NCERT solution class 9 chapter 2 Polynomials exercise 2.1 mathematics

EXERCISE 2.1

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Introduction - Polynomials, Maths Class 9th

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Q 1, Ex 2.1, Page No 32, Polynomials, NCERT Maths Class 9th

 Page No 1:

Question 1(i):

1

Answer:

We have,
$(\sqrt{x^{-3}})^{5}=\left[\left(x^{-3}\right)^{x \frac{1}{2}}\right]^{-2}$

$=\left(x^{-3 / 2}\right)^{5} x^{-\frac{3 x 5}{3}}=x^{-\frac{15}{2}}$

$(\sqrt{x-3})^{5}=\frac{1}{x^{15 / 2}}$

Question 1(ii):

1

Answer:

We have,

$\sqrt{x^{3} 4^{-2}}=\left(x^{3} y^{-2}\right)^{\frac{1}{2}}$

$=x^{3 \times \frac{1}{2}} y^{-2 \times \frac{1}{2}}=x^{\frac{3}{2}} \cdot y^{-1}$

$\sqrt{x^{3} y^{-2}}=\frac{x^{3 / 2}}{y}$

Question 1(iii):

1

Answer:

We have,

$\left(x^{-2 / 3} y^{-1 / 2}\right)^{2}=x^{-2 / 8 \times 2} y^{-1 / 2 \times 2}$

$=x^{-4 / 3} y^{-1}=\frac{1}{x^{4} / 3 y}$

Question 1(iv):

1

Answer:

We have,

$(\sqrt{x})^{2 / 3} \sqrt{y^{4}} \div \sqrt{x y^{-1 / 2}}$

$=\frac{(\sqrt{x})^{-2 / 3} \sqrt{y^{4}}}{\sqrt{x y^{-1 / 2}}}$

$=\frac{x^{1 / 2}\left(\frac{-2}{3}\right)}{x^{1 / 2} y^{(-1 / 2) x{\times (1/2)}} }$

$=\frac{x^{-1 / 3} y^{2}}{x^{1 / 2} y^{-1 / 4}}=\frac{y^{2-(-1 / 4)}}{x^{1 / 3+1 / 2}}$

$=\frac{y^{9 / 4}}{x^{5 / 6}}$

Question 1(v):

1

Answer:

We have,
$\sqrt[5]{243 x^{10} y^{5} z^{10}}=\left(3^{5} \cdot x^{10} y^{5} z^{10}\right)^{1 / 5}$

$=3^{5 \times 1 / 5} x^{10 \times 1 / 5} y^{5 \times 1 / 5} z^{10 \times 1 / 5}$

$\Rightarrow \sqrt[5]{243 x^{10} y^{5} z^{10}}=3 x^{2} y z^{2}$

Question 1(vi):

1

Answer:

We have,

$\left(\frac{x^{-4}}{y^{-10}}\right)^{5 / 4}=\frac{x^{-4 \times 5 / 4}}{y^{-10 \times 5 / 4}}$

$=\frac{x^{-5}}{y^{-25 / 2}}=\frac{y^{25 / 2}}{x^{5}}$

Q 2, Ex 2.1, Page No 32, Polynomials, NCERT Class 9th Maths

Question 2(i):

2

Answer:

We have,

(16^-1/2)^5/2
=16^-1/5×5/2
=16^-1/2
=(4²)^-1/2
$=\frac{1}{4}$

Question 2(ii):

2

Answer:

We have,

$\sqrt[3]{(343)^{-2}}=\sqrt[3]{\left(7^{3}\right)^{-2}}$

$=\left(7^{3 x-2}\right)^{1 / 2}=7^{-2}$

$=\frac{1}{49}$

Question 2(iii):

2

Answer:

We have,

$(0.001)^{1 / 3}=\left(\frac{1}{1000}\right)^{1 / 3}$

$=\frac{1}{10^{3 \times 1 / 3}}=\frac{1}{10}$

=0.1

Question 2(iv):

2

Answer:

