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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A compilation of various questions in Mathematics and Physics solved by me on GoIIT
log_{7} 10 = 1 + log_{7} (10/7) = 1 + log_{7} 1.428
log_{11} 13 = 1 + log_{11} (13/11) = 1 + log_{11 } 1.18
Let A = log_{7} 1.428
and B = log_{11} 1.18
So, 7^{A} = 1.428
and 11^{B} = 1.18
Clearly A and B will both be fractional, and B will be smaller than A; (Since, in case of A, 7 is being reduced to 1.428 but with B, a larger number 11 is being reduced to a correspondingly less value than for A. )
Thus, 1 + log_{7} 1.428 > 1 + log_{11} 1.18
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