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AP EAMCET Engineering 2018 Apr 23 Shift 2 Paper

Section: Mathematics

Question : 67 of 160

Marks: +1, -0

y = x \log\left( \frac{x}{2-3x} \right),

0 < x < \frac{2}{3}

\frac{d^2y}{dx^2}

x = \frac{1}{2}

Solution:

Consider the expression,

y = x \log\left( \frac{x}{2-3x} \right),

for

0 < x < \frac{2}{3}

Differentiate w.r.t x ,

\frac{dy}{dx} = x \left( \frac{1}{x} - \frac{-3}{2-3x} \right) + \log\left( \frac{x}{2-3x} \right)

\frac{dy}{dx} = \frac{2}{2-3x} + \log\left( \frac{x}{2-3x} \right)

Again differentiate the above,

\frac{d^2y}{dx^2} = \frac{-2}{(2-3x)^2} (-3)

+ \left( \frac{1}{x} - \frac{-3}{2-3x} \right)

= \frac{6}{(2-3x)^2} + \frac{2}{x(2-3x)}

So,

\frac{d^2y}{dx^2} \text{ at } x = \frac{1}{2} = \frac{6}{\left( \frac{1}{2} \right)^2} + \frac{2}{\frac{1}{2} \cdot \frac{1}{2}}

= 24 + 8

= 32

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