We have,
$\frac{(25)^{5 / 4} \times(243)^{3 / 5}}{(16)^{5 / 4} \times(8)^{4 / 3}}$

$=\frac{5^{2 \times 3 / 2} \times 3^{3 \times 3 / 5}}{2^{4 \times 5 / 4} \times 2^{3 \times 4 / 4}}$

$=\frac{5^{3} \times 3^{3}}{2^{5} \times 2^{4}}$


⇒$\frac{(25)^{3 / 2} \times(243)^{3 / 3}}{(16)^{5 / 4} \times(8)^{4 / 3}}$

$=\frac{125 \times 27}{2^{5+4}}=\frac{125 \times 27}{2^{9}}$

$=\frac{3375}{512}$

Question 2(v):

2

Answer:

We have,
$\left(\frac{\sqrt{2}}{5}\right)^{8} \div\left(\frac{\sqrt{2}}{5}\right)^{13}$

$=\left(\frac{\sqrt{2}}{5}\right)^{8-13}\left(\frac{\sqrt{2}}{5}\right)^{-5}$

$=\left(\frac{5}{\sqrt{2}}\right)^{5}=\frac{3125}{4 \sqrt{2}}$

Question 2(vi):

2

Answer:

We have,

$\left(\frac{5^{-1} \times 7^{2}}{5^{2} \times 7^{-4}}\right)^{7 / 2} \times\left(\frac{5^{-2} \times 7^{3}}{5^{3} \times 7^{-5}}\right)^{-5 / 2}$

$=\left(5^{-1-2} \times 7^{2-(-4)}\right)^{7 / 2} \times\left(5^{-2-3} \times 7^{3-(-5)}\right)^{-5 / 2}$

$=\left(5^{-3} \times 7^{6}\right)^{7 / 2}\left(5^{-5} \times 7^{8}\right)^{-5 / 2}$

$=\left(5^{-2 1 / 2} \times 7^{21}\right)\left(5^{25 / 2} \times 7^{-20}\right)$

$=5^{-21 / 2+25 / 2} \times 7^{21-20}=5^{2} \times 7$

=175

Q 3, Ex 2.1, Page No 32, Polynomials, Class 9th Maths NCERT

Question 3(ix):

3

Answer:

We have,
$\frac{(0.6)^{0}-(0.1)^{-1}}{\left(\frac{3}{8}\right)^{-1}\left(\frac{3}{2}\right)^{3}+\left(-\frac{1}{3}\right)^{-1}}$

$=\frac{1-\left(\frac{1}{10}\right)-1}{\left(\frac{23}{3}\right) \cdot\left(\frac{3}{2}\right)^{3}+(-3)}$

$=\frac{1-\left(\frac{1}{10}\right)-1}{\left(\frac{23}{3}\right) \cdot\left(\frac{3}{2}\right)^{3}+(-3)}$

$=\frac{1-10}{3^{3-3} \cdot 3^{3-1}-3}=\frac{-9}{3^{2}-3}$

$=\frac{3^{2}}{3(3-1)}=\frac{-3^{2-1}}{2}=\frac{-3}{2}$

Question 3(viii) :

3

Answer:

We have,
$\frac{3^{-3} \times 6^{2} \times \sqrt{98}}{5^{2} \times \sqrt[3]{1 / 25} \times(15)^{-4 / 3 }\times 3^{1/3}}$

$=\frac{3^{-3} \times 3^{2} \times 2^{2} \times 2^{1 / 2} \cdot 7^{2 \times 1 / 2}}{5^{2} \times\left(\frac{1}{5}\right)^{2 \times \frac{1}{3}} \times 3^{-4 / 3} \times 5^{-4 / 3} \times 3^{1 / 3}}$

$=\frac{3^{-3+2+\frac{4}{3}-\frac{1}{3}} \times 2^{2+\frac{1}{2} \times 7}}{5^{2-\frac{2}{3}-\frac{4}{3}} }$

$=\frac{3^{0} \times 2^{5 / 2} \times 7}{5^{0}}$

$=7 \times \sqrt{25}=7 \times 4 \sqrt{2}=28 \sqrt{2}$

Question 3(vii):

