Question

Let $f$ be a differential function such that $f\left(x\right)=f(4-x)$ and $g\left(x\right)=f(2+x)$ for all $x\in R,$ then

• A. graph of $f\left(x\right)$ is symmetric about the line $x=2$
• B. $f^{\prime }\left(2\right)=0$
• C. graph of $g\left(x\right)$ is symmetric about x-axis
• D. $g^{\prime }\left(0\right)=0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A graph of $f\left(x\right)$ is symmetric about the line $x=2$
B graph of $g\left(x\right)$ is symmetric about x-axis
C $f^{\prime }\left(2\right)=0$
D $g^{\prime }\left(0\right)=0$

Given $f\left(x\right)=f(4-x)$ ....(1)

And $g\left(x\right)=f(2+x)$ ,,,(2) Putting 2+x in place of $x$ in (1),

We get $f(2+x)=f(2-x)$........3

$\Rightarrow$ graph of $f\left(x\right)$ is symmetric about the line $x=2$

From (2) and (3),

$g\left(x\right)=f(2+x)$........4

Putting $(-x)$ in place of $x$ in (2),

we get $g(-x)=f(2-x)$ .......5

From (3), (4) and (5) $g(-x)=g\left(x\right),$

Therefore $g\left(x\right)$ is an even funcction

Hence gragh of $g\left(x\right)$ is symmetric about y-axis

Differentiating both sides of (3) w.r.t $x$, we get

$f^{\prime }(2+x)=-f^{\prime }(2-x)$

Putting $x=0,$ we get $f^{\prime }\left(2\right)=0$

Again $g\left(x\right)=g(-x)$

Differentiating w.r.t $x$ we get $g^{\prime }\left(x\right)=-g^{\prime }(-x)$

Putting $x=0,$ we get $2g^{\prime }\left(0\right)=0\Rightarrow g^{\prime }\left(0\right)=0$