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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

Short Answer

Expert verified
Based on the given function properties and the step-by-step solution, the value of f(3) is found to be 3. However, this value doesn't match with any of the given choices. Thus, the correct answer is (D) none.

Step by step solution

01

Prove that the function f(x) is uniquely determined by its properties

Here we need to show that there is only one function that satisfies the given condition. Let's set \(x=y=1\): $$f(1) . f(1) = f(1) + f(1) + f(1) - 2$$ $$f(1)^2 - 2f(1) + 1 = 0$$ $$(f(1) - 1)^2 = 0$$ Therefore, \(f(1) = 1\). Now, let's set \(y = 1\): $$f(x)f(1) = f(x) + f(1) + f(x) - 2$$ $$f(x) = f(x) + 1 + f(x) - 2$$ This simplifies to \(f(x) = 1\). This shows that the function has the form f(x) = x.
02

Find an expression for the derivative of f(x)

Since we have found that \(f(x) = x\), we can now differentiate f(x) with respect to x: $$f'(x) = \frac{d}{dx} f(x) = \frac{d}{dx} x = 1 $$
03

Utilize the given information to find f(3)

Since f(x) = x, we can simply find f(3) as follows: $$f(3) = 3$$ However, this value doesn't match with any of the given choices. Hence, the correct answer is (D) none.

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Most popular questions from this chapter

Let $f(x)=\left\\{\begin{array}{cll}x^{2} & \text { if } & x \leq x_{0} \\ a x+b & \text { if } & x>x_{0}\end{array}\right.$ 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

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