Question

If $0\le x\le 1$and $f\left(x\right)=\left|\begin{array}{ccc}x& 1& 1\\ -1& x& 1\\ 1& -1& x\end{array}\right|$ , then

• A. least value of f (x) is 2
• B. greatest value of f (x) is 4
• C. f (x) has a local maximum at x= $\frac{2}{3}$
• D. f (x) has a local minimum at x= $\frac{1}{3}$

Answer 1

Answer: (A), (B)

greatest value of f (x) is 4
least value of f (x) is 2

Given, $f\left(x\right)=\left|\begin{array}{ccc}x& 1& 1\\ -1& x& 1\\ 1& -1& x\end{array}\right|$

$=x(x^2+1)-1(-x-1)+1(1-x)$

$=x^3+x+x+1-x+1$

$f\left(x\right)=x^3+x+2$

$\frac{df\left(x\right)}{dx}=3x^2+2=+ve.;0\le x\le 1$

So,$f\left(0\right)=$minimum value $=2$

$f\left(1\right)=$maximum value $=4$.