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$f(x)=\left\{\begin{array}{1 1}\large\frac{1-\cos 4x}{x^2}&x < 0\\a&x=0\\\large\frac{\sqrt x}{\sqrt{[16+\sqrt x]-4}}&x > 0\end{array}\right.$If the function be continuous at $x=0$ then $a=$

$(a)\;4\qquad(b)\;6\qquad(c)\;8\qquad(d)\;10$

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$F(0^-)=\lim\limits_{x\to 0}\large\frac{1-\cos 4x}{x^2}$
$\qquad=\lim\limits_{x\to 0}\large\frac{2\sin^22x}{(2x)^2}$$\times 4=8$
$F(0^+)=\lim\limits_{x\to 0}\large\frac{\sqrt{x}}{\sqrt{16+\sqrt x-4}}$
$\qquad\;=\lim\limits_{x\to 0}\large\frac{\sqrt x[\sqrt{16+\sqrt x+4}}{16+\sqrt x-16}$
$4+4=8$
$f(0)=a$
$a=8$
Hence (c) is the correct answer.
answered Jan 6, 2014 by sreemathi.v
 

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