Class:9-10( অনুশীলনি:৩.১,৩.২,৩.৩,৩.৪-এর সৃজনশীল(2)।

(1)
i) \displaystyle xএর চতুর্ঘাত থেকে \displaystyle x এর বর্গের বিয়োগফল \displaystyle -4

ii) \displaystyle m এবং \displaystyle m গুনাত্তক বিপরীত সংখ্যার যোগফল \displaystyle 5

ক) \displaystyle \sqrt{m} +\frac{1}{\sqrt{m}} এর মান নির্নয় কর।
খ) প্রমান কর যে, \displaystyle \left(\sqrt{m} -\frac{1}{\sqrt{m}}\right)\left( m^{2} +\frac{1}{m^{2}}\right) =23\sqrt{3}
গ) দেখাও যে, \displaystyle \frac{x^{4}}{x^{8} -x^{4} +16} =-\frac{1}{8}

ক)

উদ্দীপকের (ii) নংএর তথ্যানুসারে,,

\displaystyle m+\frac{1}{m} =5

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

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

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

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

\displaystyle \Longrightarrow \left(\sqrt{m} +\frac{1}{\sqrt{m}}\right)^{2} =7

\displaystyle \Longrightarrow \sqrt{\left(\sqrt{m} +\frac{1}{\sqrt{m}}\right)^{2}} =\pm \sqrt{7}

\displaystyle \Longrightarrow \sqrt{m} +\frac{1}{\sqrt{m}} =\pm \sqrt{7}

কিন্তু \displaystyle \sqrt{m} +\frac{1}{\sqrt{m}} এর মান ঋনাক্তক হতে পারে না ।

সুতরাং, \displaystyle \sqrt{m} +\frac{1}{\sqrt{m}} =\sqrt{7}

খ)

উদ্দীপক অনুসারে,
\displaystyle m+\frac{1}{m} =5

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

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

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

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

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

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

\displaystyle \Longrightarrow \sqrt{m} -\frac{1}{\sqrt{m}} =\sqrt{3}

[ \displaystyle \sqrt{m} -\frac{1}{\sqrt{m}} এর মান ঋনাত্তক হতে পারে না ]
আবার, \displaystyle m^{2} +\frac{1}{m^{2}} =\left( m+\frac{1}{m}\right)^{2} 2.m.\frac{1}{m}
\displaystyle =( 5)^{2} -2

\displaystyle =25-2

\displaystyle =23

গ)

উদ্দীপকের (i) নং তথ্যঅনুযায়ী,
\displaystyle x^{4} -x^{2} =-4

\displaystyle \Longrightarrow x^{4} +4=x^{2}

\displaystyle \Longrightarrow \frac{x^{4}}{x^{2}} +\frac{4}{x^{2}} =\frac{x^{2}}{x^{2}} [উভয়পক্ষ \displaystyle x^{2} দ্বারা ভাগ করে পাইা ]

\displaystyle \Longrightarrow x^{2} +\frac{4}{x^{2}} =1

\displaystyle \Longrightarrow \left( x^{2} +\frac{4}{x^{2}}\right)^{2} =1^{2} [ বর্গ করে পাই ]

\displaystyle \Longrightarrow \left( x^{2}\right)^{2} +2.x^{2} .\frac{4}{x^{2}} +\left(\frac{4}{x^{2}}\right)^{2} =1

\displaystyle \Longrightarrow x^{4} +8+\frac{16}{x^{4}} =1

\displaystyle \Longrightarrow x^{4} +\frac{16}{x^{4}} =1-8

\displaystyle \Longrightarrow x^{4} +\frac{16}{x^{4}} =-7
এখন,
\displaystyle \frac{x^{4}}{x^{8} -x^{4} +16}

\displaystyle =\frac{x^{4}}{x^{4}\left( x^{4} -1+\frac{16}{x^{4}}\right)}

\displaystyle =\frac{1}{x^{4} +\frac{16}{x^{4}} -1}

\displaystyle =\frac{1}{-7-1}

\displaystyle =-\frac{1}{8} ( দেখানো হলো )

