Question:

For a twice continuously differentiable function 𝑔: ℝ β†’ ℝ, define
\(u_g(x,y)=\frac{1}{y}\int^y_{-y}g(x+t)dt\ \ \ \text{for}(x,y)\in \R^2, \ \ \ y \gt0.\)
Which one of the following holds for all such 𝑔 ?

Updated On: Dec 1, 2024
  • \(\frac{βˆ‚^2u_g}{βˆ‚x^2}=\frac{2}{y}\frac{βˆ‚u_g}{βˆ‚y}+\frac{βˆ‚^2u_g}{βˆ‚y^2}\)
  • \(\frac{βˆ‚^2u_g}{βˆ‚x^2}=\frac{1}{y}\frac{βˆ‚u_g}{βˆ‚y}+\frac{βˆ‚^2u_g}{βˆ‚y^2}\)
  • \(\frac{βˆ‚^2u_g}{βˆ‚x^2}=\frac{2}{y}\frac{βˆ‚u_g}{βˆ‚y}-\frac{βˆ‚^2u_g}{βˆ‚y^2}\)
  • \(\frac{βˆ‚^2u_g}{βˆ‚x^2}=\frac{1}{y}\frac{βˆ‚u_g}{βˆ‚y}-\frac{βˆ‚^2u_g}{βˆ‚y^2}\)
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The Correct Option is A

Solution and Explanation

The correct option is (A) : \(\frac{βˆ‚^2u_g}{βˆ‚x^2}=\frac{2}{y}\frac{βˆ‚u_g}{βˆ‚y}+\frac{βˆ‚^2u_g}{βˆ‚y^2}\).
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