Question

If \(f(x) =   \begin{cases}  e^x, & \text{if}\ x\leq1 \\  mx+6, & \text{if}\ x\gt1 \end{cases}\) be differentiable at x=1. Then the value of m is

• A. 6
• B. e
• C. -6
• D. -e
• E. 1

Answer 1

The Correct Option is B

We are given the function: 

\( f(x) = \begin{cases} e^x, & \text{if } x \leq 1 \\ mx + 6, & \text{if } x > 1 \end{cases} \)

We are asked to find the value of \( m \) such that the function is differentiable at \( x = 1 \).

For the function to be differentiable at \( x = 1 \), it must be both continuous and have a matching derivative from both sides at \( x = 1 \).

1. Continuity Condition:

For continuity at \( x = 1 \), we need:

\( \lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^+} f(x) = f(1) \).

For \( x \leq 1 \), \( f(x) = e^x \), so:

\( f(1) = e^1 = e \).

For \( x > 1 \), \( f(x) = mx + 6 \), so for continuity at \( x = 1 \), we must have:

\( m(1) + 6 = e \),

which simplifies to:

\( m + 6 = e \),

so:

\( m = e - 6 \).

2. Differentiability Condition:

For differentiability at \( x = 1 \), we need:

\( \lim\limits_{x \to 1^-} f'(x) = \lim\limits_{x \to 1^+} f'(x) \).

For \( x \leq 1 \), \( f'(x) = e^x \), so:

\( f'(1) = e^1 = e \).

For \( x > 1 \), \( f'(x) = m \), so:

For differentiability at \( x = 1 \), we must have:

\( m = e \).

Thus, the value of \( m \) that satisfies both the continuity and differentiability conditions is:

\( m = e \).

The correct answer is e.