Let M and m respectively be the maximum and the minimum values of f(x)=|\beginarrayccc1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \sin ^2 x & 1+\cos ^2 x & 4 \sin 4 x \sin ^2 x & \cos ^2 x & 1+4 \sin 4 x\endarray|, x \in R Then M^4 - m^4 is equal to :

Matrices and Determinants PYQ — JEE Main 2025 · 29 Jan · Shift 1 · Q44 with Solution

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Correct answer: A

Let $s=\sin^2 x,\quad c=\cos^2 x,$ so that $s+c=1.$ Also let $t=4\sin 4x.$

Then the determinant becomes

1+s & c & t\\ s & 1+c & t\\ s & c & 1+t \end{vmatrix}.$$ 2. **Use row operations to simplify** Apply: - $R_1 \to R_1-R_3$ - $R_2 \to R_2-R_3$ Then $$R_1=(1,0,-1),\qquad R_2=(0,1,-1),\qquad R_3=(s,c,1+t).$$ So $$f(x)=\begin{vmatrix} 1 & 0 & -1\\ 0 & 1 & -1\\ s & c & 1+t \end{vmatrix}.$$ 3. **Evaluate the determinant** Expand along the first row: $$f(x)=1\cdot\begin{vmatrix}1 & -1\\ c & 1+t\end{vmatrix}+(-1)\cdot\begin{vmatrix}0 & 1\\ s & c\end{vmatrix}.$$ Now, $$\begin{vmatrix}1 & -1\\ c & 1+t\end{vmatrix}=1+t+c,$$ and $$\begin{vmatrix}0 & 1\\ s & c\end{vmatrix}=-s.$$ Hence $$f(x)=1+t+c+s=1+t+(s+c)=1+t+1=2+t.$$ Since $t=4\sin 4x$, $$f(x)=2+4\sin 4x.$$ 4. **Find maximum and minimum values** Because $$-1\le \sin 4x \le 1,$$ we get $$2-4\le f(x)\le 2+4,$$ so $$-2\le f(x)\le 6.$$ Thus, $$M=6,\qquad m=-2.$$ 5. **Compute $M^4-m^4$** $$M^4-m^4=6^4-(-2)^4=1296-16=1280.$$ 6. **Option check** The correct option is: $$\boxed{\text{A: }1280}$$

View written solutionFree

Correct answer: A

Let $s=\sin^2 x,\quad c=\cos^2 x,$ so that $s+c=1.$ Also let $t=4\sin 4x.$

Then the determinant becomes

1+s & c & t\\ s & 1+c & t\\ s & c & 1+t \end{vmatrix}.$$ 2. **Use row operations to simplify** Apply: - $R_1 \to R_1-R_3$ - $R_2 \to R_2-R_3$ Then $$R_1=(1,0,-1),\qquad R_2=(0,1,-1),\qquad R_3=(s,c,1+t).$$ So $$f(x)=\begin{vmatrix} 1 & 0 & -1\\ 0 & 1 & -1\\ s & c & 1+t \end{vmatrix}.$$ 3. **Evaluate the determinant** Expand along the first row: $$f(x)=1\cdot\begin{vmatrix}1 & -1\\ c & 1+t\end{vmatrix}+(-1)\cdot\begin{vmatrix}0 & 1\\ s & c\end{vmatrix}.$$ Now, $$\begin{vmatrix}1 & -1\\ c & 1+t\end{vmatrix}=1+t+c,$$ and $$\begin{vmatrix}0 & 1\\ s & c\end{vmatrix}=-s.$$ Hence $$f(x)=1+t+c+s=1+t+(s+c)=1+t+1=2+t.$$ Since $t=4\sin 4x$, $$f(x)=2+4\sin 4x.$$ 4. **Find maximum and minimum values** Because $$-1\le \sin 4x \le 1,$$ we get $$2-4\le f(x)\le 2+4,$$ so $$-2\le f(x)\le 6.$$ Thus, $$M=6,\qquad m=-2.$$ 5. **Compute $M^4-m^4$** $$M^4-m^4=6^4-(-2)^4=1296-16=1280.$$ 6. **Option check** The correct option is: $$\boxed{\text{A: }1280}$$