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: Use continuity of limits at \(x=1\).
For \[ |x|\gt 1, \] we have \[ f(x)=\frac1{|x|} \] As \[ x\to1^+, \] \[ \lim_{x\to1^+}f(x)=1 \] For \[ |x|\leq1, \] \[ f(x)=ax^2+b \] Hence, \[ \lim_{x\to1^-}f(x)=a+b \] Since \[ \lim_{x\to1}f(x) \] exists, \[ a+b=1 \]

Step 2: Use continuity of limits at \(x=-1\).
As \[ x\to-1^-, \] again \[ f(x)=\frac1{|x|} \] Since \[ |-1|=1, \] \[ \lim_{x\to-1^-}f(x)=1 \] For \[ x\to-1^+, \] \[ f(x)=ax^2+b \] Thus, \[ \lim_{x\to-1^+}f(x)=a(-1)^2+b \] \[ =a+b \] Since \[ \lim_{x\to-1}f(x) \] exists, \[ a+b=1 \] So the required condition is \[ a+b=1 \]

Step 3: Check the given options.
Option (1): \[ 1+1=2\neq1 \] Option (2): \[ -\frac12-\frac32=-2\neq1 \] Option (3): \[ \frac32-\frac12=1 \] This satisfies the condition.
Option (4): \[ \frac12-\frac32=-1\neq1 \]

Step 4: Final conclusion.
Hence, \[ \boxed{ a=\frac32,\qquad b=-\frac12 } \] which corresponds to option (3).