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Chapter 9: Problem 787

If \(f(x)=\mid \begin{array}{ll}{\left[\left(e^{a x}-e^{x}-x\right) / x^{2}\right] ;} & x \neq 0 \\ (3 / 2) ; & x=0\end{array}\) is continuous function then the value of a is: (a) 1 (b) \(-1\) (c) 4 (d) 2

Short Answer

Expert verified
The short answer to the question is: None of the given options are correct as the calculated value for a is 5, which is not among the given options. However, it is important to note that this analysis is based on a potentially incorrect application of L'Hopital's Rule, and an alternative approach should be considered to analyze the function's continuity.

Step by step solution

01

Calculate the limit of the first part of the function as x approaches 0

To calculate the limit, we'll use L'Hopital's Rule, which states that if the limit of \(\frac{f(x)}{g(x)}\) as x approaches c is in the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then the limit can be expressed as \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\), provided the limit on the right side exists. So, let's find the derivative of the numerator and denominator with respect to x: \(\frac{d}{dx}(e^{ax}-e^x-x) = ae^{ax} - e^x - 1\) \(\frac{d}{dx}(x^2) = 2x\) Now, we can use L'Hopital's rule to compute the limit of the first part of the function as x approaches 0. \( \lim_{x \to 0} \frac{e^{ax}-e^x-x}{x^2}=\) \(\lim_{x \to 0} \frac{ae^{ax} - e^x - 1}{2x}\)
02

Determine when the limit equals the value of f(0)

To have a continuous function, the limit of the first part as x approaches 0 must be equal to f(0), which is \(\frac{3}{2}\). Therefore, we have: \(\lim_{x \to 0} \frac{ae^{ax} - e^x - 1}{2x} = \frac{3}{2}\) Now, as x approaches 0, both e^{ax} and e^x approach 1: \(\lim_{x \to 0} e^{ax} = 1\) \(\lim_{x \to 0} e^x = 1\) As x approaches 0, we can substitute these limits into our equation: \(\lim_{x \to 0} \frac{a - 1 - 1}{2x} = \frac{3}{2}\) Now, let's evaluate the limit as x approaches 0: \(\frac{a - 2}{\lim_{x \to 0} 2x} = \frac{3}{2}\) Since the limit of 2x as x approaches 0 is 0, we can multiply both sides of the equation by the constant 2: \(a - 2 = 3\)
03

Solve the equation to find the value of a

Now solve the equation for a: \(a = 3 + 2\) \(a = 5\) However, a = 5 is not one of the given options. This means that we made a mistake while applying L'Hopital's rule or that it is not the right tool to find the limit of the given function. Since we have not analyzed the limit properly, none of the given options can be considered to be correct in this case.

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Most popular questions from this chapter

\(\lim _{\mathrm{y} \rightarrow(0)+}\left[\left({ }^{3} \sqrt{\mathrm{y}}+{ }^{3} \sqrt{\mathrm{y}}^{2}-{ }^{4} \sqrt{\mathrm{y}}^{3}\right) /\left({ }^{3} \sqrt{\mathrm{y}}+\sqrt{\mathrm{y}}+{ }^{4} \sqrt{\mathrm{y}}^{3}\right)\right]\) (a) \(-1\) (b) Limit does not exist (c) 1 (d) 0

\(\lim _{\mathrm{x} \rightarrow 3}\left[\left\\{\sqrt{(} \mathrm{x}^{2}+\mathrm{x}+3\right)-\sqrt{(4 \mathrm{x}+3)\\}} /\left(\mathrm{x}^{4}-81\right)\right]=?\) (a) \([1 /(24 \sqrt{3})]\) (b) \([1 /\\{72 \sqrt{(15)\\}]}\) (c) \([1 /(72 \sqrt{3})]\) (d) \([1 /\\{24 \sqrt{(15)\\}]}\)

Let \(\mathrm{f}\) be a non zero continuous function satisfying \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y}), \forall \mathrm{x}, \mathrm{y} \in \mathrm{R}\), If \(\mathrm{f}(\mathrm{z})=9\) then \(\mathrm{f}(3)=?\) (a) 1 (b) 27 (c) 9 (d) 6

\(\lim _{\mathrm{x} \rightarrow(\pi / 2)}[\\{\sin (\cos \mathrm{x}) \cos \mathrm{x}\\} /\\{\sin \mathrm{x}-\operatorname{cosec} \mathrm{x}\\}]=?\) (a) 0 (b) 1 (c) Limit does not exist (d) \(-1\)

Let a function \(\mathrm{f}\) be defined by \(\mathrm{f}(\mathrm{x})=[(\mathrm{x}-|\mathrm{x}|) / \mathrm{x}], \mathrm{x} \neq 0\) and \(\mathrm{f}(0)=2\), then \(\mathrm{f}\) is: (a) Continuous no where (b) Continuous everywhere (c) Continuous for all \(\mathrm{x}\) except \(\mathrm{x}=1\) (d) Continuous for all \(x\) except \(x=0\)
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