class:9-10( অনুশীলনী:৩.১,৩.২,৩.৩,৩.৪-বহুনির্বাচনী(1)

Class:9-10_math_img

1) \displaystyle x^{2} +\frac{1}{x^{2}} =18 হলে, \displaystyle x+\frac{1}{x} =কত?
দেওয়া আছে,
\displaystyle x^{2} +\frac{1}{x^{2}} =18
\displaystyle \Longrightarrow \left( x+\frac{1}{x}\right)^{2} -2.x.\frac{1}{x} =18
\displaystyle \Longrightarrow \left( x+\frac{1}{x}\right)^{2} =18+2
\displaystyle \Longrightarrow \left( x+\frac{1}{x}\right)^{2} =20
\displaystyle \Longrightarrow \sqrt{\left( x+\frac{1}{x}\right)^{2}} =\pm \sqrt{4.5}
\displaystyle \Longrightarrow x+\frac{1}{x} =\pm 2\sqrt{5}

Class:9-10_math_img

2) \displaystyle x^{2} =3+2\sqrt{2} হলে, \displaystyle \frac{1}{x} এর মান কত?
দেওয়া আছে,
\displaystyle x^{2} =3+2\sqrt{2}
\displaystyle \Longrightarrow x^{2} =2+2\sqrt{2} +1
\displaystyle \Longrightarrow x^{2} =\left(\sqrt{2}\right)^{2} +2.\sqrt{2} .1+1^{2}
\displaystyle \Longrightarrow x^{2} =\left(\sqrt{2} +1\right)^{2}
\displaystyle \Longrightarrow \sqrt{x^{2}} =\sqrt{\left(\sqrt{2} +1\right)^{2}}
\displaystyle \Longrightarrow \ x=\sqrt{2} +1
এখন,
\displaystyle \frac{1}{x} =\frac{1}{\sqrt{2} +1}
\displaystyle =\frac{1.\left(\sqrt{2} -1\right)}{\left(\sqrt{2} +1\right)\left(\sqrt{2} -1\right)}
\displaystyle =\frac{\sqrt{2} -1}{\left(\sqrt{2}\right)^{2} -( 1)^{2}}
\displaystyle =\frac{\sqrt{2} -1}{2-1}
\displaystyle \frac{1}{x} =\sqrt{2} -1

3) \displaystyle a+\frac{1}{a} =2 হলে, a এর মান নির্নয় কর।
দেওয়া আছে,,
\displaystyle a+\frac{1}{a} =2
\displaystyle \Longrightarrow \frac{a^{2} +1}{a} =2
\displaystyle \Longrightarrow a^{2} +1=2a
\displaystyle \Longrightarrow a^{2} -2a+1=0
\displaystyle \Longrightarrow ( a-1)^{2} =0
\displaystyle \Longrightarrow a-1=0
\displaystyle \Longrightarrow a=1

Class:9-10_math_img

4) \displaystyle a^{2} -5a-1=0 হলে,
i) \displaystyle a^{2} +\frac{1}{a^{2}} এর মান কত?
ii) \displaystyle a+\frac{1}{a} এর মান কত?
দেওয়া আছে,
\displaystyle a^{2} -5a-1=0
\displaystyle \Longrightarrow a^{2} -1=5a
\displaystyle \Longrightarrow a\left( a-\frac{1}{a}\right) =5a
\displaystyle \Longrightarrow a-\frac{1}{a} =5
এখন, \displaystyle a^{2} +\frac{1}{a^{2}} =\left( a-\frac{1}{a}\right)^{2} +2.a.\frac{1}{a}
\displaystyle =( 5)^{2} +2
\displaystyle =25+2
\displaystyle =27
আবার,
\displaystyle \left( a+\frac{1}{a}\right)^{2} =\left( a-\frac{1}{a}\right)^{2} +4.a.\frac{1}{a}
\displaystyle \Longrightarrow \sqrt{\left( a+\frac{1}{a}\right)^{2}} =\sqrt{\left( a-\frac{1}{a}\right)^{2} +4.a.\frac{1}{a}}
\displaystyle \Longrightarrow a+\frac{1}{a} =\sqrt{( 5)^{2} +4}
\displaystyle =\sqrt{25+4}
\displaystyle =\sqrt{29}

Class:9-10_math_img.

