9. দ্বিঘাত করণী | Exercise 9.3 all solution | Ganit Prakash Class X math solution | WBBSE Class 10 Math Solution in Bengali |
1. (a) \(m+\frac{1}{m}=\sqrt3\) হলে,
(i) \(m^2+\frac{1}{m^2}\) এবং (ii) \(m^3+\frac{1}{m^3}\) এদের সরলতম মান নির্ণয় করি।
সমাধানঃ
(i) \(m^2+\frac{1}{m^2}\)
=\(\left(m+\frac{1}{m}\right)^2-2.m.\frac{1}{m}\)
= \(\left(\sqrt3\right)^2-2\)
= 3-2 = 1
(ii) \(m^3+\frac{1}{m^3}\)
=\(\left(m+\frac{1}{m}\right)^3-3.m.\frac{1}{m}\left(m+\frac{1}{m}\right)\)
= \(\left(\sqrt3\right)^3-3.\sqrt3\)
= \(3\sqrt3-3\sqrt3\)
= 0
1. (b) দেখাই যে, \(\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}-\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}=2\sqrt{15}\)
সমাধানঃ
\(\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}-\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}\)
=\(\frac{\left(\sqrt5+\sqrt3\right)^2-\left(\sqrt5-\sqrt3\right)^2}{\left(\sqrt5-\sqrt3\right)\left(\sqrt5+\sqrt3\right)}\)
= \(\frac{4.\sqrt5.\sqrt3}{\left(\sqrt5\right)^2-\left(\sqrt3\right)^2}\)
= \(\frac{4\sqrt{15}}{5-3}\)
= \(\frac{4\sqrt{15}}{2}\)
=\(2\sqrt{15}\) [প্রমাণিত]
2. সরল করিঃ
(a) \(\frac{\sqrt2\left(2+\sqrt3\right)}{\sqrt3\left(\sqrt3+1\right)}-\frac{\sqrt2\left(2-\sqrt3\right)}{\sqrt3\left(\sqrt3-1\right)}\)
সমাধানঃ
\(\frac{\sqrt2\left(2+\sqrt3\right)}{\sqrt3\left(\sqrt3+1\right)}-\frac{\sqrt2\left(2-\sqrt3\right)}{\sqrt3\left(\sqrt3-1\right)}\)
= \(\frac{\sqrt2\left(2+\sqrt3\right)\left(\sqrt3-1\right)-\sqrt2\left(2-\sqrt3\right)\left(\sqrt3+1\right)}{\sqrt3\left(\sqrt3+1\right)\left(\sqrt3-1\right)}\)
= \(\frac{\sqrt2\left[\left(2\sqrt3+3-2-\sqrt3\right)-\left(2\sqrt3-3+2-\sqrt3\right)\right]}{\sqrt3\left\{\left(\sqrt3\right)^2-\left(1\right)^2\right\}}\)
= \(\frac{\sqrt2\left[2\sqrt3+3-2-\sqrt3-2\sqrt3+3-2+\sqrt3\right]}{\sqrt3\left(3-1\right)}\)
= \(\frac{\sqrt2.2}{\sqrt3.2}\)
= \(\frac{\sqrt2.\sqrt3}{\sqrt3.\sqrt3}\)
= \(\frac{\sqrt6}{3}\)
2. সরল করিঃ
(b) \(\frac{3\sqrt7}{\sqrt5+\sqrt2}-\frac{5\sqrt5}{\sqrt2+\sqrt7}+\frac{2\sqrt2}{\sqrt7+\sqrt5}\)
সমাধানঃ
\(\frac{3\sqrt7}{\sqrt5+\sqrt2}-\frac{5\sqrt5}{\sqrt2+\sqrt7}+\frac{2\sqrt2}{\sqrt7+\sqrt5}\)
= \(\frac{3\sqrt7\left(\sqrt5-\sqrt2\right)}{\left(\sqrt5+\sqrt2\right)\left(\sqrt5-\sqrt2\right)}-\frac{5\sqrt5\left(\sqrt2-\sqrt7\right)}{\left(\sqrt2+\sqrt7\right)\left(\sqrt2-\sqrt7\right)}\)
+\(\frac{2\sqrt2\left(\sqrt7-\sqrt5\right)}{\left(\sqrt7+\sqrt5\right)\left(\sqrt7-\sqrt5\right)}\)
= \(\frac{3\left(\sqrt{35}-\sqrt{14}\right)}{\left(\sqrt5\right)^2-\left(\sqrt2\right)^2}-\frac{5\left(\sqrt{10}-\sqrt{35}\right)}{\left(\sqrt2\right)^2-\left(\sqrt7\right)^2}\)
+\(\frac{2\left(\sqrt{14}-\sqrt{10}\right)}{\left(\sqrt7\right)^2-\left(\sqrt5\right)^2}\)
