Question

Let $y=\frac{6x^3-45x^2+108x+2}{2x^3-15x^2+36x+1}$ and $x\in (0,10)$

The maxima point for $y$ is:

• A. $x=3$
• B. $x=7$
• C. $x=\frac{3}{4}$
• D. $x=9$

Answer 1

The correct option is A $x=3$

The given information is:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\Rightarrow y^{\prime \prime }\left(2\right)=\frac{-6}{(29{)}^2}$

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

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

$\Rightarrow y^{\prime \prime }\left(3\right)=\frac{6}{(28{)}^2}$

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