Question

The value of $p$ for which the function

f(x) = \\{ \dfrac{{{{({4^x} - 1)}^3}}}{{\sin \dfrac{x}{p}\log (1 + \dfrac{{{x^2}}}{3})}};x! = 0 \\\ \\{ 12{(\log 4)^3};x = 0 \\\ $$ is continuous at $$x = 0$$ , A.$4$ B.$2$ C.$3$ D.$1$

Answer 1

Hint : Use the formulas of limits and derivatives. Simplify the function into the smallest fraction or ratio and then put the values. Check for limits for both the sides and then equate the value at which the function is continuous. For example, check at $f(x) = 0$ here.

Complete step-by-step answer :
Since, according to the question it is saying that the functions $f(x)$ for both the functions are continuous at $f(x) = 0$ .
That implies:
$f(x) = \dfrac{{{{({4^x} - 1)}^3}}}{{\sin \dfrac{x}{p}\log (1 + \dfrac{{{x^2}}}{3})}} = 12{(\log 4)^3} = f(0)$ at $\mathop {\lim }\limits_{x \to 0}$ .
To find the value of $p$ solve both the left and right side and then equate with $f(0)$ ;
Let’s check for left side first:
Simplifying $f(x) = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{({4^x} - 1)}^3}}}{{\sin \dfrac{x}{p}\log (1 + \dfrac{{{x^2}}}{3})}}$ step by step:
Dividing the numerator and denominator by ${x^3}$ ,we get:

$$f(x) = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{{{({4^x} - 1)}^3}}}{{{x^3}}}}}{{\dfrac{{\sin \dfrac{x}{p}\log (1 + \dfrac{{{x^2}}}{3})}}{{{x^3}}}}} \\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{(\dfrac{{{4^x} - 1}}{x})}^3}}}{{\dfrac{{\sin \dfrac{x}{p}\log (1 + \dfrac{{{x^2}}}{3})}}{{{x^3}}}}} \\\ = \dfrac{{\mathop {\lim }\limits_{x \to 0} {{(\dfrac{{{4^x} - 1}}{x})}^3}}}{{\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \dfrac{x}{p}\log (1 + \dfrac{{{x^2}}}{3})}}{{{x^3}}}}} \\\ 3 \;$$

According to the formula of limits we know that $\mathop {\lim }\limits_{x \to 0} (\dfrac{{{a^x} - 1}}{x}) = \log a$
So, $f(x) = \dfrac{{{{(\log 4)}^3}}}{{\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \dfrac{x}{p}\log (1 + \dfrac{{{x^2}}}{3})}}{{{x^3}}}}}$
Now Divide the denominator of denominator by $3p$ and then split the terms according to their needs and simplify them as follows:

$$f(x) = \dfrac{{{{(\log 4)}^3}}}{{\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \dfrac{x}{p}\log (1 + \dfrac{{{x^2}}}{3})}}{{{x^3}}}}} \\\ = \dfrac{{{{(\log 4)}^3}}}{{\dfrac{{\mathop {\lim }\limits_{x \to 0} \sin \dfrac{x}{p}\log (1 + \dfrac{{{x^2}}}{3})}}{{\dfrac{x}{p} \times \dfrac{{{x^2}}}{3} \times 3p}}}} \\\ = \dfrac{{{{(\log 4)}^3} \times 3p}}{{\mathop {\lim }\limits_{x \to 0} (\dfrac{{\sin \dfrac{x}{p}}}{{\dfrac{x}{p}}})\mathop {\lim }\limits_{x \to 0} (\dfrac{{\log (1 + \dfrac{{{x^2}}}{3})}}{{\dfrac{{{x^2}}}{3}}})}} \;$$

Since, we know that $\mathop {\lim }\limits_{x \to 0} (\dfrac{{\sin x}}{x}) = 1$ and $\mathop {\lim }\limits_{x \to 0} (\dfrac{{\log (1 + x)}}{x}) = 1$
Use these formulas and solved further and we get:

$$\\\ = \dfrac{{{{(\log 4)}^3} \times 3p}}{{\mathop {\lim }\limits_{x \to 0} (\dfrac{{\sin \dfrac{x}{p}}}{{\dfrac{x}{p}}})\mathop {\lim }\limits_{x \to 0} (\dfrac{{\log (1 + \dfrac{{{x^2}}}{3})}}{{\dfrac{{{x^2}}}{3}}})}} \\\ \\\ = \dfrac{{{{(\log 4)}^3} \times 3p}}{1} = 3p{(\log 4)^3} \;$$

Now, comparing these values with the right-hand side and we get:

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

Therefore, value of $p = 4$ , for which the function

f(x) = \\{ \dfrac{{{{({4^x} - 1)}^3}}}{{\sin \dfrac{x}{p}\log (1 + \dfrac{{{x^2}}}{3})}};x! = 0 \\\ \\{ 12{(\log 4)^3};x = 0 \; $$ is continuous at $$x = 0$$ . So, the correct option is option (A) i.e $p = 4$ **So, the correct answer is “Option A”.** **Note** : Always remember the formulas of limits and derivatives. Always check the limit part before applying the formula as we know that $$\mathop {\lim }\limits_{x \to 0} (\dfrac{{\sin x}}{x}) = 1$$ but if it was $$1,2$$ or something else then, it would have been wrong. Before putting the value, always reduce the function into its smallest function and also have a look at its domain and range.