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Chapter 7: Problem 4

If \(f(x)=\frac{1+x}{[x]} x \in[1,3]\) then which of following statements about \(\mathrm{f}(\mathrm{x})\) is correct (A) \(f(x)\) has local maxima at \(x=2\) (B) \(f(x)\) has no point of local maxima (C) \(\mathrm{f}(\mathrm{x})\) hasneither maxima nor minima at \(\mathrm{x}=2\) (D) \(f(x)\) has 1 point of local minimum and 2 points of local maxima

Short Answer

Expert verified
Answer: Neither

Step by step solution

01

Rewrite the function for clarity

We are given the function, \(f(x) = \frac{1+x}{[x]} x \in [1,3]\). Let's rewrite it for more clarity. As \([x]\) represents the greatest integer function, where \([x]\) is the largest integer less than or equal to x, we can say that: \(f(x) = \frac{1+x}{n}\), where n is an integer and \( n \leq x < n+1\) Since \(x \in [1,3]\), we can have two cases: 1. When x is in the interval \([1,2)\), \(f(x) = \frac{1+x}{1} = 1+x\) 2. When x is in the interval \([2,3]\) , \(f(x) = \frac{1+x}{2}\)
02

Find the first derivatives for each case

We now find the first derivative of \(f(x)\) for both cases. 1. For \(1 \leq x < 2\), the first derivative is: \(f'(x) = \frac{d}{dx}(1+x) = 1\) 2. For \(2 \leq x \leq 3\), the first derivative is: \(f'(x) = \frac{d}{dx}(\frac{1+x}{2}) = \frac{1}{2}\)
03

Analyze the critical points

We now analyze the critical points for both cases. In both cases, \(f'(x)\) is constant and non-zero; this means there are no critical points in these intervals. However, the boundary between these two cases is at x = 2. Since the function is piecewise-defined and the first derivative is not continuous at x = 2, we must consider this point specifically.
04

Evaluate the function and its first derivative at x = 2

We now evaluate \(f(x)\) and its first derivative at \(x = 2\): 1. For \(x = 2\) and \(1 \leq x < 2\), \(f(x) = 1 + x = 3\) and \(f'(x) = 1\) 2. For \(x = 2\) and \(2 \leq x \leq 3\), \(f(x) = \frac{1+x}{2}=\frac{3}{2}\) and \(f'(x)=\frac{1}{2}\) At \(x=2\), there is a discontinuity in the first derivative, but the function itself is continuous and well-defined.
05

Make a conclusion based on the analysis

Since at \(x=2\), there is a discontinuity in the first derivative, we can't talk about minimum or maximum there. Thus, the function has no point of local maxima or local minima on the given interval. Therefore, statement (C) is correct: \(\mathrm{f}(\mathrm{x})\) has neither maxima nor minima at \(\mathrm{x}=2\).

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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 of the following is are true for the function \(\mathrm{y}=\mathrm{g}(\mathrm{x})\) ? (A) \(g(x)\) monotonically decreases in $\left(-\infty, 2-\frac{1}{\sqrt{3}}\right)\( and \)\left(2+\frac{1}{\sqrt{3}}, \infty\right)$ (B) g(x) monotonically increases in $\left(2-\frac{1}{\sqrt{3}}, 2+\frac{1}{\sqrt{3}}\right)$ (C) There exists exactly one tangent to \(\mathrm{y}=\mathrm{g}(\mathrm{x})\) which is parallel to the chord joining the points \((1, g(1))\) and \((3, g(3))\) (D) There exists exactly two distinct Lagrange's Mean Value in \((0,4)\) for the function \(\mathrm{y}=\mathrm{g}(\mathrm{x})\).

Assertion \((\mathbf{A}):\) Let \(f(x)=5-4(x-2)^{21}\), then at \(x=2\) the function \(f(x)\) attains neither the least value nor the greatest value. Reason \((\mathbf{R}):\) At \(x=2\), the first derivative does not exist.

If \(a>b>0\) and $f(\theta)=\frac{\left(a^{2}-b^{2}\right) \cos \theta}{a-b \sin \theta}$, then the maximum value of \(f(\theta)\) is (A) \(\sqrt{a^{2}+b^{2}}\) (B) \(\sqrt{a^{2}-b^{2}}\) (C) \(a^{2}-b^{2}\) (D) \(\frac{a-b}{a+b}\)

For any real \(\theta\), the maximum value of \(\cos ^{2}(\cos \theta)+\) $\sin ^{2}(\sin \theta)$ is (A) 1 (B) \(1+\sin ^{2} 1\) (C) \(\left.1+\cos ^{2}\right]\) (D) Does not exist

Let \(f(x)=a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x\), where \(a, b_{1} c, d, e \in R\) and \(f(x)=0\) has a positive root \(\alpha\), then (A) \(\mathrm{f}(\mathrm{x})=0\) has root al such that \(0<\alpha_{1}<\alpha\) (B) \(\mathrm{f}^{\prime}(\mathrm{x})=0\) has at least one real root (C) \(\mathrm{f}^{\prime}(\mathrm{x})=0\) has at least two real roots (D) All of the above
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