Sunday, June 24, 2012

CBSE Class 9 Maths Polynomials Exercise 2.5

Algebraic Identities


1.  (x + y)2 = x2 + 2xy + y2

2.  (x – y)2  = x2 – 2xy + y2 

3.  x2  – y2  = (x + y) (x – y)

4.  (x + a) (x + b) =  x2 + (a+b)x + ab

5.  (x - a)(x + b) = x2 + (b-a)x - ab


6.  (x + a) (x - b) =  x2 + (a-b)x - ab

7. (x - a)(x - b) =  x2 - (a+b)x + ab

8.  (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

9.  (x + y)3 = x3 + y3 + 3xy(x + y) = x3 + 3x2y + 3xy2 + y3

10. (x - y)3 = x3 - y3 - 3xy(x - y)  = x3 - 3x2y + 3xy2 - y3

11.  x3 + y3 + z3- 3xyz = (x + y + z)(x2 + y2 + z2- xy - yz -zx)

12.   x2 + y2 = ½[ (x + y)2 +  (x – y)2 ]

13.  xy = ¼[ (x + y)2 -  (x – y)2 ]

14.  x2 + y2  =  (x + y)2- 2xy

15.  (x – y)2 = (x + y)2- 4xy

16.  x2 + y2 =  (x – y)2 + 2xy

17.  (x + y)2 = (x – y)2 + 4xy

18. (x + a)(x + b)(x + c) = x3  + (a + b + c)x2 + (ab + bc + ca)x + abc

19.  x3 + y3 = (x + y) (x2- xy + y2)

20.  x3 - y3 = (x - y) (x2+ xy + y2)

21.  x2 + y2 + z2 -  xy - yz -zx = ½[ (x - y)2 +  (y – z)2 + (z – x)2]

Exercise 2.5

Q1: Use suitable identities to find the following products:
(i) (x + 4) (x + 10) 

(ii) (x + 8) (x – 10) 

(iii) (3x + 4) (3x – 5)
 
(iv) (y2 + 3/2)(y2 – 3/2) 

(v) (3 – 2x) (3 + 2x) 

Answer:  
(i) Using identity, (x + a) (x + b) =  x2 + (a+b)x + ab
(x + 4)(x + 10) =  x2 + (4+10)x + (4)(10)
  =  x2 + 14x + 40

(ii)  Using identity, (x + a) (x + b) =  x2 + (a+b)x + ab
(x + 8) (x – 10) = x2 + (8+(-10))x + (8)(-10)
 =  x2 +(8-10)x - 80
 =  x2 -2x - 80
 (Note, you may use identity, (x + a) (x - b) =  x2 + (a-b)x - ab directly here)


(iii) Using identity, (x + a) (x + b) =  x2 + (a+b)x + ab
Here x = 3x, a = 4 and b = -5
(3x + 4) (3x – 5) = (3x)2 + (4+(-5))(3x) + (4)(-5)
= 9x2 + (4-5)(3x) + (-20)
= 9x2 -3x -20

(iv) Using identity,  x2  – y2  = (x + y) (x – y)







(v)  Using identity, (x + y) (x – y) =  x2  – y2 
(3 - 2x)(3 + 2x) = (3)2  – (2x)2  = 9 - 4x2


Q2: Evaluate the following products without multiplying directly:
(i) 103 × 107 

(ii) 95 × 96 
(iii) 104 × 96

Answer:
(i) 103  ✕ 107 = (100 + 3)(100 + 7)
Using identity, (x + a) (x + b) =  x2 + (a+b)x + ab
= (100)2 + (3 + 7)100 + (3)(7)
= 10000 + (10)(100) + 21 
= 10000 + 1000 + 21
= 11021

(ii) 95 ✕ 96 = (100 - 5)(100 - 4)
Using identity, (x - a)(x - b) =  x2 - (a+b)x + ab
=  (100)2 - (5 + 4)(100) + (5)(4)
= 10000 - 900 + 20
= 9120