= \(\frac{3\left(\sqrt{35}-\sqrt{14}\right)}{5-2}-\frac{5\left(\sqrt{10}-\sqrt{35}\right)}{2-7}\)
+\(\frac{2\left(\sqrt{14}-\sqrt{10}\right)}{7-5}\)
= \(\frac{3\left(\sqrt{35}-\sqrt{14}\right)}{3}-\frac{5\left(\sqrt{10}-\sqrt{35}\right)}{-5}\)
+\(\frac{2\left(\sqrt{14}-\sqrt{10}\right)}{2}\)
= \(\left(\sqrt{35}-\sqrt{14}\right)+\left(\sqrt{10}-\sqrt{35}\right)\)
+\(\left(\sqrt{14}-\sqrt{10}\right)\)
= \(\sqrt{35}-\sqrt{14}+\sqrt{10}-\sqrt{35}+\sqrt{14}-\sqrt{10}\)
= 0
2. সরল করিঃ
(c) \(\frac{4\sqrt3}{2-\sqrt2}-\frac{30}{4\sqrt3-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}\)
সমাধানঃ
\(\frac{4\sqrt3}{2-\sqrt2}-\frac{30}{4\sqrt3-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}\)
= \(\frac{4\sqrt3\left(2+\sqrt2\right)}{\left(2-\sqrt2\right)\left(2+\sqrt2\right)}-\frac{30\left(4\sqrt3+\sqrt{18}\right)}{\left(4\sqrt3-\sqrt{18}\right)\left(4\sqrt3+\sqrt{18}\right)}\)
\(-\frac{3\sqrt2\left(3+\sqrt{12}\right)}{\left(3-\sqrt{12}\right)\left(3+\sqrt{12}\right)}\)
= \(\frac{4\left(2\sqrt3+\sqrt6\right)}{\left(2\right)^2-\left(\sqrt2\right)^2}-\frac{30\left(4\sqrt3+\sqrt{18}\right)}{\left(4\sqrt3\right)^2-\left(\sqrt{18}\right)^2}\)
\(-\frac{3\left(3\sqrt2+\sqrt{24}\right)}{\left(3\right)^2-\left(\sqrt{12}\right)^2}\)
= \(\frac{4\left(2\sqrt3+\sqrt6\right)}{4-2}-\frac{30\left(4\sqrt3+3\sqrt2\right)}{48-18}\)
\(-\frac{3\left(3\sqrt2+2\sqrt6\right)}{9-12}\)
= \(\frac{4\left(2\sqrt3+\sqrt6\right)}{2}-\frac{30\left(4\sqrt3+3\sqrt2\right)}{30}\)
\(-\frac{3\left(3\sqrt2+2\sqrt6\right)}{-3}\)
= \(2\left(2\sqrt3+\sqrt6\right)-\left(4\sqrt3+3\sqrt2\right)\)
\(+\left(3\sqrt2+2\sqrt6\right)\)
= \(4\sqrt3+2\sqrt6-4\sqrt3-3\sqrt2+3\sqrt2+2\sqrt6\)
=\(4\sqrt{6}\)
2. সরল করিঃ
(d) \(\frac{3\sqrt2}{\sqrt3+\sqrt6}-\frac{4\sqrt3}{\sqrt6+\sqrt2}+\frac{\sqrt6}{\sqrt2+\sqrt3}\)
সমাধানঃ
\(\frac{3\sqrt2}{\sqrt3+\sqrt6}-\frac{4\sqrt3}{\sqrt6+\sqrt2}+\frac{\sqrt6}{\sqrt2+\sqrt3}\)
= \(\frac{3\sqrt2\left(\sqrt3-\sqrt6\right)}{\left(\sqrt3+\sqrt6\right)\left(\sqrt3-\sqrt6\right)}-\frac{4\sqrt3\left(\sqrt6-\sqrt2\right)}{\left(\sqrt6+\sqrt2\right)\left(\sqrt6-\sqrt2\right)}\)
\(+\frac{\sqrt6\left(\sqrt2-\sqrt3\right)}{\left(\sqrt2+\sqrt3\right)\left(\sqrt2-\sqrt3\right)}\)
= \(\frac{3\sqrt2\left(\sqrt3-\sqrt6\right)}{\left(\sqrt3\right)^2-\left(\sqrt6\right)^2}-\frac{4\sqrt3\left(\sqrt6-\sqrt2\right)}{\left(\sqrt6\right)^2-\left(\sqrt2\right)^2}+\frac{\sqrt6\left(\sqrt2-\sqrt3\right)}{\left(\sqrt2\right)^2-\left(\sqrt3\right)^2}\)
= \(\frac{3\left(\sqrt6-\sqrt{12}\right)}{\left(\sqrt3\right)^2-\left(\sqrt6\right)^2}-\frac{4\left(\sqrt{18}-\sqrt6\right)}{\left(\sqrt6\right)^2-\left(\sqrt2\right)^2}+\frac{\left(\sqrt{12}-\sqrt{18}\right)}{\left(\sqrt2\right)^2-\left(\sqrt3\right)^2}\)