3

Answer:

We have,

$\left(\frac{64}{125}\right)^{-2 / 3}+\frac{1}{\left(\frac{256}{625}\right)^{1 / 4}}+\frac{\sqrt{25}}{\sqrt[3]{64}}$

$=\left(\frac{4}{5}\right)^{3 \times \frac{-2}{3}}=+\frac{1}{\left(\frac{4}{5}\right)^{4 \times \frac{1}{4}}}+\frac{\left(5^{2}\right)^{1 / 2}}{\left(4^{3}\right)^{1 / 3}}$

$=\left(\frac{4}{5}\right)^{-2}+\frac{1}{\left(\frac{4}{5}\right)}+\frac{5}{4}$

$=\left(\frac{4}{5}\right)^{2}+\frac{5}{4}+\frac{5}{4}$

$=\frac{25+20+20}{16}=\frac{65}{16}$

Question 3(vi):

3

Answer:

We have,
$\frac{2^{n}+2^{n-1}}{2^{n+1}-2^{n}}$

$=\frac{2^{n-1}(2+1)}{2^{n-1}\left(2^{2}-2\right)}=\frac{3}{4-2}=\frac{3}{2}$

Question 3(v):

3

Answer:

Prove that:
$\sqrt{\frac{1}{4}}+(0.01)^{-1 / 2}(27)^{2 / 3}=\frac{3}{2}$

We have,
$=\left(\frac{1}{2^{2}}\right)^{\frac{1}{2}}+\left(\frac{1}{100}\right)^{-1 / 2}-(3)^{3 \times \frac{2}{3}}$

$=\left(\frac{1}{2^{2 \times \frac{1}{2}}}\right)+\left(\frac{1}{10}\right)^{2 \times-\frac{1}{2}}-3^{2}$

$=\frac{1}{2}+\left(\frac{1}{10}\right)^{-1}-9$

$=\frac{1}{2}+10-9=\frac{1}{2}+1=\frac{3}{2}$

Question 3(iv):

3

Answer:

We have,
$\frac{2^{1 / 2} \times 3^{1 / 3} \times 4^{1 / 4}}{10^{-1 / 5} \times 5^{3 / 5}} \div \frac{3^{4 / 3} \times 5^{-7 / 5}}{4^{-3 / 5} \times 6}$

$=\frac{2^{1 / 2} \times 3^{1 / 3} \times 2^{2 \times 1 / 3}}{2^{-1 / 5} \times 5^{-1 / 5} \times 5^{3 / 5}} \div \frac{4^{4 / 3} \times 5^{-7 / 5}}{2^{2 x-3 / 5} \times 3 \times 2}$

$=\frac{2^{1 / 2+1 / 5+1 / 2} \times 3^{1 / 3}}{5^{-1 / 5+3 / 5}} \div \frac{3^{4 / 3-1} \times 5^{-7 / 5}}{2^{-6 / 5+1}}$

$=\frac{2^{12 / 10} \times 3^{1 / 3}}{5^{2 / 5}} \div \frac{3^{1 / 3} \times 5^{-7 / 5}}{2^{-1 / 5}}$

$=\frac{2^{12\times 10} \times 3^{1 / 3}}{5^{2 / 5}} \times \frac{2^{-1 / 5}}{3^{1 / 3} \times 5^{-7 / 5}}$

$=\frac{2^{\frac{10}{10}-\frac{1}{5}} \times 3^{1 / 3-1 / 3}}{5^{2 / 5-7 / 5}}=\frac{2^{1}}{5^{-1}}$

=2×5

=10

Question 3(iii):

3

Answer:

We have,
$\left(\frac{1}{4}\right)^{-2}-3 \times 8^{2 / 3} \times 4^{0}+\left(\frac{9}{16}\right)^{-1 / 2}$

$=\left(\frac{1}{2^{2}}\right)^{-2}-3 \times 2^{3 \times 2 / 3} \times 1+\left(\frac{3}{4}\right)^{2 \times-1 / 2}$

