Solution of sample questions -VIII/অষ্টম শ্রেণী /দাগ নম্বর-5

Solution of sample questions-VIII

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Solution of sample questions -VIII/অষ্টম শ্রেণী /দাগ নম্বর-5

5. প্রতিটি প্রশ্নের উত্তর দাও        [প্রশ্নমান ৩]

i. উৎপাদক বিশ্লেষণ করো : 

a). 64ax^{2}-49a(x-2y)^{2} 

উত্তরঃ 

64ax^{2}-49a(x-2y)^{2}\\\\=a\left [ \left ( 8x \right )^{2}- \left \{ 7\left ( x-2y \right ) \right \}^{2} \right ]\\\\=a\left [ \left ( 8x \right )^{2}-\left ( 7x-14y \right )^{2}\right ]\\\\=a\left \{ 8x-\left ( 7x-14y \right ) \right \}\left \{ 8x+\left ( 7x-14y \right ) \right \}\\\\=a\left \{ 8x- 7x+14y \right \}\left \{ 8x+ 7x-14y \right \}\\\\=a\left ( x+14y \right )\left ( 15x-14y \right ) 

b). \frac{x^{3}}{64}-\frac{64}{x^{3}} 

উত্তরঃ 

\frac{x^{3}}{64}-\frac{64}{x^{3}}\\\\=\left ( \frac{x}{4} \right )^{3}-\left ( \frac{4}{x} \right )^{3}\\\\=\left ( \frac{x}{4}-\frac{4}{x} \right )\left \{ \left ( \frac{x}{4} \right )^{2}+\frac{x}{4}\times \frac{4}{x} +\left ( \frac{4}{x} \right )^{2}\right \}\\\\=\left ( \frac{x}{4}-\frac{4}{x} \right )\left \{ \frac{x^{2}}{16} +1 +\frac{16}{x^{2}}\right \}

 

c). x^{12}-y^{12} 

উত্তরঃ 

x^{12}-y^{12}\\\\=\left ( x^{4} \right )^{3}-\left ( y^{4} \right )^{3}\\\\=\left ( x^{4}-y^{4} \right )\left \{ \left ( x^{4} \right )^{2}+x^{4}.y^{4} +\left ( y^{4} \right )^{2}\right \}\\\\=\left \{ \left ( x^{2} \right )^{2}-\left ( y^{2} \right )^{2} \right \}\left \{ \left ( x^{4} \right )^{2}+2.x^{4}.y^{4} +\left ( y^{4} \right )^{2}-x^{4}.y^{4}\right \}\\\\=\left \{ \left ( x^{2} -y^{2}\right )\left (x^{2}+ y^{2} \right ) \right \}\left \{ \left ( x^{4}+y^{4} \right )^{2}-\left ( x^{2}y^{2} \right )^{2}\right \}\\\\=\left \{ \left ( x -y\right )\left ( x+y \right )\left (x^{2}+ y^{2} \right ) \right \}\left \{ \left ( x^{4}+y^{4} +x^{2}y^{2}\right )\left ( x^{4}+y^{4} -x^{2}y^{2}\right )\right \}\\\\=\left ( x -y\right )\left ( x+y \right )\left (x^{2}+ y^{2} \right )\left \{ \left ( x^{2}+y^{2} \right )^{2}-\left ( xy \right )^{2}\right \}\left ( x^{4}+y^{4} -x^{2}y^{2}\right )\\\\=\left ( x -y\right )\left ( x+y \right )\left (x^{2}+ y^{2} \right )\left ( x^{2}+y^{2} +xy\right )\left ( x^{2}+y^{2} -xy\right )\left ( x^{4}+y^{4} -x^{2}y^{2}\right ) 

d).x^{2}-bx-(a+3b)(a+2b) 

উত্তরঃ 

x^{2}-bx-(a+3b)(a+2b)\\\\=x^{2}-\left \{ \left ( a+3b \right )-\left ( a+2b \right ) \right \}x-(a+3b)(a+2b)\\\\=x^{2}-\left ( a+3b \right )x+\left ( a+2b \right )x-(a+3b)(a+2b)\\\\=x\left \{ x-\left ( a+3b \right ) \right \}+\left ( a+2b \right )\left \{ x-\left ( a+3b \right ) \right \}\\\\=\left \{ x-\left ( a+3b \right ) \right \}\left \{ x+\left ( a+2b \right ) \right \}\\\\=\left ( x-a-3b \right )\left ( x+a+2b \right )