= \(\frac{3\left(\sqrt6-2\sqrt3\right)}{3-6}-\frac{4\left(3\sqrt2-\sqrt6\right)}{6-2}+\frac{\left(2\sqrt3-3\sqrt2\right)}{2-3}\)
= \(\frac{3\left(\sqrt6-2\sqrt3\right)}{-3}-\frac{4\left(3\sqrt2-\sqrt6\right)}{4}+\frac{\left(2\sqrt3-3\sqrt2\right)}{-1}\)
= \(-\left(\sqrt6-2\sqrt3\right)-\left(3\sqrt2-\sqrt6\right)-\left(2\sqrt3-3\sqrt2\right)\)
= \(-\sqrt6+2\sqrt3-3\sqrt2+\sqrt6-2\sqrt3+3\sqrt2\)
= 0
3. যদি x=2, y=3 এবং z=6 হয় তবে,
\(\frac{3\sqrt x}{\sqrt y+\sqrt z}-\frac{4\sqrt y}{\sqrt z+\sqrt x}+\frac{\sqrt z}{\sqrt x+\sqrt y}\) –এর মান হিসাব করে লিখি।
সমাধানঃ
\(\frac{3\sqrt x}{\sqrt y+\sqrt z}-\frac{4\sqrt y}{\sqrt z+\sqrt x}+\frac{\sqrt z}{\sqrt x+\sqrt y}\)
=\(\frac{3\sqrt2}{\sqrt3+\sqrt6}-\frac{4\sqrt3}{\sqrt6+\sqrt2}+\frac{\sqrt6}{\sqrt2+\sqrt3}\)
[∵x=2, y=3 এবং z=6]
= \(\frac{3\sqrt2\left(\sqrt3-\sqrt6\right)}{\left(\sqrt3+\sqrt6\right)\left(\sqrt3-\sqrt6\right)}-\frac{4\sqrt3\left(\sqrt6-\sqrt2\right)}{\left(\sqrt6+\sqrt2\right)\left(\sqrt6-\sqrt2\right)}\)
\(+\frac{\sqrt6\left(\sqrt2-\sqrt3\right)}{\left(\sqrt2+\sqrt3\right)\left(\sqrt2-\sqrt3\right)}\)
= \(\frac{3\left(\sqrt6-\sqrt{12}\right)}{\left(\sqrt3\right)^2-\left(\sqrt6\right)^2}-\frac{4\left(\sqrt{18}-\sqrt6\right)}{\left(\sqrt6\right)^2-\left(\sqrt2\right)^2}\)
\(+\frac{\left(\sqrt{12}-\sqrt{18}\right)}{\left(\sqrt2\right)^2-\left(\sqrt3\right)^2}\)
= \(\frac{3\left(\sqrt6-\sqrt{12}\right)}{3-6}-\frac{4\left(\sqrt{18}-\sqrt6\right)}{6-2}+\frac{\left(\sqrt{12}-\sqrt{18}\right)}{2-3}\)
= \(\frac{3\left(\sqrt6-\sqrt{12}\right)}{-3}-\frac{4\left(\sqrt{18}-\sqrt6\right)}{4}+\frac{\left(\sqrt{12}-\sqrt{18}\right)}{-1}\)
= \(-\left(\sqrt6-\sqrt{12}\right)-\left(\sqrt{18}-\sqrt6\right)\)
\(-\left(\sqrt{12}-\sqrt{18}\right)\)
= \(-\sqrt6+\sqrt{12}-\sqrt{18}+\sqrt6-\sqrt{12}+\sqrt{18}\)
= 0
4. \(x=\sqrt7+\sqrt6\) হলে
(i) \(x-\frac{1}{x}\) (ii) \(x+\frac{1}{x}\) (iii) \(x^2+\frac{1}{x^2}\)
এবং (iv) \(x^3+\frac{1}{x^3}\) –এদের সরলতম মান নির্ণয় করি।
সমাধানঃ
\(x=\sqrt7+\sqrt6\)
∴ \(\frac{1}{x}=\frac{1}{\sqrt7+\sqrt6}\)
\(=\frac{\left(\sqrt7-\sqrt6\right)}{\left(\sqrt7+\sqrt6\right)\left(\sqrt7-\sqrt6\right)}\)
\(=\frac{\left(\sqrt7-\sqrt6\right)}{\left(\sqrt7\right)^2+\left(\sqrt6\right)^2}\)
\(=\frac{\left(\sqrt7-\sqrt6\right)}{7-6} = \left(\sqrt7-\sqrt6\right)\)
(i) \(x-\frac{1}{x}=\sqrt7+\sqrt6-(\sqrt7-\sqrt6)\)
\(= \sqrt7+\sqrt6-\sqrt7+\sqrt6 = 2\sqrt6\)
(ii) \(x+\frac{1}{x} =\sqrt7+\sqrt6+(\sqrt7-\sqrt6)\)
\(= \sqrt7+\sqrt6+\sqrt7-\sqrt6 = 2\sqrt7\)
(iii) \(x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2.x.\frac{1}{x}\)