2) \displaystyle i) \ 2a=\sqrt{5} +\sqrt{3} এবং \displaystyle 2b=\sqrt{5} -\sqrt{3}

\displaystyle ii) \ p=\sqrt{\frac{5p}{2} +\frac{1}{6}}

ক) দেখাও যে, \displaystyle \frac{1}{2a} =b

খ) \displaystyle 9p^{2} +\frac{1}{4p^{2}} এর মান নির্নয় কর ।

গ) প্রমান কর যে, \displaystyle ab\left( a^{4} -b^{4}\right) =2\sqrt{15}

ক)

দেওয়া আছে,,

\displaystyle 2a=\sqrt{5} +\sqrt{3} এবং \displaystyle 2b=\sqrt{5} -\sqrt{3}

এখন, \displaystyle \frac{1}{2a}

\displaystyle =\frac{1}{\sqrt{5} +\sqrt{3}}

\displaystyle =\frac{\left(\sqrt{5} -\sqrt{3}\right)}{\left(\sqrt{5} +\sqrt{3}\right)\left(\sqrt{5} -\sqrt{3}\right)}

\displaystyle =\frac{\left(\sqrt{5} -\sqrt{3}\right)}{\left(\sqrt{5}\right)^{2} -\left(\sqrt{3}\right)^{2}}

\displaystyle =\frac{\sqrt{5} -\sqrt{3}}{5-3}

\displaystyle =\frac{\sqrt{5} -\sqrt{3}}{2}

\displaystyle =\frac{2b}{2}

\displaystyle =b

সুতরং, \displaystyle \frac{1}{2a} =b (দেখানো হলো)

খ)

দেওয়া আছে,

\displaystyle p=\sqrt{\frac{5p}{2} +\frac{1}{6}}

\displaystyle \Longrightarrow p^{2} =\frac{5p}{2} +\frac{1}{6} [ বর্গ করে পাই ]

\displaystyle \Longrightarrow p^{2} =\frac{15p+1}{6}

\displaystyle \Longrightarrow 6p^{2} =15p+1

\displaystyle \Longrightarrow 6p^{2} -1=15p

\displaystyle \Longrightarrow \frac{6p^{2}}{2p} -\frac{1}{2p} =\frac{15p}{2p} [উভয় পক্ষ \displaystyle 2p দ্বারা ভাগ করে পাই ]

\displaystyle \Longrightarrow 3p-\frac{1}{2p} =\frac{15}{2}

\displaystyle \Longrightarrow \left( 3p-\frac{1}{2p}\right)^{2} =\left(\frac{15}{2}\right)^{2} [উভয় পক্ষ বর্গ করে পাই ]

\displaystyle \Longrightarrow ( 3p)^{2} -2.3p.\frac{1}{2p} +\left(\frac{1}{2p}\right)^{2} =\frac{( 15)^{2}}{2^{2}}

\displaystyle \Longrightarrow 9p^{2} -3+\frac{1}{4p^{2}} =\frac{225}{4}

\displaystyle \Longrightarrow 9p^{2} +\frac{1}{4p^{2}} =\frac{225}{4} +3

\displaystyle \Longrightarrow 9p^{2} +\frac{1}{4p^{2}} =\frac{225+12}{4}

\displaystyle \Longrightarrow 9p^{2} +\frac{1}{4p^{2}} =\frac{237}{4}

গ)


দেওয়া আছে,

\displaystyle 2a=\sqrt{5} +\sqrt{3} এবং \displaystyle 2b=\sqrt{5} -\sqrt{3}

এখন,
\displaystyle 2a+2b=\sqrt{5} +\sqrt{3} +\sqrt{5} -\sqrt{3}

\displaystyle \Longrightarrow 2( a+b) =2\sqrt{5}

\displaystyle \Longrightarrow a+b=\frac{2\sqrt{5}}{2}

\displaystyle \Longrightarrow a+b=\sqrt{5}

আবার,

\displaystyle 2a-2b=\left(\sqrt{5} +\sqrt{3}\right) -\left(\sqrt{5} -\sqrt{3}\right)

