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Maths 2 Marks Questions

Algebraic Identities [NEW] · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 41 questions.

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Question 12 Marks
Write the following in the expanded form:
(a2 + b2 + c2)2
Answer
We have,
(a2 + b2 + c2)2 = (a2)2 + (b2)2 + (c2)2 + 2a2b2 + 2b2c2 + 2a2c2
$\big[\therefore$(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz$\big]$
(a2 + b2 + c2)2 = a4 + b4 + c4 + 2a2b2 + 2b2c2 + 2c2a2
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Question 22 Marks
Write the following in the expanded form:
(2 + x - 2y)2
Answer
We have,
(2 + x - 2y)2 = 22 + x2 + (-2y)2 + 2(2)(x) + 2(x)(-2y) + 2(2)(-2y)
$\big[\therefore$(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz$\big]$
= 4 + x2 + 4y2 + 4x − 4xy - 8y
(2 + x - 2y)2 = 4 + x2 + 4y2 + 4x - 4xy - 8y
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Question 32 Marks
Write the following in the expanded form:
(x + 2y + 4z)2
Answer
We have,
(x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 + 2x × 2y + 2 × 2y × 4z + 2x × 4z
$\big[\therefore$(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz$\big]$
(x + 2y + 4z)2 = x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz
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Question 42 Marks
Write the following in the expanded form:
$\Big(\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{z}}+\frac{\text{z}}{\text{x}}\Big)^2$
Answer
We have,
$\Big(\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{z}}+\frac{\text{z}}{\text{x}}\Big)^2\\=\Big(\frac{\text{x}}{\text{y}}\Big)^2+\Big(\frac{\text{y}}{\text{z}}\Big)+\Big(\frac{\text{z}}{\text{x}}\Big)^2\\+2\frac{\text{x}}{\text{y}}\frac{\text{y}}{\text{z}}+2\frac{\text{y}}{\text{z}}\frac{\text{z}}{\text{x}}+2\frac{\text{z}}{\text{x}}\frac{\text{x}}{\text{y}}$
$\big[\therefore\big(\text{x}+\text{y}+\text{z}\big)\\=\text{x}^2+\text{y}^2+\text{z}^2+2\text{xy}+2\text{yz}+2\text{xz}\big]$
$\therefore\Big(\frac{\text{x}}{\text{y}}+\frac{\text{y}}{\text{z}}+\frac{\text{z}}{\text{x}}\Big)^2\\=\Big(\frac{\text{x}^2}{\text{y}^2}\Big)+\Big(\frac{\text{y}^2}{\text{z}^2}\Big)+\Big(\frac{\text{z}^2}{\text{x}^2}\Big)+2\frac{\text{x}}{\text{z}}+2\frac{\text{y}}{\text{x}}+2\frac{\text{x}}{\text{y}}$
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Question 52 Marks
Write the following in the expanded form:
$\Big(\frac{\text{a}}{\text{bc}}+\frac{\text{b}}{\text{ca}}+\frac{\text{c}}{\text{ab}}\Big)^2$
Answer
We have,
$\Big(\frac{\text{a}}{\text{bc}}+\frac{\text{b}}{\text{ca}}+\frac{\text{c}}{\text{ab}}\Big)^2=\Big(\frac{\text{a}}{\text{bc}}\Big)^2+\Big(\frac{\text{b}}{\text{ca}}\Big)^2+\Big(\frac{\text{c}}{\text{ab}}\Big)^2\\+2\Big(\frac{\text{a}}{\text{bc}}\Big)\Big(\frac{\text{b}}{\text{ca}}\Big)+2\Big(\frac{\text{b}}{\text{ca}}\Big)\Big(\frac{\text{c}}{\text{ab}}\Big)+2\Big(\frac{\text{a}}{\text{bc}}\Big)\Big(\frac{\text{c}}{\text{ab}}\Big)$
$\big[\therefore\big(\text{x}+\text{y}+\text{z}\big)=\text{x}^2+\text{y}^2+\text{z}^2+2\text{xy}+2\text{yz}+2\text{xz}\big]$
$\Big(\frac{\text{a}}{\text{bc}}+\frac{\text{b}}{\text{ca}}+\frac{\text{c}}{\text{ab}}\Big)^2\Big(\frac{\text{a}}{\text{bc}}+\frac{\text{b}}{\text{ca}}+\frac{\text{c}}{\text{ab}}\Big)^2\\=\Big(\frac{\text{a}^2}{\text{b}^2\text{c}^2}\Big)+\Big(\frac{\text{b}^2}{\text{c}^2\text{a}^2}\Big)+\Big(\frac{\text{c}^2}{\text{a}^2\text{b}^2}\Big)+2\frac{2}{\text{a}^2}+\frac{2}{\text{b}^2}+\frac{2}{\text{c}^2}$
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Question 62 Marks
Write the following in the expanded form:

