Question

If the extreme values of the function \( f(x) = (2\sqrt{6}+1)\cos x + (2\sqrt{2}-\sqrt{3})\sin x - 6 \) are m and M, then \( \sqrt{|M^2-m^2|} = \)

• A. 6
• B. 12
• C. \( 6\sqrt{2} \)
• D. \( 12\sqrt{3} \)

Hint:

For any function of the form \(f(x) = a\cos x + b\sin x + c\), the range is \( [c - \sqrt{a^2+b^2}, c + \sqrt{a^2+b^2}] \). The term \( \sqrt{a^2+b^2} \) represents the amplitude of the sinusoidal part.

Answer 1

The Correct Option is B

The function is of the form \( f(x) = A\cos x + B\sin x + C \).
The maximum value (M) is \( C + \sqrt{A^2+B^2} \) and the minimum value (m) is \( C - \sqrt{A^2+B^2} \).
Here, \( A = 2\sqrt{6}+1 \), \( B = 2\sqrt{2}-\sqrt{3} \), and \( C = -6 \).
Let's calculate \( A^2+B^2 \).
\( A^2 = (2\sqrt{6}+1)^2 = (2\sqrt{6})^2 + 2(2\sqrt{6})(1) + 1^2 = 24 + 4\sqrt{6} + 1 = 25 + 4\sqrt{6} \).
\( B^2 = (2\sqrt{2}-\sqrt{3})^2 = (2\sqrt{2})^2 - 2(2\sqrt{2})(\sqrt{3}) + (\sqrt{3})^2 = 8 - 4\sqrt{6} + 3 = 11 - 4\sqrt{6} \).
\( A^2 + B^2 = (25 + 4\sqrt{6}) + (11 - 4\sqrt{6}) = 36 \).
So, \( \sqrt{A^2+B^2} = \sqrt{36} = 6 \).
The maximum value is \( M = C + 6 = -6 + 6 = 0 \).
The minimum value is \( m = C - 6 = -6 - 6 = -12 \).
Now, we need to calculate \( \sqrt{|M^2-m^2|} \).
\( M^2 = 0^2 = 0 \).
\( m^2 = (-12)^2 = 144 \).
\( \sqrt{|M^2-m^2|} = \sqrt{|0 - 144|} = \sqrt{|-144|} = \sqrt{144} = 12 \).