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Chapter 9: Problem 766

If \(f(x)=\mid \begin{array}{ll}n\left(1-x^{2}\right), & x>3 \\ 3 x+1, & x \leq 3\end{array}\) is continuous at \(\mathrm{x}=3\) then value of \(\mathrm{n}\) is: (a) \(2.25\) (b) \(1.25\) (c) \(-2.25\) (d) \(-1.25\)

Short Answer

Expert verified
The value of n that makes the given function continuous at x=3 is \(-1.25\).

Step by step solution

01

Evaluating the function at x=3

For the first case, we have x>3, and for the second case, x≤3. Since we need the value of n when x=3, the second case is relevant. So, we'll evaluate the function when x=3: f(3) = 3(3) + 1 = 10
02

Finding the limit as x approaches 3

Next, we'll find the limit of the function as x approaches 3 from the left and the right. For x≤3, the limit is the same as f(3) because the function is continuous for x≤3: \[\lim_{x\to 3^{-}} f(x) = 10\] For x>3, we'll find the limit as x approaches 3 from the right: \[\lim_{x\to 3^{+}} f(x) = \lim_{x\to 3^{+}} n(1-x^2)\] Since we're interested in the value as x approaches 3, we can substitute x=3: \[\lim_{x\to 3^{+}} f(x) = n(1-3^2)\] \[\lim_{x\to 3^{+}} f(x) = n(-8)\]
03

Equating the limits and solving for n

Now, we have the limit as x approaches 3 from the left and the right. If the function is continuous at x=3, these limits must be equal: 10 = n(-8) We can now solve the equation for n: \[n = -\frac{10}{8}\] \[n = -1.25\] Since n = -1.25, the correct answer is: (d) \(-1.25\)

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Most popular questions from this chapter

\(\lim _{x \rightarrow \sqrt{2}}\left[\left(x^{9}-3 x^{8}+x^{6}-9 x^{4}-4 x^{2}-16 x+84\right)\right.\) \(\left./\left(x^{5}-3 x^{4}-4 x+12\right)\right]=?\) (a) \(11+\sqrt{2}\) (b) \(11-\sqrt{2}\) (c) \(\sqrt{2}-11\) (b) \(11+2 \sqrt{2}\)

If \(\mathrm{f}(\mathrm{x})=[\\{\tan [(\pi / 6)-\mathrm{x}]\\} /(\operatorname{Cot} 3 \mathrm{x})] ; \mathrm{x} \neq(\pi / 6)\), is continuous at \(\mathrm{x}=(\pi / 6)\) then \(\mathrm{f}(\pi / 6)=\ldots \ldots\) (a) \([1 /(3 \sqrt{3})]\) (b) \((\sqrt{3} / 2)\) (c) \((1 / 3)\) (d) \([1 /(6 \sqrt{3})]\)

If \(\mathrm{f}(\mathrm{x})=\mid \begin{array}{ll}\mathrm{m}+3 \mathrm{nx}, & \mathrm{x}>1 \\ 11, & \mathrm{x}=1 \\ 5 \mathrm{nx}-2 \mathrm{~m}, & \mathrm{x}<1\end{array}\) is continuous at \(\mathrm{x}=1\) then \(\mathrm{m}=\ldots .\) and \(\mathrm{n}=\ldots . .\) ? (a) \(\mathrm{m}=2, \mathrm{n}=-3\) (b) \(\mathrm{m}=-2, \mathrm{n}=3\) (c) \(\mathrm{m}=2, \mathrm{n}=3\) (d) \(m=3, n=3\)

\(\lim _{\mathrm{x} \rightarrow 0}\left(3 / \mathrm{x}^{3}\right) \sin \left(\pi^{2}+2 \mathrm{x}\right)-\left(3 / \mathrm{x}^{3}\right) \sin \left(\pi^{2}+\mathrm{x}\right)\) \(-\left(1 / \mathrm{x}^{3}\right) \sin \left(\pi^{2}+3 \mathrm{x}\right)+\left(1 / \mathrm{x}^{3}\right) \sin [\pi(1+\pi)]=?\) (a) \(\operatorname{Cos} \pi^{2}\) (b) \(-\operatorname{Cos} \pi^{2}\) (c) \(-\pi\) (d) \(\pi\)

If \(\mathrm{f}(\mathrm{x})=\left[\left(\mathrm{e}^{(1 / \mathrm{x})}-\mathrm{e}^{-(1 / \mathrm{x})}\right) /\left(\mathrm{e}^{(1 / \mathrm{x})}+\mathrm{e}^{-(1 / \mathrm{x})}\right)\right], \mathrm{x} \neq 0\) and \(\lim _{\mathrm{x} \rightarrow(0+)+} \mathrm{f}(\mathrm{x})=\mathrm{a}, \lim _{\mathrm{x} \rightarrow(0)-} \mathrm{f}(\mathrm{x})=\mathrm{b}\) then the value of \(\mathrm{a}\) and \(\mathrm{b}\) are: (a) \(\mathrm{a}=1, \mathrm{~b}=-1\) (b) \(a=0, b=1\) (c) \(a=-1, b=1\) (d) \(\mathrm{a}=1, \mathrm{~b}=0\)
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