\displaystyle \Longrightarrow 2( a-b) =\left(\sqrt{5} +\sqrt{3} -\sqrt{5} +\sqrt{3}\right)

\displaystyle \Longrightarrow 2( a-b) =2\sqrt{3}

\displaystyle \Longrightarrow a-b=\frac{2\sqrt{3}}{2}

\displaystyle \Longrightarrow a-b=\sqrt{3}

বামপক্ষ\displaystyle =ab\left( a^{4} -b^{4}\right)

\displaystyle =ab\left{\left( a^{2}\right)^{2} -\left( b^{2}\right)^{2}\right}

\displaystyle =ab\left( a^{2} +b^{2}\right)\left( a^{2} -b^{2}\right)

\displaystyle =\frac{1}{8} .4ab.2\left( a^{2} +b^{2}\right)( a+b)( a-b)

\displaystyle =\frac{1}{8} .\left{( a+b)^{2} -( a-b)^{2}\right}\left{( a+b)^{2} +( a-b)^{2}\right} .\sqrt{5} .\sqrt{3}

\displaystyle =\frac{1}{8} .\left{\left(\sqrt{5}\right)^{2} -\left(\sqrt{3}\right)^{2}\right} .{\left(\sqrt{5}\right)^{2} +\left(\sqrt{3}\right)^{2} .\sqrt{15}

\displaystyle =\frac{1}{8} .( 5-3)( 5+3) .\sqrt{15}

\displaystyle =\frac{1}{8} .2.8.\sqrt{15}

\displaystyle =2\sqrt{15}
= ডানপক্ষ
( প্রমানিত )

i) একটি সংখ্যা p এর বর্গের চারগুন সংখ্যাটির দুই গুন অপেক্ষা 1 কম ।

ii) \displaystyle h=\sqrt{13} +2\sqrt{3}

ক) \displaystyle ( 5a+3b)( 3a+4c) কে দুইটি বর্গের অন্তরফলরূপে প্রকাশ কর।

খ) প্রমান কর যে, \displaystyle \frac{13h}{h^{2} -\sqrt{13} h+1} =\sqrt{13}

গ) দেখাও যে, \displaystyle 16\left( p^{4} +\frac{1}{256p^{4}}\right) =-1

ক)

ধরি, \displaystyle 5a+4b=x

\displaystyle 3a+4c=y

আমরা জানি,

\displaystyle xy=\left(\frac{x+y}{2}\right)^{2} -\left(\frac{x-y}{2}\right)^{2}

\displaystyle =\left(\frac{5a+4b+3a+4c}{2}\right)^{2} -\left(\frac{5a+4b-3a-4c}{2}\right)^{2}

\displaystyle =\left(\frac{8a+4b+4c}{2}\right)^{2} -\left(\frac{2a+4b-4c}{2}\right)^{2}

\displaystyle =\left(\frac{2( 4a+2b+2c)}{2}\right)^{2} -\left(\frac{2( a+2b-2c)}{2}\right)^{2}

\displaystyle =( 4a+2b+2c)^{2} -( a+2b-2c)^{2}

খ)

দেওয়া আছে,,

\displaystyle h=\sqrt{13} +2\sqrt{3}

\displaystyle \frac{1}{h} =\frac{1}{\sqrt{13} +2\sqrt{3}}

\displaystyle =\frac{1.\left(\sqrt{13} -2\sqrt{3}\right)}{\left(\sqrt{13} +2\sqrt{3}\right) .\left(\sqrt{13} -2\sqrt{3}\right)}

\displaystyle =\frac{\left(\sqrt{13} -2\sqrt{3}\right)}{\left(\sqrt{13}\right)^{2} -\left( 2\sqrt{3}\right)^{2}}

\displaystyle =\frac{\left(\sqrt{13} -2\sqrt{3}\right)}{13-( 2)^{2} .\left(\sqrt{3}\right)^{2}}

