$f(x)$ is a function such that $\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| \leq\left(x_{1}

Question

$f(x)$ is a function such that $\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| \leq\left(x_{1}-x_{2}\right)^{2}$ and tangent at some point on $f(x)$ passes through $(1,2)$. The equation of the tangent is y=n, value of n is___?

Professor PC Thomas Classes · Answer

Given $\left|f\left(x_{1}ight)-f\left(x_{2}ight)ight| \leq\left(x_{1}-x_{2}ight)^{2}$\ $\Rightarrow\left|\frac{f\left(x_{1}ight)-f\left(x_{2}ight)}{x_{1}-x_{2}}ight| \leq\left|x_{1}-x_{2}ight|$\ If $x_{1} ightarrow x_{2}$, then $\left|x_{1}-x_{2}ight| ightarrow 0$\ $	herefore \lim _{x_{1} ightarrow x_{2}}\left|\frac{f\left(x_{1}ight)-f\left(x_{2}ight)}{x_{1}-x_{2}}ight|=\left|f^{\prime}\left(x_{2}ight)ight|$\ $\Rightarrow\left|f^{\prime}\left(x_{2}ight)ight| \leq 0 \Rightarrow f^{\prime}\left(x_{2}ight)=0 \forall x_{2}$\ $\Rightarrow f(x)=c$, where $c$ is a constant\ The tangent passes through $(1,2)$\ $	herefore \quad f(1)=2 \Rightarrow c=2$\ Equation of the tangent is\ $y-2=f^{\prime}(1)(x-1) \Rightarrow y-2=0 \Rightarrow y=2$.\