यदि 1916x2dx=αsin1(βx)+c है, तो α+1β= क्या है?

  1. 1
  2. 712
  3. 1912
  4. 912

Answer (Detailed Solution Below)

Option 1 : 1
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संकल्पना:

1a2x2dx=sin1(xa)+c

गणना:

माना कि I = 1916x2dx है। 

=14916x2dx=141(34)2x2dx=14sin1[x(34)]+c=14sin1(4x3)+c

अब, I = 14sin1(4x3)+c=αsin1(βx)+c

इसलिए, α = 14andβ=43

निम्न का मान ज्ञात करने के लिए:

α+1β=14+34=44=1

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