$=\left(\frac{1}{2}\right)^{-4}-3 \times 2^{2}+\left(\frac{3}{4}\right)^{-1}$

$=2^{4}-12+\frac{4}{3}$

$=16-12+\frac{4}{3}$

$=\frac{12+4}{3}=\frac{16}{3}$

Question 3(ii):

3

Answer:

We have,
$9^{3 / 2}-3 \times 5^{0}-\left(\frac{1}{81}\right)^{-1 / 2}$

$=3^{3 \times 3 / 2}-3 \times 1-\left(\frac{1}{9^{2 \times-1 / 2}}\right)$

$=3^{3}-3-\frac{1}{3-2}$

$=3^{3}-3-3^{2}$

=27-12=15

Question 3(i):

3

Answer:

$\sqrt{3 \times 5^{-3}} \div \sqrt[3]{3^{-1}} \sqrt{5} \times \sqrt[6]{3 \times 5^{-4}}=\frac{3}{5}$

We have,

$3^{1 / 2} \times 5^{-3 / 2} \div 3^{-1 \times \frac{1}{3}} \cdot\left(5^{1 / 2}\right)^{1 / 3} \times 3^{1 / 6} \times\left(5^{4}\right)^{1 / 6}$

$=3^{1 / 2+1 / 3} \times 5^{-3 / 2-1 / 6} \times 3^{1 / 6} \times 5^{2 / 3}$

$=3^{5 / 6} \times 5^{-10 / 6} \times 3^{1 /6} \times 5^{2 / 3}$

$=3^{5 / 6+1 /6} \times 5^{-5 / 3+2 / 3}$

$=3^{9 / 6} \times 5^{-3 / 3}$

$=3^{1} \times 5^{-1}=\frac{3}{5}$

Q 4, Ex 2.1, Page No 32, Polynomials, Maths Class 9th

Question 4:

4

Answer:

We have,
⇒$27^{x}=\frac{9}{3^{x}}$
⇒$\left(3^{3}\right)^{x}=\frac{3^{2}}{3^{x}}$
⇒$3^{3 x}=3^{2-x}$
⇒3x=2-x
⇒4x=2

Q 5, Ex 2.1, Page No 32, Polynomials, Maths Class 9th

Question 5(i):

5

Answer:

We have,

$2^{5 x} \div 2^{x}=\sqrt[5]{2^{20}}$

$\Rightarrow 2^{5 x-x}=(2)^{20 \times \frac{1}{5}}$

$\Rightarrow 2^{4 x}=2^{4}$

⇒4x=4

⇒x=1

Question 5(ii):

5

Answer:

We have,

⇒(2³)²=(2²)x

⇒2¹²=2^2x
⇒2x=12
⇒x=6

Question 5(iii):

5

Answer:

We have,

$\left(\frac{3}{5}\right)^{x}\left(\frac{5}{3}\right)^{2 x}=\frac{125}{27}$

⇒$\left(\frac{5}{3}\right)^{-x}\left(\frac{5}{3}\right)^{2 x}=\left(\frac{5}{3}\right)^{3}$

⇒$\left(\frac{5}{3}\right)^{-x+2 x}=\left(\frac{5}{3}\right)^{3}$

⇒$\left(\frac{5}{3}\right)^{x}=\left(\frac{5}{3}\right)^{3} \Rightarrow x=3$

Question 5(iv):

5

Answer:

We have,

$5^{x-2} \times 3^{2 x-3}=5 \times 3^{3}$
⇒x-2=1,2x-3=3
⇒x=3,2x=6
⇒x=3

Question 5(v):

5

Answer:

We have,

⇒2^x-5.5^x-4=2⁰×5¹

⇒x-5=0 and x-4=1

⇒x=5

Question 5(vi):

5

Answer:

We have,
2^x-7.5^x-4=1250

⇒2^x-7.5^x-4=2¹×625

⇒2^x-7.5^x-4=2¹×5⁴

⇒x-7=1 and x-4=4

⇒x=8

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