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TS EAMCET 2017 Code A Solved Paper

Section: Mathematics

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Question : 62 of 160

Marks: +1, -0

The local maximum of

y=x^3-3x^2+5

is attained at

x = 0
x = 2
x = l
x = -l

Solution:

Consider the given equation.

y=x^{3}-3 x^{2}+5 \quad . . .

Differentiate the equation with respect to

x

\;\;\frac{dy}{dx}=3 x^{2}-6 x

For local maxima and minima,

\;\;\frac{dy}{dx}=0

3 x^{2}-6 x=0

3 x(x-2)=0

x =0,2

Differentiate equation (2) with respect to

x

\;\;\frac{d^{2}y}{dx^{2}}=6 x-6

\left(\;\;\frac{d^{2}y}{dx^{2}}\right)_{x=0}=-6<0

Therefore,

x=0

is a point of local maxima.

\left(\;\;\frac{d^{2}y}{dx^{2}}\right)_{x=2}

=6 \times 2-6

=6>0

Therefore,

x=2

is a point of local minima.

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