The maximum value of f(x) = | \beginmatrix \sin ^2x & 1 + \cos ^2x & \cos 2x 1 + \sin ^2x & \cos ^2x & \cos 2x \sin ^2x & \cos ^2x & \sin 2x \endmatrix |,x \in R is :

Application of Derivatives PYQ — JEE Main 2021 · 16 Mar · Shift 2 · Q40 with Solution

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Correct answer: A

\sin^2 x & 1+\cos^2 x & \cos 2x\\ 1+\sin^2 x & \cos^2 x & \cos 2x\\ \sin^2 x & \cos^2 x & \sin 2x \end{vmatrix}.$$ We need the maximum value of $f(x)=|D(x)|$. 2. Use row operations to simplify the determinant. Let $$R_1\to R_1-R_3,\qquad R_2\to R_2-R_3.$$ Then $$D(x)=\begin{vmatrix} 0 & 1 & \cos 2x-\sin 2x\\ 1 & 0 & \cos 2x-\sin 2x\\ \sin^2 x & \cos^2 x & \sin 2x \end{vmatrix}.$$ Set $$a=\cos 2x-\sin 2x,\qquad s=\sin^2 x,\qquad c=\cos^2 x.$$ So $$D(x)=\begin{vmatrix} 0&1&a\\ 1&0&a\\ s&c&\sin 2x \end{vmatrix}.$$ 3. Expand along the first row: $$D(x)=0-\begin{vmatrix}1&a\\ s&\sin 2x\end{vmatrix}+a\begin{vmatrix}1&0\\ s&c\end{vmatrix}.$$ Hence $$D(x)=-(\sin 2x-as)+ac.$$ So $$D(x)=-\sin 2x+a(s+c).$$ Since $$s+c=\sin^2 x+\cos^2 x=1,$$ we get $$D(x)=-\sin 2x+(\cos 2x-\sin 2x)=\cos 2x-2\sin 2x.$$ Therefore $$f(x)=|\cos 2x-2\sin 2x|.$$ 4. Now find its maximum value. For any expression of the form $$A\cos\theta+B\sin\theta,$$ the maximum absolute value is $$\sqrt{A^2+B^2}.$$ Here, $$A=1,\qquad B=-2.$$ So $$\max |\cos 2x-2\sin 2x|=\sqrt{1^2+(-2)^2}=\sqrt5.$$ 5. Thus the maximum value is $$\boxed{\sqrt5}.$$ So the correct option is **A**.

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Correct answer: A

\sin^2 x & 1+\cos^2 x & \cos 2x\\ 1+\sin^2 x & \cos^2 x & \cos 2x\\ \sin^2 x & \cos^2 x & \sin 2x \end{vmatrix}.$$ We need the maximum value of $f(x)=|D(x)|$. 2. Use row operations to simplify the determinant. Let $$R_1\to R_1-R_3,\qquad R_2\to R_2-R_3.$$ Then $$D(x)=\begin{vmatrix} 0 & 1 & \cos 2x-\sin 2x\\ 1 & 0 & \cos 2x-\sin 2x\\ \sin^2 x & \cos^2 x & \sin 2x \end{vmatrix}.$$ Set $$a=\cos 2x-\sin 2x,\qquad s=\sin^2 x,\qquad c=\cos^2 x.$$ So $$D(x)=\begin{vmatrix} 0&1&a\\ 1&0&a\\ s&c&\sin 2x \end{vmatrix}.$$ 3. Expand along the first row: $$D(x)=0-\begin{vmatrix}1&a\\ s&\sin 2x\end{vmatrix}+a\begin{vmatrix}1&0\\ s&c\end{vmatrix}.$$ Hence $$D(x)=-(\sin 2x-as)+ac.$$ So $$D(x)=-\sin 2x+a(s+c).$$ Since $$s+c=\sin^2 x+\cos^2 x=1,$$ we get $$D(x)=-\sin 2x+(\cos 2x-\sin 2x)=\cos 2x-2\sin 2x.$$ Therefore $$f(x)=|\cos 2x-2\sin 2x|.$$ 4. Now find its maximum value. For any expression of the form $$A\cos\theta+B\sin\theta,$$ the maximum absolute value is $$\sqrt{A^2+B^2}.$$ Here, $$A=1,\qquad B=-2.$$ So $$\max |\cos 2x-2\sin 2x|=\sqrt{1^2+(-2)^2}=\sqrt5.$$ 5. Thus the maximum value is $$\boxed{\sqrt5}.$$ So the correct option is **A**.