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2x - 3y - 3 এবং -4x + qy - \(\dfrac{p}{2}\) সমীকরণের সিস্টেমটি অসঙ্গতিপূর্ণ হলে নীচের কোনটি p এর মান হতে পারে না?

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DSSSB TGT Maths Female Subject Concerned - 22 Sep 2018 Shift 2
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  1. -18
  2. -24
  3. -12
  4. -36

Answer (Detailed Solution Below)

Option 3 : -12
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Detailed Solution

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ধারণা:

  • যদি একটি সিস্টেমের কোনও সমাধান না থাকে তবে এটিকে অসঙ্গতিপূর্ণ বলা হয়।
  • সমীকরণের একটি সিস্টেম যার কোনো সমাধান নেই,

a1 x + b1 y = c1 এবং a2 x + b2 y = c2

গণনা:

প্রদত্ত সমীকরণগুলি হল 2x - 3y - 3 = 0 এবং -4x + qy - = 0

সুতরাং, এখানে এই সমীকরণগুলিকে সমীকরণ (1) এবং (2) এর সাথে তুলনা করে পাই,

a1 = 2, b1 = -3, c1 = -3;

a2 = -4, b2 = q, c2 = 

এখন, একটি সিস্টেম অসঙ্গতিপূর্ণ হওয়ার জন্য,

সুতরাং, প্রদত্ত সিস্টেমটি অসঙ্গতিপূর্ণ যখন p ≠ 12

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