Question

The value of p for which the function $f\left(x\right)=\frac{(4^x-1{)}^3}{sin\left(\frac{x}{p}\right)ln(1+\frac{x^2}{3})},x\ne 0=12(ln4{)}^3,x=0$ may be continuous at x = 0 is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is D 4

$f\left(0\right)=RHL=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f\left(h\right)12(ln4{)}^3=\underset{h\to 0}{lim}\frac{(4^h-1{)}^3}{sin\left(\frac{h}{p}\right).ln(1+\frac{h^2}{3})}=\underset{h\to 0}{lim}\frac{{\frac{(4^h-1)}{h}}^3}{\frac{sin\left(h/p\right)}{\left(h/p\right)}.\frac{ln(1+h^2/3)}{\left(h^2/3\right)}.\frac{1}{\left(3p\right)}}=\underset{h\to 0}{lim}\frac{(ln4{)}^3.3p}{1.1}Hence,p=4$