Question

Let $f\left(x\right)=\{\begin{array}{ccc}\frac{1-\cos 2x}{2x^2}& :& x\ne 0\\ k& :& x=0\end{array}$.

Then the value of $k$ for which, $f\left(x\right)$ will be continuous at $x=0$ is

• A. $0$
• B. $1$
• C. $2$
• D. none of these

Answer 1

Answer: (B)

$1$

If $f\left(x\right)$ is to be continuous at $x=0$ then we must have, $\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)$.

$f\left(0\right)=k$

Now,

$\underset{x\to 0}{lim}f\left(x\right)$

$=\underset{x\to 0}{lim}\frac{1-\cos 2x}{2x^2}$

$=\underset{x\to 0}{lim}\frac{2{\sin }^{2}x}{2x^2}$

$={\left(\underset{x\to 0}{lim}\frac{\sin x}{x}\right)}^2$

$=1$.

So if $f\left(x\right)$ to be continuous at $x=0$ then $k$ should be $1$.