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Matrices and Determinants question

2021 · 27 Jul · Shift 1 · Q43
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Matrices and Determinants question

2021 · 27 Jul · Shift 1 · Q43

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
Let f(x)=∣sin⁡2x−2+cos⁡2xcos⁡2x2+sin⁡2xcos⁡2xcos⁡2xsin⁡2xcos⁡2x1+cos⁡2x∣,x∈[0,π]f(x) = \left| {\begin{matrix} {{{\sin }^2}x} & { - 2 + {{\cos }^2}x} & {\cos 2x} \\ {2 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \\ {{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \cos 2x} \\ \end{matrix} } \right|,x \in [0,\pi ]f(x)=​sin2x2+sin2xsin2x​−2+cos2xcos2xcos2x​cos2xcos2x1+cos2x​​,x∈[0,π]. Then the maximum value of f(x) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 6

  1. Let s=sin⁡2x,c=cos⁡2x,cos⁡2x=c−s.s=\sin^2 x,\quad c=\cos^2 x,\quad \cos 2x = c-s.s=sin2x,c=cos2x,cos2x=c−s. Also, s+c=1.s+c=1.s+c=1.

Then the determinant becomes

s & -2+c & c-s\\ 2+s & c & c-s\\ s & c & 1+(c-s) \end{vmatrix}.$$ Since $1+\cos 2x=1+c-s=(s+c)+c-s=2c$, the matrix is $$f(x)=\begin{vmatrix} s & c-2 & c-s\\ s+2 & c & c-s\\ s & c & 2c \end{vmatrix}.$$ 2. Now apply row operations that do not change the determinant: - $R_2 \to R_2-R_1$ - $R_3 \to R_3-R_1$ Then $$R_2=(2,2,0),\qquad R_3=(0,2,2c-(c-s))=(0,2,s+c)= (0,2,1).$$ So, $$f(x)=\begin{vmatrix} s & c-2 & c-s\\ 2 & 2 & 0\\ 0 & 2 & 1 \end{vmatrix}.$$ 3. Expand along the first row: $$f(x)=s\begin{vmatrix}2&0\\2&1\end{vmatrix}-(c-2)\begin{vmatrix}2&0\\0&1\end{vmatrix}+(c-s)\begin{vmatrix}2&2\\0&2\end{vmatrix}.$$ Compute the minors: $$\begin{vmatrix}2&0\\2&1\end{vmatrix}=2,$$ $$\begin{vmatrix}2&0\\0&1\end{vmatrix}=2,$$ $$\begin{vmatrix}2&2\\0&2\end{vmatrix}=4.$$ Hence, $$f(x)=2s-2(c-2)+4(c-s).$$ Simplify: $$f(x)=2s-2c+4+4c-4s=4+2c-2s.$$ But $$c-s=\cos 2x,$$ so $$f(x)=4+2\cos 2x.$$ 4. Since $x\in[0,\pi]$, we have $$\cos 2x\in[-1,1].$$ Therefore, $$f(x)=4+2\cos 2x$$ has maximum when $\cos 2x=1$. Thus, $$f_{\max}=4+2(1)=6.$$ 5. So the required maximum value is $$\boxed{6}.$$
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