Let f(x) is a function continuous for all `x in R` except at x = 0 such that `f'(x) lt 0, AA x in (-oo, 0) and f'(x) gt 0, AA x in (0, oo)`. If `lim_(x rarr 0^(+)) f(x) = 3, lim_(x rarr 0^(-)) f(x) = 4 and f(0) = 5`, then the image of the point (0, 1) about the line, `y.lim_(x rarr 0) f(cos^(3) x - cos^(2) x) = x. lim_(x rarr 0) f(sin^(2) x - sin^(3) x)`, is

Options

A. `((12)/(25),(-9)/(25))`
B. `((12)/(25),(9)/(25))`
C. `((16)/(25),(-8)/(25))`
D. `((24)/(25),(-7)/(25))`

Answer

To solve the given problem step by step, we will analyze the function \( f(x) \) and the limits provided, and then find the image of the point \( (0, 1) \) about the specified line.

### Step 1: Understand the behavior of \( f(x) \)

Given:
- \( f'(x) < 0 \) for \( x < 0 \) (decreasing function)
- \( f'(x) > 0 \) for \( x > 0 \) (increasing function)
- \( \lim_{x \to 0^+} f(x) = 3 \)
- \( \lim_{x \to 0^-} f(x) = 4 \)
- \( f(0) = 5 \)

From this information, we can conclude that as \( x \) approaches 0 from the left, \( f(x) \) approaches 4, and as \( x \) approaches 0 from the right, \( f(x) \) approaches 3. The function is continuous everywhere except at \( x = 0 \).

### Step 2: Calculate the limits

We need to evaluate:
1. \( \lim_{x \to 0} f(\cos^3 x - \cos^2 x) \)
2. \( \lim_{x \to 0} f(\sin^2 x - \sin^3 x) \)

#### For \( \lim_{x \to 0} f(\cos^3 x - \cos^2 x) \):

First, simplify \( \cos^3 x - \cos^2 x \):
\[ \cos^3 x - \cos^2 x = \cos^2 x (\cos x - 1) \]
As \( x \to 0 \), \( \cos x \to 1 \) and \( \cos^2 x \to 1 \), so:
\[ \cos^3 x - \cos^2 x \to 1 \cdot (1 - 1) = 0 \]
Thus, we need to find \( \lim_{x \to 0^-} f(0) \) and \( \lim_{x \to 0^+} f(0) \):
\[ \lim_{x \to 0^-} f(x) = 4 \quad \text{and} \quad \lim_{x \to 0^+} f(x) = 3 \]
So,
\[ \lim_{x \to 0} f(\cos^3 x - \cos^2 x) = 4 \]

#### For \( \lim_{x \to 0} f(\sin^2 x - \sin^3 x) \):

Now simplify \( \sin^2 x - \sin^3 x \):
\[ \sin^2 x - \sin^3 x = \sin^2 x (1 - \sin x) \]
As \( x \to 0 \), \( \sin x \to 0 \) and \( \sin^2 x \to 0 \), so:
\[ \sin^2 x (1 - \sin x) \to 0 \cdot (1 - 0) = 0 \]
Thus, we need to find:
\[ \lim_{x \to 0^-} f(0) = 4 \quad \text{and} \quad \lim_{x \to 0^+} f(0) = 3 \]
So,
\[ \lim_{x \to 0} f(\sin^2 x - \sin^3 x) = 3 \]

### Step 3: Set up the equation for the line

Now we have:
\[ y \cdot 4 = x \cdot 3 \]
This simplifies to:
\[ 4y = 3x \quad \Rightarrow \quad y = \frac{3}{4}x \]

### Step 4: Find the image of the point \( (0, 1) \)

The slope of the line is \( \frac{3}{4} \). The slope of the perpendicular line will be:
\[ m = -\frac{4}{3} \]
Using the point-slope form of a line:
\[ y - 1 = -\frac{4}{3}(x - 0) \]
This simplifies to:
\[ y - 1 = -\frac{4}{3}x \quad \Rightarrow \quad y = -\frac{4}{3}x + 1 \]

### Step 5: Find the intersection of the two lines

Set the two equations equal to each other:
\[ \frac{3}{4}x = -\frac{4}{3}x + 1 \]
Multiply through by 12 to eliminate fractions:
\[ 9x = -16x + 12 \]
Combine like terms:
\[ 25x = 12 \quad \Rightarrow \quad x = \frac{12}{25} \]

Substituting \( x \) back into \( y = \frac{3}{4}x \):
\[ y = \frac{3}{4} \cdot \frac{12}{25} = \frac{36}{100} = \frac{9}{25} \]

### Final Answer

The image of the point \( (0, 1) \) about the line is:
\[ \left( \frac{12}{25}, \frac{9}{25} \right) \]