Let be a singular matrix. Let . If M and m are respectively the maximum and the minimum values of f in [1, ], then 3(M – m) is equal to: [2026]

Question

Let $A = [\begin{matrix} 1 & 3 & - 1 \\ 2 & 1 & α \\ 0 & 1 & - 1 \end{matrix}]$ be a singular matrix. Let $f (x) = \int_{0}^{x} (t^{2} + 2 t + 3) d t, x \in [1, α]$ . If M and m are respectively the maximum and the minimum values of f in [1, $α$ ], then 3(M – m) is equal to: [2026]

Smart Achievers · Accepted Answer

(2)

Given, $A = [\begin{matrix} 1 & 3 & - 1 \\ 2 & 1 & α \\ 0 & 1 & - 1 \end{matrix}]$

Since, A is a singular matrix, |A| = 0

$\Rightarrow 1 (- 1 - α) - 3 (- 2 - 0) - 1 (2 - 2) = 0$

$\Rightarrow α = 3$

Also, given $f (x) = \int_{0}^{x} (t^{2} + 2 t + 3) d t, x \in [1, 3]$

$= {(\frac{t^{3}}{3} + t^{2} + 3 t)}_{0}^{x} = \frac{x^{3}}{3} + x^{2} + 3 x$

$f' (x) = x^{2} + 2 x + 3 = {(x + 1)}^{2} + 2$

$∴$ f(x) is strictly increasing on [1, 3]

M = f(3) = 9 + 9 + 9 = 27

and $m = f (1) = \frac{1}{3} + 1 + 3 = \frac{13}{3}$

$∴ M - m = 27 - \frac{13}{3} = \frac{68}{3}$

$\Rightarrow 3 (m - m) = 68$

Solution

Q.
Let $A = [\begin{matrix} 1 & 3 & - 1 \\ 2 & 1 & α \\ 0 & 1 & - 1 \end{matrix}]$ be a singular matrix. Let $f (x) = \int_{0}^{x} (t^{2} + 2 t + 3) d t, x \in [1, α]$ . If M and m are respectively the maximum and the minimum values of f in [1, $α$ ], then 3(M – m) is equal to: [2026]

1 64

2 68

3 72

4 76

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