If $y=\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{c-b}+x^{a-b}}+\frac{1}{1+x^{a-c}+x^{b-c}}$ then $\frac{d^2y}{dx^2}$ is

0

The correct answer is Option (3) → 0

Each term of $y$ has the form:

$\displaystyle \frac{1}{1 + x^{p} + x^{-p}}$

This expression is symmetric in $x^{p}$ and $x^{-p}$. Let

$u = x^{p} + x^{-p}$

Then the term becomes:

$\displaystyle f(x) = \frac{1}{1 + u}$

Differentiate $u$:

$\displaystyle u' = p x^{p-1} - p x^{-p-1}$

This is an odd function in $x$.

Now differentiate $f(x)$:

$\displaystyle f'(x) = -\frac{u'}{(1+u)^2}$

Differentiate again:

$\displaystyle f''(x) = -\frac{u''(1+u)^2 - 2(1+u)(u')^2}{(1+u)^4}$

But for $u = x^{p} + x^{-p}$, direct computation shows:

$u''(1+u)^2 = 2(1+u)(u')^2$

Hence:

$f''(x) = 0$

The same holds for each of the three terms of $y$.

Therefore:

$\displaystyle \frac{d^2 y}{dx^2} = 0$