Question

Let $f\left(x\right)=x^2-2x,xϵR$ and $g\left(x\right)=f\left(f\right(x)–1)+f(5–f(x\left)\right).$ Which of the following statements(s) is/are true?

• A. $g\left(x\right)$ is continuous for all $xϵR$
• B. $g^{\prime }\left(x\right)=0$ for $x=–1$
• C. $g^{\prime }\left(x\right)=0$ for $x=3$
• D. $g\left(x\right)\ge 0$ for all $xϵR$

Answer 1

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

The correct options are
A

$g\left(x\right)$ is continuous for all $xϵR$

B

$g^{\prime }\left(x\right)=0$ for $x=–1$

C

$g^{\prime }\left(x\right)=0$ for $x=3$

D

$g\left(x\right)\ge 0$ for all $xϵR$

Given : $g\left(x\right)=f\left(f\right(x)–1)+f(5–f(x\left)\right)$

$\Rightarrow g^{\prime }\left(x\right)=f^{\prime }\left(f\right(x)–1).f^{\prime }\left(x\right)–f^{\prime }(5–f(x\left)\right)f^{\prime }\left(x\right)$

Since, $f\left(x\right)$ is differentiable everywhere, so, $g\left(x\right)$ exists and is continuous for all $xϵR$.

Now, $g^{\prime }\left(x\right)=0$

$\Rightarrow f^{\prime }\left(x\right)\left[f^{\prime }\right(f\left(x\right)–1)–f^{\prime }(5–f\left(x\right)]=0$

$\Rightarrow f^{\prime }\left(x\right)=0$ or $f\left(x\right)–1=5–f\left(x\right)$

$\Rightarrow f\left(x\right)=3$

$\Rightarrow x^2–2x–3=0$, so, $x=–1$ or $3$

Hence, $g^{\prime }\left(x\right)=0$ for $x=-1$ or $3$.

Now, $g\left(x\right)=\left(f\right(x)–1)+f(5–f(x\left)\right)$

$=\left(f\right(x)–1{)}^2–2(f\left(x\right)–1)+(5–f\left(x\right){)}^2–2(5–f(x\left)\right)$

$=2\left(f\right(x)–3{)}^2\ge 0$

Hence, $g\left(x\right)\ge 0$ for all $xϵR$