Let A=[aij] be a 3×3 matrix, where aij={1, if i=j−x if|i−j|=12x+1, otherwise Let a function be defined as f(x)=det(A). Then the sum of maximum and minimum values of f on R is equal to

A=[1−x2x+1−x1−x2x+1−x1]∴|A|=4x3−4x2−4x=f(x)(let)Let f'(x)=0⇒x=1,−1/3Sum of minimum and maximum values isf(1)+f(−1/3)=−4+2027=-8827

Let A=[aij] be a 3×3 matrix, where aij={1, if i=j−x if|i−j|=12x+1, otherwise Let a function be defined as f(x)=det(A). Then the sum of maximum and minimum values of f on R is equal to

A=[1−x2x+1−x1−x2x+1−x1]∴|A|=4x3−4x2−4x=f(x)(let)Let f'(x)=0⇒x=1,−1/3Sum of minimum and maximum values isf(1)+f(−1/3)=−4+2027=-8827

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Let $A = [a_{i j}] be a 3 \times 3$ matrix, where $a_{i j} = {\begin{cases} 1, if i=j \\ - x if | i - j | = 1 \\ 2 x + 1, otherwise \end{cases}$ Let a function be defined as $f (x) = \det (A)$ . Then the sum of maximum and minimum values of f on R is equal to

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$- \frac{20}{27}$

$\frac{20}{27}$

$\frac{88}{27}$

$- \frac{88}{27}$

answer is C.

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Detailed Solution

$A = [\begin{matrix} 1 & - x & 2 x + 1 \\ - x & 1 & - x \\ 2 x + 1 & - x & 1 \end{matrix}]$
$\begin{array}{l} ∴ | A | = 4 x^{3} - 4 x^{2} - 4 x = f (x) (l e t) \\ L e t f^{'} (x) = 0 \\ \Rightarrow x = 1, - 1 / 3 \end{array}$
Sum of minimum and maximum values is
$f (1) + f (- 1 / 3) = - 4 + \frac{20}{27} = - \frac{88}{27}$