(m + 2n - 5p)2

Answer
We have,
(m + 2n - 5p)2 = m2 + (2n)2 + (-5p)2 + 2m × 2n + (2 × 2n × -5p) + 2m × -5p
$\big[\therefore$(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz$\big]$
(m + 2n - 5p)2 = m2 + 4n2 + 25p2 + 4mn − 20np - 10pm
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Question 72 Marks
Write the following in the expanded form:
(ab + bc + ca)2
Answer
We have,
(ab + bc + ca)2 = (ab)2 + (bc)2 + (ca)2 + 2(ab)(bc) + 2(bc)(ca) + 2(ab)(ca)
$\big[\therefore$(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz$\big]$
= a2b2 + b2c2 + c2a2 + 2(ac)b2 + 2(ab)(c)2 + 2(bc)(a)2
(ab + bc + ca)2 = a2b2 + b2c2 + c2a2 + 2acb2 + 2abc2 + 2bca2
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Question 82 Marks
Write the following in the expanded form:
(a + 2b + c)2
Answer
We have,
(a + 2b + c)2 = a2 + (2b)2 + c2 + 2a(2b) + 2ac + 2(2b)c
$\big[\therefore$(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz$\big]$
$\therefore$ (a + 2b + c)2 = a2 + 4b2 + c2 + 4ab + 2ac + 4bc
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Question 92 Marks
Write the following in the expanded form:
(-3x + y + z)2
Answer
We have,
(-3x + y + z)2 = [(-3x)2 + y2 + z2 + 2(-3x)y + 2yz + 2(-3x)z]
$\big[\therefore$(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz$\big]$
9x2 + y2 + z2 - 6xy + 2yz - 6xz
(-3x + y + z)2 = 9x2 + y2 + z2 - 6xy + 2xy - 6xz
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Question 102 Marks
Write the following in the expanded form:
(2x - y + z)2
Answer
We have,
(2x - y + z)2 = (2x)2 + (-y)2 + (z)2 + 2(2x)(-y) + 2(-y)(z) + 2(2x)(z)
$\big[\therefore$(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz$\big]$
(2x - y + z)2 = 4x2 + y2 + z2 - 4xy - 2yz + 4xz
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Question 112 Marks
Write the following in the expanded form:
(-2x + 3y + 2z)2
Answer
We have,
(-2x + 3y + 2z)2 = (-2x)2 + (3y)2 + (2z)2 + 2(-2x)(3y) + 2(3y)(2z) + 2(-2x)(2z)
$\big[\therefore$(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz$\big]$
(-4x + 6y + 4z)2 = 4x2 + 9y2 + 4z2 - 12xy + 12yz - 8xz
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Question 122 Marks
Write the following in the expanded form:
(2a - 3b - c)2
Answer
We have,
(2a - 3b - c)2 = [(2a) + (-3b) + (-c)2
(2a)2 +( -3b)2 + (-c)2 + 2(2a)(-3b) + 2(-3b)(−c) + 2(2a)(-c)
$\big[\therefore$(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz$\big]$
4a2 + 9b2 + c2 − 12ab + 6bc - 4ca
$\therefore$ (2a - 3b - c)2 = 4x2 + 9y2 + c2 - 12ab + 6bc - 4ca
