Applications of Derivatives
Monotonically Increasing Functions
Grade 12
Question:
<p>If \[ f(x) = \begin{cases} -e^{-x} + k, & x \leq 0 \\ e^x + 1, & 0 < x < 1 \\ ex^2 + \lambda, & x \geq 1 \end{cases} \] is one-one and monotonically increasing for all \( x \in R \), then difference of maximum value of \( k \) and minimum value of \( \lambda \) is:</p>
<p>(a) 0</p>
<p>(b) 1</p>
<p>(c) 2</p>
<p>(d) 3</p>
Step-by-Step Solution
Key Concept: For differentiability at x=0, both continuity and equality of left and right derivatives must hold. Use these two conditions to find k and m separately.
<p><strong>Step 1: Apply continuity at x = 0</strong></p><p>For f to be continuous at x = 0:</p><p>f(0⁻) = f(0⁺) ⟹ -e⁰ + k = e⁰ + 1</p><p>⟹ -1 + k = 1 + 1 ⟹ <strong>k = 3</strong></p><p><strong>Step 2: Find derivatives from each side</strong></p><p>For x < 0: f(x) = -e⁻ˣ + 3 ⟹ f'(x) = e⁻ˣ</p><p>⟹ f'(0⁻) = e⁰ = 1</p><p>For x > 0: f(x) = eˣ + 1 ⟹ f'(x) = meˣ</p><p>⟹ f'(0⁺) = m·e⁰ = m</p><p><strong>Step 3: Apply differentiability at x = 0</strong></p><p>For f to be differentiable: f'(0⁻) = f'(0⁺)</p><p>⟹ 1 = m ⟹ <strong>m = 1</strong></p><p><strong>Step 4: Calculate k + m</strong></p><p>k + m = 3 + 1 = <strong>4</strong></p><p>∴ Answer: B</p>
Correct Answer: B