 1
   
   = Using section formula,
   

   x =	5 × k + 1 × -1
   	k + 1

   
   

   ⇒ x =	5k - 1
   	k + 1

   
   

   y =	7 × k + 1 × 1
   	k + 1

   
   

   ⇒ y =	7k + 1
   	k + 1

   
   Putting the value of x and y in the equation of line, we get
   

   5k - 1	+	7k + 1	= 4
   k + 1		k + 1

   
   12k = 4(k + 1)
   → 12k = 4k + 4
   8k = 4
   k = 1/2
   ∴ Ratio is 1 : 2.

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117. For what value of x the points (x, – 1), (2, 1) and (4, 5) are collinear ?

1. 1. x = 2

   
   
   

   2. x = – 1

   
   
   

   3. x = 4

   
   
   

   4. x = 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   When three points are collinear then area of triangle is zero.
   

   ⇒ ar∆ =			x		- 1		1
   	1		2		1		1
   	2		4		5		1

   
   = 1/2 [x (1 – 5) + 1(2 – 4) + 1(10 – 4)]
   ⇒ – 4x – 2 + 6 = 0
   ⇒ – 4x + 4 = 0
   ⇒ x = 1

   Correct Option: D

   When three points are collinear then area of triangle is zero.
   

   ⇒ ar∆ =			x		- 1		1
   	1		2		1		1
   	2		4		5		1

   
   = 1/2 [x (1 – 5) + 1(2 – 4) + 1(10 – 4)]
   ⇒ – 4x – 2 + 6 = 0
   ⇒ – 4x + 4 = 0
   ⇒ x = 1

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118. If	1	+	1	=	1	, then the value of (p³ – q³) is
     	p		q		p + q

1. 1. p – q

   
   
   

   2. pq

   
   
   

   3. 1

   
   
   

   4. 0

   
   
   

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1. View Hint View Answer Discuss in Forum

   1	+	1	=	1
   p		q		p + q

   
   ⇒ ( p + q)² = pq
   ⇒ p² + 2pq + q² = pq
   ⇒ p² + pq + q² = 0
   ∴ p³ – q³ = (p – q) (p² + pq + q²) = 0

   Correct Option: D

   1	+	1	=	1
   p		q		p + q

   
   ⇒ ( p + q)² = pq
   ⇒ p² + 2pq + q² = pq
   ⇒ p² + pq + q² = 0
   ∴ p³ – q³ = (p – q) (p² + pq + q²) = 0

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119. If x = a +	1	and y = a –	1	, then the value of x4 + y4 – 2x²y² is :
     	a		a

1. 1. 4

   
   
   

   2. 8

   
   
   

   3. 16

   
   
   

   4. 64

   
   
   

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1. View Hint View Answer Discuss in Forum

   x = a +	1	; y = a -	1
   	a		a

   
   

   ∴ x² - y² =

   a +

   1

   ²

   -

   a -

   1

   ²

   = 4a ×

   1

   = 4

   a

   a

   a

   
   = 4 [∵ (a + b)² – (a – b)² = 4ab]
   ∴ x⁴ + y⁴ – 2x²y² = (x² – y²)²
   = 4² = 16

   Correct Option: C

   x = a +	1	; y = a -	1
   	a		a

   
   

   ∴ x² - y² =

   a +

   1

   ²

   -

   a -

   1

   ²

   = 4a ×

   1

   = 4

   a

   a

   a

   
   = 4 [∵ (a + b)² – (a – b)² = 4ab]
   ∴ x⁴ + y⁴ – 2x²y² = (x² – y²)²
   = 4² = 16

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120. If a³ – b³ = 56 and a – b = 2, what is the value of (a² + b²) ?

1. 1. 12

   
   
   

   2. 20

   
   
   

   3. 28

   
   
   

   4. 32

   
   
   

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1. View Hint View Answer Discuss in Forum

   a³ – b³ = 56
   ⇒ (a – b)³ + 3ab (a – b) = 56
   ⇒ (2)³ + 3ab × 2 = 56
   ⇒ 6ab = 56 – 8 = 48
   ⇒ ab = 48 ÷ 6 = 8
   ∴ a² + b² = (a – b)² + 2ab
   = 2²+ 2 × 8
   = 4 + 16 = 20

   Correct Option: B

   a³ – b³ = 56
   ⇒ (a – b)³ + 3ab (a – b) = 56
   ⇒ (2)³ + 3ab × 2 = 56
   ⇒ 6ab = 56 – 8 = 48
   ⇒ ab = 48 ÷ 6 = 8
   ∴ a² + b² = (a – b)² + 2ab
   = 2²+ 2 × 8
   = 4 + 16 = 20

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