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Question 132 Marks
Simplify the following:
$\frac{7.83\times7.83-1.17\times1.17}{6.66}$
Answer
We have,
$\frac{7.83\times7.83-1.17\times1.17}{6.66}$
$=\frac{(7.83+1.17)(7.83-1.17)}{6.66}$ $\big[\because(\text{a}-\text{b})^2=(\text{a}+\text{b})(\text{a}-\text{b})\big]$
$=\frac{(9.00)(6.66)}{6.66}=9$
$\therefore\frac{7.83\times7.83-1.17\times1.17}{6.66}=9$
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Question 142 Marks
Simplify the following:
322 × 322 - 2 × 322 × 22 + 22 × 22
Answer
We have,
322 × 322 - 2 × 322 × 22 + 22 × 22
= (322-22)2 [a2 + b2 - 2ab = (a – b)2]
= (300)2 [ Where a = 322 and b = 22]
= 90000
Therefore, 322 × 322 - 2 × 322 × 22 + 22 × 22 = 90000.
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Question 152 Marks
Simplify the following:
175 × 175 + 2 × 175 × 25 + 25 × 25
Answer
We have,
175 × 175 + 2 × 175 × 25 + 25 × 25 = (175)2 + 2(175)(25) + (25)2
= (175 + 25)2 [a2 + b2 + 2ab = (a + b)2]
= (200)2 [Where a = 175 and b = 25]
= 40000
Therefore, 175 × 175 + 2 × 175 × 25 + 25 × 25 = 40000.
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Question 162 Marks
Simplify the following:
0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24
Answer
We have,
0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24
= (0.76 + 0.24)2 [a2 + b2 + 2ab = (a + b)2]
= (1.00)2 [Where a = 0.76 and b = 0.24]
= 1
Therefore, 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × 0.24 = 1
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Question 172 Marks
Simplify the following expressions:
(x + y - 2z)2 - x2 - y2 - 3z2 + 4xy
Answer
Expanding, we get
(x + y - 2z)2 - x2 - y2 - 3z2 + 4xy
= [x2 + y2 + 4z2 + 2xy + 2y(−2z) + 2a(-2c)] - x2 - y2 - 3z2 + 4xy
= z2 + 6xy - 4yz - 4zx
(x + y - 2z)2 - x2 - y2 - 3z2 + 4xy = z2 + 6xy - 4yz - 4zx
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Question 182 Marks
If  $\text{x}-\frac{1}{\text{x}}=-1$ find the value of $\text{x}^2+\frac{1}{\text{x}^2}.$
Answer
We have,
 $\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\big(\frac{1}{\text{x}}\big)^2-2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$\Rightarrow(-1)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$ $\big[\because\text{x}-\frac{1}{\text{x}}=-1\big]$
$\Rightarrow2+1=\text{x}^2+\frac{1}{\text{x}^2}$
$\therefore\ \text{x}^2+\frac{1}{\text{x}^2}=3.$
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Question 192 Marks
If  $\text{x}+\frac{1}{\text{x}}=11$ find the value of $\text{x}^2+\frac{1}{\text{x}^2}.$
