Question

If \( f(x) = \begin{cases} ax^2 - bx + 2, & x<3 \\ bx - 3, & x \geq 3 \end{cases} \) is differentiable at every \( x \in \mathbb{R} \), then the area (in sq units) of the triangle formed by the line \(\frac{x}{a} + \frac{y}{b} = 1\) with the coordinate axes is:

• A. \(\frac{175}{81}\)
• B. \(\frac{175}{27}\)
• C. \(\frac{35}{27}\)
• D. \(\frac{125}{27}\)

Hint:

For a line \(\frac{x}{a} + \frac{y}{b} = 1\), the triangle formed with the coordinate axes has vertices \((0, 0)\), \((a, 0)\), \((0, b)\), and area \(\frac{1}{2}ab\).

Answer 1

The Correct Option is B

Step 1: Ensure differentiability of \( f(x) \) at \( x = 3 \).
For \( f(x) \) to be differentiable at \( x = 3 \), it must be continuous and the derivatives from both sides must be equal.
Continuity at \( x = 3 \):
Left-hand limit (\( x<3 \)): \[ f(3^-) = a(3)^2 - b(3) + 2 = 9a - 3b + 2 \] Right-hand limit (\( x \geq 3 \)): \[ f(3^+) = b(3) - 3 = 3b - 3 \] Equate for continuity: \[ 9a - 3b + 2 = 3b - 3 \implies 9a - 6b + 5 = 0 \quad (1) \] Differentiability at \( x = 3 \):
Left-hand derivative: \[ f'(x) = \frac{d}{dx} (ax^2 - bx + 2) = 2ax - b \implies f'(3^-) = 2a(3) - b = 6a - b \] Right-hand derivative: \[ f'(x) = \frac{d}{dx} (bx - 3) = b \implies f'(3^+) = b \] Equate the derivatives: \[ 6a - b = b \implies 6a - 2b = 0 \implies 3a - b = 0 \implies b = 3a \quad (2) \] Solve equations (1) and (2): Substitute \( b = 3a \) into (1): \[ 9a - 6(3a) + 5 = 0 \implies 9a - 18a + 5 = 0 \implies -9a + 5 = 0 \implies a = \frac{5}{9} \] Then: \[ b = 3a = 3 \cdot \frac{5}{9} = \frac{15}{9} = \frac{5}{3} \] So, \( a = \frac{5}{9} \), \( b = \frac{5}{3} \).
Step 2: Find the area of the triangle formed by the line \(\frac{x}{a} + \frac{y}{b} = 1\).
The line intersects the x-axis (\( y = 0 \)) at: \[ \frac{x}{a} = 1 \implies x = a \] It intersects the y-axis (\( x = 0 \)) at: \[ \frac{y}{b} = 1 \implies y = b \] The vertices of the triangle are \((0, 0)\), \((a, 0)\), and \((0, b)\). The area of a right triangle with legs \( a \) and \( b \) is: \[ \text{Area} = \frac{1}{2} \cdot a \cdot b \] Substitute \( a = \frac{5}{9} \), \( b = \frac{5}{3} \): \[ \text{Area} = \frac{1}{2} \cdot \frac{5}{9} \cdot \frac{5}{3} = \frac{1}{2} \cdot \frac{25}{27} = \frac{25}{54} \] Final Answer: \[ \boxed{2} \]