MHT CET · Maths · Continuity and Differentiability
If \(f(x)=\frac{e^{x^2}-\cos x}{x^2}\) if \(x \neq 0\) is continuous at \(x=0\), then \(f(0)=\).
- A \(\frac{1}{2}\)
- B \(\frac{3}{2}\)
- C \(\frac{2}{3}\)
- D \(\frac{-3}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{2}\)
Step-by-step Solution
Detailed explanation
For continuity at \(x=0, \lim _{x \rightarrow 0} f(x)=f(0)\)
\(
\begin{aligned}
& \Rightarrow f(0)=\lim _{x \rightarrow 0} \frac{e^{x^2}-\cos x}{x^2}=\lim _{x \rightarrow 0} \\
& \frac{\left(1+x^2+\frac{x^4}{2 !}+\ldots \ldots\right)-\left(1-\frac{x^2}{2 !}+\ldots . . .\right)}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{\frac{3}{2} x^2+\ldots . .}{x^2}=\frac{3}{2}
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow f(0)=\lim _{x \rightarrow 0} \frac{e^{x^2}-\cos x}{x^2}=\lim _{x \rightarrow 0} \\
& \frac{\left(1+x^2+\frac{x^4}{2 !}+\ldots \ldots\right)-\left(1-\frac{x^2}{2 !}+\ldots . . .\right)}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{\frac{3}{2} x^2+\ldots . .}{x^2}=\frac{3}{2}
\end{aligned}
\)
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