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)
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Answer:
($y-2 = 0$)
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.
