Question

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

• A. \(64\)
• B. \(68\)
• C. \(72\)
• D. \(76\)

Answer 1

The Correct Option is C

Concept: A matrix is singular if its determinant is zero. Also, \[ f(x)=\int_0^x g(t)dt \Rightarrow f'(x)=g(x) \] Thus maxima and minima in a closed interval occur at endpoints or critical points.

Step 1: {Find \( \alpha \) using determinant condition.} \[ |A|= \begin{vmatrix} 1&3&-1\\ 2&1&\alpha\\ 0&1&-1 \end{vmatrix} =0 \] Expanding: \[ =1 \begin{vmatrix} 1&\alpha\\ 1&-1 \end{vmatrix} -3 \begin{vmatrix} 2&\alpha\\ 0&-1 \end{vmatrix} -1 \begin{vmatrix} 2&1\\ 0&1 \end{vmatrix} \] \[ =(-1-\alpha)-3(-2)-2 \] \[ =-1-\alpha+6-2 \] \[ =3-\alpha \] Thus \[ 3-\alpha=0 \] \[ \alpha=3 \]

 Step 2: {Find \(f(x)\).} \[ f(x)=\int_0^x(t^2+2t+3)dt \] \[ =\frac{x^3}{3}+x^2+3x \] 

Step 3: {Find critical points.} \[ f'(x)=x^2+2x+3 \] Discriminant: \[ 4-12<0 \] Thus \(f'(x)>0\), so \(f(x)\) is increasing. 

Step 4: {Find maximum and minimum.} Minimum at \(x=1\): \[ f(1)=\frac13+1+3=\frac{13}{3} \] Maximum at \(x=3\): \[ f(3)=9+9+9=27 \]

 Step 5: {Compute required value.} \[ M-m=27-\frac{13}{3} \] \[ =\frac{68}{3} \] \[ 3(M-m)=68 \] Thus value is \(68\).