SATHEE: Application of Derivatives 1 Question 17

Application of Derivatives 1 Question 17

####17. If $| f (x_{1}) - f (x_{2}) | \leq {(x_{1} - x_{2})}^{2}, \forall x_{1}, x_{2} \in R$ . 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 $∣ f (x_{1)} - f (x_{2}) ∣\leq {(x_{1} - x_{2})}^{2}, \forall x_{1}, x_{2} \in R$

$\Rightarrow | f (x_{1}) - f (x_{2}) | \leq {| x_{1} - x_{2} |}^{2}$

[as $x^{2} = | x |^{2}]$

$\begin{aligned} ∴ | \frac{f (x_{1}) - f (x_{2})}{x_{1} - x_{2}} | \leq | x_{1} - x_{2} | \\ \Rightarrow lim_{x_{1} \to x_{2}} \frac{f (x_{1}) - f (x_{2})}{x_{1} - x_{2}} \leq lim_{x_{1} \to x_{2}} | x_{1} - x_{2} | \\ \Rightarrow | f^{'} (x_{1}) | \leq 0, \forall x_{1} \in R \\ ∴ | f^{'} (x) | \leq 0, which shows | f^{'} (x) | = 0 \end{aligned}$

[as modulus is non negative or $| f^{'} (x) | \geq 0$ ]

$∴ f^{'} (x) = 0$ or $f (x)$ is constant function.

$\Rightarrow$ Equation of tangent at $(1, 2)$ is

$\begin{aligned} \frac{y - 2}{x - 1} = f^{'} (x) \\ or [∵ as f^{'} (x) = 0] \end{aligned}$

$\Rightarrow y - 2 = 0$ is required equation of tangent.

Application of Derivatives 1 Question 17

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Application of Derivatives 1 Question 17

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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