(iii)  104 ✕ 96 = (100 + 4)(100 - 4)
Using identity, (x + y) (x – y) =  x2  – y2 

Here x = 100, y = 4
=  (100)2  – (4)2 = 10000 - 16 = 9984

Q3: Factorise the  following using appropriate identities:
(i) 9x2 + 6xy + y2 
(ii) 4y2 – 4y + 1 
(iii) x2 - y2/100

Answer:
(i)  9x2 + 6xy + y2
  =  (3x)2 + 2(3x)(y) + (y)2
(a + b)2 = a2 + 2ab + b2
∴ = (3x + y)2
 ∴ = (3x + y)(3x + y)

(ii)  4y2 – 4y + 1 
= (2y)2 – 2(2y)(1) + 12 
(x – y)2  = x2 – 2xy + y2 
= (2y – 1)2 = (2y - 1)(2y - 1)







Q4: Expand each of the following, using suitable identities:
(i) (x + 2y + 4z)2 
(ii) (2x – y + z)2
(iii) (–2x + 3y + 2z)2
(iv) (3a – 7b – c)2
(v)  (–2x + 5y – 3z)2
(vi) [¼a - ½b + 1]2

Answer: Using identity (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca here,

(i)  (x + 2y + 4z)2
Here a = x, b = 2y and x = 4z
 =  x2 + (2y)2 + (4z)2 + 2x(2y) + 2(2y)(4z) + 2(4z)x
 =  x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx


(ii) (2x – y + z)2
Here a = 2x, b = -y and c = z
=  (2x)2 + (-y)2 + (z)2 + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)
=  4x2 + y2 + z2 - 4xy -2yz + 4xz

(iii) (–2x + 3y + 2z)2
Here a = -2x, b= 3y and c = 2z
= (-2x)2 + (3y)2 + (2z)2 + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)
= 4x2 + 9y2 + 4z2 -12xy +12yz -8zx

(iv)  (3a – 7b – c)2
Here  a= 3a, b = -7b and c = -c
= (3a)2 + (-7b)2 + (-c)2 + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)
= 9a2 + 49b2 + c2- 42ab + 14bc - 6ac

(v) (–2x + 5y – 3z)2
Here a = -2x, b = 5y and c = -3z
= (-2x)2 + (5y)2 + (-3z)2 + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x)
= 4x2 + 25y2 + 9z2 - 20xy -30yz + 12zx

(vi)
 









Q5: Factorise:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
(ii) 2x2 + y2  + 8z2  – 2√2 xy + 4√2 yz – 8xz


Answer:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
= (2x)2 + (3y)2 + (4z)2 + 2(2x)(3y) – 2(3y)(4z) – 2(2x)(4z)
= (2x)2 + (3y)2 + (-4z)2 + 2(2x)(3y) + 2(3y)(-4z) + 2(2x)(-4z)
Using identity, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
= (2x + 3y -4z)2
=  (2x + 3y -4z)(2x + 3y -4z)

(ii) 2x2 + y2  + 8z2  – 2√2 xy + 4√2 yz – 8xz
 = (√2x)2 + y2  + (2√2z)2  – 2(√2 x)(y) + 2(y)(2√2z) – 2(√2x)(2√2z)
 = (-√2x)2 + y2  + (2√2z)2  + 2(-√2 x)(y) + 2(y)(2√2z) + 2(-√2x)(2√2z)
 Using identity, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
 = (-√2x + y + 2√2z )2
 = (-√2x + y + 2√2z )(-√2x + y + 2√2z )

Q6: Write the following cubes in expanded form:
(i)  (2x + 1)3
(ii) (2a – 3b)3
(iii) (3x/2 + 1)3
(iv) (x - 2y/3)3

Answer:
(i)  (2x + 1)3
Using identity (x + y)3 = x3 + y3 + 3xy(x + y)
=  (2x)3 + (1)3 + 3(2x)(1)(2x + 1)
=  8x3 + 1 + 6x(2x + 1)
=  8x3 + 1 + 12x2 + 6x

(ii) (2a – 3b)3
Using identity (x - y)3 = x3 - y3 - 3xy(x - y) 
= (2a)3 - (3b)3 - 3(2a)(3b)(2a - 3b)
= 8a3 - 27b3 - 18ab(2a - 3b)
= 8a3 - 27b3 -36a2b + 54ab2

