Question

The value of $y^{\prime \prime }\left(1\right)$ if $x^3-2x^2{y}^2+5x+y-5=0$ when $y\left(1\right)>1,$ is equal to

• A. $\frac{22}{7}$
• B. $-7\frac{21}{28}$
• C. $8$
• D. $-8\frac{22}{27}$

Answer 1

The correct option is D $-8\frac{22}{27}$

Given expression is $x^3-2x^2{y}^2+5x+y-5=0$ ...(1)

Differentiating the given expression, we get

$3x^2-4x{y}^2-4x^{2}y{y}^{\prime }+5+y^{\prime }=0$ ....(2)

Putting $x=1$ in the given expression, we get

$3-2y^2+5+y-5=0\Rightarrow 2y^2-y-1=0$

$\Rightarrow y=1$ or $y=-\frac{1}{2}$

Now $\because y\left(1\right)>0\Rightarrow y\left(1\right)=1$

Differentiating again we get

$6x-4y^2-8xy{y}^{\prime }-8xy{y}^{\prime }-4x2{\left(y^{\prime }\right)}^2-4x^{2}y{y}^{\prime \prime }+y^{\prime \prime }=0$

Putting $x=1,y=1$ and $y^{\prime }\left(1\right)=\frac{4}{3},$ we get

$6-4-8\left(\frac{4}{3}\right)-8\left(\frac{4}{3}\right)-4\left(\frac{16}{9}\right)-3y^{\prime \prime }\left(1\right)=0$

$\Rightarrow y^{\prime \prime }\left(1\right)=-8\frac{22}{27}$