\(= \left(2\sqrt7\right)^2-2 =28-2 = 26\)
(iv) \(x^3+\frac{1}{x^3} = \left(x+\frac{1}{x}\right)^3-3.x.\frac{1}{x}\left(x+\frac{1}{x}\right)\)
\(= \left(2\sqrt7\right)^3-3\left(2\sqrt7\right)\)
\(= 56\sqrt7-6\sqrt7 = 50\sqrt7\)
5. সরল করিঃ \(\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}\)
সরলফল 14 হলে, x এর মান কী কী হবে হিসাব করে লিখি।
সমাধানঃ
\(\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}\)
\(=\frac{\left(x+\sqrt{x^2-1}\right)^2+\left(x-\sqrt{x^2-1}\right)^2}{\left(x-\sqrt{x^2-1}\right)\left(x+\sqrt{x^2-1}\right)}\)
\(=\frac{2\left\{\left(x\right)^2+\left(\sqrt{x^2-1}\right)^2\right\}}{\left(x\right)^2-\left(\sqrt{x^2-1}\right)^2}\)
\(=\frac{2\left(x^2+x^2-1\right)}{x^2-(x^2-1)}\)
\(=\frac{2\left({2x}^2-1\right)}{x^2-x^2+1} = 2\left({2x}^2-1\right)\)
∴\(2\left({2x}^2-1\right)=14\)
বা, \(\left({2x}^2-1\right)=\frac{14}{2}\)
বা, \({2x}^2=7+1\)
বা, \(x^2=\frac{8}{2}\)
বা, \(x^2=4\)
∴ \(x=\pm2\)
6. যদি \(a=\frac{\sqrt5+1}{\sqrt5-1}\) ও \(b=\frac{\sqrt5-1}{\sqrt5+1}\) হয়, তবে নীচের মানগুলি নির্ণয় করি।
(i) \(\frac{a^2+ab+b^2}{a^2-ab+b^2}\)
সমাধানঃ
\(a+b=\frac{\sqrt5+1}{\sqrt5-1}+\frac{\sqrt5-1}{\sqrt5+1}\)
\(=\frac{\left(\sqrt5+1\right)^2+\left(\sqrt5-1\right)^2}{\left(\sqrt5+1\right)\left(\sqrt5-1\right)}\)
\(=\frac{2\left\{\left(\sqrt5\right)^2+(1)^2\right\}}{\left(\sqrt5\right)^2-(1)^2}\)
\(=\frac{2\left(5+1\right)}{5-1} = \frac{2\times6}{4}\) = 3
ab= \(\frac{\sqrt5+1}{\sqrt5-1}\times\frac{\sqrt5-1}{\sqrt5+1}\)=1
∴\(\frac{a^2+ab+b^2}{a^2-ab+b^2} = \frac{a^2+b^2+ab}{a^2+b^2-ab}\)
\(=\frac{\left(a+b\right)^2-2ab+ab}{\left(a+b\right)^2-2ab-ab}\)
\(=\frac{\left(3\right)^2-2.1+1}{\left(3\right)^2-2.1-1}\)
\(=\frac{9-2+1}{9-2-1} = \frac{8}{6} = \frac{4}{3}\)
(ii) \(\frac{{(a-b)}^3}{{(a+b)}^3}\)
সমাধানঃ
a+b=\(\frac{\sqrt5+1}{\sqrt5-1}+\frac{\sqrt5-1}{\sqrt5+1}\)
= \(\frac{\left(\sqrt5+1\right)^2+\left(\sqrt5-1\right)^2}{\left(\sqrt5+1\right)\left(\sqrt5-1\right)}\)
= \(\frac{2\left\{\left(\sqrt5\right)^2+(1)^2\right\}}{\left(\sqrt5\right)^2-(1)^2}\)
= \(\frac{2\left(5+1\right)}{5-1} = \frac{2\times6}{4} = 3\)
a-b=\(\frac{\sqrt5+1}{\sqrt5-1}-\frac{\sqrt5-1}{\sqrt5+1}\)
= \(\frac{\left(\sqrt5+1\right)^2-\left(\sqrt5-1\right)^2}{\left(\sqrt5+1\right)\left(\sqrt5-1\right)}\)
= \(\frac{4.\sqrt5.1}{\left(\sqrt5\right)^2-(1)^2} = \frac{4\sqrt5}{5-1} = \frac{4\sqrt5}{4} = \sqrt5\)
∴ \(\frac{{(a-b)}^3}{{(a+b)}^3}=\frac{{(\sqrt5)}^3}{{(3)}^3} = \frac{5\sqrt5}{27}\)
(iii) \(\frac{{3a}^2+5ab+{3b}^2}{{3a}^2-5ab+3b^2}\)
সমাধানঃ
a+b=\(\frac{\sqrt5+1}{\sqrt5-1}+\frac{\sqrt5-1}{\sqrt5+1}\)
= \(\frac{\left(\sqrt5+1\right)^2+\left(\sqrt5-1\right)^2}{\left(\sqrt5+1\right)\left(\sqrt5-1\right)}\)
