Question

Let F and G be inverse functions respectively of

$f(x) = (2x-3\pi)^5 + \frac{4x}{3} + \cos x$ and $g(x) = e^x + 2x$.

Then $7F'(2\pi) + 25G'(\ln 3)$ is equal to

• A. 10
• B. 7
• C. 9
• D. 8

Answer 1

Answer: (D)

8

Answer 2

Let F and G be inverse functions of $f(x)$ and $g(x)$ respectively. The derivative of an inverse function is given by the formula $F'(y) = \frac{1}{f'(x)}$ where $y = f(x)$. Similarly, $G'(y) = \frac{1}{g'(x)}$ where $y = g(x)$.

First, we need to find $F'(2\pi)$. Let $y = 2\pi$. We need to find $x_0$ such that $f(x_0) = 2\pi$. $f(x) = (2x-3\pi)^5 + \frac{4x}{3} + \cos x$. We check if there is a simple value of $x$ for which $f(x) = 2\pi$. Let's try $x = \frac{3\pi}{2}$. $f(\frac{3\pi}{2}) = (2(\frac{3\pi}{2})-3\pi)^5 + \frac{4(\frac{3\pi}{2})}{3} + \cos(\frac{3\pi}{2})$ $f(\frac{3\pi}{2}) = (3\pi-3\pi)^5 + \frac{6\pi}{3} + 0$ $f(\frac{3\pi}{2}) = 0^5 + 2\pi + 0 = 2\pi$. So, $x_0 = \frac{3\pi}{2}$ is the value such that $f(x_0) = 2\pi$.

Now we need to find $f'(x)$. $f'(x) = \frac{d}{dx}\left((2x-3\pi)^5 + \frac{4x}{3} + \cos x\right)$ $f'(x) = 5(2x-3\pi)^4 \cdot \frac{d}{dx}(2x-3\pi) + \frac{4}{3} \cdot \frac{d}{dx}(x) + \frac{d}{dx}(\cos x)$ $f'(x) = 5(2x-3\pi)^4 \cdot 2 + \frac{4}{3} \cdot 1 - \sin x$ $f'(x) = 10(2x-3\pi)^4 + \frac{4}{3} - \sin x$.

Now evaluate $f'(x_0)$ at $x_0 = \frac{3\pi}{2}$. $f'(\frac{3\pi}{2}) = 10(2(\frac{3\pi}{2})-3\pi)^4 + \frac{4}{3} - \sin(\frac{3\pi}{2})$ $f'(\frac{3\pi}{2}) = 10(3\pi-3\pi)^4 + \frac{4}{3} - (-1)$ $f'(\frac{3\pi}{2}) = 10(0)^4 + \frac{4}{3} + 1 = 0 + \frac{4}{3} + \frac{3}{3} = \frac{7}{3}$.

Using the inverse function derivative formula, $F'(2\pi) = \frac{1}{f'(3\pi/2)} = \frac{1}{7/3} = \frac{3}{7}$. So, $7F'(2\pi) = 7 \cdot \frac{3}{7} = 3$.

Next, we need to find $G'(\ln 3)$. Let $y = \ln 3$. We need to find $x_1$ such that $g(x_1) = \ln 3$. $g(x) = e^x + 2x$. We need to solve $e^{x_1} + 2x_1 = \ln 3$ for $x_1$.

It seems there is a high probability of a typo in the question. Assuming the question intended the second term to be $25G'(y_1)$ where $y_1 = g(\ln 3)$, the result is 8. Let's assume the typo is in the argument of G'. Let the argument be $y_1$ such that the corresponding $x_1$ is simple, say $x_1=0$. If $x_1=0$, $g(0)=1$. $g'(0)=3$. $G'(1) = 1/3$. $25G'(1) = 25/3$. $3+25/3 \neq$ any option. If $x_1=1$, $g(1)=e+2$. $g'(1)=e+2$. $G'(e+2) = 1/(e+2)$. $25G'(e+2) = 25/(e+2)$. $3+25/(e+2) \neq$ any option.

Given the strong indication from the first term's calculation and the options, it is most probable that the second term $25G'(\ln 3)$ was intended to be $25G'(g(\ln 3))$. If $y_1 = g(\ln 3)$, then $G'(y_1) = \frac{1}{g'(\ln 3)}$. $g'(x) = e^x + 2$. $g'(\ln 3) = e^{\ln 3} + 2 = 3 + 2 = 5$. So $G'(g(\ln 3)) = \frac{1}{5}$. Then $25G'(g(\ln 3)) = 25 \cdot \frac{1}{5} = 5$. The expression becomes $7F'(2\pi) + 25G'(g(\ln 3)) = 3 + 5 = 8$. This matches option (D). Assuming this intended meaning, the answer is 8.