Algebraic Function MCQ Quiz in বাংলা - Objective Question with Answer for Algebraic Function - বিনামূল্যে ডাউনলোড করুন [PDF]

Last updated on Mar 27, 2025

পাওয়া Algebraic Function उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). এই বিনামূল্যে ডাউনলোড করুন Algebraic Function MCQ কুইজ পিডিএফ এবং আপনার আসন্ন পরীক্ষার জন্য প্রস্তুত করুন যেমন ব্যাঙ্কিং, এসএসসি, রেলওয়ে, ইউপিএসসি, রাজ্য পিএসসি।

Latest Algebraic Function MCQ Objective Questions

Algebraic Function Question 1:

\(\sqrt{3 + 2 \sqrt{2}}\) - \(\sqrt{3 - 2 \sqrt{2}}\)  সমান:

  1. 1
  2. \(\sqrt{3} + 1\)
  3. \(\sqrt{3} - 1\)
  4. 2

Answer (Detailed Solution Below)

Option 4 : 2

Algebraic Function Question 1 Detailed Solution

প্রদত্ত:

\(X=\sqrt{3 + 2 \sqrt{2}}-\sqrt{3 - 2 \sqrt{2}}\)

\(=\sqrt {{(2 + 1) + 2} \sqrt 2}-\sqrt {{(2 + 1) - 2} \sqrt 2}\)

\(=\sqrt {{{{\left( {\sqrt 2 } \right)}^2} + {1^2}} + 2\sqrt 2}-\sqrt {{{{\left( {\sqrt 2 } \right)}^2} + {1^2}} - 2\sqrt 2}\)

\(=\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}}-\sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}}\)

= √2 + 1 – (√2 – 1)

= 2

Top Algebraic Function MCQ Objective Questions

\(\sqrt{3 + 2 \sqrt{2}}\) - \(\sqrt{3 - 2 \sqrt{2}}\)  সমান:

  1. 1
  2. \(\sqrt{3} + 1\)
  3. \(\sqrt{3} - 1\)
  4. 2

Answer (Detailed Solution Below)

Option 4 : 2

Algebraic Function Question 2 Detailed Solution

Download Solution PDF

প্রদত্ত:

\(X=\sqrt{3 + 2 \sqrt{2}}-\sqrt{3 - 2 \sqrt{2}}\)

\(=\sqrt {{(2 + 1) + 2} \sqrt 2}-\sqrt {{(2 + 1) - 2} \sqrt 2}\)

\(=\sqrt {{{{\left( {\sqrt 2 } \right)}^2} + {1^2}} + 2\sqrt 2}-\sqrt {{{{\left( {\sqrt 2 } \right)}^2} + {1^2}} - 2\sqrt 2}\)

\(=\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}}-\sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}}\)

= √2 + 1 – (√2 – 1)

= 2

Algebraic Function Question 3:

\(\sqrt{3 + 2 \sqrt{2}}\) - \(\sqrt{3 - 2 \sqrt{2}}\)  সমান:

  1. 1
  2. \(\sqrt{3} + 1\)
  3. \(\sqrt{3} - 1\)
  4. 2

Answer (Detailed Solution Below)

Option 4 : 2

Algebraic Function Question 3 Detailed Solution

প্রদত্ত:

\(X=\sqrt{3 + 2 \sqrt{2}}-\sqrt{3 - 2 \sqrt{2}}\)

\(=\sqrt {{(2 + 1) + 2} \sqrt 2}-\sqrt {{(2 + 1) - 2} \sqrt 2}\)

\(=\sqrt {{{{\left( {\sqrt 2 } \right)}^2} + {1^2}} + 2\sqrt 2}-\sqrt {{{{\left( {\sqrt 2 } \right)}^2} + {1^2}} - 2\sqrt 2}\)

\(=\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}}-\sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}}\)

= √2 + 1 – (√2 – 1)

= 2

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