= \(\frac{2\left\{\left(\sqrt5\right)^2+(1)^2\right\}}{\left(\sqrt5\right)^2-(1)^2}\)
= \(\frac{2\left(5+1\right)}{5-1} = \frac{2\times6}{4} = 3\)
ab=\(\frac{\sqrt5+1}{\sqrt5-1}\times\frac{\sqrt5-1}{\sqrt5+1}\)=1
∴ \(\frac{{3a}^2+5ab+{3b}^2}{{3a}^2-5ab+3b^2}=\frac{{3a}^2+5ab+{3b}^2}{{3a}^2-5ab+3b^2}\)
= \(\frac{{3(a}^2+b^2)+5ab}{3(a^2+b^2)-5ab}\)
= \(\frac{3\left\{\left(a+b\right)^2-2ab\right\}+5ab}{3\left\{\left(a+b\right)^2-2ab\right\}-5ab}\)
= \(\frac{3\left\{\left(3\right)^2-2.1\right\}+5.1}{3\left\{\left(3\right)^2-2.1\right\}-5.1}\)
= \(\frac{3\left(9-2\right)+5}{3\left(9-2\right)-5}\)
= \(\frac{21+5}{21-5} = \frac{26}{16} = \frac{13}{8} =1\frac{5}{8}\)
(iv) \(\frac{a^3+b^3}{a^3-b^3}\)
সমাধানঃ
a+b=\(\frac{\sqrt5+1}{\sqrt5-1}+\frac{\sqrt5-1}{\sqrt5+1}\)
= \(\frac{\left(\sqrt5+1\right)^2+\left(\sqrt5-1\right)^2}{\left(\sqrt5+1\right)\left(\sqrt5-1\right)}\)
= \(\frac{2\left\{\left(\sqrt5\right)^2+(1)^2\right\}}{\left(\sqrt5\right)^2-(1)^2}\)
= \(\frac{2\left(5+1\right)}{5-1}\)
= \(\frac{2\times6}{4}\) = 3
a-b=\(\frac{\sqrt5+1}{\sqrt5-1}-\frac{\sqrt5-1}{\sqrt5+1}\)
= \(\frac{\left(\sqrt5+1\right)^2-\left(\sqrt5-1\right)^2}{\left(\sqrt5+1\right)\left(\sqrt5-1\right)}\)
= \(\frac{4.\sqrt5.1}{\left(\sqrt5\right)^2-(1)^2}\)
= \(\frac{4\sqrt5}{5-1}\)
= \(\frac{4\sqrt5}{4}\)
= \(\sqrt5\)
ab= \(\frac{\sqrt5+1}{\sqrt5-1}\times\frac{\sqrt5-1}{\sqrt5+1}\)=1
∴ \(\frac{a^3+b^3}{a^3-b^3}=\frac{\left(a+b\right)^3-3ab(a+b)}{\left(a-b\right)^3+3ab(a-b)}\)
= \(\frac{\left(3\right)^3-3.1.3}{\left(\sqrt5\right)^3+3.1.(\sqrt5)}\)
= \(\frac{27-9}{5\sqrt5+3\sqrt5}\)
= \(\frac{18}{8\sqrt5}\)
= \(\frac{18\sqrt5}{8\sqrt5.\sqrt5}\)
=\(\frac{18\sqrt5}{8.5} = \frac{9\sqrt5}{20}\)
7. যদি \(x=2+\sqrt3\), \(y=2-\sqrt3\) হয়, তবে নিম্নলিখিত সরলতম মান নির্ণয় করি।
(a) (i) \(x-\frac{1}{x}\)
সমাধানঃ
\(x=2+\sqrt3\)
∴\(\frac{1}{x}=\frac{1}{2+\sqrt3}\)
= \(\frac{\left(2-\sqrt3\right)}{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}\)
= \(\frac{2-\sqrt3}{\left(2\right)^2-\left(\sqrt3\right)^2}\)
=\(\frac{\left(2-\sqrt3\right)}{4-3}\)
= \(2-\sqrt3\)
∴ \(x-\frac{1}{x}=2+\sqrt3-(2-\sqrt3)\)
= \(2+\sqrt3-2+\sqrt3 = 2\sqrt3\)
(a) (ii) \(y^2+\frac{1}{y^2}\)
সমাধানঃ
\(y=2-\sqrt3\)
∴\(\frac{1}{y}=\frac{1}{2-\sqrt3}\)
=\(\frac{\left(2+\sqrt3\right)}{\left(2-\sqrt3\right)\left(2+\sqrt3\right)}\)
= \(\frac{2+\sqrt3}{\left(2\right)^2-\left(\sqrt3\right)^2}\)
=\(\frac{\left(2+\sqrt3\right)}{4-3}\)
= \(2+\sqrt3\)
∴\(y^2+\frac{1}{y^2}=\left(y-\frac{1}{y}\right)^2+2.y.\frac{1}{y}\)
= \(\left(2-\sqrt3-2+\sqrt3\right)^2+2.1\)
= \(\left(2\sqrt3\right)^2+2\) = 12+2 = 14
(a) (iii) \(x^3-\frac{1}{x^3}\)
সমাধানঃ
\(x=2+\sqrt3\)
∴\(\frac{1}{x}=\frac{1}{2+\sqrt3}\)
= \(\frac{\left(2-\sqrt3\right)}{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}\)
= \(\frac{2-\sqrt3}{\left(2\right)^2-\left(\sqrt3\right)^2}\)
=\(\frac{\left(2-\sqrt3\right)}{4-3}\)
= \(2-\sqrt3\)
\(x-\frac{1}{x}=2+\sqrt3-2+\sqrt3=2\sqrt3\)
∴\(x^3-\frac{1}{x^3}=\left(x-\frac{1}{x}\right)^3+2.x.\frac{1}{x}.\left(x-\frac{1}{x}\right)\)
= \({(2\sqrt3)}^3+2.2\sqrt3\)
= \(24\sqrt3+4\sqrt3\)
= \(28\sqrt3\)
(a) (iv) \(xy+\frac{1}{xy}\)
সমাধানঃ
xy=\(\left(2+\sqrt3\right)\left(2-\sqrt3\right)\)
= \(\left(2\right)^2-\left(\sqrt3\right)^2\)
= 4-3=1
∴ \(xy+\frac{1}{xy}=1+\frac{1}{1}\)=2
7. যদি \(x=2+\sqrt3, y=2-\sqrt3\) হয়, তবে নিম্নলিখিত সরলতম মান নির্ণয় করি।
(b) \(3x^2-5xy+3y^2\)
সমাধানঃ
x+y=\(\left(2+\sqrt3\right)+\left(2-\sqrt3\right)\)
=\(2+\sqrt3+2-\sqrt3\)=4
xy=\(\left(2+\sqrt3\right)\left(2-\sqrt3\right)\)
= \(\left(2\right)^2-\left(\sqrt3\right)^2\)
=4-3=1
\(3x^2-5xy+3y^2\)
= \(3(x^2+y^2)-5xy\)
= \(3\left\{\left(x+y\right)^2-2xy\right\}-5xy\)
= \(3\left\{\left(4\right)^2-2.1\right\}-5.1\)
= 3(16-2)-5
= 42-5
= 37
8.\(x=\frac{\sqrt7+\sqrt3}{\sqrt7-\sqrt3}\) এবং xy=1 হলে,
দেখাই যে, \(\frac{x^2+xy+y^2}{x^2-xy+y^2}=\frac{12}{11}\)
সমাধানঃ
\(x=\frac{\sqrt7+\sqrt3}{\sqrt7-\sqrt3}\) এবং xy=1
∴ \(y=\frac{1}{x}=\frac{\sqrt7-\sqrt3}{\sqrt7+\sqrt3}\)
x+y=\(\frac{\sqrt7+\sqrt3}{\sqrt7-\sqrt3}+\frac{\sqrt7-\sqrt3}{\sqrt7+\sqrt3}\)
= \(\frac{\left(\sqrt7+\sqrt3\right)^2+\left(\sqrt7-\sqrt3\right)^2}{\left(\sqrt7-\sqrt3\right)\left(\sqrt7+\sqrt3\right)}\)
= \(\frac{2\left\{\left(\sqrt7\right)^2+\left(\sqrt3\right)^2\right\}}{\left(\sqrt7\right)^2-\left(\sqrt3\right)^2}\)
= \(\frac{2\left(7+3\right)}{7-3}\)
= \(\frac{2\times10}{4} = 5\)
\(\frac{x^2+xy+y^2}{x^2-xy+y^2} = \frac{x^2+y^2+xy}{x^2+y^2-xy}\)
= \(\frac{\left(x+y\right)^2-2xy+xy}{\left(x+y\right)^2-2xy-xy}\)
= \(\frac{\left(5\right)^2-2.1+1}{\left(5\right)^2-2.1-1}\)
= \(\frac{\left(5\right)^2-2.1+1}{\left(5\right)^2-2.1-1}\)
= \(\frac{25-2+1}{25-2-1}\)
= \(\frac{24}{22}\)
= \(\frac{12}{11}\) [প্রমাণিত]
9.\(\left(\sqrt7+1\right)\) এবং \(\left(\sqrt5+\sqrt3\right)\)-এর মধ্যে কোনটি বড়ো লিখি।
সমাধানঃ
\(\left(\sqrt7+1\right)^2=\left\{\left(\sqrt7\right)^2+2.\sqrt7.1+1^2\right\}\)
= \(7+2\sqrt7+1\)
= \(8+2\sqrt7\)
\(\left(\sqrt5+\sqrt3\right)^2=\left\{\left(\sqrt5\right)^2+2.\sqrt5.\sqrt3+\left(\sqrt3\right)^2\right\}\)
= \((5+2\sqrt{15}+3)\)
= \(8+2\sqrt{15}\)
যেহেতু, \(\sqrt{15}>\sqrt7\)
∴ \(\left(\sqrt5+\sqrt3\right)^2>\left(\sqrt7+1\right)^2\)
∴ \(\left(\sqrt5+\sqrt3\right)>\left(\sqrt7+1\right)\)
10. অতিসংক্ষিপ্ত উত্তরধর্মী প্রশ্ন(V.S.A.)