(iii)  (3x/2 + 1)3
Using identity (x + y)3 = x3 + y3 + 3xy(x + y)
=  (3x/2)3 + (1)3 + 3(3x/2)(1)(3x/2 + 1)
= 27x3/8 + 1 + (9x/2)(3x/2 + 1)
= 27x3/8 + 27x2/4 - 9x/2 + 1

(iv) (x - 2y/3)3
Using identity (x - y)3 = x3 - y3 - 3xy(x - y)
= x3 - (2y/3)3 - 3x(2y/3)(x - 2y/3)
= x3 - 8y3/27 - 2xy/3(x - 2y/3)
=  x3 - 8y3/27 - 2x2y + 4xy2/3

Q7: Evaluate the following using suitable identities:
(i) (99)3
(ii) (102)3
(iii) (998)3


Answer:
(i) (99)3
= (100-1)3
Using identity (x - y)3 = x3 - y3 - 3xy(x - y) 
= (100)3 - (1)3 - 3(100)(1)(100 - 1)
= 1000000 - 1 -300(99)
= 1000000 − 1 − 29700
= 970299

(ii) (102)3
= (100 + 2)3
Using identity (x + y)3 = x3 + y3 + 3xy(x + y)
= (100)3 + (2)3 + 3(100)(2)(100+2)
= 1000000 + 8 + 600(102)
= 1000000 + 8 + 61200
= 1061208

(iii) (998)3
= (1000 - 2)3
Using identity (x - y)3 = x3 - y3 - 3xy(x - y)
= (1000)3 + (2)3 + 3(100)(2)(1000-2)
= 1000000000 − 8 − 6000(998)
= 1000000000 − 8 − 5988000
= 1000000000 − 5988008
= 994011992

Q8: Factorise each of the following:
(i)  8a3 + b3 + 12a2b + 6ab2
(ii) 8a3 – b3 – 12a2b + 6ab2
(iii) 27 – 125a3  – 135a + 225a2
(iv) 64a3 – 27b3 – 144a2b + 108ab2





Answer:
(i) 8a3 + b3 + 12a2b + 6ab2
= (2a)3 + b3 + 3(2a)2b + 3(2a)b2

(x + y)3 =  x3 + 3x2y + 3xy2 + y3
= (2a + b)3

(ii) 8a3 – b3 – 12a2b + 6ab2
=  (2a)3 - b3 - 3(2a)2b + 3(2a)b2
∵  (x - y)3 = x3 - y3 - 3xy(x - y)  = x3 - 3x2y + 3xy2 - y3

=  (2a - b)3

(iii) 27 – 125a3  – 135a + 225a2
 = 33 – (5a)3 – 3(3)2(5a) + 3(3)(5a)2
 ∵  (x - y)3 = x3 - y3 - 3xy(x - y)  = x3 - 3x2y + 3xy2 - y3
 = (3 - 5a)3

(iv) 64a3 – 27b3 – 144a2b + 108ab2
= (4a)3 - (3b)3 –3(4a)2(3b) + 3(4a)(3b)2
= (4a - 3b)3                            [∵  (x - y)3 = x3 - 3x2y + 3xy2 - y3]








Q9. Verify :
(i) x3 + y3 = (x + y) (x2  – xy + y2)
(ii) x3 –  y3 = (x – y) (x2 + xy +  y2)

Answer:
(i) x3 + y3 = (x + y) (x2  – xy + y2)
∵  (x + y)3 = x3 + y3 + 3xy(x + y)
⇒  x3 + y3 = (x + y)3 - 3xy(x + y)
⇒  x3 + y3 = (x+y) [(x + y)2 - 3xy]
⇒  x3 + y3 = (x+y) [x2 + y2 + 2xy - 3xy]
⇒  x3 + y3 = (x+y) (x2 + y2 -xy)                      ... (answer)


