JEE MAIN - Mathematics (2021 - 20th July Morning Shift - No. 9)
Let $$A = [{a_{ij}}]$$ be a 3 $$\times$$ 3 matrix, where $${a_{ij}} = \left\{ {\matrix{
1 & , & {if\,i = j} \cr
{ - x} & , & {if\,\left| {i - j} \right| = 1} \cr
{2x + 1} & , & {otherwise.} \cr
} } \right.$$
Let a function f : R $$\to$$ R be defined as f(x) = det(A). Then the sum of maximum and minimum values of f on R is equal to:
Let a function f : R $$\to$$ R be defined as f(x) = det(A). Then the sum of maximum and minimum values of f on R is equal to:
$$ - {{20} \over {27}}$$
$${{88} \over {27}}$$
$${{20} \over {27}}$$
$$ - {{88} \over {27}}$$
Explanation
$$A = \left[ {\matrix{
1 & { - x} & {2x + 1} \cr
{ - x} & 1 & { - x} \cr
{2x + 1} & { - x} & 1 \cr
} } \right]$$
$$\left| A \right| = 4{x^3} - 4{x^2} - 4x = f(x)$$
$$f'(x) = 4(3{x^2} - 2x - 1) = 0$$
$$ \Rightarrow x = 1;x = {{ - 1} \over 3}$$
$$\therefore$$ $$f(1) = - 4;f\left( { - {1 \over 3}} \right) = {{20} \over {27}}$$
Sum $$ = - 4 + {{20} \over 7} = - {{88} \over {27}}$$
$$\left| A \right| = 4{x^3} - 4{x^2} - 4x = f(x)$$
$$f'(x) = 4(3{x^2} - 2x - 1) = 0$$
$$ \Rightarrow x = 1;x = {{ - 1} \over 3}$$
$$\therefore$$ $$f(1) = - 4;f\left( { - {1 \over 3}} \right) = {{20} \over {27}}$$
Sum $$ = - 4 + {{20} \over 7} = - {{88} \over {27}}$$
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