JEE Mains · Maths · STD 12 - 5. continuity and differentiation
For \(\mathrm{a}, \mathrm{b}>0\), let \(f(x)=\left\{\begin{array}{l}\frac{\tan ((a+1) x)+b \tan x}{x}, x<0 \\ \frac{\sqrt{a x+b^2 x^2}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}, x>0\end{array}\right.\) be a continous function at \(x=0\). Then \(\frac{b}{a}\) is equal to
- A \(5\)
- B \(4\)
- C \(8\)
- D \(6\)
Answer & Solution
Correct Answer
(D) \(6\)
Step-by-step Solution
Detailed explanation
\( \lim _{x \rightarrow 0} f(x)=f(0)=3 \) \( \lim _{x \rightarrow 0^{+}} \frac{\sqrt{a x+b^2 x^2}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}=3 \) \( \lim _{x \rightarrow 0^{+}} \frac{a x+b^2 x^2-a x}{b \sqrt{a} x^{3 / 2}\left(\sqrt{a x+b^2 x^2}+\sqrt{a x}\right)} \)…
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