Solution of sample questions -VII

by

5. নিম্ন লিখিত প্রশ্নগুলির উত্তর দাও [প্রশ্নমান 3]

i). \left ( 5+9x \right ) এবং \left ( 6-7x+4x^{2} \right )  যোগফল থেকে \left ( x^{2}-9x \right ) এবং \left ( -2x^{2}+3x+5 \right )
এর যোগফল বিয়োগ করো। 

উত্তরঃ 

\left ( 5+9x \right ) এবং \left ( 6-7x+4x^{2} \right )  যোগফল =\left ( 5+9x \right ) +\left ( 6-7x+4x^{2} \right )\\\\= 5+9x +6-7x+4x^{2} \\\\=4x^{2}+9x -7x+5+6\\\\=4x^{2}+2x+11.......(i) 
এবং \left ( x^{2}-9x \right )

এবং \left ( -2x^{2}+3x+5 \right ) এর যোগফল =\\\left ( x^{2}-9x \right )+ \left ( -2x^{2}+3x+5 \right )\\\\= x^{2}-9x -2x^{2}+3x+5\\\\= x^{2}-2x^{2}-9x+3x+5\\\\=-x^{2}-6x+5.............(ii) 
(i) থেকে (ii) বিয়োগ করে পাই, 
\\\left ( 4x^{2}+2x+11 \right )-\left (-x^{2}-6x+5 \right )\\\\=4x^{2}+2x+11+x^{2}+6x-5\\\\= 5x^{2}+8x+6 

ii). যদি A=2a+3b-5c,B=2b+3c-7a,C=2c+3a-4b হয়, তবে \left ( A-B+C+2 \right ) এর মান নির্ণয় করো।  

উত্তরঃ 

\because

A=2a+3b-5c,B=2b+3c-7a,C=2c+3a-4b
\therefore \left ( A-B+C+2 \right )\\\\=\left ( 2a+3b-5c \right )-\left ( 2b+3c-7a \right )+\left ( 2c+3a-4b \right )+2\\\\=2a+3b-5c - 2b-3c+7a + 2c+3a-4b +2\\\\=2a+7a+3a+3b-2b-4b-5c-3c+2c+2\\\\=12a-3b-6c+2 

iii).  প্রথম বীজগাণিতিক সংখ্যামালাকে দ্বিতীয় বীজগাণিতিক সংখ্যামালা দিয়ে ভাগ করো।

 m^{2}n^{4}+m^{3}n^{3}-m^{4}n^{2}, -m^{4}n^{4} 

উত্তরঃ 

\therefore \left (m^{2}n^{4}+m^{3}n^{3}-m^{4}n^{2} \right )\div \left (-m^{4}n^{4} \right )\\\\=\frac{m^{2}n^{4}+m^{3}n^{3}-m^{4}n^{2}}{-m^{4}n^{4} }\\\\=\frac{m^{2}n^{4}}{-m^{4}n^{4}}+\frac{m^{3}n^{3}}{-m^{4}n^{4}}-\frac{m^{4}n^{2}}{-m^{4}n^{4} }\\\\=-m^{2-4}n^{4-4}-m^{3-4}n^{3-4}+m^{4-4}n^{2-4}\\\\=-m^{-2}-m^{-1}n^{-1}+n^{-2}\\\\=-\frac{1}{m^{2}}-\frac{1}{mn}+\frac{1}{n^{2}} 

iv). সরল করো : \left ( b-c \right )\left ( b^{2}+bc+c^{2} \right )+\left ( c-a \right )\left ( c^{2}+ca+a^{2} \right )\\\\+\left ( a-b \right )\left ( a^{2}+ab+b^{2} \right )
 

উত্তরঃ 

\\\left ( b-c \right )\left ( b^{2}+bc+c^{2} \right )+\left ( c-a \right )\left ( c^{2}+ca+a^{2} \right )+\left ( a-b \right )\left ( a^{2}+ab+b^{2} \right )\\\\=b\left ( b^{2}+bc+c^{2} \right )-c\left ( b^{2}+bc+c^{2} \right )+c\left ( c^{2}+ca+a^{2} \right )-a\left ( c^{2}+ca+a^{2} \right )\\\ +a\left ( a^{2}+ab+b^{2} \right )-b\left ( a^{2}+ab+b^{2} \right )\\\\=b^{3}+b^{2}c+bc^{2}-cb^{2}-bc^{2}-c^{3}+c^{3}+ac^{2}+a^{2}c-ac^{2}-ca^{2}-a^{3}\\\\+a^{3}+a^{2}b+ab^{2}-a^{2}b-ab^{2}-b^{3}\\\\=0 

v). গুনফল নির্ণয় করো : 

