Question

For a certain curve $\frac{d^{2}y}{d{x}^2}=6x-4$ and curve has local minimum value $5$ at $x=1$. Let the global maximum and global minimum values, where $0\le x\le 2$ ; are $M$ and $m$. Then the vaule of $(M-m)$ equals to :

• A. $-2$
• B. $2$
• C. $12$
• D. $-12$

Answer 1

The correct option is B $2$

given that $\frac{d^{2}y}{d{x}^2}=6x-4$
by integrating $\frac{dy}{dx}=3x^2-4x+a$
Since, $x=1$ is a point of local minimum, so $\frac{dy}{dx}$ at $x=1$
$\Rightarrow 0=3-4+a$ $\Rightarrow a=1$
$\frac{dy}{dx}=3x^2-4x+1$
by integrating above equation $y=x^3-2x^2+x+b$
at $x=1$ local minimum value of $f$ is $5$
$y\left(1\right)=5$
$\Rightarrow 5=1-2+1+b$ $\Rightarrow b=5$
for local max. & min. $\frac{dy}{dx}=0$
$3x^2-4x+1=0$ $\Rightarrow x=\frac{1}{3},1$
at $x=\frac{1}{3}$, $\frac{d^{2}y}{d{x}^2}=2-4=-2<0$
Therefore, $x=\frac{1}{3}$ is a point of local maximum.
thus $y\left(\frac{1}{3}\right)={\left(\frac{1}{3}\right)}^3-2{\left(\frac{1}{3}\right)}^2+\left(\frac{1}{3}\right)+5=\frac{139}{27}$
now $y\left(0\right)=5$, $y\left(\frac{1}{3}\right)=\frac{139}{27}$, $y\left(1\right)=5$ & $y\left(2\right)=7$
Therefore, global max.$=M=7$
and global min.$=m=5$
thus $M-m=2$
Ans: B