Let A = [1 3 -1; 2 1 alpha; 0 1 -1] be a singular matrix. Let f(x) = integral from 0 to x of (t^2 + 2t + 3) dt, 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 :

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q17 – JEE-MAIN 2026

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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

[1, α]

, then

3 (M - m)

is equal to :

Clue 1 of 3: Matrix Singularity and Function Extrema

First, use the property of a singular matrix to find the unknown parameter $α$ . Then, analyze the given function $f (x)$ to find its maximum and minimum values over the specified interval $[1, α]$ .

🧠 Full Solution Path

Step 1: Determine the value of

α

✦ Active

The matrix $A$ is singular, which means its determinant is zero. Calculate the determinant of $A$ :

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

Simplify the expression and set it to zero:

- 1 - α + 6 - 2 = 0 ⟹ 3 - α = 0 ⟹ α = 3

Thus, the interval for $x$ is $[1, 3]$ .

Step 2: Find the function

f (x)

and analyze its monotonicity○ Expand

Integrate the given expression to find $f (x)$ :

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

Now, find the derivative $f^{'} (x)$ to determine the function's behavior:

f^{'} (x) = \frac{d}{d x} (\frac{x^{3}}{3} + x^{2} + 3 x) = x^{2} + 2 x + 3

Calculate the discriminant of $f^{'} (x)$ : $Δ = b^{2} - 4 a c = 2^{2} - 4 (1) (3) = 4 - 12 = - 8$ . Since $Δ < 0$ and the leading coefficient (1) is positive, $f^{'} (x) > 0$ for all real $x$ . This implies that $f (x)$ is a strictly increasing function on its domain, including the interval $[1, 3]$ .

💡 Teacher's Secret Hint

Remember that for a strictly increasing function on an interval $[a, b]$ , the minimum value is at $x = a$ and the maximum value is at $x = b$ .

Step 3: Calculate M, m, and

3 (M - m)

○ Expand

Since $f (x)$ is strictly increasing on $[1, 3]$ , the minimum value $m$ occurs at $x = 1$ and the maximum value $M$ occurs at $x = 3$ .

m = f (1) = \frac{1^{3}}{3} + 1^{2} + 3 (1) = \frac{1}{3} + 1 + 3 = \frac{13}{3}

M = f (3) = \frac{3^{3}}{3} + 3^{2} + 3 (3) = 9 + 9 + 9 = 27

Finally, calculate $3 (M - m)$ :

3 (M - m) = 3 (27 - \frac{13}{3}) = 3 (\frac{81 - 13}{3}) = 3 (\frac{68}{3}) = 68

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Maths Question 17 – JEE-MAIN 2026

Clue 1 of 3: Matrix Singularity and Function Extrema

🧠 Full Solution Path