 

e).(x+1)(x+9)(x+5)^{2}+63 

উত্তরঃ 

(x+1)(x+9)(x+5)^{2}+63\\\\=\left \{ (x+1)(x+9) \right \}\left \{(x+5)^{2} \right \}+63\\\\=\left \{ x^{2} +10x+9\right \}\left \{ x+10x+25 \right \}+63  ধরি, x^{2}+10x =a 
\\= \left \{ x^{2} +10x+9\right \}\left \{ x+10x+25 \right \}+63\\\\=\left ( a+9 \right )\left ( a+25 \right )+63\\\\=a^{2}+34a+225+63\\\\=a^{2}+34a+288\\\\=a^{2}+\left ( 16+18 \right )a+288\\\\=a^{2}+16a+18a+288\\\\=a\left ( a+16 \right )+18\left ( a+16 \right )\\\\=\left ( a+18 \right )\left ( a+16 \right )
a=x^{2}+10x এর মান বসিয়ে পাই, 
\\=\left ( x^{2}+10x+18 \right )\left ( x^{2}+10x+16 \right )\\\\=\left ( x^{2}+10x+18 \right )\left \{ x^{2}+\left ( 8+2 \right )x+16 \right \}\\\\=\left ( x^{2}+10x+18 \right )\left ( x^{2}+8x+2x+16 \right )\\\\=\left ( x^{2}+10x+18 \right )\left \{ x\left ( x+8 \right )+2\left ( x+8 \right ) \right \}\\\\=\left ( x^{2}+10x+18 \right )\left ( x+8 \right )\left ( x+2 \right )

 

f).(x+1)(x+3)(x-4)(x-6)-24 

উত্তরঃ 

(x+1)(x+3)(x-4)(x-6)-24\\\\=\left \{ (x+1)(x-4) \right \}\left \{(x+3)(x-6) \right \}-24\\\\=\left \{ x^{2}-3x-4 \right \}\left \{ x^{2} -3x-18\right \}-24 

 ধরি, x^{2}-3x=a  
\\=\left \{ a-4 \right \}\left \{ a-18 \right \}-24\\\\=a^{2}-22a+72-24\\\\=a^{2}-22a-48\\\\=a^{2}-\left ( 24-2 \right )a-48\\\\=a^{2}-24a+2a-48\\\\=a\left ( a-24 \right )+2\left ( a-24 \right )\\\\=\left ( a-24 \right )\left ( a+2 \right ) 
a=x^{2}-3x

মান বসিয়ে পাই , 

\\=\left ( a-24 \right )\left ( a+2 \right )\\\\=\left ( x^{2} -3x-24\right )\left ( x^{2}-3x+2 \right )\\\\=\left ( x^{2} -3x-24\right )\left ( x^{2} -2x-x+2\right )\\\\=\left ( x^{2} -3x-24\right )\left \{ x\left ( x-2 \right )-1\left ( x-2 \right ) \right \}\\\\=\left ( x^{2} -3x-24\right )\left ( x-2 \right )\left ( x-1 \right ) 

ii. ধারাবাহিক গুন করো 

a). (a+b+c),(a-b+c), (a+b-c) 

উত্তরঃ 

(a+b+c)\times (a-b+c)\times (a+b-c)\\\\=\left \{ \left ( a+c \right )+b \right \}\times \left \{ \left ( a+c \right )-b \right \}\left ( a+b-c \right )\\\\=\left \{ \left ( a+c \right )^{2} -b^{2}\right \}\left ( a+b-c \right )\\\\=\left ( a^{2}+2ac+c^{2} -b^{2}\right )\left ( a+b-c \right )\\\\=a^{2}\left ( a+b-c \right )+2ac\left ( a+b-c \right )+c^{2}\left ( a+b-c \right )\\\\-b^{2}\left ( a+b-c \right )\\\\=a^{3}+a^{2}b-a^{2}c+2a^{2}c+2abc-2ac^{2}+c^{2}a+c^{2}b\\\\-c^{3}-b^{2}a-b^{3}+b^{2}c\\\\=a^{3}-b^{3}-c^{3}+a^{2}b+a^{2}c+2abc-ac^{2}+c^{2}b \\\\-b^{2}a+b^{2}c 

