The equation of the line passing through origin which is parallel to the tangent of the curve \(y=\dfrac{x-2}{x-3}\) at \(x=4\) is
\(y=2x \)
\(y=-2x+1\)
\( y=-x\)
\( y=x+2 \)
\( y=4x\)
The Correct Option is C
Solution and Explanation
\(y=\dfrac{x-3}{x-2}\)
\(⇒\dfrac{dy}{dx}=\dfrac{(x-3).1-(x-2).1}{(x-3)^2}\)
\(⇒\dfrac{dy}{dx}=\dfrac{-1}{(x-3)^2}\)
Then ⇒\(\dfrac{dy}{dx}\) at \(x=4\) \(=\dfrac{-1}{(4-3)^2}\)
\(=-1\)
means the slope\( =-1\)
Therefore,the equation of the line passing through origin which is parallel to the tangent of the curve \(y=\dfrac{x-2}{x-3}\) at \(x=4\) is
\(y=-x\) (_Ans)
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Concepts Used:
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Various Applications of Derivatives-
Rate of Change of Quantities:
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
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Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
- If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
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