(A) বহুবিকল্পীয় প্রশ্ন(M.C.Q):
(i) \(x=2+\sqrt3\) হলে, \(x+\frac{1}{x}\) এর মান
(a)2 (b)\(2\sqrt3\)
(c)4 (d)\(2-\sqrt3\)
সমাধানঃ
\(x=2+\sqrt3\)
∴ \(\frac{1}{x}=\frac{1}{2+\sqrt3}\)
=\(\frac{\left(2-\sqrt3\right)}{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}\)
=\(\frac{\left(2-\sqrt3\right)}{\left(2\right)^2-\left(\sqrt3\right)^2}\)
= \(\frac{\left(2-\sqrt3\right)}{4-3}\)
= \(\left(2-\sqrt3\right)\)
∴ \(x+\frac{1}{x}=\ 2+\sqrt3+2-\sqrt3\) = 4
উত্তরঃ (c) 4
(ii) যদি \(p+q=\sqrt{13}\) এবং \(p-q=\sqrt5\) হয়, তাহলে pq এর মান
(a) 2 (b) 18 (c) 9 (d) 8
সমাধানঃ
\(pq=\left(p+q\right)^2-\left(p-q\right)^2\)
= \(\left(\sqrt{13}\right)^2-\left(\sqrt5\right)^2\)
= 13-5= 8
উত্তরঃ (d) 8
(iii) যদি \(a+b=\sqrt5\) এবং \(a-b=\sqrt3\) হয়, তাহলে \(\left(a^2+b^2\right)\) এর মান
(a) 8 (b) 4 (c) 2 (d) 1
সমাধানঃ
\(2\left(a^2+b^2\right)=\left(a+b\right)^2-\left(a-b\right)^2\)
বা, 2\(\left(a^2+b^2\right)=\left(\sqrt5\right)^2-\left(\sqrt3\right)^2\)
বা, 2\(\left(a^2+b^2\right)=5-3\)
বা,\(\left(a^2+b^2\right)=\frac{2}{2}\)
∴ \((a^2+b^2)=1\)
উত্তরঃ (d) 1
(iv) \(\sqrt{125}\) থেকে \(\sqrt5\) বিয়োগ করলে বিয়োগফল হবে
(a) \(\sqrt{80}\) (b) \(\sqrt{120}\)
(c) \(\sqrt{100}\) (d) কোনটিই নয়
সমাধানঃ
\(\sqrt{125}-\sqrt5\)
= \(\sqrt{25\times5}-\sqrt5\)
= 5\(\sqrt5-\sqrt5\)
= 4\(\sqrt5\)
= \(\sqrt{4\times4\times5}\)
= \(\sqrt{80}\)
উত্তরঃ (a) \(\sqrt{80}\)
(v) \(\left(5-\sqrt3\right)\left(\sqrt3-1\right)\left(5+\sqrt3\right)\left(\sqrt3+1\right)\) এর গুনফল
(a) 22 (b) 44 (c) 2 (d) 11
সমাধানঃ
\(\left(5-\sqrt3\right)\left(\sqrt3-1\right)\left(5+\sqrt3\right)\left(\sqrt3+1\right)\)
= \(\left(5-\sqrt3\right)\left(5+\sqrt3\right)\left(\sqrt3-1\right)\left(\sqrt3+1\right)\)
= \(\left\{\left(5\right)^2-\left(\sqrt3\right)^2\right\}\left\{{\left(\sqrt3\right)^2-\left(1\right)}^2\right\}\)
= \(\left(25-3\right)\left(3-1\right)\)
= 22×2 = 44
উত্তরঃ (b) 44
(B)নীচের বিবৃতিগুলি সত্য না মিথ্যা লিখিঃ
(i) \(\sqrt{75}\) এবং \(\sqrt{147}\) সদৃশ্য করণী।
সমাধানঃ
\(\sqrt{75}=\sqrt{25\times3}=5\sqrt3\)
\(\sqrt{147}=\sqrt{49\times3}=7\sqrt3\)
5\(\sqrt3\) এবং 7\(\sqrt3\) পরস্পর সদৃশ্য করণী
উত্তরঃ সত্য
(ii) \(\sqrt\pi\) একটি দ্বিঘাত করণী।
সমাধানঃ
এখানে, \(\pi\) একটি অমূলদ সংখ্যা
তাই \(\sqrt\pi\) দ্বিঘাত করণী নয়।
উত্তরঃ মিথ্যা
(C)শূন্যস্থান পূরণ করিঃ
(i) 5\(\sqrt{11}\) একটি _________সংখ্যা।(মূলদ/অমূলদ)
উত্তরঃ অমূলদ
(ii) \(\left(\sqrt3-5\right)\) –এর অনুবন্ধী করণী_________।
উত্তরঃ \(\left(-\sqrt3-5\right)\)
(iii) দুটি দ্বিঘাত করণীর যোগফল ও গুনফল একটি মূলদ সংখ্যা হলে করণীদ্বয় _______করণী।
উত্তরঃ অনুবন্ধী
11. সংক্ষিপ্ত উত্তরধর্মী প্রশ্ন(S.A.)
