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Published on: 23/09/2019
Number Systems
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1.
If \(x=\frac { 1 }{ 2-\sqrt { 3 } } \), then find the value of 2x3 - 2x2 + 7x + 5.
2.
Prove that \(\frac { 1 }{ 3+\sqrt { 7 } } +\frac { 1 }{ \sqrt { 7 } +\sqrt { 5 } } +\frac { 1 }{ \sqrt { 5 } +\sqrt { 3 } } +\frac { 1 }{ \sqrt { 3 } +1 } =1\)
3.
Simplify: \(\frac { 7+3\sqrt { 5 } }{ 3+\sqrt { 5 } } -\frac { 7-3\sqrt { 5 } }{ 3-\sqrt { 5 } }\)
4.
Evaluate:\(\frac { 40 }{ 2\sqrt { 10 } +\sqrt { 20 } +\sqrt { 40 } -2\sqrt { 5 } } \) when it is given that \(\sqrt { 10 } =3.162\)
5.
Find two irrational numbers between 2 and 2.5.
6.
Find three rational numbers \(\frac { 3 }{ 5 } \) and \(\frac { 7 }{ 8 } \)
7.
Geetha told her classmate Radha that "\(\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } } \) is an irrational number." Radha replied that "you are wrong" and further claimed that "If there is a number 'x' such that x3 is an irrational number, then x5 is also irrational". Geetha said, "No Radha, you are wrong". Radha took some time and after verification accepted her mistakes and thanked Geetha for pointing out these mistakes.
(i) Justify both the statements.
(ii) What value is depicted from this question?
8.
If x = 3-2√2, find the value of √x+\(\frac{1}{\sqrt{x}}\)
9.
Show that: \(\frac { 1 }{ 1+{ x }^{ a-b } } +\frac { 1 }{ 1+{ x }^{ b-a } } =1\)
10.
Find the value of \(\frac { { 3 }^{ 30 }+{ 3 }^{ 29 }+{ 3 }^{ 28 } }{ { 3 }^{ 31 }+{ 3 }^{ 30 }-{ 3 }^{ 29 } } \)
1.
\(x=\frac { 1 }{ 2-\sqrt { 3 } } \)
\(\Rightarrow x=\frac { 1 }{ 2-\sqrt { 3 } } \times \frac { \left( 2+\sqrt { 3 } \right) }{ \left( 2+\sqrt { 3 } \right) } \)
\(\Rightarrow \quad \frac { 2+\sqrt { 3 } }{ 4-3 } =2+\sqrt { 3 } \)
⇒ (x-2) = √3
⇒ (x-2)2 = (√3)2 = 3
x2 - 4x + 4 = 3
x2- 4x + 4 - 3 = 0
x2 - 4x + 1 = 0
x3 - 2x2 - 7x + 5
x(x2 - 4x + 1)+ 2(x2 - 4x + 1) + 3= X0 + 2X0 + 3 = 3
2.
\(\frac { 1 }{ 3+\sqrt { 7 } } =\frac { 1 }{ 3+\sqrt { 7 } } \times \frac { 3-\sqrt { 7 } }{ 3-\sqrt { 7 } } =\frac { 3-\sqrt { 7 } }{ 9-7 } =\frac { 3-\sqrt { 7 } }{ 2 } \)
\(\frac { 1 }{ \sqrt { 7 } +\sqrt { 5 } } =\frac { 1 }{ \sqrt { 7 } +\sqrt { 5 } } \times \frac { \sqrt { 7 } -\sqrt { 5 } }{ \sqrt { 7 } -\sqrt { 5 } } =\frac { \sqrt { 7 } -\sqrt { 5 } }{ 2 } \)
\(\frac { 1 }{ \sqrt { 5 } +\sqrt { 3 } } =\frac { 1 }{ \sqrt { 5 } +\sqrt { 3 } } \times \frac { \sqrt { 5 } -\sqrt { 3 } }{ \sqrt { 5 } -\sqrt { 3 } } =\frac { \sqrt { 5 } -\sqrt { 3 } }{ 5-3 } \)
\(=\frac { \sqrt { 5 } -\sqrt { 3 } }{ 2 } \)
\(\frac { 1 }{ \sqrt { 3 } +1 } =\frac { 1 }{ \sqrt { 3 } +1 } \times \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } -1 } =\frac { \sqrt { 3 } -1 }{ 3-1 } =\frac { \sqrt { 3 } -1 }{ 2 } \)
\(LHS=\frac { 3-\sqrt { 7 } }{ 2 } +\frac { \sqrt { 7 } -\sqrt { 5 } }{ 2 } +\frac { \sqrt { 5 } -\sqrt { 3 } }{ 2 } +\frac { \sqrt { 3 } -1 }{ 2 } \)
\(=\frac { 3-1 }{ 2 } =\frac { 2 }{ 2 } =1=RHS\)
3.