(ii) ∵ (x - y)3 = x3 - y3 - 3xy(x - y)
⇒  x3 - y3 = (x - y)3 + 3xy(x - y)
⇒  x3 - y3 = (x - y)[(x - y)2 + 3xy]
⇒  x3 - y3 = (x - y)(x2 + y2 - 2xy+ 3xy)
⇒  x3 - y3 = (x - y)(x2 + y2 + xy)                     ... (answer)


Q10: Factorise each of the following
(i) 27y3 + 125z3
(ii) 64m3 - 343n3

Answer:
(i) 27y3 + 125z3
=  (3y)3 + (5z)3
Using identity x3 + y3 = (x+y) (x2 + y2 -xy)
= (3y + 5z)((3y)2 + (5z)2 -(3y)(5z))
= (3y +5z)(9y2 + 25z2 -15yz)

(ii) 64m3 - 343n3
=  (4m)3 - (7n)3
Using identity x3 - y3 = (x - y)(x2 + y2 + xy)
= (4m - 7n)((4m)2 + (7n)2 + (4m)(7n))
= (4m - 7n)(16m2 + 49n2 + 28mn)

Q11: Factorise 27x3 + y3 + z3 - 9xyz

Answer: ∵ x3 + y3 + z3- 3xyz = (x + y + z)(x2 + y2 + z2- xy - yz -zx)
∴ = (3x)3 + y3 + z3 - 3(3x)yz
   = (3x + y + z)((3x)2 + y2 + z2- 3xy - yz -3zx)
   = (3x + y + z)(9x2 + y2 + z2- 3xy - yz -3zx)


Q12: Verify that 
 x3 + y3 + z3- 3xyz = ½((x + y + z)[(x-y)2 + (y-z)2 + (z - x)2]

Answer: ∵ x3 + y3 + z3- 3xyz = (x + y + z)(x2 + y2 + z2- xy - yz -zx)
∴ RHS =  ½(x + y + z)(2x2 + 2y2 + 2z2- 2xy -2yz -2zx)
      =  ½(x + y + z)(x2 + x2 +y2 + y2 +z2 + z2 - 2xy -2yz -2zx)
      =  ½(x + y + z)[x2 + y2 - 2xy + y2 +z2 -2yz + z2 + x2 -2zx]
      = ½(x + y + z)[(x2 + y2 - 2xy) + (y2 +z2 -2yz) + (z2 + x2 -2zx)]
      = ½(x + y + z)[(x - y)2 + (y - z)2 + (z - x)2 ]    ... (answer)


Q13: If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.

Answer: ∵ x3 + y3 + z3- 3xyz = (x + y + z)(x2 + y2 + z2- xy - yz -zx)
Also it is given  x + y + z = 0
⇒  x3 + y3 + z3- 3xyz = (0)(x2 + y2 + z2- xy - yz -zx)
⇒  x3 + y3 + z3- 3xyz = 0
⇒  x3 + y3 + z3= 3xyz

Q14. Without actually calculating the cubes, find the value of each of the following:
(i) (–12)3 + (7)3 + (5)3
(ii) (28) x3 + (–15) x3 + (–13) x3




Answer:
(i) (–12)3 + (7)3 + (5)3

∵ (-12) + (7) + (5) = 0
Using identity,  if x + y + z = 0, then x3 + y3 + z3 = 3xyz.
 = 3(-12)(7)(5) = -1260      ...(answer)

(ii) (28) x3 + (–15) x3 + (–13) x3
∵  (28) + (-15) + (-13) = 0
Using identity,  if x + y + z = 0, then x3 + y3 + z3 = 3xyz.
 = 3(28)(-15)(-13) = 16380     ...(answer)

Q15: Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:

(i)  Area : 25a2 – 35a + 12
(ii) Area : 35y2 + 13y –12

Answer: Since area of rectangle = length ✕ breadth. Let us factorize the following equations into two terms.
(i)  Area =  25a2 – 35a + 12
   =  25a2 – 15a  - 20a + 12             (Using splitting method)
   =  5a(5a -3) - 4(5a - 3)
   =  (5a - 3)(5a -4)
∴ Possible length = (5a - 3)
and Possible width =  (5a -4)

