Question

Find number of $x$ for which $\frac{40}{3x^4+8x^3-18x^2+60}$ is minimum.

• A. $1$
• B. $0$
• C. $2$
• D. $3$

Answer 1

The correct option is A $1$

Let $z=\frac{1}{y}=\frac{1}{40}(3x^4+8x^3-18x^2+60)$
Now, $y$ is minimum when $z$ is maximum.
$\frac{dz}{dy}=\frac{1}{40}(12x^3+24x^2-36x)$
$\Rightarrow \frac{dz}{dy}=\frac{12}{40}(x^3+2x^2-3x)$
For maximum or minimum,
$x(x+3)(x-1)=0$
$\Rightarrow x=0,-3,1$
Now, $\frac{d^{2}z}{d{y}^2}=\frac{12}{40}(3x^2+4x-3)$
At $x=0$ , $\frac{d^{2}z}{d{y}^2}=0$
Here, second derivative test fails.
$\frac{d^{3}z}{d{y}^3}=\frac{12}{40}(6x+4)$
At $x=0$, $\frac{d^{3}z}{d{y}^3}>0$
Hence, $z$ has maximum at $x=0$
At $x=1$ , $\frac{d^{2}z}{d{y}^2}>0$
Hence, $z$ has a minimum at $x=1$
At $x=-3$ , $\frac{d^{2}z}{d{y}^2}>0$
Hence, $z$ has a minimum at $x=-3$
So, y has 2 point of maximum and 1 point of minimum.