AP EAMCET · Maths · Limits
If a function \(f(x)=\left\{\begin{array}{cc}\frac{\sqrt[3]{1+a x^2+b x^3}-\sqrt[3]{1-a x^2-b x^3}}{x^2}, & x < 0 \\ 5, & x=0 \\ \frac{\tan 3 x-\sin 3 x}{b x^3}, & x>0\end{array}\right.\)
is continuous at \(\mathrm{x}=0\), then the geometric mean of a and b is
- A \(\frac{3}{2}\)
- B \(\frac{9}{2}\)
- C \(\frac{81}{4}\)
- D \(\frac{9}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{9}{2}\)
Step-by-step Solution
Detailed explanation
\(\lim_{x \to 0^-} \frac{\sqrt[3]{1+a x^2+b x^3}-\sqrt[3]{1-a x^2-b x^3}}{x^2} = \lim_{x \to 0^-} \frac{(1+\frac{1}{3}(ax^2+bx^3)) - (1-\frac{1}{3}(ax^2+bx^3))}{x^2} = \lim_{x \to 0^-} \frac{\frac{2}{3}(ax^2+bx^3)}{x^2} = \frac{2a}{3}\)…
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