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
Answer & explanation
Correct answer: option 3
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:
$\displaystyle \frac{d^2 y}{dx^2} = 0$
$0$