Answer
We have $\text{x}+\frac{1}{\text{x}}=11$
Now, $\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\big(\frac{1}{\text{x}}\big)^2+2\times\text{x}\times\frac{1}{\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow(11)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow121=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=119.$
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Question 202 Marks
Find the following:
(x3 + 1)(x6 - x3 + 1)
Answer
We have,
(x3 + 1)(x6 - x3 + 1)
= (x3 + 1)$\big[$(x3)2 - 1 × x3 + 12$\big]$
= (x3)3 + (1)3 $\big[\because$ a3 + b3 = (a + b)(a2 - ab + b2)$\big]$
= x9 + 1
$\therefore$ (x3 + 1)(x6 - x3 + 1) = x9 + 1
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Question 212 Marks
Find the following:
(x2 - 1)(x4 + x2 + 1)
Answer
We have,
(x2 - 1)(x4 + x2 + 1)
= (x2 - 1)$\big[$(x2)2 + 1 × x2 + 12$\big]$
= (x2)3 - (1)3 $\big[\because$ a3 - b3 = (a - b)(a2 + ab + b2)$\big]$
= x6 - 1
$\therefore$ (x2 - 1)(x4 + x2 + 1) = x6 - 1
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Question 222 Marks
Find the following products:
$\Big(\frac{\text{x}}{2}+2\text{y}\Big)\Big(\frac{\text{x}^2}{4}-\text{xy}+4\text{y}^2\Big)$
Answer
We have,
$\Big(\frac{\text{x}}{2}+2\text{y}\Big)\Big(\frac{\text{x}^2}{4}-\text{xy}+4\text{y}^2\Big)$
$=\Big(\frac{\text{x}}{2}+2\text{y}\Big)\bigg[\Big(\frac{\text{x}}{2}\Big)^2-\text{xy}-(2\text{y})^2\bigg]$
$=\Big(\frac{\text{x}}{2}\Big)^3+(2\text{y})^3$ $\big[\therefore\text{a}^3+\text{b}^3=(\text{a}+\text{b})(\text{a}^2-\text{ab}+\text{b}^2)\big]$
$=\frac{\text{x}^3}{8}+8\text{y}^3$
$\therefore\Big(\frac{\text{x}}{2}+2\text{y}\Big)\Big(\frac{\text{x}^2}{4}-\text{xy}+4\text{y}^2\Big)=\frac{\text{x}^3}{8}+8\text{y}^3$
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Question 232 Marks
Find the following products:
$\Big(\frac{3}{\text{x}}-\frac{5}{\text{y}}\Big)\Big(\frac{9}{\text{x}^2}+\frac{25}{\text{y}^2}+\frac{15}{\text{xy}}\Big)$
Answer
We have,
$\Big(\frac{3}{\text{x}}-\frac{5}{\text{y}}\Big)\Big(\frac{9}{\text{x}^2}+\frac{25}{\text{y}^2}+\frac{15}{\text{xy}}\Big)$
$=\Big(\frac{3}{\text{x}}-\frac{5}{\text{y}}\Big)\bigg[\Big(\frac{3}{\text{x}}\Big)^2+\Big(\frac{5}{\text{y}}\Big)^2+\frac{3}{\text{x}}\times\frac{5}{\text{y}}\Big)\bigg]$
$=\Big(\frac{3}{\text{x}}\Big)^3-\Big(\frac{5}{\text{y}}\Big)^3$ $\big[\therefore\text{a}^3-\text{b}^3=(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)\big]$
$=\frac{27}{\text{x}^3}-\frac{125}{\text{y}^3}$
