Limits, Continuity & Differentiability
Monotonicity and Continuity
Grade 12
Question:
<p><strong>55.</strong> 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 f to be continuous at x=0, left-hand limit must equal right-hand limit equals f(0). The key is evaluating both one-sided limits and using the continuity condition to solve for k.
<p><strong>Step 1:</strong> Find the left-hand limit as x→0⁻ using f(x) = -e^(-x) + k:</p><p>lim(x→0⁻) f(x) = -e^0 + k = -1 + k</p><p><strong>Step 2:</strong> Find the right-hand limit as x→0⁺ using f(x) = e^x + 1:</p><p>lim(x→0⁺) f(x) = e^0 + 1 = 1 + 1 = 2</p><p><strong>Step 3:</strong> Find f(0) using the left piece (since 0 ≤ 0):</p><p>f(0) = -e^0 + k = -1 + k</p><p><strong>Step 4:</strong> For continuity at x=0, equate left and right limits:</p><p>lim(x→0⁻) f(x) = lim(x→0⁺) f(x)</p><p>-1 + k = 2</p><p>k = 3</p><p>∴ Answer: B</p>
Correct Answer: B