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Q. 1. Evaluate each of the following using identities:
Solution:
Given,
[?(a−b)^{2} = a^{2}+b^{2}−2ab]
Where, a = 2x, b = 1/x
(ii) (2x+y) (2xy)
Solution:
Given, (2x+y) (2xy)
= (2x)^{2}−(y)^{2}[?(a+b)(a−b) = a^{2}−b^{2}]
= 4x^{2}−y^{2}
?(2x+y)(2x−y) = 4x^{2}−y^{2}
(iii) (a^{2}b−ab^{2})^{2}
Given, (a^{2}b−ab^{2})^{2}
= (a^{2}b)^{2}+(ab^{2})^{2}−2∗a^{2}b∗ab^{2 }[?(a−b)^{2} = a^{2}+b^{2}−2ab]
Where, a = a^{2}b,b = ab^{2}
= a^{4}b^{2}+b^{4}a^{2}−2a^{3}b^{3}
?(a^{2}b−ab^{2})2 = a^{4}b^{2}+b^{4}a^{2}−2a^{3}b^{3}
(iv) (a0.1) (a+0.1)
Solution:
Given, (a0.1) (a+0.1)
= a^{2}−(0.1)^{2}[?(a+b)(a−b) = a^{2}−b^{2}]
Where, a = a and b = 0.1
= a^{2}−0.01
?(a−0.1)(a+0.1) = a^{2}−0.01
(v) (1.5x^{2}−0.3y^{2})(1.5x^{2}+0.3y^{2})
Solution:
Given, (1.5x^{2}−0.3y^{2})(1.5x^{2}+0.3y^{2})
= (1.5x^{2})^{2}−(0.3y^{2})^{2 }[?(a+b)(a−b) = a^{2}−b^{2}]
Where, a = 1.5x^{2},b = 0.3y^{2}
= 2.25x^{4}−0.09y^{4}
?(1.5x^{2}−0.3y^{2})(1.5x^{2}+0.3y^{2}) = 2.25x^{4}−0.09y^{4}
Q.2. Evaluate each of the following using identities:
(i) (399)^{2}
Solution:
We have,
399^{2 }= (4001)^{2}
= (400)^{2}+(1)^{2} – 2x400x1 [ (ab)^{2} = a^{2}+ b^{2}2ab ]
Where, a = 400 and b = 1
= 160000 + 1 – 8000
= 159201
Therefore, (399)^{2} = 159201.
(ii) (0.98)^{2}
Solution:
We have,
(0.98)^{2} = (10.02)^{2}
= (1)^{2 }+ (0.02)^{2} – 2x1x0.02
= 1 + 0.0004 – 0.04 [ Where, a=1 and b=0.02 ]
= 1.0004 – 0.04
= 0.9604
Therefore, (0.98)^{2} = 0.9604
(iii) 991x1009
Solution:
We have,
991x1009
= (10009) (1000+9)
= (1000)^{2} – (9)^{2} [ (a+b) (ab) = a^{2} – b^{2} ]
= 1000000 – 81 [ Where a=1000 and b=9 ]
= 999919
Therefore, 991x1009 = 999919
(iv) 117x83
Solution:
We have,
117x83
= (100+17) (10017)
= (100)^{2} – (17)^{2} [ (a+b) (ab) = a^{2} – b^{2} ]
= 10000 – 289 [ Where a=100 and b=17 ]
= 9711
Therefore, 117x83 = 9711
Q.3. Simplify each of the following:
(i) 175 x 175 +2 x 175 x 25 + 25 x 25
Solution:
We have,
175 x 175 +2 x 175 x 25 + 25 x 25 = (175)^{2} + 2 (175) (25) + (25)^{2}
= (175+25)^{2} [ a^{2}+ b^{2}+2ab = (a+b)^{2} ]
= (200)^{2} [ Where a=175 and b=25 ]
= 40000
Therefore, 175 x 175 +2 x 175 x 25 + 25 x 25 = 40000.
(ii) 322 x 322 – 2 x 322 x 22 + 22 x 22
Solution:
We have,
322 x 322 – 2 x 322 x 22 + 22 x 22
= (32222)^{2} [ a^{2}+ b^{2}2ab = (ab)^{2} ]
= (300)^{2} [ Where a=322 and b=22 ]
= 90000
Therefore, 322 x 322 – 2 x 322 x 22 + 22 x 22= 90000.
(iii) 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24
Solution:
We have,
0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24
= (0.76+0.24)^{2} [ a^{2}+ b^{2}+2ab = (a+b)^{2} ]
= (1.00)^{2} [ Where a=0.76 and b=0.24 ]
= 1
Therefore, 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24 = 1.
