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Chapter 7: Problem 7

The total number of values of \(x\), where \(f(x)=2^{-x}\) (cos \(x\) $+\cos \sqrt{3} \mathrm{x}$ ) attains its maximum value is (A) \(\underline{1}\) (B) \(\underline{2}\) (C) 4 (D) None

Short Answer

Expert verified
Answer: There are no values of \(x\) where the function attains its maximum value, or (D) None.

Step by step solution

01

Finding the First Derivative

First, apply the product rule for derivatives: \((uv)' = u'v + uv'\). In this case, \(u = 2^{-x}\) and \(v = \cos x + \cos{\sqrt{3}x}\). We'll find \(f'(x)\): $$ u' = -x 2^{-x-1}\ln 2, $$ $$ v' = -\sin x - \sqrt{3}\sin{\sqrt{3}x}. $$ Now we'll find \(f'(x)\) by applying the product rule: $$ f'(x) = u'v + uv' = -x 2^{-x-1}\ln 2 (\cos x + \cos{\sqrt{3}x}) + 2^{-x}(-\sin x - \sqrt{3}\sin{\sqrt{3}x}). $$
02

Finding the Critical Points

Now, we'll set \(f'(x)\) to zero and solve for \(x\): $$ 0 = -x 2^{-x-1}\ln 2 (\cos x + \cos{\sqrt{3}x}) + 2^{-x}(-\sin x - \sqrt{3}\sin{\sqrt{3}x}). $$ Upon examining the equation, one may observe that \(x=0\) seems to be one of the solutions, as it causes the first term in the equation to cancel out. We will confirm this solution in the next step.
03

Finding the Second Derivative

Now, we find the second derivative of \(f(x)\) to analyze the concavity at the critical points we found: $$ f''(x) = \frac{d}{dx} \left[ -x 2^{-x-1}\ln 2 (\cos x + \cos{\sqrt{3}x}) + 2^{-x}(-\sin x - \sqrt{3}\sin{\sqrt{3}x}) \right]. $$ At \(x=0\), we have $$ f''(0) = \frac{d}{dx} \left[ -x 2^{-x-1}\ln 2 (\cos x + \cos{\sqrt{3}x}) + 2^{-x}(-\sin x - \sqrt{3}\sin{\sqrt{3}x}) \right]_{x=0}. $$
04

Evaluating Concavity and Finding Maximum

Now, we'll evaluate the second derivative at the critical point \(x=0\). If \(f''(0) < 0\), then \(f(x)\) has a local maximum at \(x=0\). If \(f''(0) > 0\), then there is no local maximum at \(x=0\). If \(f''(0) = 0\), the test is inconclusive, and we'll have to consider other methods. After evaluating \(f''(0)\) using the second derivative we found in step 3, we get the following value: $$ f''(0) = 4\ln 2 - 4 - 2\sqrt{3}. $$ As this value is positive, there is no local maximum at \(x=0\). Since this was the only critical point we derived, there is no maximum value attained by the given function. Therefore, the answer is (D) None.

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Most popular questions from this chapter

Consider \(f, g\) and \(h\) be three real valued dilierentiable functions defined on \(\mathrm{R}\). Let $g(x)=x^{3}+g^{\prime \prime}(1) x^{2}+\left(3 g^{\prime}(1)-g^{\prime \prime}(1)-1\right) x+3 g^{\prime}(1)\(, \)f(x)=x g(x)-12 x+1$ and \(f(x)=(h(x))^{2}\) where \(h(0)=1\). Which one of the following does not hold good for \(\mathrm{y}=\mathrm{h}(\mathrm{x})\) ? (A) Exactly one critical point (B) No point of inflection (C) Exactly one real zero in \((0,3)\) (D) Exactly one tangent parallel to \(\mathbf{x}\)-axis

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

If \(a, b, c d \in R\) such that \(\frac{a+2 c}{b+3 d}+\frac{4}{3}=0\), then the equation \(a x^{2}+c x+d=0\) has (A) atleast one root in \((-1,0)\) (B) atleast one root in \((0,1)\) (C) no root in \((-1,1)\) (D) no root in \((0,2)\)

The global minimum value of $\mathrm{e}^{\left(2 \mathrm{x}^{2}-2 x+1\right) \sin ^{\prime} \mathrm{x}}$ is (A) \(\mathrm{c}\) (B) \(1 / 3\) (C) 1 (D) 0

Let \(g(x)=-\frac{f(-1)}{2} x^{2}(x-1)-f(0)\left(x^{2}-1\right)\) \(+\frac{f(1)}{2} x^{2}(x+1)-f^{\prime}(0) x(x-1)(x+1)\) where \(f\) is a thrice differentiable function. Then the correct statements are (A) there exists \(x \in(-1,0)\) such that \(f^{\prime}(x)=g^{\prime}(x)\) (B) there exists \(x \in(0,1)\) such that $f^{\prime \prime}(x)=g^{\prime \prime}(x)$ (C) there exists \(x \in(-1,1)\) such that $f^{\prime \prime}(x)=g^{\prime \prime \prime}(x)$ (D) there exists \(x \in(-1,1)\) such that \(f^{\prime \prime}(x)=3 f(1)-3 f(-1)\) \(-6 \mathrm{f}^{\prime}(0)\)
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