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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Normal to the curve
$y = x^3 - 3x + 2$
at the point $(2,4)$ is:
Solution
Differentiate:$\dfrac{dy}{dx} = 3x^2 - 3$
At $x = 2$:
$\dfrac{dy}{dx} = 3(4) - 3 = 12 - 3 = 9$
So slope of tangent $m_t = 9$
Slope of normal:
$m_n = -\dfrac{1}{9}$
Equation of normal at $(2,4)$:
$y - 4 = -\dfrac{1}{9}(x - 2)$
Multiply by 9:
$9y - 36 = -(x - 2)$
$9y - 36 = -x + 2$
$ x + 9y - 38 = 0$
Answer: (c) $x + 9y - 38 = 0$
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