(iv)
Solution:
We have,
[?(a−b)^{2 }= (a+b)(a−b)]
= 9
Q.4. If ,find the value of
Solution:
Q.5. If ,find the value of
Solution: We have,
Now,
Q. 6. If find the value of
Solution:
We have,
Q. 7. If find the value of
Solution:
We have,
Q. 8. If find the value of
Solution:
We have,
Q. 9. If 9x^{2 }+ 25y^{2 }= 181 and xy = 6, find the value of 3x + 5y.
Solution:
We have,
(3x + 5y)^{2} = (3x)^{2} + (5y)^{2} + 2*3x*5y
⇒(3x + 5y)^{2} = 9x^{2} + 25y^{2} + 30xy
= 181 + 30(6) [Since, 9x^{2 }+ 25y^{2 }= 181 and xy = 6]
⇒ (3x+5y)^{2} = 1
⇒ (3x+5y)^{2} = (±1)^{2}
⇒ 3x+5y = ±1
Q.10. If 2x + 3y = 8 and xy = 2, find the value of 4x^{2 }+ 9y^{2}.
Solution:
We have,
(2x + 3y)^{2} = (2x)^{2} + (3y)^{2} + 2*2x*3y
⇒ (2x + 3y)^{2} = 4x^{2} + 9y^{2} + 12xy [Since, 2x + 3y = 8 and xy = 24 ]
⇒ (8)^{2} = 4x^{2} + 9y^{2} + 24
⇒ 64 – 24 = 4x^{2} + 9y^{2}
⇒ 4x^{2} + 9y^{2} = 40
Q. 11. If 3x  7y = 10 and xy = 1, find the value of 9x^{2 }+ 49y^{2}.
Solution:
We have,
(2  7y)^{2} = (3x)^{2} + (7y)^{2}  2*3x*7y
⇒ (3x  7y)^{2} = 9x^{2} + 49y^{2}  42xy [Since, 3x  7y = 10 and xy = 1 ]
⇒ (10)^{2} = 9x^{2} + 49y^{2} + 42
⇒ 100 – 42 = 9x^{2} + 49y^{2}
⇒ 9x^{2} + 49y^{2} = 58
Q.12. Simplify each of the following products:
(i)
Solution:
Solution:
We have,
[?(a+b)(a+b) = (a+b)^{2} and (a+b)(a−b) = a^{2}−b^{2}]
Solution:
We have,
[?(a−b)(a−b) = (a−b)^{2}]
(iv) (x^{2}+x−2)(x^{2}−x+2)
Solution:
(x^{2}+x−2)(x^{2}−x+2)
[(x)^{2}+(x−2)][(x^{2}−(x+2)]
⇒ (x^{2})^{2}−(x−^{2})^{2} [(a – b) (a + b) = a^{2}  b^{2}]
⇒ x^{4}−(x^{2}+4−4x)[?(a−b)^{2} = a^{2}+b^{2}−2ab]
⇒ x^{4}−x^{2}+4x−4
?(x^{2}+x−2)(x^{2}−x+2) = x^{4}−x^{2}+4x−4
(v) (x^{3}−3x−x)(x^{2}−3x+1)
Solution:
We have,
(x^{3}−3x−x)(x^{2}−3x+1)
⇒ x(x^{2}−3x−1)(x^{2}−3x+1)
⇒ x[(x^{2}−3x)2−(1)2][?(a+b)(a−b) = a^{2}−b^{2}]
⇒ x[(x^{2})^{2}+(−3x)^{2}−2(3x)(x^{2})−1]
⇒ x[x^{4}+9x^{2}−6x^{3}−1]
⇒ x^{5}−6x^{4}+9x^{3}−x
?(x^{3}−3x−x)(x^{2}−3x+1) = x^{5}−6x^{4}+9x^{3}−x
(vi) (2x^{4}−4x^{2}+1)(2x^{4}−4x^{2}−1)
Solution:
We have,
(2x^{4}−4x^{2}+1)(2x^{4}−4x^{2}−1)
⇒ [(2x^{4}−4x^{2})^{2}−(1)^{2}] [?(a+b)(a−b) = a^{2}−b^{2}]
⇒ [(2x^{4})^{2}+(4x^{2})^{2}−2(2x^{4})(4x^{2})−1]
⇒ 4x^{8}−16x^{6}+16x^{4}−1 [?(a−b)^{2 }= a^{2}+b^{2}−2ab]
?(2x^{4}−4x^{2}+1)(2x^{4}−4x^{2}−1) = 4x^{8}−16x^{6}+16x^{4}−1
Q.13. Prove that a^{2}+b^{2}+c^{2}−ab−bc−ca is always nonnegative for all values of a, b and c.
Solution:
We have,
a^{2}+b^{2}+c^{2}−ab−bc−ca
Multiply and divide by ‘2’
?a^{2}+b^{2}+c^{2}−ab−bc−ca ≥ 0
Hence, a^{2}+b^{2}+c^{2}−ab−bc−ca ≥ 0 is always nonnegative for all values of a, b and c.
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