Question

If $g\left(x\right)=f\left(x\right)+f(x-1)$ and $f^{\prime \prime }\left(x\right)<0,0\le x\le 1$ then find the intervals of monotonicity of g(x).

• A. $xϵ[\frac{1}{2},1]$
• B. $xϵ[\frac{-1}{2},1]$
• C. $xϵ[\frac{1}{3},1]$
• D. $xϵ[\frac{-1}{3},1]$

Answer 1

The correct option is A $xϵ[\frac{1}{2},1]$

$g^{\prime }\left(x\right)=f^{\prime }\left(x\right)-f^{\prime }(1-x)$ Now $f^{\prime \prime }\left(x\right)<0$ when $0\le x\le 1$

Hence f'(x) is a decreasing function when $0\le x\le 1.$

Now g(x) is increasing or decreasing according $g^{\prime }\left(x\right)=f^{\prime }\left(x\right)-f(1-x)>0$ or $<0$ or $f^{\prime }\left(x\right)>f^{\prime }(1-x)$ for increasing i.e.$f^{\prime }(1-x)<f^{\prime }\left(x\right)...\left(1\right)$

Since f' is a decreasing function then by definition later value is less than the earlier value i.e.$f^{\prime }\left(a\right)<f^{\prime }\left(b\right)$ when $a>b.$ or $f^{\prime }\left(x\right)<f^{\prime }(1-x)$ for decreasing ...(2)

Hence $\left(1\right)\Rightarrow (1-x)>x$ or $1>2x$ or $x<\frac{1}{2}$

$\therefore xϵ[0,\frac{1}{2}]$ for $g\left(x\right)$ to be increasing and $\Rightarrow x>(1-x)$ or $2x>1$ or $x>\frac{1}{2}$ $\therefore xϵ[\frac{1}{2},1]$ for $g\left(x\right)$ to be decreasing.