Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{1-cos4x}{x^2}& whenx<0\\ a,& whenx=0\\ \frac{\sqrt{x}}{\sqrt{(16+\sqrt{x})}-4},& whenx>0\end{array}$ , is continuous at x = 0, then the value of 'a' will be

• A. 8
• B. -8
• C. 4
• D. None of these

Answer 1

The correct option is A 8

${lim}_{x\to 0-}f\left(x\right)={lim}_{x\to 0-}\left(\frac{2si{n}^{2}2x}{(2x{)}^2}\right)4=8$
${lim}_{x\to 0+}f\left(x\right)={lim}_{x\to 0+}\sqrt{16+\sqrt{x}+4}=8.$ Hence a = 8