\left ( -\frac{3}{5}s^{2}t \right )\times \left ( \frac{15}{7}st^{2}u \right )\times \left ( \frac{7}{9}su^{2} \right ) 

উত্তরঃ 

\left ( -\frac{3}{5}s^{2}t \right )\times \left ( \frac{15}{7}st^{2}u \right )\times \left ( \frac{7}{9}su^{2} \right )\\\\=\left (-\frac{3}{5}\times \frac{15}{7}\times \frac{7}{9} \right )\times s^{2}t \times st^{2}u \times su^{2} \\\\=-s^{2+1+1}t^{1+2}u^{1+2}=-s^{4}t^{3}u^{3} 

vi). সরল করো : 6a-2b-ab-\left ( 3a+b-ab \right )+2ab-b+a 

উত্তরঃ 

6a-2b-ab-\left ( 3a+b-ab \right )+2ab-b+a\\\\=6a-2b-ab- 3a-b+ab+2ab-b+a\\\\=6a-3a+a-2b-b-b-ab+ab+2ab\\\\=3a+a-4b+2ab\\\\=4a-4b+2ab 

vii). বিস্তার করে সংখ্যাটি লেখো :

 9\times 10^{4}+5\times 10^{3}+6\times 10^{2}+7\times 10 

উত্তরঃ 

\\9\times 10^{4}+5\times 10^{3}+6\times 10^{2}+7\times 10\\\\=90000+5000+600+70\\\\=95670 

viii). পূর্ণবর্গাকারে প্রকাশ করো : 

p^{2}q^{2}+10pqr+25r^{2} যখন p=2, q=-1, r=3 

উত্তরঃ 

\\p^{2}q^{2}+10pqr+25r^{2}\\\\=\left ( pq \right )^{2}+2.pq.5r+\left ( 5r \right )^{2} \\\\=\left ( pq+5r \right )^{2} \because \left ( a+b \right )^{2}=a^{2}+2.a.b+b^{2}যখন p=2, q=-1, r=3
\\\therefore \left ( pq+5r \right )^{2}=\left ( 2\times -1+5\times 3 \right )^{2}\\\\=\left ( -2+15 \right )^{2}=\left ( 13 \right )^{2}=169 

ix). 8xy\left ( x^{2}+y^{2} \right )  এর মান লিখি যখন, \left ( x+y \right )=5, \left ( x-y \right )=1 

উত্তরঃ 

আমরা জানি যে, 4ab=\left ( a+b \right )^{2}-\left ( a-b \right )^{2}  এবং 2\left ( a^{2} +b^{2}\right )=\left ( a+b \right )^{2}+\left ( a-b \right )^{2} 
8xy\left ( x^{2}+y^{2} \right ) \\\\=4xy\times 2\left ( x^{2}+y^{2} \right )\\\\=\left \{ \left ( x+y \right )^{2}-\left ( x-y \right )^{2} \right \}\times \left \{ \left ( x+y \right )^{2}+\left ( x-y \right )^{2} \right \}\\\\=\left \{ \left ( 5 \right )^{2}-\left ( 1 \right )^{2} \right \}\times \left \{ \left ( 5 \right )^{2}+\left ( 1 \right )^{2} \right \}\\\\=\left \{ 25 - 1 \right \}\times \left \{ 25+ 1 \right \}\\\\=24\times 26=624 

x). \frac{x^{2}+y^{2}}{2xy} এর মান লিখি যখন \left ( x+y \right )=9 এবং \left ( x-y \right )=5 

উত্তরঃ 

\frac{x^{2}+y^{2}}{2xy}=\frac{2\left (x^{2}+y^{2} \right )}{4xy} (লব ও হর কে 2 দিয়ে গুন করে পাই )
আমরা জানি যে, 4ab=\left ( a+b \right )^{2}-\left ( a-b \right )^{2} \\\\2\left ( a^{2} +b^{2}\right )=\left ( a+b \right )^{2}+\left ( a-b \right )^{2} 
\\\therefore \frac{2\left (x^{2}+y^{2} \right )}{4xy}=\frac{\left \{ \left ( x+y \right )^{2}+\left ( x-y \right )^{2} \right \}}{\left \{ \left ( x+y \right )^{2}-\left ( x-y \right )^{2} \right \}}\\\\=\frac{\left \{ \left ( 9 \right )^{2}+\left (5 \right )^{2} \right \}}{\left \{ \left ( 9 \right )^{2}-\left ( 5 \right )^{2} \right \}}=\frac{81+25 }{ 81-25}\left \{ \because \left ( x+y \right )=9; \left ( x-y \right )=5 \right \} 
=\frac{106}{56}=\frac{53}{28} 

