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Chapter 3: Problem 37

The function \(f(x)=\) maximum \(\\{\sqrt{x(2-x)}, 2-x\\}\) is nondifferentiable at \(x\) equal to (A) 1 (B) 0,2 (C) 0,1 (D) 1,2

Short Answer

Expert verified
Answer: The function f(x) is non-differentiable at x = 0 and x = 2.

Step by step solution

01

Identify the Piecewise Functions

First, we can identify that the given function is a piecewise function of two functions: 1. \(g(x) = \sqrt{x(2-x)}\) which is valid for \(0\le x\le 2\) 2. \(h(x) = 2-x\) which is valid for \(x \le 2\)
02

Check for Discontinuity or Undefined Points

Both \(g(x)\) and \(h(x)\) are continuous in their respective domains. The function \(f(x)\) should be continuous in the domains of \(g(x)\) and \(h(x)\). Thus, there are no discontinuous points.
03

Check for Corners

Next, we need to check for corner points where there is a change in the slope of the function. These will occur at the points where \(g(x) = h(x)\). To find such points, we will equate the two functions and solve for x: \(\sqrt{x(2-x)} = 2 - x\) Squaring both sides, we get: \(x(2-x) = (2-x)^2\) Expanding and simplifying the equation, we get: \(x^2-2x = 0\) \(x(x-2) = 0\) This gives us two possible points: 1. \(x = 0\) 2. \(x = 2\) These are the points where \(g(x)\) and \(h(x)\) intersect, so these will be the corner points.
04

Differentiate the Piecewise Functions

Next, we will differentiate both \(g(x)\) and \(h(x)\) in their respective domains to check their differentiability. For \(g(x)\): \(\frac{d}{dx}\sqrt{x(2-x)} = \frac{d}{dx}\sqrt{x(2-x)} = \frac{2-2x}{2\sqrt{x(2-x)}}\) which is valid for \(0\le x\le 2\) For \(h(x)\): \(\frac{d}{dx}(2 - x) = -1\) which is valid for \(x\le2\)
05

Check for Differentiability at the Intersection Points

Finally, we will check the differentiability of \(f(x)\) at the intersection points \(x = 0\) and \(x = 2\). At \(x = 0\), \(g'(x)\) is not defined as there is division by 0, and \(h'(0) = -1\). Since the derivatives are not equal, f(x) is non-differentiable at \(x = 0\). At \(x=2\), \(g'(x)\) is not defined as there is division by 0, and \(h'(2) = -1\). Since the derivatives are not equal, f(x) is also non-differentiable at \(x = 2\). From our analysis, we find that the function \(f(x)\) is non-differentiable at \(x = 0\) and \(x = 2\). Therefore, the correct answer is (B) 0, 2.

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Most popular questions from this chapter

If $f(x)=\left\\{\begin{array}{ll}{[x]+\sqrt{\\{x\\}}} & x<1 \\\ \frac{1}{[x]+\\{x\\}^{2}} & x \geq 1\end{array}\right.\(, then [where \)[.]$ and \(\\{.\) represents greatest integer part and fractional part respectively.] (A) \(f(x)\) is continuous at \(x=1\) but not differentiable (B) \(f(x)\) is not continuous at \(x=1\) (C) \(\mathrm{f}(\mathrm{x})\) is differentiable at \(\mathrm{x}=1\) (D) \(\lim _{x \rightarrow 1} f(x)\) does not exist

Let \(f(x)=\cos x\) and \(g(x)=\) $g(x)= \begin{cases}\text { minimum }\\{f(t): 0 \leq t \leq x\\}, x \in[0, \pi] \\ \sin x-1, & x>\pi\end{cases}$ then (A) \(g(x)\) is discontinuous at \(x=\pi\) (B) \(g(x)\) is continuous for \(x \in[0, \infty)\) (C) \(\mathrm{g}(\mathrm{x})\) is differentiable at \(\mathrm{x}=\pi\) (D) \(g(x)\) is differentiable for \(x[0, \infty)\)

If \(f(x)\) has isolated point discontinuity at \(x=a\) such that \(|f(x)|\) is continuous at \(x=\) a then (A) \(|\mathrm{f}(\mathrm{x})|\) must be differentiable at \(\mathrm{x}=\mathrm{a}\) (B) \(\lim _{x \rightarrow a} f(x)\) does not exist (C) \(\lim _{x \rightarrow a} f(x)+f(a)=0\) (D) \(\mathrm{f}(\mathrm{a})=0\)

Let \(f: R \rightarrow R, f(x-f(y))=f(f(y))+x f(y)+f(x)-1 \forall x\), \(y \in R\), if \(f(0)=1\) and \(f^{\prime}(0)=0\), then (A) \(\mathrm{f}(\mathrm{x})=1-\frac{\mathrm{x}^{2}}{2}\) (B) \(f(x)=x^{2}+1\) (C) \(f(x)=\left(\frac{2 x+1}{x+1}\right)\) (D) none of these

If $f(x)=\left\\{\begin{array}{ll}{[x]+\sqrt{\\{x\\}}} & x<1 \\\ \frac{1}{[x]+\\{x\\}^{2}} & x \geq 1\end{array}\right.$, then [where [. ] and \\{ - \(\\}\) represent greatest integer and fractional part functions respectively] (A) \(\mathrm{f}(\mathrm{x})=\) is continuous at \(\mathrm{x}=1\) but not differentiable (B) \(\mathrm{f}(\mathrm{x})\) is not continuous at \(\mathrm{x}=1\)
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