$\therefore\Big(\frac{3}{\text{x}}-\frac{5}{\text{y}}\Big)\Big(\frac{9}{\text{x}^2}+\frac{25}{\text{y}^2}+\frac{15}{\text{xy}}\Big)=\frac{27}{\text{x}^3}-\frac{125}{\text{y}^3}$
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Question 242 Marks
Find the following products:
$\Big(\frac{3}{\text{x}}-2\text{x}^2\Big)\Big(\frac{9}{\text{x}^2}+4\text{x}^2-6\text{x}\Big)$
Answer
We have,
$\Big(\frac{3}{\text{x}}-2\text{x}^2\Big)\Big(\frac{9}{\text{x}^2}+4\text{x}^2-6\text{x}\Big)$
$=\Big(\frac{3}{\text{x}}-2\text{x}^2\Big)\bigg[\Big(\frac{3}{\text{x}}\Big)^2+(2\text{x})^2+\frac{3}{\text{x}}-\times2\text{x}^2\bigg]$
$=\Big(\frac{3}{\text{x}}\Big)^3-(2\text{x}^2)^3$ $\big[\therefore\text{a}^3-\text{b}^3=(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)\big]$
$=\frac{27}{\text{x}^3}-8\text{x}^6$
$\therefore\Big(\frac{3}{\text{x}}-2\text{x}^2\Big)\Big(\frac{9}{\text{x}^2}+4\text{x}^2-6\text{x}\Big)=\frac{27}{\text{x}^3}-8\text{x}^6$
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Question 252 Marks
Find the following products:
$\Big(\frac{2}{\text{x}}+3\text{x}\Big)\Big(\frac{4}{\text{x}^2}+9\text{x}^2-6\Big)$
Answer
We have,
$\Big(\frac{2}{\text{x}}+3\text{x}\Big)\Big(\frac{4}{\text{x}^2}+9\text{x}^2-6\Big)$
$\Big(\frac{2}{\text{x}}+3\text{x}\Big)\bigg[\Big(\frac{2}{\text{x}}\Big)^2+(3\text{x})^2-\frac{2}{\text{x}}\times3\text{x}\bigg]$
$=\Big(\frac{2}{\text{x}}\Big)^3+(3\text{x})^3$ $\big[\therefore\text{a}^3+\text{b}^3=(\text{a}+\text{b})(\text{a}^2-\text{ab}+\text{b}^2)\big]$
$=\frac{8}{\text{x}^3}+27\text{x}^3$
$\therefore\Big(\frac{2}{\text{x}}+3\text{x}\Big)\Big(\frac{4}{\text{x}^2}+9\text{x}^2-6\Big)=\frac{8}{\text{x}^3}+27\text{x}^3$
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Question 262 Marks
Find the following products:
$\Big(3+\frac{5}{\text{x}}\Big)\Big(9-\frac{15}{\text{x}}+\frac{25}{\text{x}^2}\Big)$
Answer
We have,
$\Big(3+\frac{5}{\text{x}}\Big)\Big(9-\frac{15}{\text{x}}+\frac{25}{\text{x}^2}\Big)$
$\Big(3+\frac{5}{\text{x}}\Big)\Big[(3)^2-3\times\frac{5}{\text{x}}+\Big(\frac{5}{\text{x}}\Big)^2\Big]$
$=(3)^3+\Big(\frac{5}{\text{x}}\Big)^2$ $\big[\therefore\text{a}^3+\text{b}^3=(\text{a}+\text{b})(\text{a}^2-\text{ab}+\text{b}^2)\big]$
$=27+\frac{125}{\text{x}^3}$
$\therefore\Big(3+\frac{5}{\text{x}}\Big)\Big(9-\frac{15}{\text{x}}+\frac{25}{\text{x}^2}\Big)=27+\frac{125}{\text{x}^3}$
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Question 272 Marks
Find the following products:
(7p4 + q)(49p8 - 7p4q + q2)
Answer
we have,

