Limits, Continuity & Differentiability
Continuity — Rationalization
nta_pyq_2024_apr
Grade 12
Question:
For $a,b>0$, let $f(x)=\begin{cases}\dfrac{\tan((a+1)x)+b\tan x}{x}, & x<0\\ 3, & x=0\\ \dfrac{\sqrt{ax+b^2x^2}-\sqrt{ax}}{b\sqrt{ax}\sqrt{x}}, & x>0\end{cases}$ be a continuous function at $x=0$. Then $\dfrac{b}{a}$ is equal to:
Step-by-Step Solution
Key Concept: LHL: $\lim_{x\to0^-}\frac{\tan((a+1)x)+b\tan x}{x}=(a+1)+b=a+b+1=3$. RHL: $\lim_{x\to0^+}\frac{\sqrt{ax+b^2x^2}-\sqrt{ax}}{b\sqrt{ax}\cdot x}=\frac{b}{2a}\cdot\frac{1}{... }$: rationalize to get $\frac{b}{2\sqrt{a}\cdot\sqrt{a}}=\frac{b}{2a}=3$.
RHL gives $b/(2a)=3\Rightarrow b=6a$. $b/a=6$.
Correct Answer: 1