Question:

For a,b>0a, b > 0, let f(x)={tan((a+1)x)+btanxx,x<0,x3,x=0,ax+b2x2axbaxx,x>0 f(x) = \begin{cases} \frac{\tan((a+1)x) + b \tan x}{x}, & x < 0, \\ \frac{x}{3}, & x = 0, \\ \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b\sqrt{a x \sqrt{x}}}, & x > 0 \end{cases} be a continuous function at x=0x = 0. Then ba\frac{b}{a} is equal to:

Updated On: Mar 20, 2025
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The Correct Option is D

Solution and Explanation

For f(x) f(x) to be continuous at x=0 x = 0 , we require:
limx0f(x)=f(0)=limx0+f(x)\lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x).
Given: f(0)=03=0f(0) = \frac{0}{3} = 0.
Right-hand limit: 

Consider:
limx0+ax+b2x2axbax\lim_{x \to 0^+} \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b \sqrt{ax}}.
To simplify, multiply the numerator and the denominator by the conjugate of the numerator:
limx0+(ax+b2x2ax)×(ax+b2x2+ax)bax×(ax+b2x2+ax)\lim_{x \to 0^+} \frac{(\sqrt{ax + b^2x^2} - \sqrt{ax}) \times (\sqrt{ax + b^2x^2} + \sqrt{ax})}{b \sqrt{ax} \times (\sqrt{ax + b^2x^2} + \sqrt{ax})}.
This simplifies to:
limx0+ax+b2x2axbax×(ax+b2x2+ax)\lim_{x \to 0^+} \frac{ax + b^2x^2 - ax}{b \sqrt{ax} \times (\sqrt{ax + b^2x^2} + \sqrt{ax})}.
Simplifying further:
limx0+b2x2bax×(ax+b2x2+ax)\lim_{x \to 0^+} \frac{b^2x^2}{b \sqrt{ax} \times (\sqrt{ax + b^2x^2} + \sqrt{ax})}.
Canceling terms and simplifying:
limx0+bxa×2ax=b2a\lim_{x \to 0^+} \frac{bx}{\sqrt{a} \times 2 \sqrt{ax}} = \frac{b}{2a}.
For continuity, we equate this limit to the value of f(0) f(0) :
b2a=3    ba=6\frac{b}{2a} = 3 \implies \frac{b}{a} = 6.
Therefore:
6\boxed{6}.

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