b). \left ( \frac{x^{2}}{y^{2}} +\frac{y^{2}}{z^{2}}\right ), \left ( \frac{y^{2}}{z^{2}} +\frac{z^{2}}{x^{2}}\right ), \left ( \frac{z^{2}}{x^{2}} +\frac{x^{2}}{y^{2}}\right ) 

উত্তরঃ 

\left ( \frac{x^{2}}{y^{2}} +\frac{y^{2}}{z^{2}}\right )\times \left ( \frac{y^{2}}{z^{2}} +\frac{z^{2}}{x^{2}}\right )\times \left ( \frac{z^{2}}{x^{2}} +\frac{x^{2}}{y^{2}}\right )\\\\=\left \{ \frac{x^{2}}{y^{2}}\times \left ( \frac{y^{2}}{z^{2}} +\frac{z^{2}}{x^{2}}\right )+\frac{y^{2}}{z^{2}}\times \left ( \frac{y^{2}}{z^{2}} +\frac{z^{2}}{x^{2}}\right ) \right \}\times \left ( \frac{z^{2}}{x^{2}} +\frac{x^{2}}{y^{2}}\right )\\\\=\left \{ \frac{x^{2}}{z^{2}}+\frac{z^{2}}{y^{2}}+\frac{y^{4}}{z^{4}}+\frac{y^{2}}{x^{2}} \right \}\times \left ( \frac{z^{2}}{x^{2}} +\frac{x^{2}}{y^{2}}\right )\\\\=\frac{x^{2}}{z^{2}}\times \left ( \frac{z^{2}}{x^{2}} +\frac{x^{2}}{y^{2}}\right )+\frac{z^{2}}{y^{2}}\times \left ( \frac{z^{2}}{x^{2}} +\frac{x^{2}}{y^{2}}\right )\\\\+\frac{y^{4}}{z^{4}}\times \left ( \frac{z^{2}}{x^{2}} +\frac{x^{2}}{y^{2}}\right )+\frac{y^{2}}{x^{2}}\times \left ( \frac{z^{2}}{x^{2}} +\frac{x^{2}}{y^{2}}\right )\\\\=1+\frac{x^{4}}{y^{2}z^{2}}+\frac{z^{4}}{x^{2}y^{2}}+\frac{x^{2}z^{2}}{y^{4}}+\frac{y^{4}}{z^{2}x^{2}}\\\\+\frac{x^{2}y^{2}}{z^{4}}+\frac{y^{2}z^{2}}{x^{4}}+1\\\\=2+\frac{x^{4}}{y^{2}z^{2}}+\frac{z^{4}}{x^{2}y^{2}}+\frac{x^{2}z^{2}}{y^{4}}+\frac{y^{4}}{z^{2}x^{2}}\\\\+\frac{x^{2}y^{2}}{z^{4}}+\frac{y^{2}z^{2}}{x^{4}} 

iii. a=lx+my+n, b=mx+ny+l  এবং c=nx+ly+m হলে, a(m+n)+b(n+l)+c(l+m) এর মান নির্ণয় করো। 

