Question

Let $f(x) = \begin{cases} \dfrac{1}{|x|}, & |x|>1 \\ ax^2 + b, & |x| \leq 1 \end{cases}$. If $\lim_{x \to 1} f(x)$ and $\lim_{x \to -1} f(x)$ exist, then the possible values for $a$ and $b$ are

• A. $a = b = 1$
• B. $a = -\dfrac{1}{2}, b = -\dfrac{3}{2}$
• C. $a = \dfrac{3}{2}, b = -\dfrac{1}{2}$
• D. $a = \dfrac{1}{2}, b = -\dfrac{3}{2}$

Hint:

Check continuity and differentiability conditions on piecewise functions by equating limits and function values at boundaries.

Answer 1

The Correct Option is C

Continuity at $x=1$ implies $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$
So $a(1)^2 + b = \dfrac{1}{|1|} = 1$
Similarly, at $x = -1$, $a(1)^2 + b = 1$
From these, $a + b = 1$
Also, differentiability conditions or further limits yield $a = \dfrac{3}{2}$, $b = -\dfrac{1}{2}$