Question

Let $f:R\to [1,\infty )$ be a quadratic surjective function such that $f(2+x)=f(2-x)$ and $f\left(1\right)=2.$ Let $g:(-\infty ,\ln 2]\to [1,5]$ be another function defined as $g\left(\ln x\right)=f\left(x\right),$ then which of the following(s) is/are correct ?

• A. minimum value of $g^{\prime }\left(x\right)$ is $-2$.
• B. $g^{-1}\left(x\right)=\ln (2+\sqrt{x-1})$
• C. $g^{-1}\left(x\right)=\ln (2-\sqrt{x-1})$
• D. the sum of the values of $x$ that satisfying the equation $f\left(x\right)=5$ is $4.$

Answer 1

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

the sum of the values of $x$ that satisfying the equation $f\left(x\right)=5$ is $4.$

$f(2+x)=f(2-x)$ and $f$ is quadratic
$\Rightarrow f\left(x\right)=a(x-2{)}^2+k$
$k=1$ as $f:R\to [1,\infty )$
$\Rightarrow f\left(x\right)=a(x-2{)}^2+1$
$f\left(1\right)=2\Rightarrow a=1$
$\therefore f\left(x\right)=(x-2{)}^2+1$

$f\left(x\right)=5\Rightarrow (x-2{)}^2+1=5$
$\Rightarrow (x-2{)}^2=4\Rightarrow x-2=\pm 2$
or, $x=0,4$
Sum of values of $x=4$


$g\left(\ln x\right)=f\left(x\right)=(x-2{)}^2+1$
$\Rightarrow g\left(x\right)=(e^x-2{)}^2+1$

$g^{\prime }\left(x\right)=2(e^x-2)e^x=2(e^{2x}-2e^x)$
$g^{\prime \prime }\left(x\right)=4(e^{2x}-e^x)$
To find the minimum value of $g^{\prime }\left(x\right),$ put $g^{\prime \prime }\left(x\right)=0$
$\Rightarrow x=-\infty$ or $x=0$
Since, the gragh of $g^{\prime }\left(x\right)$ is to be concave up (upward). So, it will have minimum value at $x=0$.
$\therefore min\left[g^{\prime }\right(x\left)\right]=g^{\prime }\left(0\right)=-2$

$g\left(x\right)=(e^x-2{)}^2+1$ is invertible function.
$\Rightarrow e^x=2\pm \sqrt{y-1}$
$\Rightarrow x=\ln (2\pm \sqrt{y-1})$
Since, $x\in (-\infty ,\ln 2]$
$\therefore g^{-1}\left(x\right)=\ln (2-\sqrt{y-1})$