Chapter 3: Problem 17
Given \(\mathrm{f}(\mathrm{x})\) is a differentiable function of \(\mathrm{x}\), satisfying \(f(x) . f(y)=f(x)+f(y)+f(x y)-2\) and that \(f(2)=5\). Then \(\mathrm{f}(3)\) is equal to (A) 10 (B) 24 (C) 15 (D) none
Chapter 3: Problem 17
Given \(\mathrm{f}(\mathrm{x})\) is a differentiable function of \(\mathrm{x}\), satisfying \(f(x) . f(y)=f(x)+f(y)+f(x y)-2\) and that \(f(2)=5\). Then \(\mathrm{f}(3)\) is equal to (A) 10 (B) 24 (C) 15 (D) none
All the tools & learning materials you need for study success - in one app.
Get started for freeLet $f(x)=\left\\{\begin{array}{cll}x^{2} & \text { if } & x \leq x_{0} \\ a x+b & \text { if } & x>x_{0}\end{array}\right.$ The values of the coefficients a and \(\mathrm{b}\) for which the function is continuous and has a derivative at \(\mathrm{x}_{0}\), are (A) \(a=x_{0}, b=-x_{0}\) (B) \(a=2 x_{0}, b=-x_{0}^{2}\) (C) $\mathrm{a}=\mathrm{a}=\mathrm{x}_{\mathrm{e}}^{2}, \mathrm{~b}=-\mathrm{x}_{0}$ (D) \(a=x_{0}, b=-x_{0}^{2}\)
The set of values of \(x\) for which the function defined as $f(x)=\left[\begin{array}{ll}1-x \quad x<1 & \\ (1-x)(2-x) & 1 \leq x \leq 2 \\\ 3-x & x>2\end{array}\right.$ fails to be continuous or differentiable, is (A) \(\\{1\\}\) (B) \(\\{2\\}\) (C) \(\\{1,2\\}\) (D) \(\phi\)
Let \(\mathrm{f}^{\prime \prime}(\mathrm{x})\) be continuous at \(\mathrm{x}=0\) and \(\mathrm{f}^{\prime \prime}(0)=4\) then value of $\lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^{2}}$ is (A) 11 (B) 2 (C) 12 (D) none
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 \(\mathrm{x}=1\)
If $\mathrm{f}(\mathrm{x}+\mathrm{y}+\mathrm{z})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y}) \cdot \mathrm{f}(\mathrm{z})\( for all \)\mathrm{x}, \mathrm{y}, \mathrm{z}$ and \(f(2)=4, f^{\prime}(0)=3\), then \(f^{\prime}(2)\) equals (A) 12 (B) 9 (C) 16 (D) 6
What do you think about this solution?
We value your feedback to improve our textbook solutions.