We are given the piecewise function:
\[ f(x) = \begin{cases} 2x & \text{for} \ x < 1 \\ 5a - x & \text{for} \ x \geq 1 \end{cases} \]
For \( f(x) \) to be continuous at \( x = 1 \), the values of the function from both sides must be equal at \( x = 1 \).
For \( x \to 1^- \) (as \( x \) approaches 1 from the left):
\[ \lim_{x \to 1^-} f(x) = 2(1) = 2 \]
For \( x \to 1^+ \) (as \( x \) approaches 1 from the right):
\[ \lim_{x \to 1^+} f(x) = 5a - 1 \]
For continuity at \( x = 1 \), we equate both limits:
\[ 2 = 5a - 1 \]
Solving for \( a \):
\[ 2 + 1 = 5a \] \[ 3 = 5a \] \[ a = \frac{3}{5} \]
Answer: \( \frac{3}{5} \)
Match List-I with List-II
List-I | List-II |
---|---|
(A) \( f(x) = |x| \) | (I) Not differentiable at \( x = -2 \) only |
(B) \( f(x) = |x + 2| \) | (II) Not differentiable at \( x = 0 \) only |
(C) \( f(x) = |x^2 - 4| \) | (III) Not differentiable at \( x = 2 \) only |
(D) \( f(x) = |x - 2| \) | (IV) Not differentiable at \( x = 2, -2 \) only |
Choose the correct answer from the options given below:
Match List-I with List-II
List-I | List-II |
---|---|
(A) \( f(x) = |x| \) | (I) Not differentiable at \( x = -2 \) only |
(B) \( f(x) = |x + 2| \) | (II) Not differentiable at \( x = 0 \) only |
(C) \( f(x) = |x^2 - 4| \) | (III) Not differentiable at \( x = 2 \) only |
(D) \( f(x) = |x - 2| \) | (IV) Not differentiable at \( x = 2, -2 \) only |
Choose the correct answer from the options given below: