Question

Let f : \(\R → \R\) be the function defined by
\(f(x)=\text{det}\begin{pmatrix}1+x & 9 & 9 \\ 9 & 1+x & 9 \\ 9 & 9 & 1+x \end{pmatrix}\)
Then the maximum value of f on the interval [9, 10] equals

• A. 118
• B. 112
• C. 114
• D. 116

Answer 1

The Correct Option is D

The given problem requires us to find the maximum value of the function \(f(x) = \text{det}\begin{pmatrix} 1+x & 9 & 9 \\ 9 & 1+x & 9 \\ 9 & 9 & 1+x \end{pmatrix}\) on the interval \([9, 10]\). To solve this, we need to calculate the determinant of the matrix.

The determinant of a \(3 \times 3\) matrix \(A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}\) is given by:

\(\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)\)

For the given matrix:

\(a = 1+x\)	\(b = 9\)	\(c = 9\)
\(d = 9\)	\(e = 1+x\)	\(f = 9\)
\(g = 9\)	\(h = 9\)	\(i = 1+x\)

Substitute these values into the determinant formula:

\(\text{det} = (1+x)((1+x)(1+x) - 9 \cdot 9) - 9(9(1+x) - 9 \cdot 9) + 9(9 \cdot 9 - 9(1+x))\)

Simplifying each term:

• \((1+x)(1+x)^2 = (1+x)^3\)
• \(9 \cdot 9 = 81\)

Therefore,

\(\text{det} = (1+x)(x^2 + 2x + 1 - 81) - 9(9x + 81 - 81) + 9(81 - 9x)\)

Simplify the expression:

\(= (1+x)(x^2 + 2x - 80) - 81x + 729 - 81x\)

Which results in:

\(f(x) = x^3 + 2x^2 - 80x + x^2 + 2x - 80 - 162x + 729\)

\(= x^3 + 3x^2 - 240x + 729\)

Now, to find the maximum value on the interval \([9, 10]\), evaluate at both endpoints:

• At \(x = 9\): \(f(9) = 9^3 + 3 \cdot 9^2 - 240 \cdot 9 + 729 = 729 + 243 - 2160 + 729 = 118\)
• At \(x = 10\): \(f(10) = 10^3 + 3 \cdot 10^2 - 240 \cdot 10 + 729 = 1000 + 300 - 2400 + 729 = 116\)

Thus, the maximum value of \(f(x)\) on the interval \([9, 10]\) is 116.