Question

Let two curves $y = f(x)$ with $f(0) = 1$ and $g(x) = \int_{-\infty}^{x} f(t) dt$ with $g(0) = 4$ are such that, their tangents at the points with equal abscissa intersects on x-axis, then

• A. $\int_{-1}^{1} \frac{x^{2} + 4}{4 + g(x)} dx = \frac{13}{3}$
• B. Area bounded by $y = f(x)$, $y = g(x)$, $x = 1$ and $y$ axis is $12 \left(e^{\frac{1}{4}}-1\right)$ sq. unit
• C. $f(x) - g(x)$ is decreasing in [-1, 1]
• D. Fundamental period of $f\left(\left\{\frac{x}{2}\right\}\right)$ is 8 ({.} denotes fractional part of x)

Answer 1

Answer: (B), (C)

(B), (C)

Answer 2

The condition that tangents at points with equal abscissa intersect on the x-axis implies: $x - \frac{f(x)}{f'(x)} = x - \frac{g(x)}{g'(x)}$ $\frac{f(x)}{f'(x)} = \frac{g(x)}{f(x)}$ since $g'(x) = f(x)$. This leads to $g(x) = 4f(x)$. Given $f(0)=1$ and $g(0)=4$, we find $f(x) = e^{x/4}$ and $g(x) = 4e^{x/4}$.

(A) The integral evaluates to $\frac{13}{12}$, not $\frac{13}{3}$. (B) The area is $\int_{0}^{1} (g(x) - f(x)) dx = \int_{0}^{1} 3e^{x/4} dx = 12(e^{1/4} - 1)$. This is correct. (C) $f(x) - g(x) = -3e^{x/4}$. The derivative is $-\frac{3}{4}e^{x/4} < 0$, so $f(x) - g(x)$ is decreasing. This is correct. (D) The period of $\{x/2\}$ is 2. The period of $f(\{x/2\})$ is also 2, not 8.