(i) \(x=3+2\sqrt2\) হলে, \(x+\frac{1}{x}\) –এর মান লিখি।
সমাধানঃ
\(x=3+2\sqrt2\)
∴\(\frac{1}{x}=\frac{1}{3+2\sqrt2}\)
=\(\frac{\left(3-2\sqrt2\right)}{\left(3+2\sqrt2\right)\left(3-2\sqrt2\right)}\)
= \(\frac{\left(3-2\sqrt2\right)}{\left(3\right)^2-\left(2\sqrt2\right)^2}\)
= \(\frac{\left(3-2\sqrt2\right)}{9-8}=\left(3-2\sqrt2\right)\)
∴\(x+\frac{1}{x}=\left(3+2\sqrt2\right)-\left(3-2\sqrt2\right)\)
= \(3+2\sqrt2-3+2\sqrt2\)
= \(4\sqrt2\)
(ii) \(\left(\sqrt{15}+\sqrt3\right)\) এবং \(\left(\sqrt{10}+\sqrt8\right)\) এর মধ্যে কোনটি বড়ো লিখি।
সমাধানঃ
\(\left(\sqrt{15}+\sqrt3\right)^2=\left(\sqrt{15}\right)^2+2.\sqrt{15}.\sqrt3+\left(\sqrt3\right)^2\)
= \(15+2\sqrt{5\times3\times3}+3\)
= \(18+2\times3\sqrt5\)
= \(18+6\sqrt5\)
\(\left(\sqrt{10}+\sqrt8\right)^2=\left(\sqrt{10}\right)^2+2.\sqrt{10}.\sqrt8+\left(\sqrt8\right)^2\)
= \(10+2\sqrt{5\times2\times2\times2\times2}+8\)
= \(18+2\times2\times2\sqrt5\)
= \(18+8\sqrt5\)
যেহেতু, \(18+8\sqrt5>18+6\sqrt5\)
∴ \(\left(\sqrt{10}+\sqrt8\right)^2>\left(\sqrt{15}+\sqrt3\right)^2\)
সুতরাং, \(\left(\sqrt{10}+\sqrt8\right)>\left(\sqrt{15}+\sqrt3\right)\)
(iii) দুটি মিশ্র দ্বিঘাত করণী লিখি যাদের গুনফল একটি মূলদ সংখ্যা।
সমাধানঃ
দুটি মিশ্র দ্বিঘাত করণী হল \(2+\sqrt3\) ও \(2-\sqrt3\)
দ্বিঘাত করণী দুটির গুনফল একটি মূলদ সংখ্যা।
[∵ দ্বিঘাত করণী দুটি পরস্পর অনুবন্ধী করণী]
(iv) \(\sqrt{72}\) থেকে কত বিয়োগ করলে \(\sqrt{32}\) হবে তা লিখি।
সমাধানঃ
ধরি, \(\sqrt{72}\) থেকে x বিয়োগ করলে \(\sqrt{32}\) হবে
∴ \(\sqrt{72}-x = \sqrt{32}\)
বা, \(x = \sqrt{72}-\sqrt{32}\)
বা, \(x = \sqrt{36\times2}-\sqrt{16\times2}\)
বা, \(x = 6\sqrt2-4\sqrt2\)
∴\(x = 2\sqrt2\)
∴ \(\sqrt{72}\) থেকে \(2\sqrt2\) বিয়োগ করলে \(\sqrt{32}\) হবে।
(v) \(\left(\frac{1}{\sqrt2+1}+\frac{1}{\sqrt3+\sqrt2}+\frac{1}{\sqrt4+\sqrt3}\right)\) এর সরলতম মান লিখি।
সমাধানঃ
\(\left(\frac{1}{\sqrt2+1}+\frac{1}{\sqrt3+\sqrt2}+\frac{1}{\sqrt4+\sqrt3}\right)\)
= \(\frac{\sqrt2-1}{\left(\sqrt2+1\right)\left(\sqrt2-1\right)}+\frac{\left(\sqrt3-\sqrt2\right)}{\left(\sqrt3+\sqrt2\right)\left(\sqrt3-\sqrt2\right)}\)
+\(\frac{\left(\sqrt4-\sqrt3\right)}{\left(\sqrt4+\sqrt3\right)\left(\sqrt4-\sqrt3\right)}\)
= \(\frac{\sqrt2-1}{\left(\sqrt2\right)^2-\left(1\right)^2}+\frac{\left(\sqrt3-\sqrt2\right)}{\left(\sqrt3\right)^2-\left(\sqrt2\right)^2}\)
+\(\frac{\left(\sqrt4-\sqrt3\right)}{\left(\sqrt4\right)^2-\left(\sqrt3\right)^2}\)
= \(\frac{\sqrt2-1}{2-1}+\frac{\left(\sqrt3-\sqrt2\right)}{3-2}+\frac{\left(\sqrt4-\sqrt3\right)}{4-3}\)
= \(\frac{\sqrt2-1}{1}+\frac{\left(\sqrt3-\sqrt2\right)}{1}+\frac{\left(2-\sqrt3\right)}{1}\)
= \(\sqrt2-1+\sqrt3-\sqrt2+2-\sqrt3\)
= 1
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