Question

If f\left( x \right) = \left\\{ \dfrac{{1 - \cos 4x}}{{{x^2}}},when\ x < 0 \\\ a,when\ x = 0 \\\ \dfrac{{\sqrt x }}{{\sqrt {\left( {16 + \sqrt x } \right)} - 4}},when\ x > 0 \\\ \right.
is continuous at $x = 0$, then the value of $a$ will be
A. $8$
B. $- 8$
C. $4$
D. None of these

Answer 1

Here to find the value of $a$, as it is continuous at $x = 0$ let us apply the definition of continuity.
$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = f\left( a \right)$
As the function is continuous at $x = 0$ apply this equation.
In which we need to find $\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)$ and $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)$ to get the value of $a$.

Formula used:
$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( a \right)$

Complete step-by-step answer:
The given function $f\left( x \right)$ is

\dfrac{{1 - \cos 4x}}{{{x^2}}},when\ x < 0 \\\ a,when\ x = 0 \\\ \dfrac{{\sqrt x }}{{\sqrt {\left( {16 + \sqrt x } \right)} - 4}},when\ x > 0 \\\ \right.$$ Which is continuous at $$x = 0$$. Applying the continuity function for $$x < 0$$ and $$x > 0$$ for function $$f\left( x \right)$$ we get $$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = f\left( a \right)$$ In which, when $$x < 0$$, the function is $$\dfrac{{1 - \cos 4x}}{{{x^2}}}$$ We get $$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{1 - \cos 4x}}{{{x^2}}}$$ $$ = \mathop {\lim }\limits_{x \to {0^ - }} \left( {\dfrac{{2{{\sin }^2}2x}}{{{{\left( {2x} \right)}^2}}}} \right) \cdot 4$$ Hence after simplifying the terms we get $$\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = 8$$ Now let us find the terms of $$\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sqrt x }}{{\sqrt {\left( {16 + \sqrt x } \right)} - 4}}$$ After simplifying the terms of the function, we get $$ = \mathop {\lim }\limits_{x \to {0^ + }} \sqrt {\left( {16 + \sqrt x } \right)} + 4$$ Hence, $$\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 8$$ Also, $$f\left( 0 \right) = 8$$ The value of $$a$$ is $$8$$. **So, the correct answer is “Option A”.** **Additional Information:** The continuity of a real function $$\left( f \right)$$ on a subset of the real numbers is defined when the function exists at point c and is given as $$\mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)$$ A real function $$\left( f \right)$$is said to be continuous if it is continuous at every point in the domain of $$f$$. Consider a function $$f\left( x \right)$$, and the function is said to be continuous at every point in $$\left[ {a,b} \right]$$including the endpoints a and b. Continuity of “$$f$$” at a means, $$\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)$$ Continuity of “$$f$$” at b means, $$\mathop {\lim }\limits_{x \to b} f\left( x \right) = f\left( b \right)$$ **Note:** To find the value of any given term, we need to see that the given function is continuous at $$x = 0$$ or $$y = 0$$ with respect to the given function and the conditions of $$x$$. Hence based on this we need to find the values of the given function.