Question

for which value of \(\lambda\) is the function ,\(f(x) = \begin{cases} \lambda(x^2-2x) & \text{if } x \leq 0 \\ 4x+1& \text{if } x > 0 \end{cases}\) continuous at \(x=0 ?\)

• A. \(\lambda=0\)
• B. \(\lambda=1\)
• C. \(\lambda=-1\)
• D. for no value of \(\lambda\)

Answer 1

The Correct Option is D

To determine the value of \(\lambda\) for which the function \(f(x) = \begin{cases} \lambda(x^2-2x) & \text{if } x \leq 0 \\ 4x+1 & \text{if } x > 0 \end{cases}\) is continuous at \(x=0\), we need the function values to be equal from both sides of \(x=0\). For continuity at \(x=0\), it must hold that:

• The left-hand limit exists.
• The right-hand limit exists.
• The left-hand limit, right-hand limit, and \(f(0)\) are equal.

Calculate \(f(0)\): Since \(x \leq 0\), \(f(0) = \lambda(0^2 - 2\cdot0) = 0\).

Calculate the left-hand limit as \(x \to 0^-\):

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \lambda(x^2 - 2x) = \lambda(0^2 - 2\cdot0) = 0 \]

Calculate the right-hand limit as \(x \to 0^+\):

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (4x + 1) = 4(0) + 1 = 1 \]

For \(f(x)\) to be continuous at \(x=0\), the limits and function value must match:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]

This implies

\[ 0 = 1 = 0 \]

Which is a contradiction. Hence, no value of \(\lambda\) will satisfy the continuity condition at \(x=0\).

The correct answer is: for no value of \(\lambda\).