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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

माना \(f:[0,1] \rightarrow R\) इस प्रकार है कि सभी \(x , y , \varepsilon[0,1]\) के लिए \(f( xy )=f( x ) . f( y ) x , y , \varepsilon[0,1]\) है तथा \(f(0) \neq 0\) है। यदि \(y = y ( x )\) अवकल समीकरण \(\frac{ dy }{ dx }=f( x )\) को संतुष्ट करता है और \(y (0)=1\) है, तो \(y \left(\frac{1}{4}\right)+ y \left(\frac{3}{4}\right)\) बराबर है

  1. A \(4\)
  2. B \(3\)
  3. C \(5\)
  4. D \(2\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3\)

Step-by-step Solution

Detailed explanation

\(f\left( {xy} \right) = f\left( x \right).f\left( y \right)\) \(f\left( 0 \right) = 1\) as \(f\left( 0 \right) \ne 0\) \( \Rightarrow f\left( x \right) = 1\) \(\frac{{dy}}{{dx}} = f\left( x \right) = 1\) \( \Rightarrow y = x + c\) At, \(x = 0,y = 1 \Rightarrow c = 1\)…
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