If \( f(x)=\begin{cases}\displaystyle\frac{\sin(a-b)x}{x}, & x\lt 0\\ 2a+b, & x=0,\ (b\gt 0)\\ \displaystyle\frac{-\tan 12x}{bx}, & x\gt 0\end{cases} \) is continuous at \( x=0 \), then
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( a=-4,b=2 \)
Continuity at x = 0 needs left-hand limit = right-hand limit = f(0). Using \( \displaystyle\lim_{x\to 0}\frac{\sin kx}{x}=\lim_{x\to 0}\frac{\tan kx}{x}=k \):
\( \displaystyle \lim_{x\to 0^{-}}\frac{\sin(a-b)x}{x}=a-b \)
\( \displaystyle \lim_{x\to 0^{+}}\frac{-\tan 12x}{bx}=-\frac{12}{b} \)
\( f(0)=2a+b \)
Left-hand limit = f(0):
\( a-b=2a+b \)
\( a=-2b \)
f(0) = right-hand limit:
\( \displaystyle 2a+b=-\frac{12}{b} \)
\( \displaystyle -4b+b=-\frac{12}{b} \)
\( \displaystyle -3b=-\frac{12}{b} \)
\( b^{2}=4\ \Rightarrow\ b=2\ (\text{since } b\gt 0) \)
\( a=-2b=-4 \)
Check: a − b = −6 and \( -\frac{12}{b}=-6 \).
Shortcut: The condition a − b = 2a + b gives a = −2b, and b > 0. Only Option (4) satisfies both (−4 = −2 × 2); then \( -\frac{12}{2}=-6=a-b \) confirms it.