TS EAMCET · Maths · Trigonometric Ratios & Identities
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 \(\sqrt{\left|\mathrm{M}^2-\mathrm{m}^2\right|}=\)
- A \(6\)
- B \(12\)
- C \(6 \sqrt{2}\)
- D \(12 \sqrt{3}\)
Answer & Solution
Correct Answer
(B) \(12\)
Step-by-step Solution
Detailed explanation
\(A = 2\sqrt{6}+1\), \(B = 2\sqrt{2}-\sqrt{3}\), \(C = -6\) \(A^2 = (2\sqrt{6}+1)^2 = 24+4\sqrt{6}+1 = 25+4\sqrt{6}\) \(B^2 = (2\sqrt{2}-\sqrt{3})^2 = 8-4\sqrt{6}+3 = 11-4\sqrt{6}\) \(\sqrt{A^2+B^2} = \sqrt{(25+4\sqrt{6})+(11-4\sqrt{6})} = \sqrt{36} = 6\)…
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