Question

The value of $p$ for which the function
$f\left(x\right)=\{\begin{array}{cc}\frac{(4^x-1{)}^3}{\sin \frac{x}{p}\log (1+\frac{x^2}{3})}& ;x\ne 0\\ 12(\log 4{)}^3& ;x=0\end{array}$ is continuous at $x=0$, is

• A. $4$
• B. $2$
• C. $3$
• D. $1$

Answer 1

The correct option is A $4$

We know that the function is continuous at $x=0$ if $\underset{x\to 0}{lim}f\left(x\right)$ exists and is equal to $f\left(0\right)$.

Therefore, for continuity we have,

$\underset{x\to 0}{lim}\frac{{(4^x-1)}^3}{\sin \frac{x}{p}\log (1+\frac{x^2}{3})}=f\left(0\right)=12{\left(\log 4\right)}^3$

Since

$\underset{x\to 0}{lim}\frac{(a^x-1)}{x}=\log a$ for $a\ne 0,a>1$,

$\underset{x\to 0}{lim}\frac{\sin x}{x}=1$, and

$\underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1$,

the limit on LHS can be rewritten as:

$\underset{x\to 0}{lim}\frac{{\left(\frac{4^x-1}{x}\right)}^3\cdot x^3}{\frac{\sin \frac{x}{p}}{\frac{x}{p}}\cdot \frac{x}{p}\cdot \frac{\log (1+\frac{x^2}{3})}{\frac{x^2}{3}}\cdot \frac{x^2}{3}}$

$=\underset{x\to 0}{lim}\frac{{\left(\log 4\right)}^3\cdot x^3}{1\cdot \frac{x}{p}\cdot 1\cdot \frac{x^2}{3}}$ (using standard limits stated above)

$=3p{\left(\log 4\right)}^3$ .... (because $\underset{x\to 0}{lim}\frac{x^3}{x^3}=1$)

According to the condition for continuity, this must equal $f\left(0\right)=12{\left(\log 4\right)}^3$

$3p{\left(\log 4\right)}^3=12{\left(\log 4\right)}^3$

$\Rightarrow 3p=12$

$\Rightarrow p=4$