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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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The points at which the tangents to be curve $y=x^3-12x+18$ are parallel to x-axis are:

$\begin{array}{1 1} (A)\;(2,-2),(-2,-34) \\ (B)\;(2,34),(-2,0)\\(C)\;(0,34),(-2,0) \\ (D)\;(2,2),(-2,34) \end{array} $

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1 Answer

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Toolbox:
  • Slope of a line is $ ax+by+c=0$ is $ - \bigg( \large\frac{ Coefficient\: of \: x}{ Coefficient\: of \: y} \bigg)$
  • If $ y = f(x),\: then \: \bigg( \large\frac{dy}{dx} \bigg)_p = $ slope of the normal to $ y = f(x)$ at point $p$
Step 1
Equation of the curve is
$ y = x^3-12x+18$
On differentiating w.r.t $x$ we get,
$ \large\frac{dy}{dx}=3x^2-12$
If the tangent is parallel to $x$ - axis then
$ \large\frac{dy}{dx}=0$
$ \Rightarrow 3x^2-12=0$
$ x^2=4$
$ \Rightarrow x = \pm 2$
when $ x = 2, \: y=2$
when $x = -2, \: y=-4$
Hence the points are $(2,2)\: and \: (-2, +34)$
The correct answer is D
answered Aug 9, 2013 by thanvigandhi_1
 

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