Application of Derivatives 1 Question 17
17. If $\left|f\left(x _1\right)-f\left(x _2\right)\right| \leq\left(x _1-x _2\right)^{2}, \forall x _1, x _2 \in R$. Find the equation of tangent to the curve $y=f(x)$ at the point $(1,2)$.
$(2005,4$ M)
Show Answer
Solution:
- As $\mid f\left(x _{1)}-f\left(x _2\right) \mid \leq\left(x _1-x _2\right)^{2}, \forall x _1, x _2 \in R\right.$
$\Rightarrow\left|f\left(x _1\right)-f\left(x _2\right)\right| \leq\left|x _1-x _2\right|^{2}$
[as $\left.x^{2}=|x|^{2}\right]$
$$ \begin{aligned} & \therefore \quad\left|\frac{f\left(x _1\right)-f\left(x _2\right)}{x _1-x _2}\right| \leq\left|x _1-x _2\right| \\ & \Rightarrow \quad \lim _{x _1 \rightarrow x _2} \frac{f\left(x _1\right)-f\left(x _2\right)}{x _1-x _2} \leq \lim _{x _1 \rightarrow x _2}\left|x _1-x _2\right| \\ & \Rightarrow \quad\left|f^{\prime}\left(x _1\right)\right| \leq 0, \forall x _1 \in R \\ & \therefore \quad\left|f^{\prime}(x)\right| \leq 0, \text { which shows }\left|f^{\prime}(x)\right|=0 \end{aligned} $$
[as modulus is non negative or $\left|f^{\prime}(x)\right| \geq 0$ ]
$\therefore f^{\prime}(x)=0$ or $f(x)$ is constant function.
$\Rightarrow$ Equation of tangent at $(1,2)$ is
$$ \begin{aligned} & \frac{y-2}{x-1}=f^{\prime}(x) \\ & \text { or } \quad\left[\because \text { as } f^{\prime}(x)=0\right] \end{aligned} $$
$\Rightarrow y-2=0$ is required equation of tangent.