(7p4 + q)(49p8 - 7p4q + q2)

= (7p4 + q)$\big[$(7p4)2 - 7p4 × q + (q)2$\big]$

= (7p4)3 + (q)3 $\big[\because$ a3 + b3 = (a + b)(a2 - ab + b2)$\big]$

= 343p12 + q3

$\therefore$ (7p4 + q)(49p8 - 7p4q + q2) = 343p12 + q3

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Question 282 Marks
Find the following products:
(4x - 5y)(16x2 + 20xy + 25y2)
Answer
we have,

(4x - 5y)(16x2 + 20xy + 25y2)

= (4x - 5y)$\big[$(4x)2 + 4x × 5y + (5y)2$\big]$

= (4x)3 - (5y)3 $\big[\because$ a3 - b3 = (a - b)(a2 + ab + b2)$\big]$

= 64x3 - 125y3

$\therefore$ (4x - 5y)(16x2 + 20xy + 25y2) = 64x3 - 125y3

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Question 292 Marks
Find the following products:
(3x + 2y)(9x2 - 6xy + 4y2)
Answer
we have,
(3x + 2y)(9x2 - 6xy + 4y2)
= (3x + 2y)$\big[$(3x)2 - 3x × 2y + (2y)2$\big]$
= (3x)3 + (2y)3 $\big[\because$ a3 + b3 = (a + b)(a2 - ab + b2)$\big]$
= 27x3 + 8y3
$\therefore$ (3x + 2y)(9x2 - 6xy + 4y2) = 27x3 + 8y3
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Question 302 Marks
Find the following products:
(1 - x)(1 + x + x2)
Answer
We have,
(1 - x)(1 + x + x2)
 = (1 - x)$\big[$(1)2 + 1 × x + (x)2$\big]$
= 13 - x3 $\big[\because$ a3 - b3 = (a - b)(a2 + ab + b2)$\big]$
= 1 - x3
$\therefore$ (1 - x)(1 + x + x2) = 1 - x3
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Question 312 Marks
Find the following products:
(1 + x)(1 - x + x2)
Answer
We have,
(1 + x)(1 - x + x2)
 = (1 + x)$\big[$(1)2 - 1 × x + (x)2$\big]$
= 13 + x3 $\big[\because$ a3 - b3 = (a + b)(a2 - ab + b2)$\big]$
$\therefore$ (1 + x)(1 -+ x + x2) = 1 + x3
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Question 322 Marks
Evaluate the following using identities:
$\big(\text{a}^2\text{b}-\text{b}^2\text{a}\big)^2$
Answer
We have,
$\big(\text{a}^2\text{b}-\text{a}\text{b}^2\big)^2$
$=\big(\text{a}^2\text{b}\big)^2+\big(\text{ab}^2\big)^2-2\times\text{a}^2\times\text{ab}^2$ $\begin{bmatrix}\because\big(\text{a}-\text{b}\big)^2=\text{a}^2+\text{b}^2-2\text{ab}\\\text{Where}\ \text{a}=\text{a}^2\text{b}\ \text{and}\ \text{b}=\text{ab}^2\end{bmatrix}$
$=\text{a}^4\text{b}^2+\text{b}^4\text{a}^2-2\text{a}^3\text{b}^3$
$\therefore\big(\text{a}^2\text{b}-\text{b}^2\text{a}\big)^2=\text{a}^4\text{b}^2+\text{b}^4\text{a}^2-2\text{a}^3\text{b}^3$
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Question 332 Marks
Evaluate the following using identities:
$\big(\text{a}-0.1\big)\big(\text{a}+0.1\big)$
Answer
We have,
$\big(\text{a}-0.1\big)\big(\text{a}+0.1\big)=\text{a}^2-(0.1)^2$ $\begin{bmatrix}\because(\text{a}-\text{b})(\text{a}+\text{b})\\\text{a}=\text{a}:\text{b}=0.1\end{bmatrix}$
$=\text{a}^2-0.01$
$(\text{a}-0.1)(\text{a}+0.1)=\text{a}^2-0.01$
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Question 342 Marks
Evaluate the following using identities:
$\Big(2\text{x}-\frac{1}{\text{x}}\Big)^2$
Answer
We have,
$\Big(2\text{x}-\frac{1}{\text{x}}\Big)^2=(2\text{x})^2+\big(\frac{1}{\text{x}}\big)^2-2.2\text{x}.\frac{1}{\text{x}}$