5) \displaystyle x=3+2\sqrt{2} হলে,
i) \displaystyle x+\frac{1}{x} এর মান কত?
ii) \displaystyle x^{2} +\frac{1}{x^{2}} এর মান কত?
দেওয়া আছে,,
\displaystyle x=3+2\sqrt{2}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{1}{3+2\sqrt{2}}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left( 3-2\sqrt{2}\right)}{\left( 3+2\sqrt{2}\right)\left( 3-2\sqrt{2}\right)}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left( 3-2\sqrt{2}\right)}{( 3)^{2} +\left( 2\sqrt{2}\right)^{2} \ }

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left( 3-2\sqrt{2}\right)}{9+( 2)^{2} .\left(\sqrt{2}\right)^{2}}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left( 3-2\sqrt{2}\right)}{9+4.2}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left( 3-2\sqrt{2}\right)}{9+8}

\displaystyle \Longrightarrow \frac{1}{x} =3-2\sqrt{2}

এখন,

\displaystyle x+\frac{1}{x} =3+2\sqrt{2} +3-2\sqrt{2}

\displaystyle =6

Class:9-10_math_img

5) \displaystyle x=3+2\sqrt{2} হলে,
i) \displaystyle x+\frac{1}{x} এর মান কত?
ii) \displaystyle x^{2} +\frac{1}{x^{2}} এর মান কত?
দেওয়া আছে,,
\displaystyle x=3+2\sqrt{2}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{1}{3+2\sqrt{2}}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left( 3-2\sqrt{2}\right)}{\left( 3+2\sqrt{2}\right)\left( 3-2\sqrt{2}\right)}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left( 3-2\sqrt{2}\right)}{( 3)^{2} +\left( 2\sqrt{2}\right)^{2} \ }

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left( 3-2\sqrt{2}\right)}{9+( 2)^{2} .\left(\sqrt{2}\right)^{2}}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left( 3-2\sqrt{2}\right)}{9+4.2}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left( 3-2\sqrt{2}\right)}{9+8}

\displaystyle \Longrightarrow \frac{1}{x} =3-2\sqrt{2}

এখন,
\displaystyle x+\frac{1}{x} =3+2\sqrt{2} +3-2\sqrt{2}

\displaystyle =6

আবার,

\displaystyle \left( x^{2} +\frac{1}{x^{2}}\right) =\left( x+\frac{1}{x}\right)^{2} -2.x.\frac{1}{x} =( 6)^{2} -2=36-2=34

Class:9-10_math_img.

6) \displaystyle y=\sqrt{6} +\sqrt{5} হলে,
i) \displaystyle \frac{1}{y} এর মান কত?
ii) \displaystyle y^{2} এর মান কত?
দেওয়া আছে,
\displaystyle \ y=\sqrt{6} +\sqrt{5}

\displaystyle \Longrightarrow \frac{1}{y} =\frac{1}{\ \sqrt{6} +\sqrt{5}}

\displaystyle \Longrightarrow \frac{1}{y} =\frac{\left(\sqrt{6} -\sqrt{5}\right)}{\left( \ \sqrt{6} +\sqrt{5}\right)\left(\sqrt{6} -\sqrt{5}\right)}

\displaystyle \Longrightarrow \frac{1}{y} =\frac{\left(\sqrt{6} -\sqrt{5}\right)}{\left( \ \sqrt{6}\right)^{2} -\left(\sqrt{5}\right)^{2}}

\displaystyle \Longrightarrow \frac{1}{y} =\frac{\left(\sqrt{6} -\sqrt{5}\right)}{6-5}

\displaystyle \Longrightarrow \frac{1}{y} =\frac{\left(\sqrt{6} -\sqrt{5}\right)}{1} =\left(\sqrt{6} -\sqrt{5}\right)

আবার,\displaystyle \left(\sqrt{6} +\sqrt{5}\right)^{2} =\left(\sqrt{6}\right)^{2} +2.\sqrt{6} .\sqrt{5} +\left(\sqrt{5}\right)^{2}

\displaystyle =6+2\sqrt{30} +5

\displaystyle =11+2\sqrt{30}

Class:9-10_math_img

7) \displaystyle p^{2} =3p+1 হলে,
i) \displaystyle p^{2} -\frac{1}{p^{2}} এর মান কত?
\displaystyle ii) \ p^{4} -\frac{1}{p^{4}} এর মান কত?
দেওয়া আছে,,
\displaystyle p^{2} =3p+1
\displaystyle \Longrightarrow p^{2} -1=3p
\displaystyle \Longrightarrow p\left( p-\frac{1}{p}\right) =3p
\displaystyle \Longrightarrow p-\frac{1}{p} =3
\displaystyle \Longrightarrow \left( p+\frac{1}{p}\right)^{2} =\left( p-\frac{1}{p}\right)^{2} +4.p.\frac{1}{p}
\displaystyle \Longrightarrow \left( p+\frac{1}{p}\right)^{2} =( 3)^{2} +4
\displaystyle \Longrightarrow \sqrt{\left( p+\frac{1}{p}\right)^{2}} =\sqrt{( 3)^{2} +4}
\displaystyle \Longrightarrow p+\frac{1}{p} =\sqrt{9+4}
\displaystyle \Longrightarrow p+\frac{1}{p} =\sqrt{13}
এখন,\displaystyle p^{2} -\frac{1}{p^{2}} =\left( p+\frac{1}{p}\right)\left( p-\frac{1}{p}\right) =\sqrt{13} .3=3\sqrt{13}