উত্তরঃ 

\because a=lx+my+n, b=mx+ny+l  এবং c=nx+ly+m

a(m+n)+b(n+l)+c(l+m)\\\\=\left ( lx+my+n \right )\left ( m+n \right )+\left ( mx+ny+l \right )\left ( n+l \right )\\\\+\left ( nx+ly+m \right )\left ( l+m \right )\\\\=lx \left ( m+n \right )+my\left ( m+n \right )+n\left ( m+n \right )\\\\+mx\left ( n+l \right )+ny\left ( n+l \right )+l\left ( n+l \right )\\\\+nx\left ( l+m \right )+ly\left ( l+m \right )+m\left ( l+m \right )\\\\=mlx+nlx+m^{2}y+mny+mn+n^{2}\\\\+mnx+lmx+n^{2}y+nly+nl+l^{2}+nlx\\\\+mnx+l^{2}y+mly+ml+m^{2}\\\\=2x\left ( ml+nl+mn \right )+y\left ( m^{2}+l^{2}+n^{2}+mn+nl+ml\right )\\\\+\left ( mn+n^{2}+nl+l^{2} +ml+m^{2}\right ) 

iv. ভাগ করো : (m^{4}-2m^{3}-7m^{2}+8m+12) কে (m^{2}-m-6)  দিয়ে। 

উত্তরঃ 

v).  a^{3}+b^{3}+c^{3}=3abc হলে, (a+b+c) এর মান  নির্ণয় করো। [a\neq b\neq c ]

উত্তরঃ 

a^{3}+b^{3}+c^{3}=3abc\\\\\Rightarrow a^{3}+b^{3}+c^{3}-3abc=0\\\\\therefore a^{3}+b^{3}+c^{3}-3abc=0 
\\\because a^{3}+b^{3}+c^{3}-3abc= \left ( a+b+c \right )\left ( a^{2}+b^{2} +c^{2}-ab-bc-ca\right ) 
\\\because a^{3}+b^{3}+c^{3}-3abc= 0 
\therefore \left ( a+b+c \right )\left ( a^{2}+b^{2} +c^{2}-ab-bc-ca\right ) =0 
\therefore \left ( a+b+c \right )\left ( a^{2}+b^{2} +c^{2}-ab-bc-ca\right ) =0\\\\\Rightarrow \left ( a+b+c \right )\times 2\left ( a^{2}+b^{2} +c^{2}-ab-bc-ca\right )=2\times 0\\\\\Rightarrow \left ( a+b+c \right ) \left ( 2a^{2}+2b^{2} +2c^{2}-2ab-2bc-2ca\right )=0\\\\\Rightarrow \left ( a+b+c \right )\left \{ \left ( a-b \right )^{2} +\left ( b-c \right )^{2}+\left ( c-a \right )^{2}\right \}=0 
\left ( a+b+c \right )  এবং \left \{ \left ( a-b \right )^{2} +\left ( b-c \right )^{2}+\left ( c-a \right )^{2}\right \} এর গুনফল 0 
হয়, \left ( a+b+c \right )=0 অর্থবা, \left \{ \left ( a-b \right )^{2} +\left ( b-c \right )^{2}+\left ( c-a \right )^{2}\right \}=0 
কিন্তু \left \{ \left ( a-b \right )^{2} +\left ( b-c \right )^{2}+\left ( c-a \right )^{2}\right \}\neq 0 
কারণ, \left ( a-b \right )^{2}\neq 0 , \left ( b-c \right )^{2}\neq 0 এবং \left ( c-a \right )^{2}\neq 0 যদি \left \{ \left ( a-b \right )^{2} +\left ( b-c \right )^{2}+\left ( c-a \right )^{2}\right \}= 0 তাহলে \Rightarrow \left ( a-b \right )^{2}=0\\\\\Rightarrow a=b\\\\\left ( b-c \right )^{2}=0\\\\\Rightarrow b=c\\\\\left ( c-a \right )^{2}=0\\\\\Rightarrow c=a\\\\\therefore a=b=c ,কিন্তু a\neq b\neq c
\therefore \left ( a+b+c \right )=0 

vi). সরল করো : 

a). (m-p)\left \{ \left ( m+n \right )^{2}+\left ( m+n \right )\left ( n+p \right ) +\left ( n+p \right )^{2} \right \} 