xi). 8x^{2}+50y^{2} কে দুটি বর্গের সমষ্টিরূপে প্রকাশ করো। 

উত্তরঃ 

8x^{2}+50y^{2}=2\left ( 4x^{2}+25y^{2} \right )\\\\=2\left \{ (2x)^{2}+(5y)^{2} \right \}\\\\=\left \{ \left ( 2x+5y \right )^{2} +\left ( 2x-5y \right )^{2}\right \} 

xii). k  এর মান বা মানগুলির জন্য c^{2}+kc+\frac{1}{9} পূর্ণবর্গ হবে লেখো। 

উত্তরঃ 

c^{2}+kc+\frac{1}{9}\\\\=\left ( c \right )^{2}+2.c.\frac{k}{2}+\left ( \frac{k}{2} \right )^{2}-\left ( \frac{k}{2} \right )^{2}+\frac{1}{9}\\\\=\left ( c+\frac{k}{2} \right )^{2}+\left ( -\frac{k^{2}}{4}+\frac{1}{9} \right ) 
উপরের রাশিটি পূর্ণবর্গ হবে যদি \left ( -\frac{k^{2}}{4}+\frac{1}{9} \right )=0 হয় 
\therefore \left ( -\frac{k^{2}}{4}+\frac{1}{9} \right )=0\\\\\Rightarrow \frac{k^{2}}{4}=\frac{1}{9} \\\\\Rightarrow k^{2}=\frac{4}{9}\\\\\therefore k=\pm \sqrt{\frac{4}{9}}=\pm \frac{2}{3}  

xiii). \left ( x-y \right )=3, xy=28 হলে \left ( x^{2}+y^{2} \right ) মান  লেখো। 

উত্তরঃ 

\because \left ( x-y \right )=3, xy=28 
\left ( x^{2}+y^{2} \right )=\left ( x-y \right )^{2}+2xy\\\\=\left ( 3 \right )^{2}+2\times 28=9+56=65 

xiv). l^{2}+m^{2}=13 এবং l+m=5 হলে lm এর মান লেখো। 

উত্তরঃ 

\because l^{2}+m^{2}=13 এবং l+m=5 
আমরা জানি যে, l^{2}+m^{2}=\left ( l+m \right )^{2}-2lm 
l^{2}+m^{2}=\left ( l+m \right )^{2}-2lm\\\\\therefore 13=\left ( 5 \right )^{2}-2lm\\\\\Rightarrow 2lm=25-13=12\\\\\Rightarrow lm=\frac{12}{2}=6\\\\\therefore lm=6 

xv). 2x-\frac{1}{x}=4 হলে x^{2}-\frac{1}{4x^{2}} এর মান লেখো। 

উত্তরঃ 

\because 2x-\frac{1}{x}=4\\\\\Rightarrow 2\left ( x-\frac{1}{2x} \right )=4\\\\\Rightarrow \left ( x-\frac{1}{2x} \right )=2........(i) 
(i) নং সমীকরনের উভয়পক্ষকে বর্গ করে পাই, 
\left ( x-\frac{1}{2x} \right )^{2}=2^{2}\\\\\Rightarrow x^{2}+\frac{1}{4x^{2}}-2.x.\frac{1}{2x}=4\\\\\Rightarrow x^{2}+\frac{1}{4x^{2}}=4+1=5 

xvi). a^{2}+b^{2}=5ab হলে দেখাও যে \frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}=23 

উত্তরঃ 

a^{2}+b^{2}=5ab\\\\\Rightarrow \frac{a^{2}+b^{2}}{ab}=5\\\\\Rightarrow \frac{a^{2}}{ab}+\frac{b^{2}}{ab}=5\\\\\Rightarrow \frac{a}{b}+\frac{b}{a}=5.........(i) 
(i) নং সমীকরনের উভয়পক্ষকে বর্গ করে পাই, 
\frac{a}{b}+\frac{b}{a}=5\\\\\Rightarrow \left (\frac{a}{b}+\frac{b}{a} \right )^{2}=(5)^{2}\\\\\Rightarrow \left (\frac{a}{b} \right )^{2}+\left (\frac{b}{a} \right )^{2}+2.\frac{a}{b}.\frac{b}{a} =25\\\\\Rightarrow \frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}=25-2=23\\\\\therefore \frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}=23 

xvii). m-\frac{1}{m-2}=6 হলে \left ( m-2 \right )^{2}+\frac{1}{\left ( m-2 \right )^{2}} এর মান লেখো।  

