Question

Let f^{"}(x) > 0 \forall x \epsilon R and g(x) = f(2-x)+f(4+x). Then g(x) is increasing in

• A. $(-\infty ,-1)$
• B. $(-\infty ,0)$
• C. $(-1,-\infty )$
• D. $(1,\infty )$

Answer 1

The correct option is C $(-1,-\infty )$

$f^{\prime \prime }\left(x\right)>0\forall xϵR$
$\Rightarrow f^{{}^{\prime }}\left(x\right)$ is increasing $\forall xϵR$
Here, for g(x) to be increasing function $g^{{}^{\prime }}\left(x\right)>0$
$-f^{{}^{\prime }}(2-x)+f^{{}^{\prime }}(4+x)>0$
$f^{{}^{\prime }}(4+x)>f^{{}^{\prime }}(2-x)$
But $f^{{}^{\prime }}\left(x\right)$ is increasing function
$4+x>2-x$ $\forall xϵR$
$x>-1$
$xϵ(-1,\infty )$