Question

If a function \( f(x) \) is given as: \[ f(x) = \begin{cases} \frac{\sqrt{1+ax^2+bx^3}-\sqrt{1-ax^2-bx^3}}{x^2}, & x<0 \\ 5, & x = 0 \\ \frac{\tan3x-\sin3x}{bx^3}, & x>0 \end{cases} \] and is continuous at \( x = 0 \), then the geometric mean of \( a \) and \( b \) is:

• A. \( \frac{3}{2} \)
• B. \( \frac{9}{2} \)
• C. \( \frac{81}{4} \)
• D. \( \frac{9}{4} \)

Hint:

For checking continuity, ensure: \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \] Apply first-order approximations for small \( x \) when dealing with trigonometric or radical functions.

Answer 1

The Correct Option is D

Step 1: Condition for Continuity at \( x = 0 \) For \( f(x) \) to be continuous at \( x = 0 \), we require: \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \] Step 2: Evaluating \( \lim_{x \to 0^-} f(x) \) Using first-order approximations for small \( x \), \[ \sqrt{1+ax^2+bx^3} \approx 1 + \frac{ax^2}{2} + \frac{bx^3}{2} \] \[ \sqrt{1-ax^2-bx^3} \approx 1 - \frac{ax^2}{2} - \frac{bx^3}{2} \] Subtracting, \[ \sqrt{1+ax^2+bx^3} - \sqrt{1-ax^2-bx^3} = \left( \frac{ax^2}{2} + \frac{bx^3}{2} \right) - \left(-\frac{ax^2}{2} - \frac{bx^3}{2} \right) \] \[ = ax^2 + bx^3 \] Dividing by \( x^2 \), \[ \lim_{x \to 0^-} f(x) = a + bx \] Setting \( x = 0 \), \[ \lim_{x \to 0^-} f(x) = a \] Step 3: Evaluating \( \lim_{x \to 0^+} f(x) \) Using first-order approximations, \[ \tan 3x \approx 3x + \frac{9x^3}{3} \] \[ \sin 3x \approx 3x - \frac{9x^3}{6} \] Subtracting, \[ \tan 3x - \sin 3x = \left( 3x + \frac{9x^3}{3} \right) - \left( 3x - \frac{9x^3}{6} \right) \] \[ = \frac{9x^3}{3} + \frac{9x^3}{6} = 3x^3 + \frac{3x^3}{2} = \frac{9x^3}{2} \] Dividing by \( bx^3 \), \[ \lim_{x \to 0^+} f(x) = \frac{9}{2b} \] Step 4: Equating Limits for Continuity \[ a = \frac{9}{2b} \] Solving for \( ab \), \[ ab = \frac{9}{2} \] Thus, the geometric mean of \( a \) and \( b \) is: \[ \sqrt{ab} = \sqrt{\frac{9}{2}} = \frac{9}{4} \]