(ii) Area : 35y2 + 13y –12
= 35y2 + 28y -15y –12
= 7y(5y + 4)-3(5y + 4)
= (5y +4)(7y - 3)
∴ Possible length = (5y +4)
and Possible width = (7y - 3)


Q16: What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume : 3x2 – 12x
(ii) Volume : 12ky2  + 8ky – 20k

Answer:
Since, volume of cuboid = length ✕ breadth ✕ height
Let us factorise the equations into three terms.
(i) Volume : 3x2 – 12x
 = 3x(x-4)
⇒ Possible length = 3, width = x and height = (x- 4)
or Possible length = 1, width = 3x and height = (x- 4)

(ii) Volume : 12ky2  + 8ky – 20k
= 4k(3y2  + 2y – 5)
= 4k( 3y2 - 3y + 5y - 5)
= 4k( 3y(y - 1) + 5(y -1))
= 4k(y-1)(3y + 5)
⇒ Possible length = 4k, width = (y-1) and height = (3y + 5)

✪ Extra Problems ✪

Q17: If x + x-1 = 5, evaluate x3 – x-3

Answer: x + x-1 = 5
Cubing both sides, we get
 (x + x-1)3 = 53
Using identity,  (x - y)3 = x3 - y3 - 3xy(x - y) 
 x3 - x-3 -3(x)(x-1)(x - x-1) = 125

⇒ x3 - x-3 -3(x - x-1) = 125
⇒ x3 - x-3 -3(5) = 125
⇒ x3 - x-3  = 125 + 15
⇒ x3 - x-3  =140                ...(answer)


Q18: if x2 + x-2 = 102, evaluate  x + x-1

Answer: x2 + x-2 = 102
⇒  x2 + x-2 - 2 = 102 - 2
⇒  x2 + x-2 - 2(x)(x-1) = 100
⇒  (x - x-1)2 = 100
⇒  x - x-1 = 10

RD Sharma Exercise 4.1
Q19: Evaluate (a - 0.1)(a + 0.1)

Answer: Using identity, x2  – y2  = (x + y) (x – y)
(a - 0.1)(a + 0.1) = a2  – (0.1)2 = a2  – 0.01

Q20: Evaluate
(i) (399)2
(ii) (0.98)2
(iii) 991 ✕ 1009
(iv) 117 ✕ 83

Answer:
(i) (399)2 = (400 -1)
Using identity, (x – y)2  = x2 – 2xy + y2
= (400)2 - 2(400)(1) + (1)2
= 160000 - 800 + 1
= 159201

(ii) (0.98)2 = (1 - 0.02)2
Using identity,  (x – y)2  = x2 – 2xy + y2
= (1)2 – 2(1)(0.02) + (0.02)2
= 1 -0.04 + 0.0004
= 0.9604

(iii) 991 ✕ 1009 = (1000 - 9)(1000 + 9)
Using identity, x2  – y2  = (x + y) (x – y)
= (1000)2  – (9)2 
= 106 - 81
= 999919

(iv) 117 ✕ 83 = (100 + 17)(100 - 17)
Using identity, x2  – y2  = (x + y) (x – y)
= (100)2  – (17)2 = 10000 - 289 = 9711

Q21: Simplify 0.76 ✕ 0.76 + 2 ✕ 0.76 ✕ 0.24 + 0.24 ✕ 0.24

Answer: Using identity (x + y)2  = x2 + 2xy + y2
= (0.76 + 0.24)2 = (1)2 = 1

Q22: If x + x-1 = 11, evaluate x2 + x-2

Answer: x + x-1 = 11
(x + x-1)2 = (x)2 +(x-1)2 + 2(x)(x-1) = 112
⇒  x2 + x-2+ 2 = 121
⇒  x2 + x-2 = 121 - 2 = 119

Q23: Prove that a2 + b2 + c2  –ab -bc -ca is always non-negative for all values of a, b and c.

