Question

Consider two sets

$A = \{\frac{p}{q} ; p, q \text{ are coprime odd integers }\}$ and

$B = \{\frac{p+1}{q} : p, q \text{ are odd integers and } p \neq -1 \}$.

Functions $f: A \rightarrow R$ & $g: B \rightarrow R$ are both defined as $f(x) = x^x$ & $g(x) = x^{|x|}$ on respective domains. Then mark incorrect option

• A. $f(\frac{1}{3}) \cdot g(-\frac{2}{3}) = \frac{\sqrt[3]{4}}{3}$
• B. $g(x)$ is an even function.
• C. $f(1) + g(2) = 6$
• D. $[g(-\frac{6}{5})]^{5/2} - [g(\frac{4}{5})]^{15/4} - [f(\frac{3}{5})]^5 = 1$

Answer 1

Answer: (C)

C

Answer 2

Option (A): $f(\frac{1}{3}) = (\frac{1}{3})^{\frac{1}{3}}$. $g(-\frac{2}{3}) = (-\frac{2}{3})^{\frac{2}{3}} = (\frac{4}{9})^{\frac{1}{3}}$. $f(\frac{1}{3}) \cdot g(-\frac{2}{3}) = \frac{1}{3^{1/3}} \cdot \frac{4^{1/3}}{9^{1/3}} = \frac{\sqrt[3]{4}}{3}$. Correct.

Option (B): $g(x) = x^{|x|}$. $g(-x) = (-x)^{|-x|} = (-x)^{|x|}$. If $x = \frac{N}{D}$ where $N$ is even and $D$ is odd, then $|x| = \frac{|N|}{|D|} = \frac{a}{b}$ where $a$ is even. $g(x) = x^{a/b}$. For $x<0$, $x^{a/b} = (\text{sgn}(x))^a |x|^{a/b}$. Since $a$ is even, $(\text{sgn}(x))^a = 1$. So $g(x) = |x|^{a/b}$. Similarly $g(-x) = |x|^{a/b}$. Thus $g(x)$ is even. Correct.

Option (C): $f(1) = 1^1 = 1$. $g(2) = 2^{|2|} = 2^2 = 4$. $f(1) + g(2) = 1+4=5 \neq 6$. Incorrect.

Option (D): $[g(-\frac{6}{5})]^{5/2} = [g(\frac{6}{5})]^{5/2} = [(\frac{6}{5})^{\frac{6}{5}}]^{\frac{5}{2}} = (\frac{6}{5})^3 = \frac{216}{125}$. $[g(\frac{4}{5})]^{15/4} = [(\frac{4}{5})^{\frac{4}{5}}]^{\frac{15}{4}} = (\frac{4}{5})^3 = \frac{64}{125}$. $[f(\frac{3}{5})]^5 = [(\frac{3}{5})^{\frac{3}{5}}]^5 = (\frac{3}{5})^3 = \frac{27}{125}$. $\frac{216}{125} - \frac{64}{125} - \frac{27}{125} = \frac{125}{125} = 1$. Correct.

The incorrect option is (C).