Question

Let f: ℝ → ℝ be defined as
\(f(x) = \left\{   \begin{array}{ll}     [e^x] &  x < 0 \\     [a e^x + [x-1]] & 0 \leq x < 1 \\     [b + [\sin(\pi x)]] &  1 \leq x < 2 \\     [[e^{-x}] - c] &  x \geq 2 \\   \end{array} \right.\)
Where a, b, c ∈ ℝ and [t] denotes greatest integer less than or equal to t. 
Then, which of the following statements is true?

• A. There exists a, b, c ∈ ℝ such that ƒiscontinuous on ∈ ℝ .
• B. If ƒ is discontinuous at exactly one point, then a + b + c = 1
• C. If ƒ is discontinuous at exactly one point, then a + b + c ≠ 1
• D. ƒ is discontinuous at atleast two points, for any values of a, b and c

Answer 1

The Correct Option is C

To determine which statement about the function \(f\) is true, we explore its continuity based on the given piecewise definition: \( f(x) = \left\{ \begin{array}{l} [e^x] \quad & x < 0 \\ [a e^x + [x-1]] \quad & 0 \leq x < 1 \\ [b + [\sin(\pi x)]] \quad & 1 \leq x < 2 \\ [[e^{-x}] - c] \quad & x \geq 2 \\ \end{array} \right.\). Below, we analyze its continuity at the points where the function definition changes, namely \( x = 0, 1, 2 \):

1. Continuity at \( x=0 \): 

• Left-hand Limit (LHL) as \( x \to 0^- \): \([e^x] \to [e^0] = 1\).
• Right-hand Limit (RHL) as \( x \to 0^+ \): \([a \cdot 1 + [-1]] = [a - 1]\).
• Limit Condition: Continuity requires \( 1 = [a-1] \), implying \( 0 \leq a < 2 \).

2. Continuity at \( x=1 \):

• LHL as \( x \to 1^- \): \([a e^1 + [1-1]] = [ae]\).
• RHL as \( x \to 1^+ \): \([b + [\sin(\pi)] = [b]\) since \(\sin(\pi) = 0\).
• Limit Condition: Continuity at \( x=1 \) requires \( [ae] = [b] \).

3. Continuity at \( x=2 \):

• LHL as \( x \to 2^- \): \([b + [\sin(2\pi)] = [b]\).
• RHL as \( x \to 2^+ \): \([[e^{-2}]] - c = [-c]\) since \( e^{-2} \approx 0.135\), and hence \([e^{-x}] - c \) becomes \([-c]\).
• Limit Condition: Continuity at \( x=2 \) requires \( [b] = -c \).

To be discontinuous at exactly one point, one limit condition must fail. Suppose only at \( x=0 \) we have a discontinuity:

• This implies not all conditions for continuity are satisfied simultaneously.
• Solving previous conditions with one failure, \( a-1 \) differing from 1 hints \( a \neq 2 \), and \( a, b, c \) fulfilling tested continuity relation: \( a, b, c \in \mathbb{R}\) not resolving as \( a+b+c \neq 1 \) consistently.

After analyzing potential discontinuities of \( f \), if \( f \) is discontinuous at exactly one point, then the condition \( a+b+c \neq 1 \) is valid.

Answer 2

The correct answer is (C) : If ƒ is discontinuous at exactly one point, then a + b + c ≠ 1
\(f(x) = \left\{   \begin{array}{ll}     0 &  x < 0 \\     a e^{x}-1 &  0 \leq x < 1 \\     b &  x = 1 \\     b - 1 & 1 < x < 2 \\     -c &  x \geq 2 \\   \end{array} \right.\)
To be continuous at x = 0
a – 1 = 0
to be continuous at x = 1
ae – 1 = b = b – 1 ⇒ not possible
to be continuous at x = 2
b – 1 = – c
⇒ b + c = 1
If a = 1 and b + c = 1 then f(x) is discontinuous at exactly one point.