Limits, Continuity & Differentiability
Continuity And Differentiability
nta_abhyas_2025
Grade None
Question:
If $f(x) = \begin{cases} \frac{\sqrt{a+x}-\sqrt{a-x}}{x} & -1 \leq x < 0 \\ \frac{3x+2}{x-3} & 0 \leq x \leq 1 \end{cases}$ is continuous in $[-1, 1]$, then the value of $a$ is
1
-1
\frac{1}{2}
-\frac{1}{2}
Step-by-Step Solution
Key Concept: Use continuity conditions to relate function values and derivatives at specific points to find unknown constants.
Given that $f(x)$ is continuous on $[-1, 1]$, it is continuous at $x = 0$. We have $f(0) = \frac{3}{4}$. Computing $f'(0) = \lim_{x \to 0} \frac{3x}{\sqrt{x^2 + x + 4}}$ (using rationalization). This limit evaluates to $\frac{3}{4}$. By continuity and differentiability at $x = 0$, equating coefficients from the given condition $f(1) = 1$ and $f'(1) = a + b$, we solve to find $a = \frac{-1}{2}$.
Correct Answer: -1