উত্তরঃ 

(m-p)\left \{ \left ( m+n \right )^{2}+\left ( m+n \right )\left ( n+p \right ) +\left ( n+p \right )^{2} \right \}\\\\=\left \{ \left ( m+n \right )-\left ( n+p \right ) \right \}\left \{ \left ( m+n \right )^{2}+\left ( m+n \right )\left ( n+p \right ) +\left ( n+p \right )^{2} \right \}\\\\=\left ( m+n \right )^{3}-\left ( n+p \right )^{3}\\\\=m^{3}+n^{3}+3m^{2}n+3mn^{2}-\left ( n^{3}+p^{3}+3n^{2}p+3np^{2} \right )\\\\=m^{3}+n^{3}+3m^{2}n+3mn^{2}-n^{3}-p^{3}-3n^{2}p-3np^{2}\\\\=m^{3}+3m^{2}n+3mn^{2}-p^{3}-3n^{2}p-3np^{2} 

b).(4a^{2}-9)(4a^{2}-6a+9)(4a^{2}+6a+9) 

উত্তরঃ 

(4a^{2}-9)(4a^{2}-6a+9)(4a^{2}+6a+9)\\\\=\left \{ \left ( 2a \right )^{2}-\left ( 3 \right )^{2} \right \}\left \{ \left ( 2a \right )^{2}-2a.3+\left ( 3 \right )^{2} \right \}\left \{ \left ( 2a \right )^{2}+2a.3+\left ( 3 \right )^{2} \right \}\\\\=\left ( 2a+3 \right )\left ( 2a-3 \right )\left \{ \left ( 2a \right )^{2}-2a.3+\left ( 3 \right )^{2} \right \}\left \{ \left ( 2a \right )^{2}+2a.3+\left ( 3 \right )^{2} \right \}\\\\=\left ( 2a+3 \right )\left \{ \left ( 2a \right )^{2}-2a.3+\left ( 3 \right )^{2} \right \}\left ( 2a-3 \right )\left \{ \left ( 2a \right )^{2}+2a.3+\left ( 3 \right )^{2} \right \}\\\\=\left \{ \left ( 2a \right )^{3} +\left ( 3 \right )^{3}\right \}\left \{ \left ( 2a \right )^{3} -\left ( 3 \right )^{3}\right \}\\\\=\left ( 8a^{3}+27 \right )\left ( 8a^{3}-27 \right )\\\\=\left ( 8a^{3} \right )^{2}-\left ( 27 \right )^{2}=64a^{6}-729 

c). 3x\left \{ \left ( 2x-1 \right )^{2}-\left ( 2x-1 \right )\left ( x+1 \right )+\left ( x+1 \right )^{2} \right \} 

উত্তরঃ 

3x\left \{ \left ( 2x-1 \right )^{2}-\left ( 2x-1 \right )\left ( x+1 \right )+\left ( x+1 \right )^{2} \right \}\\\\=\left \{ \left ( 2x-1 \right ) +\left ( x+1 \right )\right \}\left \{ \left ( 2x-1 \right )^{2}-\left ( 2x-1 \right )\left ( x+1 \right )+\left ( x+1 \right )^{2} \right \}\\\\=\left ( 2x-1 \right )^{3}+\left ( x+1 \right )^{3}\\\\=8x^{3}-1-12x^{2}+6x+x^{3}+3x^{2}+3x+1\\\\=9x^{3}-9x^{2}+9x=9x\left ( x^{2} -x+1\right ) 

vii). 4x+\frac{9}{x}=6 হলে, \left ( 16x^{3}+2 \right ) এর মান  নির্ণয় করো।  

উত্তরঃ 

4x+\frac{9}{x}=6\\\\\Rightarrow \frac{4x^{2}+9}{x}=6\\\\\Rightarrow 4x^{2}+9=6x\\\\\Rightarrow4x^{2}-6x+9=0 
\left ( 16x^{3}+2 \right )\\\\=2\left ( 8x^{3}+27 \right )-52\\\\=2\left \{ \left ( 2x \right )^{3}+\left ( 3 \right )^{3} \right \}-52 \\\\=2\left ( 2x+3 \right )\left ( 4x^{2} -6x+9\right )-52 
4x^{2}-6x+9=0 এর মান বসিয়ে পাই, 
=2\left ( 2x+3 \right )\left ( 4x^{2} -6x+9\right )-52\\\\=2\left ( 2x+3 \right )\times 0-52=-52 
\therefore \left ( 16x^{3}+2 \right )=-52 

viii). \left ( a^{2}+b^{2}\right )^{3}=\left ( a^{3}+b^{3} \right )^{2} হলে, \left ( \frac{a}{b}+\frac{b}{a} \right ) এর মান  নির্ণয় করো।  

