Question

If $f\left(x\right)=4x^3-x^2-2x+1$ and $g\left(x\right)=\left[\begin{array}{cc}Minf\left(t\right):0\le t\le x;& 0\le x\le 1\\ 3-x;& 1<x\le 2\end{array}\right]$ then $g(\frac{1}{4})+g(\frac{3}{4})+g(\frac{5}{4})$ has the value equal to

• A. $\frac{7}{4}$
• B. $\frac{9}{4}$
• C. $\frac{13}{4}$
• D. $\frac{5}{2}$

Answer 1

The correct option is D $\frac{5}{2}$

Given, $f\left(x\right)=4x^3-x^2-2x+1$

Taking first derivative of the function f(x) we get
$f^{{}^{\prime }}\left(x\right)=2(6x^2-x-1)$
Putting this equals to zero we get, $x=-\frac{1}{3}$, $\frac{1}{2}$.
Taking second derivative of f(x) we get, $f^{{}^{\prime \prime }}\left(x\right)=2(12x-1)$ this shows that f(x) has a local maxima at $x=-\frac{1}{3}$ and a local minima at $x=\frac{1}{2}$.
Which means in between $0\le x\le 1$ f(x) has a minima at 1/2 and $f\left(x\right)=1$ at $x=0$ and $f\left(x\right)=2$ at $x=1$, also $f\left(x\right)=1/4$ at $x=1/2$.
As per question we have $g\left(x\right)=$ Min $f\left(t\right):0\le t\le x$,
$g\left(\frac{1}{4}\right)$ = $f\left(\frac{1}{4}\right)$ as min of f(x) in between 0 to 1/4 is at 1/4
$f\left(\frac{1}{4}\right)$ = 1/2
$g\left(\frac{3}{4}\right)$ = $f\left(\frac{1}{2}\right)$ because in between 0 to 3/4, we have minima at x= 1/2 which is the local minima of function as well.
$f\left(\frac{1}{2}\right)=1/4$
$g\left(\frac{5}{4}\right)$ = $\frac{7}{4}$
So $g\left(\frac{1}{4}\right)$ + $g\left(\frac{3}{4}\right)$ + $g\left(\frac{5}{4}\right)$ = $\frac{5}{2}$