Given that: f (x) = , if f(x) is continous at x = 0, the value of a is

JEE MainMathematicsToughContinuity and differentiability

Given that: f (x) = $\left\{ \begin{matrix}
\frac{1 - cos4x}{x^{2}}, & \ if\ x < 0 \\
a &,\ if\ x = 0 \\
\frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4} &,\ if\ x > 0
\end{matrix} \right.\ $, if f(x) is continous at x = 0, the value of a
is

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For f(x) to be continous at x = 0, we have
f (0^-) = f(0) = f(0^+)......... (1)
here f(0) = a

f(0^{-}) = \lim_{x \rightarrow 0^{-}}\mspace{2mu}\frac{1 - \cos{4x}}{x^{2}} = \lim_{x \rightarrow 0^{-}}\mspace{2mu}\frac{2\sin^{2}{2x}}{x^{2}}

= \lim_{x \rightarrow 0 -}\mspace{2mu} 8\left( \frac{sin2x}{2x} \right)^{2} = 8 \cdot 1^{2} = 8

.....(2)

f\left( 0^{+} \right) = \lim_{x \rightarrow 0^{+}}\mspace{2mu}\frac{\sqrt{x}}{\sqrt{16 + \sqrt{x} - 4}} = \lim_{x \rightarrow 0^{+}}\mspace{2mu}\frac{\sqrt{x}}{\sqrt{16 + \sqrt{x} - 4}} \times \frac{\sqrt{16 + \sqrt{x}} + 4}{\sqrt{16 + \sqrt{x}} + 4}

= \lim_{x \rightarrow 0^{+}}\mspace{2mu}\frac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{(16 + \sqrt{x}) - 16} = \lim_{x \rightarrow 0^{+}}\mspace{2mu}(\sqrt{16 + \sqrt{x}} + 4) = 8

...........(3)
Hence from (1), (2) & (3) we get a = 8.

Source: Grass-G / M-19 Mathematics Paper 28 Nov Fourth.docx

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