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Chapter 3: Problem 35

The number of points where the function \(f(x)=\left(x^{2}-1\right)\left|x^{2}-x-2\right|+\sin (|x|)\) is not differentiable is (A) 0 (B) \(I\) (C) 2 (D) 3

Short Answer

Expert verified
Answer: (C) 2

Step by step solution

01

Identify critical points for each term separately

First, we need to identify the critical points for each term, i.e., the points where the function might be non-differentiable. We have three terms in the function: a quadratic term, an absolute value term, and a sine term. In the function \(f(x)\), - Quadratic term: \(x^2 - 1\) is differentiable everywhere - Sine term: \(\sin(|x|)\) is differentiable everywhere For the absolute value term, let's break it down further: - Absolute value term: \(\left|x^2 - x - 2\right|\) is not differentiable at the points where \(x^2 - x - 2 = 0\)
02

Find the roots of the polynomial x^2 - x - 2

To find the points where the absolute value term might be non-differentiable, we need to find the roots of the polynomial \(x^2 - x - 2\). We can do this by using the quadratic formula: \(x =\frac{-b \pm \sqrt{b^2-4ac}}{2a}\), where \(a=1, b=-1\) and \(c=-2\) \(x =\frac{1 \pm \sqrt{1+8}}{2} = \frac{1 \pm \sqrt{9}}{2} = \frac{1 \pm 3}{2}\) The roots are \(x=2\) and \(x=-1\)
03

Check if the function is non-differentiable at the critical points

Now, we have two critical points: \(x=2\) and \(x=-1\). We need to check if the function is indeed non-differentiable at these points. We form the complete function: \(f(x)=\left(x^{2}-1\right)\left|x^{2}-x-2\right|+\sin (|x|)\) We observe that at these points, the function is continuous, but its derivative changes as it crosses these points due to the presence of \(\left|x^2-x-2\right|\). Hence, at \(x=2\) and \(x=-1\), the function has non-differentiable points. Therefore, there are 2 non-differentiable points in the function \(f(x)\)
04

Compare the result with the provided options

Our result is that there are 2 non-differentiable points in the function \(f(x)=\left(x^{2}-1\right)\left|x^{2}-x-2\right|+\sin (|x|)\). Comparing it with the given options, we find out that the correct answer is: (C) 2

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Most popular questions from this chapter

Let \(\mathrm{f}(\mathrm{x})\) be differentiable at \(\mathrm{x}=\mathrm{h}\) then \(\lim _{x \rightarrow h} \frac{(x+h) f(x)-2 h f(h)}{x-h}\) is equal to (A) \(f(h)+2 h f^{\prime}(h)\) (B) \(2 \mathrm{f}(\mathrm{h})+\mathrm{hf}^{\prime}(\mathrm{h})\) (C) \(\mathrm{hf}(\mathrm{h})+2 \mathrm{f}^{\prime}(\mathrm{h})\) (D) \(h f(h)-2 f^{\prime}(h)\)

Let \(f(x)\) be defined for all \(x \in R\) and the continuous. Let $\mathrm{f}(\mathrm{x}+\mathrm{y})-\mathrm{f}(\mathrm{x}-\mathrm{y})=4 \mathrm{xy} \forall \mathrm{x}, \mathrm{y}=\in \mathrm{R}$ and \(f(0)=0\) then (A) \(\mathrm{f}(\mathrm{x})\) is bounded (B) \(f(x)+f\left(\frac{1}{x}\right)=f\left(x+\frac{1}{x}\right)+2\) (C) \(\mathrm{f}(\mathrm{x})+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\mathrm{f}\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)+2\) (D) none of these

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 $f(x)=\left\\{\begin{array}{ll}{[x]+\sqrt{\\{x\\}}} & x<1 \\\ \frac{1}{[x]+\\{x\\}^{2}} & x \geq 1\end{array}\right.$, 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 \(\mathrm{x}=1\)

Suppose \(f\) is a differentiable function such that \(f(x+y)=f(x)+f(y)+5 x y\) for all \(x, y\) and \(f^{\prime}(0)=3\). The minimum value of \(f(x)\) is (A) \(-1\) (B) \(-9 / 10\) (C) \(-9 / 25\) (D) None
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