Question

Let \[ f(x)= \begin{cases} \dfrac{1}{|x|}, & \text{for } |x|\gt 1 \\[0.3cm] ax^2+b, & \text{for } |x|\leq1 \end{cases} \] If \[ \lim_{x\to1}f(x) \quad \text{and} \quad \lim_{x\to-1}f(x) \] exist, then the possible values of \(a\) and \(b\) are

• A. \(a=b=1\)
• B. \(a=-\dfrac12,\; b=-\dfrac32\)
• C. \(a=\dfrac32,\; b=-\dfrac12\)
• D. \(a=\dfrac12,\; b=-\dfrac32\)

Hint:

For existence of a limit at a point: \[ \text{LHL}=\text{RHL} \] Always equate the left-hand and right-hand limits of piecewise functions.

Answer 1

The Correct Option is C

Step 1: Set up the matching at $x=1$.
For $|x|>1$, $f(x)=\dfrac{1}{|x|}$, so as $x\to 1^+$, $f(x)\to 1$. For $|x|\le 1$, $f(x)=ax^2+b$, so as $x\to 1^-$, $f(x)\to a+b$.
Step 2: Require the limit at $x=1$ to exist.
For $\lim_{x\to1}f(x)$ to exist, the two side limits must agree: $a+b = 1$.
Step 3: Set up the matching at $x=-1$.
As $x\to -1^-$, $f(x)=\dfrac{1}{|x|}\to 1$. As $x\to -1^+$, $f(x)=ax^2+b \to a(1)+b = a+b$.
Step 4: Note this gives the same condition.
Existence at $x=-1$ again requires $a+b = 1$, so the symmetry produces no new equation. We pick the option satisfying this.
Step 5: Test the candidate $a=\frac32,\,b=-\frac12$.
Check $a+b = \frac32 - \frac12 = 1$. This satisfies the matching condition exactly.
Step 6: Confirm against the others.
Among the options, $a=\frac32,\,b=-\frac12$ is the pair giving $a+b=1$ (note $a=b=1$ gives $a+b=2\neq1$), so it is the answer.
\[ \boxed{a=\dfrac32,\ b=-\dfrac12} \]