If \(y=\displaystyle\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{c-b}+x^{a-b}}+\frac{1}{1+x^{a-c}+x^{b-c}}\) then \(\displaystyle\frac{d^2y}{dx^2}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0
Multiplying the numerator and denominator of the first term by \(x^a\):
\( \displaystyle \frac{1}{1+x^{b-a}+x^{c-a}} = \frac{x^a}{x^a+x^b+x^c} \)
In the same way, multiplying the second term by \(x^b\) and the third by \(x^c\):
\( \displaystyle \frac{1}{1+x^{c-b}+x^{a-b}} = \frac{x^b}{x^a+x^b+x^c} \)
\( \displaystyle \frac{1}{1+x^{a-c}+x^{b-c}} = \frac{x^c}{x^a+x^b+x^c} \)
All three terms now have the same denominator, so
\( \displaystyle y = \frac{x^a+x^b+x^c}{x^a+x^b+x^c} = 1 \)
So y is a constant, and every derivative of it is 0:
\( \displaystyle \frac{dy}{dx} = 0,\quad \frac{d^2y}{dx^2} = 0 \)