AP EAMCET · Maths · Application of Derivatives
The constant \(c\) of Lagrange's mean value theorem for \(f(x)=\cos x-\sin 2 x\) in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) is
- A 0
- B \(\sin ^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right)\)
- C \(\cos ^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right)\)
- D \(\pm \frac{\pi}{4}\)
Answer & Solution
Correct Answer
(B) \(\sin ^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right)\)
Step-by-step Solution
Detailed explanation
The given function \(f(x)=\cos x-\sin 2 x\), is a continuous function in interval \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and differentiable in interval \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). Now, according to Lagrange's mean value theorem, there exist…
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