Direction: Study the following questions carefully and choose the right answer.
Important for :
1
 If x + 1 = 3, then the value of 3x2 – 4x + 3 is x x2 – x + 1
» Explain it
C
 Given,  x + 1 = 3 x

 So, 3x2 – 4x + 3 x2 – x + 1

⇒
 3x ( x – 4 + 1 ) 3 x
=
 3 [( x + 1 ) – 4 ] x 3

 x ( x – 1 + 1 ) x
 ( x + 1 ) – 1 x

 =
 3 ( 3 – 4 ) 3
 =
 3 ( 5 ) 3
3 – 1 2

 = 5 . 2

Hence, option C is  correct.

2
If x = 2015, y = 2014 and z = 2013, then value of x2 + y2 + z2 – xy – yz – zx is
» Explain it
A
x – y = 2015 – 2014 = 1
y – z = 2014 – 2013 = 1
z – x = 2013 – 2015 = –2

x2 + y2 + z2 – xy – yz – zx

Numerator & denominator multiplied by 2

 = 1 (2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx) 2

 = 1 (x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx) 2

 = 1 [(x– y)2 + (y – z)2 + (z – x)2] 2

 = 1 [1 + 1 + 4] = 1 × 6 = 3 2 2

Hence, option A is correct.

3
If p3 + 3p2 + 3p = 7, then the value of p2 + 2p is
» Explain it
B
p3 + 3p2 + 3p = 7

⇒   p3 + 3p2 + 3p + 1 = 7 + 1 = 8

⇒   (p + 1)3 = (2)3

⇒    p + 1= 2 ⇒ p = 1

∴    p2 + 2p = 1 + 2 × 1 = 1 + 2 = 3.

Hence, option B is correct.

4
 If x = p + 1 and y = p – 1 then the value of x4 – 2x2y2 + y4 is p p
» Explain it
C
 x = p + 1 and y = p – 1 p p

 ∴    x + y = p + 1 + p – 1 = 2p p p

 x – y = p + 1 – p + 1 = 2 p p p

x4 – 2x2y2 + y4  =  (x2 – y2)2  =  [(x – y)(x + y)]2     ......(i)

By putting the value of x and y in equation (i), we get

 = ( 2p × 2 ) 2 = (4)2 = 16. p

Hence, option C is correct.

5
If a + b + c = 0, then the value of (a + b – c)2 + (b + c – a)2 + (c + a – b)2 is
» Explain it
C
Given, a + b + c = 0

∴    a + b = – c,   b + c = – a,   c + a = – b

∴   (a + b – c)2 + (b + c – a)2 + (c + a – b)2

⇒   (– c – c)2 + (– a – a)2 + (– b – b)2

⇒   (– 2c)2 + (– 2a)2 + (–2b)2

⇒   4c2 + 4a2 + 4b2 = 4(a2 + b2 + c 2)

Hence, option C is correct.