Question

Let a function $f\left(x\right)$ satisfies $f^2\left(x\right)-f^2\left(y\right)=4(x-y)$ and $f\left(0\right)=2(f\left(x\right)\ge 0)$whose domain is $[a,\infty )$ and it is differentiable on $(a,\infty )$ The number of points where $g\left(x\right)=max\left\{f\right(x),6,7,\left|x\right|\}$ is non derivable when $xϵ[a,10]$

• A. $2$
• B. $3$
• C. $6$
• D. $8$

Answer 1

The correct option is A $2$

$f^2\left(x\right)-f^2\left(y\right)={\left(2\sqrt{x}\right)}^2-{\left(2\sqrt{y}\right)}^2$

$\Rightarrow f\left(x\right)=2\sqrt{x}+k$

Given $f\left(0\right)=2$

$\Rightarrow 2=2\times =0+k\Rightarrow k=2$

$f\left(x\right)=2(\sqrt{x}+1)$

x should be non negative

Since we have root function

$\Rightarrow x\ge 0$

$\Rightarrow x\in (0,\infty )$

$g\left(x\right)=max\{f\left(x\right),6,7,\left|x\right|\},x\in [0,10]$

$2\sqrt{x}+2>x(\left|x\right|=x,x\ge 0)$

$4x>x^2+4-4x$

$x^2-8x+4<0$

${(x-4)}^2-12<0\Rightarrow 0<x<2(2+\sqrt{3})$

$x<7.4$

$2\sqrt{x}+2>7$

$\sqrt{x}>2.5$

$x>6.25$

$\left|x\right|>7$

$x>7$

$\Rightarrow g\left(x\right)=\{\begin{array}{c}7;0\le x<6.25\\ 2\sqrt{x}+2;6.25\le x<2(2+\sqrt{3})\\ \left|x\right|;x\ge 2(2+\sqrt{3})\end{array}$

$g^{\prime }\left(x\right)=\{\begin{array}{c}0;0\le x<6.25\\ 1/\sqrt{x};6.25\le x<2(2+\sqrt{3})\\ \left|x\right|/x;x\ge 2(2+\sqrt{3})\end{array}$

at $x=6.25$

${lim}_{x\to {6.25}^{+}}{g}^{\prime }\left(x\right)={lim}_{x\to {6.25}^{+}}0=0$

${lim}_{x\to {6.25}^{-}}{g}^{\prime }\left(x\right)={lim}_{x\to {6.25}^{-}}\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{6.25}}=\frac{1}{2.5}$

LHL$\ne$RHL $\Rightarrow$Non differentiable

$x=2(2+\sqrt{3})$

${lim}_{x\to {(4+2\sqrt{3})}^{+}}{g}^{\prime }\left(x\right)=\frac{|4+2\sqrt{3}|}{4+2\sqrt{3}}=1$

${lim}_{x\to {(4+2\sqrt{3})}^{-}}{g}^{\prime }\left(x\right)=\frac{1}{\sqrt{4+2\sqrt{3}}}\ne 1$

LHL$\ne$RHL $\Rightarrow$Non differentiable

$\Rightarrow$At $2$ points the function is non derivable