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

જો વિધેય \(f : [0,1]\,\to R\) આપેલ છે કે જેથી  \(f\,(xy) = f\,(x)\,f\,(y)\) દરેક \(x,y\,\in [0,1]\) માટે શક્ય થાય  અને \(f \,(0)\,\ne 0.\) જો \(y=y\,(x)\) એ વિકલ સમીકરણ \(\frac{{dy}}{{dx}} = f(x)\) નો ઉકેલ છે અને \(y(0) = 1\) તો \(y\left( {\frac{1}{4}} \right) + y\left( {\frac{3}{4}} \right)\) ની કિમંત મેળવો.

  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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