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

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Determinants

Let

A = 120311 - 1 α - 1

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

[1, α]

then

3 (M - m)

is equal to :

Answer: (B)

Singularity gives

∣ A ∣ = 0

: expanding,

1 (- 1 - α) - 3 (- 2) + (- 1) (2) = 3 - α = 0

, so

α = 3

. Then

f (x) = \int_{0}^{x} (t^{2} + 2 t + 3) d t = \frac{x ^{3}}{3} + x^{2} + 3 x

[1, 3]

, and

f^{'} (x) = t^{2} + 2 t + 3 > 0

throughout (negative discriminant), so

f

increases on

[1, 3]

. Thus

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

and

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

, giving

3 (M - m) = 81 - 13 = 68

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Single correct option