$\Big(2\text{x}-\frac{1}{\text{x}}\Big)^2=4\text{x}^2+\frac{1}{\text{x}^2}-4$ $\begin{bmatrix}\because(\text{a}-\text{b})^2=\text{a}^2+\text{b}^2-2\text{ab},\\\text{Where}\ \text{a}=2\text{x}\ \text{and}\ \text{b}=\frac{1}{\text{x}}\end{bmatrix}$
$\therefore\Big(2\text{x}-\frac{1}{\text{x}}\Big)^2=4\text{x}^2+\frac{1}{\text{x}^2}-4$
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Question 352 Marks
Evaluate the following using identities:
$\big(1.5\text{x}^2-0.3\text{y}^2\big)\big(1.5\text{x}+0.3\text{y}^2\big)$
Answer
We have,
$\big(1.5\text{x}^2-0.3\text{y}^2\big)\big(1.5\text{x}+0.3\text{y}^2\big)$
$=\big[1.5\text{x}^2\big]^2-\big[0.3\text{y}^2\big]^2$ $\begin{bmatrix}\because(\text{a}+\text{b})(\text{a}-\text{b})=\text{a}^2-\text{b}^2\\\text{a}=1.5\text{x}^2\ \text{and}\ \text{b}=0.3\text{y}^2\end{bmatrix}$
$=2.25\text{x}^4-0.09\text{y}^4$
$=\big[1.5\text{x}^2-0.3\text{y}^2\big]\big[1.5\text{x}^2+0.3\text{y}^2\big]\\=2.25\text{x}^4-0.09\text{y}^4.$
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Question 362 Marks
Evaluate the following using identities:
$(2\text{x} +\text{y})(2\text{x} - \text{y})$
Answer
We have,
$(2\text{x} +\text{y})(2\text{x} - \text{y})$
$= (2\text{x})^2 - (\text{y})^2 $ $\begin{bmatrix}\because(\text{a}+\text{b})(\text{a}-\text{b})=\text{a}^2-\text{b}^2\\\text{Where}\ \text{a}=2\text{x}\ \text{and}\ \text{b}=\text{y}\end{bmatrix}$
$=4\text{x}^2-\text{y}^2$
$(2\text{x}+\text{y})(2\text{x}-\text{y})=4\text{x}^2-\text{y}^2$
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Question 372 Marks
Evaluate the following using identities:
(399)2
Answer
We have,
3992 = (400 - 1)2
= (400)2 + (1)2 - 2 × 400 × 1 [(a - b)2 = a2 + b2 - 2ab]
Where, a = 400 and b = 1
= 160000 + 1 - 8000
= 159201
Therefore, (399)2 = 159201
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Question 382 Marks
Evaluate the following using identities:
(0.98)2
Answer
We have,
(0.98)2 = (1 - 0.02)2
= (1)2 + (0.02)2 - 2 × 1 × 0.02
= 1 + 0.0004 - 0.04 [Where, a = 1 and b = 0.02]
= 1.0004 - 0.04
= 0.9604
Therefore, (0.98)2 = 0.96041
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Question 392 Marks
Evaluate:
483 - 303 - 183
Answer
Let a = 48, b = -30 and c = -18
Then,
a + b + c = 48 - 30 - 18
= 0
$\therefore$ a3 + b3 + c3 = 3abc
⇒ (48)3 + (-30)3 + (-18)3 = 3 × (48) × (-30) × (-18)
= 144 × 540
= 77760
$\therefore$ 483 - 303 - 183 = 77760
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Question 402 Marks
Evaluate:
253 - 753 + 503
Answer
Let a = 25, b = -75 and c = 50
Then,
a + b + c = 25 - 75 + 50
= 0
$\therefore$ a3 + b3 + c3 = 3abc
⇒ (25)3 + (-75)3 + (50)3 = 3 × 25 × (-75) × 50
= -75 × 75 × 50
= -5625 × 50
= -281250
$\therefore$ 253 - 753 + 503 = -281250
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Question 412 Marks
Evaluate:
(0.2)3 - (0.3)3 + (0.1)3
Answer
Let a = 0.2, b = -0.3 and c = 0.1
Then,
a + b + c = 0.2 - 0.3 + 0.1
= 0.3 - 0.3
⇒ a + b + c = 0
$\therefore$ a3 + b3 + c3 = 3abc
⇒ (0.2)3 + (-0.3)3 + (0.1)3 = 3 × (0.2) × (-0.3) × (0.1)
= -0.018
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2 Marks Questions - Maths STD 9 Questions - Vidyadip