The minimum value of $y$ is:

\frac{83}{28}

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\frac{77}{29}

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3

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2

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Solution

The correct option is A $\frac{83}{28}$
The given information is:

$y = \frac{6 x^{3} - 45 x^{2} + 108 x + 2}{2 x^{3} - 15 x^{2} + 36 x + 1}$

$\Rightarrow y^{'} = \frac{(18 x^{2} - 90 x + 108) (2 x^{3} - 15 x^{2} + 36 x + 1) - (6 x^{2} - 30 x + 36) (6 x^{3} - 45 x^{2} + 108 x + 2)}{(2 x^{3} - 15 x^{2} + 36 x + 1)^{2}}$

$\Rightarrow y^{'} = \frac{18 (x^{2} - 5 x + 6) (2 x^{3} - 15 x^{2} + 36 x + 1) - 6 (x^{2} - 5 x + 6) (6 x^{3} - 45 x^{2} + 108 x + 2)}{(2 x^{3} - 15 x^{2} + 36 x + 1)^{2}}$

$\Rightarrow y^{'} = \frac{6 (x^{2} - 5 x + 6) (6 x^{3} - 45 x^{2} - 108 x + 3 - 6 x^{3} + 45 x^{2} - 108 x - 2)}{(2 x^{3} - 15 x^{2} + 36 x + 1)^{2}}$

$\Rightarrow y^{'} = \frac{6 (x^{2} - 5 x + 6)}{(2 x^{3} - 15 x^{2} + 36 x + 1)^{2}}$

Now the extremum points are where $y^{'} = 0$

$\Rightarrow \frac{6 (x^{2} - 5 x + 6)}{(2 x^{3} - 15 x^{2} + 36 x + 1)^{2}} = 0$

$\Rightarrow 6 (x^{2} - 5 x + 6) = 0$

The extremum points are $x = 2$ and $x = 3$ .

Again differentiating w.r.t to x we get,

$\Rightarrow y^{''} = 6 \times \frac{(2 x - 5) (2 x^{3} - 15 x^{2} + 36 x + 1)^{2} - 2 (2 x^{3} - 15 x^{2} + 36 x + 1) (6 x^{2} - 30 x + 36) (x^{2} - 5 x + 6)}{(2 x^{3} - 15 x^{2} + 36 x + 1)^{4}}$

$\Rightarrow y^{''} = 6 \times \frac{(2 x - 5) (2 x^{3} - 15 x^{2} + 36 x + 1)^{2} - 12 (2 x^{3} - 15 x^{2} + 36 x + 1) (x^{2} - 5 x + 6) (x^{2} - 5 x + 6)}{(2 x^{3} - 15 x^{2} + 36 x + 1)^{4}}$

Finding the value of $y^{''}$ at the extremum points we get,

$\Rightarrow y^{''} (2) = 6 \times \frac{(- 1) (29)^{2} - 12 \times 29 \times 0}{(29)^{4}}$

$\Rightarrow y^{''} (2) = \frac{- 6}{(29)^{2}}$

$\Rightarrow y^{''} (2) < 0$ . Hence the point $x = 2$ is a point of maxima. .....Answer

$\Rightarrow y^{''} (3) = 6 \times \frac{(1) (28)^{2} - 12 \times 28 \times 0}{(28)^{4}}$

$\Rightarrow y^{''} (3) = \frac{6}{(28)^{2}}$

$\Rightarrow y^{''} (3) > 0$ . Hence the point $x = 3$ is a point of minima.

Thus the minimum value of the function is $y (3)$ .

$\Rightarrow y (3) = \frac{{6.3}^{3} - {45.3}^{2} + 108.3 + 2}{{2.3}^{3} - {15.3}^{2} + 36.3 + 1}$

$\Rightarrow y (3) = \frac{162 - 405 + 324 + 2}{54 - 135 + 108 + 1}$

$\Rightarrow y (3) = \frac{83}{28}$

Thus the minimum value of the function is $\frac{83}{28}$ . .....Answer

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