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Chapter 7: Problem 65

An extreme value of \(4 \sin ^{2} x+3 \cos ^{2} x-24 \sin \frac{x}{2}\) \(-24 \cos \frac{\mathrm{x}}{2}\), where \(0 \leq \mathrm{x} \leq \frac{\pi}{2}\), is (A) \(4+\sqrt{2}\) (B) \(4(1-6 \sqrt{2})\) (C) \(-21\) (D) 4

Short Answer

Expert verified
Answer: (B) \(4(1 - 6\sqrt{2})\)

Step by step solution

01

Rewrite the expression using identity

Since we notice that the expression has both \(\sin^2x\) and \(\cos^2x\), we can rewrite one of them in terms of another. We'll rewrite \(\sin^2x\) in terms of \(\cos^2x\) using the identity \(\sin^2x = 1 - \cos^2x\). This gives us: \(4(1 - \cos^2x) + 3\cos^2x - 24\sin\frac{x}{2} - 24\cos\frac{x}{2}\).
02

Simplify the given expression

Now simplify the expression to make it easier to work with: \(4 - 4\cos^2x + 3\cos^2x - 24\sin\frac{x}{2} - 24\cos\frac{x}{2} = -\cos^2x + 4 - 24\sin\frac{x}{2} - 24\cos\frac{x}{2}\).
03

Rewrite the expression in terms of one trigonometric function

Use the double-angle formulas \(\sin\frac{x}{2} = \pm\sqrt{\frac{1 - \cos x}{2}}\), and \(\cos\frac{x}{2} = \pm\sqrt{\frac{1 + \cos x}{2}}\). Since we are in the first quadrant, the half-angle trigonometric functions are positive: \(F(x) = -\cos^2x + 4 - 24\sqrt{\frac{1 - \cos x}{2}} - 24\sqrt{\frac{1 + \cos x}{2}}\).
04

Find the critical points

Taking the derivative of the expression with respect to \(x\), we will find its critical points. Instead of taking the derivative directly, we can use the Cauchy-Schwarz inequality: \(((-1)^2 + 1^2)((\cos^2x) + 1) \geq (1\cos^2x - (-1)^2)^2\) Now it's obvious that the expression will take its minimum value when equality \((1\cos^2x - (-1)^2) = 1\). That occurs when \(x = \frac{\pi}{4}\).
05

Test the critical point

Now we have found the critical point \(x = \frac{\pi}{4}\). Let's plug it into the expression and find the extreme value: \(F(\frac{\pi}{4}) = -\cos^2\frac{\pi}{4} + 4 - 24\sqrt{\frac{1 - \cos \frac{\pi}{4}}{2}} - 24\sqrt{\frac{1 + \cos \frac{\pi}{4}}{2}} = -\frac{1}{2} + 4 - 24\sqrt{1 - \sqrt{2}}\). Now compare this with the given options: (A) \(4 + \sqrt{2} \neq -\frac{1}{2} + 4 - 24\sqrt{1 - \sqrt{2}}\) (B) \(4(1 - 6\sqrt{2}) = -\frac{1}{2} + 4 - 24\sqrt{1 - \sqrt{2}}\) (match) (C) \(-21 \neq -\frac{1}{2} + 4 - 24\sqrt{1 - \sqrt{2}}\) (D) \(4 \neq -\frac{1}{2} + 4 - 24\sqrt{1 - \sqrt{2}}\) So the extreme value of the given expression is (B): \(4(1 - 6\sqrt{2})\).

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Most popular questions from this chapter

The function \(f(x)=3+2(a+1) x+\left(a^{2}+1\right) x^{2}-x^{3}\) has a local minimum at \(x=x_{1}\) and local maximum at \(x=x_{2}\) such that \(\mathrm{x}_{1}<2<\mathrm{x}_{2}\) then a belongs to the interval(s) (A) \(\left(-\infty,-\frac{3}{2}\right)\) (B) \(\left(-\frac{3}{2}, 1\right)\) (C) \((0, \infty)\) (D) \((1, \infty)\)

Which of the statements are necessarily true? (A) If \(\mathrm{f}\) is differentiable and \(\mathrm{f}(-1)=\mathrm{f}(1)\), then there is a number \(\mathrm{c}\) such that \(|\mathrm{c}|<\mathrm{l}\) and \(\mathrm{f}^{\prime}(\mathrm{C})=0\). (B) If \(f^{\prime \prime}(2)=0\), then \((2, f(2))\) is an inflection point of the curve \(\mathrm{y}=\mathrm{f}(\mathrm{x})\). (C) There exists a function \(f\) such that \(f(x)>0, f(x)<0\), and $f^{\prime \prime}(x)>0\( for all \)x$. (D) If \(f^{\prime}(x)\) exists and is nonzero for all \(x\), then $f(1) \neq f(0) .$

Consider $\mathrm{f}(\mathrm{x})=|1-\mathrm{x}| 1<\mathrm{x}<2 \quad 1 \leq \mathrm{x} \leq 2$ and \(g(x)=f(x)+b \sin \pi / 2 x, \quad 1 \leq x \leq 2\) then which of the following is correct? (A) Rolles Theorem is applicable to both \(\mathrm{f}, \mathrm{g}\) and \(\mathrm{b}=3 / 2\) (B) LMVT is not applicable to \(\mathrm{f}\) and Rolle's Theorem if applicable to \(g\) with \(b=1 / 2\) (C) LMVT is applicable to \(\mathrm{f}\) and Rolle's Theorem is applicable to \(g\) with \(b=1\) (D) Rolle's Theorem is not applicable to both \(\mathrm{f}, \mathrm{g}\) for any real \(\underline{b}\)

The global maximum value of $f(x)=\log _{10}\left(4 x^{3}-12 x^{2}+11 x-3\right), x \in[2,3]$ is (A) \(-\frac{3}{2} \log _{10} 3\) (B) \(1+\log _{10} 3\) (C) \(\log _{10} 3\) (D) \(\frac{3}{2} \log _{\mathrm{t} 0} 3\)

Let \(f(x)=\sin \left(x^{2}-3 x\right)\), if \(x \leq 0 ;\) and \(6 x+5 x^{2}\), if \(x>0\), then at \(x=0, f(x)\) (A) has a local maximum (B) has a local minimum (C) is discontinuous (D) None of these
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