Class:9-10_math_img

7) \displaystyle p^{2} =3p+1 হলে,
i) \displaystyle p^{2} -\frac{1}{p^{2}} এর মান কত?
\displaystyle ii) \ p^{4} -\frac{1}{p^{4}} এর মান কত?
ii)
i) হইতে পাই,
\displaystyle p+\frac{1}{p} =\sqrt{13} \ এবং\ p^{2} -\frac{1}{p^{2}} =3\sqrt{13} \
সুতরাং, \displaystyle p^{2} +\frac{1}{p^{2}} =\left( p+\frac{1}{p}\right)^{2} -2.p.\frac{1}{p}
\displaystyle =\left(\sqrt{13}\right)^{2} -2=13-2=11

এখন,
\displaystyle p^{4} -\frac{1}{p^{4}} =\left( p^{2} +\frac{1}{p^{2}}\right) .\left( p^{2} -\frac{1}{p^{2}}\right)
\displaystyle =11.3\sqrt{13}
\displaystyle =33\sqrt{13}

Class:9-10_math_img

8) \displaystyle x^{2} -5-2\sqrt{6} =0 হলে,
\displaystyle i) \ x\ এর\ মান\ কত?
\displaystyle ii) \ x-\frac{1}{x} \ এর\ মান\ কত?
দেওয়া আছে,
\displaystyle i) \ x^{2} -5-2\sqrt{6} =0

\displaystyle \Longrightarrow x^{2} =5+2\sqrt{6}

\displaystyle \Longrightarrow x^{2} =3+2\sqrt{6} +2

\displaystyle \Longrightarrow x^{2} =\left(\sqrt{3}\right)^{2} +2.\sqrt{3} .\sqrt{2} +\left(\sqrt{2}\right)^{2}

\displaystyle \Longrightarrow x^{2} =\left(\sqrt{3} +\sqrt{2}\right)^{2}

\displaystyle \Longrightarrow \sqrt{x^{2}} =\sqrt{\left(\sqrt{3} +\sqrt{2}\right)^{2} \ } \ \ \displaystyle উভয়\ পক্ষ\sqrt{} নিয়ে\ \ পাই

\displaystyle \Longrightarrow x=\sqrt{3} +\sqrt{2}

Class:9-10_math_img

8) \displaystyle x^{2} -5-2\sqrt{6} =0 হলে,
\displaystyle i) \ x\ এর\ মান\ কত?
\displaystyle ii) \ x-\frac{1}{x} \ এর\ মান\ কত?
i) নং হইতে পাই,
\displaystyle x=\sqrt{3} +\sqrt{2}
\displaystyle \Longrightarrow \frac{1}{x} =\frac{1}{\sqrt{3} +\sqrt{2}}
\displaystyle \Longrightarrow \frac{1}{x} =\frac{1.\left(\sqrt{3} -\sqrt{2}\right)}{\left(\sqrt{3} +\sqrt{2}\right)\left(\sqrt{3} -\sqrt{2}\right)}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left(\sqrt{3} -\sqrt{2}\right)}{\left(\sqrt{3}\right)^{2} -\left(\sqrt{2}\right)^{2}}

\displaystyle \Longrightarrow \frac{1}{x} =\frac{\left(\sqrt{3} -\sqrt{2}\right)}{3-2}

\displaystyle \Longrightarrow \frac{1}{x} =\sqrt{3} -\sqrt{2}

এখন,\displaystyle x-\frac{1}{x} =\left(\sqrt{3} +\sqrt{2}\right) -\left(\sqrt{3} -\sqrt{2}\right)

\displaystyle =\sqrt{3} +\sqrt{2} -\sqrt{3} +\sqrt{2}

\displaystyle =2\sqrt{2}

Class:9-10_math_img

9 ) \displaystyle a^{2} -3a+1=0 হলে,
\displaystyle i) \ a^{2} +\frac{1}{a^{2}} \ এর মান নির্নয় কর।
\displaystyle ii) \ a^{2} -\frac{1}{a^{2}} \ =কত?
i) দেওয়া আছে,

