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TN XII Math
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Objective type Questions and Answers
The gradient of the tangent to the curve $y=8+4x-2x^{2}$ at the point where the curve cuts the $y$-axis is
\[\begin{array}{1 1}(1)8&(2)4\\(3)0&(4)-4\end{array}\]
tnstate
class12
bookproblem
p231
objective
q20
sec-a
easy
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asked
May 16, 2013
by
poojasapani_1
edited
May 16, 2013
by
poojasapani_1
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1 Answer
$y= 8+4x-2x^2$
To obtain the point of intersection with $y-axis$ put $x= 0$
$y=8$
The point of intersection is (0,8)
$y= 8+4x-2x^2$
$\large\frac{dy}{dx} $$=4-4x$
$\large\frac{dy}{dx} $$/(0,8) =4-0=4$
Hence 2 is the correct answer.
answered
May 19, 2014
by
meena.p
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