If \( f(x)=\begin{cases}\displaystyle\frac{1}{|x|} & ,\ x\ge 1\\ ax^2+b & ,\ x\lt 1\end{cases} \) is differentiable at \( x=1 \) then
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( a=-\displaystyle\frac{1}{2},b=\displaystyle\frac{3}{2} \)
For x ≥ 1, \( |x|=x \), so \( f(x)=\frac{1}{x} \). A differentiable function must also be continuous.
Continuity at x = 1:
\( \displaystyle a(1)^2+b=\frac{1}{1} \)
\( \displaystyle a+b=1 \)
Equal derivatives at x = 1:
\( \displaystyle \text{Left: }2ax\big|_{x=1}=2a,\qquad\text{Right: }-\frac{1}{x^2}\Big|_{x=1}=-1 \)
\( \displaystyle 2a=-1\;\Rightarrow\;a=-\frac{1}{2} \)
\( \displaystyle b=1-a=\frac{3}{2} \)