\displaystyle a^{2} -3a+1=0

\displaystyle \Longrightarrow a^{2} +1=3a

\displaystyle \Longrightarrow a\left( a+\frac{1}{a}\right) =3a

\displaystyle \Longrightarrow a+\frac{1}{a} =3

এখন,
\displaystyle a^{2} +\frac{1}{a^{2}}

\displaystyle =\left( a+\frac{1}{a}\right)^{2} -2.a.\frac{1}{a}

\displaystyle =( 3)^{2} -2

\displaystyle =9-2

\displaystyle =7

Class:9-10_math_img

9 ) \displaystyle a^{2} -3a+1=0 হলে,
\displaystyle i) \ a^{2} +\frac{1}{a^{2}} \ এর মান নির্নয় কর।
\displaystyle ii) \ a^{2} -\frac{1}{a^{2}} \ =কত?
(ii)
i) নং হতে পাই,
\displaystyle a+\frac{1}{a} =3

এখন,\displaystyle \left( a-\frac{1}{a}\right)^{2} =\left( a+\frac{1}{a}\right)^{2} -4.a.\frac{1}{a}

\displaystyle \Longrightarrow \left( a-\frac{1}{a}\right)^{2} =( 3)^{2} -4

\displaystyle \Longrightarrow \left( a-\frac{1}{a}\right)^{2} =9-4

\displaystyle \Longrightarrow \left( a-\frac{1}{a}\right)^{2} =5

\displaystyle \Longrightarrow \sqrt{\left( a-\frac{1}{a}\right)^{2}} =\sqrt{5}

\displaystyle \Longrightarrow a-\frac{1}{a} =\sqrt{5}

সুতরং, \displaystyle a^{2} -\frac{1}{a^{2}} =\left( a+\frac{1}{a}\right) .\left( a-\frac{1}{a}\right) =3\sqrt{5}

Class:9-10_math_img

10) \displaystyle p+q=\sqrt{3} এবং \displaystyle p^{2} -q^{2} =\sqrt{6} হলে,
i) \displaystyle pq এর মান বের কর।
\displaystyle ii) \ p^{2} +q^{2} এর মান বের কর।
i) দেওয়া আছে,
\displaystyle p^{2} -q^{2} =\sqrt{6}

\displaystyle \Longrightarrow ( p+q)( p-q) =\sqrt{6}

\displaystyle \Longrightarrow \sqrt{3}( p-q) =\sqrt{6}

\displaystyle \Longrightarrow p-q=\frac{\sqrt{6}}{\sqrt{3}}

\displaystyle \Longrightarrow p-q=\frac{\sqrt{2} .\sqrt{3}}{\sqrt{3}}

\displaystyle \Longrightarrow p-q=\sqrt{2}

এখন,
\displaystyle pq=\left(\frac{p+q}{2}\right)^{2} -\left(\frac{p-q}{2}\right)^{2}

\displaystyle =\left(\frac{\sqrt{3}}{2}\right)^{2} -\left(\frac{\sqrt{2}}{2}\right)^{2}

\displaystyle =\frac{\left(\sqrt{3}\right)^{2}}{2^{2}} -\frac{\left(\sqrt{2}\right)^{2}}{2^{2}}

\displaystyle =\frac{3}{4} -\frac{2}{4}

\displaystyle =\frac{3-2}{4}

\displaystyle =\frac{1}{4}

Class:9-10_math_img

10) \displaystyle p+q=\sqrt{3} এবং \displaystyle p^{2} -q^{2} =\sqrt{6} হলে,
i) \displaystyle pq এর মান বের কর।
\displaystyle ii) \ p^{2} +q^{2} এর মান বের কর।

(ii) i) নং হতে পাই,
\displaystyle p+q=\sqrt{3}
\displaystyle p-q=\sqrt{2}
আমরা জানি,
\displaystyle 2\left( p^{2} +q^{2}\right) =( p+q)^{2} +( p-q)^{2}

\displaystyle \Longrightarrow p^{2} +q^{2} =\frac{( p+q)^{2} +( p-q)^{2}}{2}

\displaystyle \Longrightarrow p^{2} +q^{2} =\frac{\left(\sqrt{3}\right)^{2} +\left(\sqrt{2}\right)^{2}}{2}

\displaystyle \Longrightarrow p^{2} +q^{2} =\frac{3+2}{2}

\displaystyle \Longrightarrow p^{2} +q^{2} =\frac{5}{2}

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