A)
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By Applying partial fractions -
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BY using the statement that f(0) = 0 ........We get C = 0
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Compilation of various questions in Mathematics and Physics I previously solved on GoIIT. For what I'm working on now, see www.profillic.com
x^2 + [x^2/1+x^2 = 3 x^4 - x^2 = 3 x^2(x^2 - 1) = 3 Let x^2 = y .....thus x = root y y(y-1)=3 y^2 - y - 3 = 0 y = 1+- root(1 + 12) / 2 y = 1 +-rt(13) /2 now x = root y so x = root (1 + rt(13) / 2) other value of y is truncated since x cannot be negative due to root. x = root [ (1 + rt(13) ) / 2] |