Question

If $\left|f\right(x_1)-f(x_2\left)\right|<(x_1-x_2{)}^2$ for all $x_1$ $x_2$ $\in$ R. Find the equation of tangent to the curve y = f(x) at the point (1, 2).

• A. $x=2$
• B. $y=2$
• C. $y=1$
• D. $x=1$

Answer 1

Answer: (B)

$y=2$

As $|f\left(x_1\right)-f\left(x_2\right)|\le {(x_1-x_2)}^2,\forall x_1,x_2\in R$
$\Rightarrow |f\left(x_1\right)-f\left(x_2\right)|\le {|x_1-x_2|}^2$ $(x^2={\left|x\right|}^2)$
$\therefore \left|\frac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}\right|\le |x_1-x_2|\Rightarrow \underset{x_1\to x_2}{lim}\left|\frac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}\right|\le \underset{x_1\to x_2}{lim}|x_1-x_2|$
$\Rightarrow \left|f^{\prime }\left(x_1\right)\right|\le 0,\forall x_1\in R$
$\therefore \left|f^{\prime }\left(x\right)\right|\le 0$, which showsw $\left|f^{\prime }\left(x\right)\right|=0$ (as modulus is non negative or $\left|f^{\prime }\left(x\right)\right|\ge 0$)
$\therefore f^{\prime }\left(x\right)=0$ or $f\left(x\right)$ is a constant function.
$\Rightarrow$ Equation of tangent at $(1,2)$ is
$\frac{y-2}{x-1}=f^{\prime }\left(x\right)$ or $y-2=0$ [$\because$ as $f^{\prime }\left(x\right)=0$]
$\Rightarrow y-2=0$ is required equation of tangent.