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Chapter 9: Problem 768

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) \(m=3, n=3\)

Short Answer

Expert verified
The solution is \(m = \frac{11}{4}\) and \(n = \frac{11}{4}\). None of the given choices satisfy the condition of continuity at x=1.

Step by step solution

01

Calculate the left-hand limit.

To find the left-hand limit, evaluate the limit of the function as x approaches 1 from the left: \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (5nx - 2m).\)
02

Calculate the right-hand limit.

To find the right-hand limit, evaluate the limit of the function as x approaches 1 from the right: \(\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (m + 3nx).\)
03

Set the left-hand limit and the right-hand limit equal.

For the function to be continuous at x=1, the left-hand limit must be equal to the right-hand limit: \(5n - 2m = m + 3n.\)
04

Solve for m in terms of n.

Solve the equation obtained in step 3 for m, so we can then find n: \(m = 3n - 2m \Rightarrow 3m = 3n \Rightarrow m = n.\)
05

Check the value of the function at x=1.

The value of the function at x=1 is given as: \(f(1) = 11.\)
06

Set the right-hand limit equal to the function value at x=1.

For the function to be continuous at x=1, the right-hand limit must be equal to the function value at x=1: \(m(1) + 3n(1) = 11 \Rightarrow m + 3n = 11.\)
07

Substitute m=n and solve for n.

Use the relation m=n from step 4 and substitute it into the equation from step 6 to solve for n: \(n + 3n = 11 \Rightarrow 4n = 11 \Rightarrow n = \frac{11}{4}.\)
08

Find the value of m.

Substitute the value of n back into the relation m=n: \(m = n \Rightarrow m = \frac{11}{4}.\) The solution is \(m = \frac{11}{4}\) and \(n = \frac{11}{4}\). None of the choices given in the problem matches this solution, and therefore none of the choices satisfy the condition of continuity at x=1.

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Most popular questions from this chapter

\(\lim _{\mathrm{x} \rightarrow(0)+}\left[\left(\tan \mathrm{x}+\tan ^{2} \mathrm{x}+\tan ^{3} \mathrm{x}+\tan ^{4} \mathrm{x}+\ldots \ldots \infty\right) / \pi \mathrm{x}\right]\) [where \(0<|\mathrm{x}|<(\pi / 4)]\) (a) \((1 / \pi)\) (b) \(\left(1 / \pi^{\mathrm{n}}\right)\) (c) \(\pi\) (d) 0

\(\lim _{\mathrm{x} \rightarrow(\pi / 4)}\left[\left\\{16 \sqrt{2}-(\operatorname{Sin} \mathrm{x}+\cos \mathrm{x})^{9}\right\\} /(1-\operatorname{Sin} 2 \mathrm{x})\right]=?\) (a) \(9 \sqrt{2}\) (b) \(18 \sqrt{2}\) (c) \(36 \sqrt{2}\) (d) \(16 \sqrt{2}\)

\(\lim _{(1 / \mathrm{x}) \rightarrow 0}\left[\left\\{(2+3 \mathrm{x})^{40}(4+3 \mathrm{x})^{5}\right\\} /(2-3 \mathrm{x})^{45}\right]=?\) (a) \((40 / 9)\) (b) \(-35\) (c) \(-1\) (d) \((8 / 9)\)

The value of \(\lim _{\mathrm{x} \rightarrow(\pi / 4)}\left[\left\\{(\tan \mathrm{x})^{\tan \mathrm{x}}-\tan \mathrm{x}\right\\} /\\{\ln (\tan \mathrm{x})-\tan \mathrm{x}+1\\}\right]\) is (a) 0 (b) 1 (c) \(-2\) (d) none of these

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