উত্তরঃ 

m-\frac{1}{m-2}=6\\\\\Rightarrow m-\frac{1}{m-2}=4+2\\\\\Rightarrow m-2-\frac{1}{m-2}=4\\\\\Rightarrow \left ( m-2 \right )-\frac{1}{\left (m-2 \right )}=4..........(i) 
(i) নং সমীকরণের উভয়পক্ষকে বর্গ করে পাই, 
\left \{ \left ( m-2 \right )-\frac{1}{\left (m-2 \right )} \right \}^{2}=\left (4 \right )^{2}\\\\\Rightarrow \left ( m-2 \right )^{2}+\frac{1}{\left (m-2 \right )^{2}}-2.\left ( m-2 \right ).\frac{1}{\left (m-2 \right )}=16\\\\\Rightarrow \left ( m-2 \right )^{2}+\frac{1}{\left (m-2 \right )^{2}}-2=16\\\\\Rightarrow \left ( m-2 \right )^{2}+\frac{1}{\left (m-2 \right )^{2}}=16+2=18\\\\\therefore\left ( m-2 \right )^{2}+\frac{1}{\left (m-2 \right )^{2}}=18 

xviii). 2x+\frac{1}{x}=5 হলে 4x^{2}+\frac{1}{x^{2}} এর মান  লেখো। 

উত্তরঃ 

2x+\frac{1}{x}=5......(i) 
(i) নং সমীকরণের উভয়পক্ষকে বর্গ করে পাই, 
\left (2x+\frac{1}{x} \right )^{2}=\left (5 \right )^{2}\\\\\Rightarrow \left ( 2x \right )^{2}+\left ( \frac{1}{x} \right )^{2}+2.2x.\frac{1}{x}=25\\\\\Rightarrow 4x^{2}+\frac{1}{x^{2}}+4=25\\\\\Rightarrow 4x^{2}+\frac{1}{x^{2}}=25-4=21\\\\\therefore 4x^{2}+\frac{1}{x^{2}}=21 

xix). 

xx). \left ( x^{2}+5x \right )\left ( x-1 \right )-\left ( x^{2}+2x \right )\left ( x-1 \right )+3x\left ( x+1 \right ) 

উত্তরঃ 

\\\left ( x^{2}+5x \right )\left ( x-1 \right )-\left ( x^{2}+2x \right )\left ( x-1 \right )+3x\left ( x+1 \right )\\\\=x^{2} \left ( x-1 \right )+5x\left ( x-1 \right )-x^{2}\left ( x-1 \right )-2x\left ( x-1 \right )+3x\left ( x+1 \right )\\\\= x^{3}-x^{2}+5x^{2}-5x-x^{3}+x^{2}-2x^{2}+2x+3x^{2}+3x\\\\=x^{3}-x^{3}-x^{2}+5x^{2}+x^{2}-2x^{2}+3x^{2}-5x+2x+3x\\\\=6x^{2} 

  শ্রেণী সপ্তম/-কষে দেখি-12.2

xxi). m+\frac{1}{m}=p-2 হলে দেখাও যে, m^{2}+\frac{1}{m^{2}}=p^{2}-4p+2 

উত্তরঃ 

\because m+\frac{1}{m}=p-2 

উভয়পক্ষকে বর্গকরে পাই, 
\left (m+\frac{1}{m} \right )^{2}=\left (p-2 \right )^{2}\\\\\Rightarrow \left ( m \right )^{2}+\left ( \frac{1}{m} \right )^{2}+2.m.\frac{1}{m} =p^{2}-4p+4\\\\\Rightarrow m^{2}+\frac{1}{m^{2}}+2=p^{2}-4p+4\\\\\Rightarrow m^{2}+\frac{1}{m^{2}}=p^{2}-4p+2\\\\\therefore m^{2}+\frac{1}{m^{2}}=p^{2}-4p+2 

Class-VII-Factorization-WB:

http://sdtutoronline.com/class-vii-factorization-wb/

http://sdtutoronline.com/class-vii-kosedekhi-12-2-wb-math-7-algebra/

CLASS-VII-দাগ নাম্বার -C -Annual Examination—2021/MODEL QUESTION/Mathematics

Spread/ share this post

Leave a Comment