Question:
Let
\(A=[a_{ij}]\)
be a \(3\times3\) matrix, where
\(a_{ij}=\left\{\begin{matrix}1&,&\text{if } i=j\\-x&,&\text{if } |i-j|=1\\2x+1&,&\text{otherwise.}\end{matrix}\right.\)
Let a function \(f:\mathbb{R}\to\mathbb{R}\) be defined as \(f(x)=\det(A)\). Then the sum of maximum and minimum values of \(f\) on \(\mathbb{R}\) is equal to:
Let
\(A=[a_{ij}]\)
be a \(3\times3\) matrix, where
\(a_{ij}=\left\{\begin{matrix}1&,&\text{if } i=j\\-x&,&\text{if } |i-j|=1\\2x+1&,&\text{otherwise.}\end{matrix}\right.\)
Let a function \(f:\mathbb{R}\to\mathbb{R}\) be defined as \(f(x)=\det(A)\). Then the sum of maximum and minimum values of \(f\) on \(\mathbb{R}\) is equal to:
\(A=[a_{ij}]\)
be a \(3\times3\) matrix, where
\(a_{ij}=\left\{\begin{matrix}1&,&\text{if } i=j\\-x&,&\text{if } |i-j|=1\\2x+1&,&\text{otherwise.}\end{matrix}\right.\)
Let a function \(f:\mathbb{R}\to\mathbb{R}\) be defined as \(f(x)=\det(A)\). Then the sum of maximum and minimum values of \(f\) on \(\mathbb{R}\) is equal to:
Show Hint
Expand \(\det A\) to get a cubic in \(x\); the "maximum and minimum values" are its local maximum and local minimum values (found from \(f^{\prime}(x)=0\)).
Updated On: Nov 4, 2026
- \(-\frac{20}{27}\)
- \(\frac{88}{27}\)
- \(\frac{20}{27}\)
- \(-\frac{88}{27}\)
Show Solution
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The Correct Option is D
Solution and Explanation
\(A=\left[\begin{matrix}1&-x&2x+1\\-x&1&-x\\2x+1&-x&1\end{matrix}\right]\). Let \(c=2x+1\).
\(\det A=1\cdot(1-x^2)-(-x)(-x+xc)+c(x^2-c)=1-x^2+x^2(c-1)+cx^2-c^2\).
With \(c-1=2x\) and \(c^2=4x^2+4x+1\): \(\det A=1-x^2+2x^3+2x^3+x^2-4x^2-4x-1=4x^3-4x^2-4x\).
\(f(x)=4x^3-4x^2-4x\), \(f^{\prime}(x)=12x^2-8x-4=4(3x+1)(x-1)\). Critical points \(x=-\frac13\) (local maximum) and \(x=1\) (local minimum).
\(f\left(-\frac13\right)=-\frac{4}{27}-\frac{4}{9}+\frac43=\frac{-4-12+36}{27}=\frac{20}{27}\), \(f(1)=4-4-4=-4\).
Sum \(=\frac{20}{27}-4=-\frac{88}{27}\).
Answer: (4) \(-\frac{88}{27}\)
\(\det A=1\cdot(1-x^2)-(-x)(-x+xc)+c(x^2-c)=1-x^2+x^2(c-1)+cx^2-c^2\).
With \(c-1=2x\) and \(c^2=4x^2+4x+1\): \(\det A=1-x^2+2x^3+2x^3+x^2-4x^2-4x-1=4x^3-4x^2-4x\).
\(f(x)=4x^3-4x^2-4x\), \(f^{\prime}(x)=12x^2-8x-4=4(3x+1)(x-1)\). Critical points \(x=-\frac13\) (local maximum) and \(x=1\) (local minimum).
\(f\left(-\frac13\right)=-\frac{4}{27}-\frac{4}{9}+\frac43=\frac{-4-12+36}{27}=\frac{20}{27}\), \(f(1)=4-4-4=-4\).
Sum \(=\frac{20}{27}-4=-\frac{88}{27}\).
Answer: (4) \(-\frac{88}{27}\)
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