\displaystyle =\frac{\sqrt{13} -2\sqrt{3}}{13-4.3}

\displaystyle =\frac{\sqrt{13} -2\sqrt{3}}{13-12}

\displaystyle =\frac{\sqrt{13} -2\sqrt{3}}{1}

\displaystyle \frac{1}{h} =\sqrt{13} -2\sqrt{3}

আবার, \displaystyle h+\frac{1}{h} =\sqrt{13} +2\sqrt{3} +\sqrt{13} -2\sqrt{3}

\displaystyle \Longrightarrow \frac{h^{2} +1}{h} =2\sqrt{13}

\displaystyle \Longrightarrow h^{2} +1=2\sqrt{13} h

বামপক্ষ\displaystyle =\frac{13h}{h^{2} -\sqrt{13} h+1}

\displaystyle =\frac{13h}{h^{2} +1-\sqrt{13} h}

\displaystyle =\frac{13h}{2\sqrt{13} h-\sqrt{13} h}

\displaystyle =\frac{13h}{\sqrt{13} h}

\displaystyle =\frac{13}{\sqrt{13}}

\displaystyle =\frac{\sqrt{13} .\sqrt{13}}{\sqrt{13}}

\displaystyle =\sqrt{13}

= ডানপক্ষ

সুতরাং, \displaystyle \frac{13h}{h^{2} -\sqrt{13} h+1} =\sqrt{13} ( প্রমানিত )

গ)

উদ্দীপকের (i) নংএর তথ্য অনুসারে,

\displaystyle 4p^{2} =2p-1

\displaystyle \Longrightarrow 4p^{2} +1=2p

\displaystyle \Longrightarrow \frac{4p^{2}}{2p} +\frac{1}{2p} =\frac{2p}{2p} [ উভয়পক্ষ \displaystyle 2p দ্বারা ভাগ করে পাই ]

\displaystyle \Longrightarrow 2p+\frac{1}{2p} =1

\displaystyle \Longrightarrow \ 2\left( p+\frac{1}{4p}\right) =1

\displaystyle \Longrightarrow \left( p+\frac{1}{4p}\right) =\frac{1}{2}

\displaystyle \Longrightarrow \left( p+\frac{1}{4p}\right)^{2} =\left(\frac{1}{2}\right)^{2} [ উভয়পক্ষ বর্গ করে পাই ]

\displaystyle \Longrightarrow p^{2} +2.p.\frac{1}{4p} +\left(\frac{1}{4p}\right)^{2} =\frac{1}{4}

\displaystyle \Longrightarrow p^{2} +\frac{1}{2} +\frac{1}{16p^{2}} =\frac{1}{4}

\displaystyle \Longrightarrow p^{2} +\frac{1}{16p^{2}} =\frac{1}{4} -\frac{1}{2}

\displaystyle \Longrightarrow p^{2} +\frac{1}{16p^{2}} =\frac{1-2}{4}

\displaystyle \Longrightarrow p^{2} +\frac{1}{16p^{2}} =-\frac{1}{4}

\displaystyle \Longrightarrow \left( p^{2} +\frac{1}{16p^{2}}\right)^{2} =\left( -\frac{1}{4}\right)^{2} [ উভয়পক্ষ বর্গ করে পাই ]

\displaystyle \Longrightarrow \left( p^{2}\right)^{2} +2.p^{2} .\frac{1}{16p^{2}} +\left(\frac{1}{16p^{2}}\right)^{2} =\frac{1}{16}

\displaystyle \Longrightarrow p^{4} +\frac{1}{8} +\frac{1}{256p^{4}} =\frac{1}{16}

\displaystyle \Longrightarrow p^{4} +\frac{1}{256p^{4}} =\frac{1}{16} -\frac{1}{8}

\displaystyle \Longrightarrow p^{4} +\frac{1}{256p^{4}} =\frac{1-2}{16}

\displaystyle \Longrightarrow p^{4} +\frac{1}{256p^{4}} =\frac{-1}{16}

\displaystyle \Longrightarrow 16( p^{4} +\frac{1}{256p^{4}} =-1 ( দেখানো হলো )

Post Author: showrob

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