Let gx-log fx where fx is a twice differentiable positive function on 0- - such that fx-1-x fx- Then- for N- 1-2-3- - - - g- N -1-2- g -1-2-41-1-9-1-25- -1-2 N-12-41-1-9-1-25- - -1-2 N-12C- 41-1-9-1-25- -1-2 N-12D- 41-1-9-1-25- - -1-2 N-12

The correct option is A g”(x+1) g”(x)= 1/x^2 Replace x by (x 1/2) g^" ( x+1/2 ) g^" ( x 1/2 )= 4/(2x 1)^2 Put x = 1, 2, 3, ....... N and then add ...

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Discriminant

Let gx=log fx...

Question

Let $g (x) = l o g f (x)$ where f(x) is a twice differentiable positive function on $0, \infty$ such that $f (x + 1) = x f (x)$ . Then, for $N = 1, 2, 3, . . . . . g^{''} (N + \frac{1}{2}) - g^{''} (\frac{1}{2})$ =

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g (x) = log (f (x))

f (x)

(0, \infty)

f (x + 1) = x f (x)

N = 1, 2, 3, \dots g^{''} (N + \frac{1}{2}) - g^{''} (\frac{1}{2}) =

$(1 + \frac{3}{1}) (1 + \frac{5}{4}) (1 + \frac{7}{9}) \dots (1 + \frac{(2 n + 1)}{n^{2}}) = (n + 1)^{2}$

(1 + \frac{3}{1}) (1 + \frac{5}{4}) (1 + \frac{7}{9}) . . . (1 + \frac{(2 n + 1)}{n^{2}}) = {(n + 1)}^{2}

(1 + \frac{3}{1}) (1 + \frac{5}{4}) (1 + \frac{7}{9}) . . . (1 + \frac{(2 n + 1)}{n^{2}}) = {a (n + 1)}^{2}

Prove the following by using the principle of mathematical induction for all n ∈ N:

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