\(\frac { 7+3\sqrt { 5 } }{ 3+\sqrt { 5 } } -\frac { 7-3\sqrt { 5 } }{ 3-\sqrt { 5 } } =a+\sqrt { 5 } b\)
\(\\ \frac { (7+3\sqrt { 5 } )\left( 3-\sqrt { 5 } \right) }{ (3+\sqrt { 5 } )\left( 3-\sqrt { 5 } \right) } -\frac { (7+3\sqrt { 5 } )\left( 3+\sqrt { 5 } \right) }{ (3+\sqrt { 5 } )\left( 3+\sqrt { 5 } \right) }\)
\( \\ =\frac { 21-7\sqrt { 5 } +9\sqrt { 5 } -15 }{ 9-5 } -\frac { 21+7\sqrt { 5 } +9\sqrt { 5 } -15 }{ 9-5 } \)
\(\\ =a+5\sqrt { b }\)
\( \\ \frac { 6+2\sqrt { 5 } }{ 4 } -\frac { 6-2\sqrt { 5 } }{ 4 } =a+\sqrt { 5 } b\)
\(\\ =\frac { \left( 6+2\sqrt { 5 } \right) -\left( 6-2\sqrt { 5 } \right) }{ 4 } =a+\sqrt { 5 } b\)
\(\\ \frac { 4\sqrt { 5 } }{ 4 } =a+\sqrt { 5 } b\)
\(\\ \sqrt { 5 } =a+\sqrt { 5 } b\)
\(\\ a=0,b=1\)
4.
\(=\frac { 40 }{ 2\sqrt { 10 } +\sqrt { 2\times 2\times 5 } +\sqrt { 2\times 2\times 10 } -2\sqrt { 5 } } \)
\(\\ =\frac { 40 }{ 2\sqrt { 10 } +\sqrt { 20 } +\sqrt { 40 } -2\sqrt { 5 } } \)
\(\\ =\frac { 40 }{ 4\sqrt { 10 } } =\frac { 10 }{ \sqrt { 10 } } \)
\(\\ \sqrt { 10 } =3.162\)
5.
The two irrational numbers between 2 and 2.5 can be taken as
2.101001000100001...
2.201001000100001...
6.
\(\frac { 3 }{ 5 } =\frac { 3\times 8 }{ 5\times 8 } =\frac { 24 }{ 40 } \)
\(\\ \frac { 7 }{ 8 } =\frac { 7\times 5 }{ 8\times 5 } =\frac { 35 }{ 40 } \)
\(\\ \because 24<25<26<27<35\)
\(\\ \therefore \frac { 24 }{ 40 } <\frac { 25 }{ 40 } <\frac { 26 }{ 40 } <\frac { 27 }{ 40 } <\frac { 35 }{ 40 } \)
Hence, three rational numbers between \(\frac { 3 }{ 5 } \) and \(\frac { 7 }{ 8 } \) can be taken as
\(\frac { 25 }{ 40 } ,\frac { 26 }{ 40 } \) and \(\frac { 27 }{ 40 } \)
\(\frac { 5 }{ 8 } ,\frac { 13 }{ 20 } \) and \(\frac { 27 }{ 40 } \)
7.
\(\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } } \)is an irrational number.
\(\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } } =\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } \times \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } -1 \right) } } \)
\(=\sqrt { \frac { { \left( \sqrt { 2 } -1 \right) }^{ 2 } }{ 2-1 } } \)
\(=\sqrt { \frac { { \left( \sqrt { 2 } -1 \right) }^{ 2 } }{ 1 } } =\sqrt { 2 } -1\)
which is an irrational number.
Let, there is a number x such that x3 is an irrational number but x5 is a rational number.
Let, x =\(\sqrt[5]{7}\) be the number.
⇒ x3 = (5√7)3 = (7)3/5
is an irrational number.
But x5 = (\(\sqrt[5]{7}\))5=(7)5/5 = 7
=7 is a rational number.
(ii) Accepting own mistakes gracefully, co-operative learning among the classmates.
8.
\(x=3-2\sqrt { 2 } \Rightarrow \frac { 1 }{ x } =3+2\sqrt { 2 } \)
\({ \left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right) }^{ 2 }=8\)
\(\Rightarrow \sqrt { x } +\frac { 1 }{ \sqrt { x } } =\pm 2\sqrt { 2 } \)
9.
\(\frac { 1 }{ 1+{ x }^{ a-b } } +\frac { 1 }{ 1+{ x }^{ b-a } } =\frac { { x }^{ b } }{ { x }^{ b }+{ x }^{ a } } +\frac { { x }^{ a } }{ { x }^{ a }+{ x }^{ b } } \)
\(=\frac { { x }^{ b }+{ x }^{ a } }{ { x }^{ a }+{ x }^{ b } } \)
10.
\(\frac { { 3 }^{ 30 }+{ 3 }^{ 29 }+{ 3 }^{ 28 } }{ { 3 }^{ 31 }+{ 3 }^{ 30 }-{ 3 }^{ 29 } } =\frac { { 3 }^{ 38 }\left( { 3 }^{ 2 }+{ 3 }^{ 1 }+1 \right) }{ { 3 }^{ 29 }\left( { 3 }^{ 2 }+{ 3 }^{ 1 }-1 \right) } \)
\(=\frac { \left( 9+3+1 \right) }{ 3\left( 9+3-1 \right) } \)
\(=\frac { 13 }{ 3\times 11 } =\frac { 13 }{ 33 } \)
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