Question

solve it ..I came to the conclusion the question was wrong ...do it guys if you can​

Answer 1

CORRECT QUESTION:

The maximum value of

$\left|\begin{array}{c c c} 1 + \sin^2x&\cos^2x&4\sin 2x \\ \\ \sin^2x&1 + \cos^2x&4\sin 2x \\ \\\sin^2x& \cos^2x&1 + 4\sin 2x \end{array} \right|$

ANSWER:

• The maximum value = 6

GIVEN:

$\left|\begin{array}{c c c} 1 + \sin^2x&\cos^2x&4\sin 2x \\ \\ \sin^2x&1 + \cos^2x&4\sin 2x \\ \\\sin^2x& \cos^2x&1 + 4\sin 2x \end{array} \right|$

TO FIND:

The maximum value of

$\left|\begin{array}{c c c} 1 + \sin^2x&\cos^2x&4\sin 2x \\ \\ \sin^2x&1 + \cos^2x&4\sin 2x \\ \\\sin^2x& \cos^2x&1 + 4\sin 2x \end{array} \right|$

EXPLANATION:

$\left|\begin{array}{c c c} 1 + \sin^2x&\cos^2x&4\sin 2x \\ \\ \sin^2x&1 + \cos^2x&4\sin 2x \\ \\\sin^2x& \cos^2x&1 + 4\sin 2x \end{array} \right|$

$R_1\implies R_1 - R_2$

$\left|\begin{array}{c c c} 1 + \sin^2x-sin^2x&\cos^2x-\cos^2x-1&4\sin 2x -4\sin2x \\ \\ \sin^2x&1 + \cos^2x&4\sin 2x \\ \\\sin^2x& \cos^2x&1 + 4\sin 2x \end{array} \right|$

$\left|\begin{array}{c c c} 1& - 1&0 \\ \\ \sin^2x&1 +\cos^2x&4\sin 2x \\ \\\sin^2x& \cos^2x&1 + 4\sin 2x \end{array} \right|$

$R_2\implies R_2 - R_3$

$\left|\begin{array}{c c c} 1& - 1&0 \\ \\ \sin^2x-\sin^2x&1 +\cos^2x-\cos^2x&4\sin 2x-1 - 4\sin 2x \\ \\\sin^2x& \cos^2x&1 + 4\sin 2x \end{array} \right|$

$\left|\begin{array}{c c c} 1& - 1&0 \\ \\0&1& - 1 \\ \\\sin^2x& \cos^2x&1 + 4\sin 2x \end{array} \right|$

$C_2\implies C_2+C_1$

$\left|\begin{array}{c c c} 1&-1+1&0 \\ \\0&0+1& - 1 \\ \\\sin^2x& \cos^2x + \sin^{2} x&1 + 4\sin 2x \end{array} \right|$

$\left|\begin{array}{c c c} 1& 0&0 \\ \\0&1& - 1 \\ \\\sin^2x&1&1 + 4\sin 2x \end{array} \right|$

$\Delta= 1(1(1 + 4\sin x) - 1(-1)) - 0+0$

$\Delta = 1 + 4 \sin 2x + 1$

$\Delta = 2+ 4 \sin 2x$

MAXIMUM VALUE OF SIN 2X = 1

$\Delta = 2+ 4 (1)$

$\Delta = 6$

HENCE THE MAXIMUNM VALUE IS EQUAL TO 6

Answer 2

Answer:

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