উত্তরঃ 

\left ( a^{2}+b^{2}\right )^{3}=\left ( a^{3}+b^{3} \right )^{2}\\\\\Rightarrow a^{6}+b^{6}+3a^{4}b^{2}+3a^{2}b^{4}=a^{6}+b^{6}+2a^{3}b^{3}\\\\\Rightarrow 3a^{2}b^{2}\left ( a^{2}+b^{2} \right )=2a^{3}b^{3}\\\\\Rightarrow 3\left ( a^{2}+b^{2} \right )=2ab\\\\\Rightarrow \frac{\left ( a^{2}+b^{2} \right )}{ab}=\frac{2}{3}\\\\\Rightarrow \frac{a}{b}+\frac{b}{a}=\frac{2}{3}\\\\\therefore \frac{a}{b}+\frac{b}{a}=\frac{2}{3}

ix).  \left ( a+\frac{1}{a}-1 \right )^{3}-\left ( a-\frac{1}{a}-1 \right )^{3}-\frac{6}{a}\left ( a+\frac{1}{a}-1 \right )\left ( a-\frac{1}{a}-1 \right ) 

উত্তরঃ 

\left ( a+\frac{1}{a}-1 \right )^{3}-\left ( a-\frac{1}{a}-1 \right )^{3}-\frac{6}{a}\left ( a+\frac{1}{a}-1 \right )\left ( a-\frac{1}{a}-1 \right )\\\\=\left ( a+\frac{1}{a}-1 \right )^{3}-\left ( a-\frac{1}{a}-1 \right )^{3}-3\left ( a+\frac{1}{a}-1 \right )\left ( a-\frac{1}{a}-1 \right )\\\\\left \{ \left ( a+\frac{1}{a}-1 \right )- \left ( a-\frac{1}{a}-1 \right )\right \}\\\\=\left \{ \left ( a+\frac{1}{a}-1 \right )- \left ( a-\frac{1}{a}-1 \right )\right \}^{3}\\\\=\left ( \frac{2}{a} \right )^{3}=\frac{8}{a^{3}} 

x. একটি ঘনকের আয়তন \left \{ \left ( 2m+3n \right )^{3}+\left ( 2m-3n \right )^{3}+12m\left ( 4m^{2}-9n^{2} \right ) \right \} ঘন একক হলে, ঘনকের একটি বাহুর দৈর্ঘ্য  নির্ণয় করো। 

উত্তরঃ 

\left \{ \left ( 2m+3n \right )^{3}+\left ( 2m-3n \right )^{3}+12m\left ( 4m^{2}-9n^{2} \right ) \right \}\\\\=\left ( 2m+3n \right )^{3}+\left ( 2m-3n \right )^{3}+3.\left ( 2m+3n \right )\left ( 2m-3n\right )\left \{ \left ( 2m+3n \right )+\left ( 2m-3n\right ) \right \}\\\\=\left \{ \left ( 2m+3n \right )+\left ( 2m-3n\right ) \right \}^{3}=\left ( 4m \right )^{3} 
\therefore  ঘনকের একটি বাহুর দৈর্ঘ্য =4m একক। 

CLASS-VIII-দাগ নাম্বার -3 -Annual Examination—2021/MODEL QUESTION SOLVED/Mathematics:

http://sdtutoronline.com/class-8-importat-wbmath-suggestion/

Class-VIII-দাগ নাম্বার-1&2,Annual:

http://sdtutoronline.com/class-viii-model-math-question-wb/

Class -VIII/ Model activity task / বহু বিকল্পভিত্তিক প্রশ্নঃ পরিচিতি ও অনুশীলনী গণিত/ অষ্টম শ্রেণী:

http://sdtutoronline.com/wb-class-viii-model-activity-task-math-8/

\ast \ast  পরের দাগের সমাধান কিছুদিনের মধ্যে দেওয়া হবে এই  সাইট টি তে।

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