Question

Given $f:[0,\infty )\to R$ be a strictly increasing function such that the functions $g\left(x\right)=f\left(x\right)-3x$ and $h\left(x\right)=f\left(x\right)-x^3$ are both strictly increasing function. Then the function $F\left(x\right)=f\left(x\right)-x^2-x$ is

• A. increasing in $(0,1)$ and decreasing in $(1,\infty )$
• B. decreasing in $(0,1)$ and increasing in $(1,\infty )$
• C. increasing throughout $(0,\infty )$
• D. decreasing throughout $(0,\infty )$

Answer 1

The correct option is C increasing throughout $(0,\infty )$

$g\left(x\right)=f\left(x\right)-3x\Rightarrow g^{\prime }\left(x\right)=f^{\prime }\left(x\right)-3...\left(1\right)$
$h\left(x\right)=f\left(x\right)-x^3\Rightarrow h^{\prime }\left(x\right)=f^{\prime }\left(x\right)-3x^2...\left(2\right)$
$F\left(x\right)=f\left(x\right)-x^2-x\Rightarrow F^{\prime }\left(x\right)=f^{\prime }\left(x\right)-2x-1$
$\Rightarrow F^{\prime }\left(x\right)=\frac{1}{2}\left(f^{\prime }\right(x)+f^{\prime }(x\left)\right)-2x-1$
Now using (1) and (2)
$F^{\prime }\left(x\right)=\frac{1}{2}\left(g^{\prime }\right(x)+h^{\prime }(x)+3x^2-4x-1)$
$F^{\prime }\left(x\right)=\frac{1}{2}\left(g^{\prime }\right(x)+h^{\prime }(x)+3((x-2/3{)}^2+1/6)$
Since $h^{\prime }\left(x\right),g^{\prime }\left(x\right)>0$
$\Rightarrow F^{\prime }\left(x\right)>0$
Hence $F\left(x\right)$ is increasing in $(0,\infty )$