Answer: To prove a2 + b2 + c2  –ab -bc -ca ≥ 0.
We know that square of any number (+ve or -ve) is always +ve.
a2 + b2 + c2- ab - bc -ca
= ½[2a2 + 2b2 + 2c2 -2ab -2bc -2ca]
⇒ = ½[a2 + b2 -2ab + b2 + c2 -2bc + a2 + c2 -2ca]
⇒ = ½[(a2 + b2 -2ab) + (b2 + c2 -2bc) + (a2 + c2 -2ca)]
⇒ = ½[(a - b)2 + (b - c)2 + (c - a)2 ]
Here, all the terms are always be positive,
⇒ a2 + b2 + c2  –ab -bc -ca ≥ 0.

Q24: If x + x-1 = √5, evaluate x2 + x-2 and x4 + x-4

Answer:  x + x-1 = √5
  (x + x-1 )2 = (√5)2
⇒  x2 + x-2 + 2 = 5
⇒  x2 + x-2 = 5 -3 = 3
(x2 + x-2)2 = (3)2
⇒  x4 + x-4 + 2 = 9
⇒  x4 + x-4 = 9 -2 = 7



Q25: If 9x2+ 25y2 = 181 and xy = -6. Find the value of 3x + 5y

Answer:  9x2+ 25y2 = 181
       (3x)2 + (5y)2 = 181
⇒ (3x)2 + (5y)2 + 30xy - 30xy = 181
⇒ (3x)2 + (5y)2 + 2(5x)(6y) = 181 + 30xy
⇒  (3x + y)2 = 181 + 30(-6) = 181 - 180
⇒ (3x + y)2 =1
⇒ 3x + y = ∓1

Q26: Simplify (x3- 3x2 - x)(x2- 3x + 1)

Answer:  (x3- 3x2 - x)(x2- 3x + 1)
⇒ = x3(x2- 3x + 1) - 3x2(x2- 3x + 1) -x(x2- 3x + 1)
    =  x5 - 3x4 + x3- 3x4+ 9x3- 3x2- x3 + 3x2- x
    = x5 - 6x4 + 9x3 - x

miscellaneous problems

Q27:  Factorise (x - y)3 + (y - z)3 + (z - x)3


Answer: Let (x - y) = a, (y - z) = b and (z - x) = c
⇒ = a3 + b3 + c3
∵  a + b + c = (x - y) + (y -z) + (z -x) = 0
Using identity, p3 + q3 + r3 = 3xyz (if p + q + r = 0)
⇒= 3abc
   = 3(x -y)(y - z)(z - x)

Q28: Factorise: abx + aby - bcx - bcy

Answer: abx + aby - bcx - bcy
= ab(x + y) -bc(x + y)
= (x + y)(ab -ac)
= a(b - c)(x + y)

See here the video tutorial by Khanacademy.org on Factorization of sum of cubes:



Q29(CBSE 2011):  Show that (xa-b)a+b + (xb-c)b+c + (xc-a)c+a = 1

Answer: (xa-b)a+b + (xb-c)b+c + (xc-a)c+a
 Using identity  (x + y) (x – y) = x2  – y2

⇒ = xa2-b2 + xb2-c2 + xc2-a2 
 Using identity ap.ap = ap+q
⇒ = xa2- b2+ b2- c2 + c2-a2
⇒ = x0 = 1

Q30: Evaluate 5252- 4752
(a) 100
(b) 10000
(c) 50000
(d) 100000

Answer: (c) 50000
x2  – y2  = (x + y) (x – y)
⇒ (525 + 475)(525 - 475) = (1000)(50) = 50000

Q31: If a+ b + c = 0, then (a3 + b3 + c3 ) = ?

(a) abc
(b) 2abc
(c) 3abc
(d) 4abc

Answer: (c) 3abc [Hint: See Q 13 above]

10 comments:

  1. very very thanks and now im good in maths thanks very very much

    ReplyDelete
  2. there is a mistake in exercise 2.5 Q16 (i) it is 3x(x-4) not 3x(x-12). pls correct

    ReplyDelete
  3. you forgot for both the heights in the same question it should also be (x-4)

    ReplyDelete
  4. helped me to clear all my confusions

    ReplyDelete
  5. can give me answer of any another question not of ncert

    ReplyDelete
  6. in question 24 ,it should be 5-2=3 not 5-3=3.
    please correct the mistake

    ReplyDelete