If $f:\mathbb{R} \to \mathbb{R}$ is defined as:
$f(x) = \left\{ {\begin{array}{*{20}{c}}
  {\dfrac{{\cos 3x - \cos x}}{{{x^2}}}}&{x \ne 0} \\
  \lambda &{x = 0}
\end{array}} \right.$
And if $f$ is continuous at $x = 0$ then $\lambda $ is equal to:
A) $ - 2$
B) $ - 4$
C) $ - 6$
D) $ - 8$

Hint:Given function is defined differently at two different points. It is also given that the function is continuous at that point. So, we will use the basic definition of continuity and we will take the limit at that point. Then we will equate the limiting value of the function to the given value of the function.

Complete step-by-step answer:
The given definition of the function is:
$f(x) = \left\{ {\begin{array}{*{20}{c}}
  {\dfrac{{\cos 3x - \cos x}}{{{x^2}}}}&{x \ne 0} \\
  \lambda &{x = 0}
\end{array}} \right.$
It is also given that the function $f$ is continuous at point $x = 0$ .
We will use the definition of continuity.
We say that the function $f(x)$ is continuous at the point $x = a$ if $\mathop {\lim }\limits_{x \to a} f(x) = f(a)$ that is limiting value of the function is same as the functional value.
In the above case as the function is continuous at $x = 0$ , we get:
$\mathop {\lim }\limits_{x \to 0} f(x) = f(0)$
But the value of the function at $x = 0$ is $\lambda $ .
Therefore,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\cos 3x - \cos x}}{{{x^2}}} = \lambda….. (1)$
Consider the left-hand side.
If we directly substitute $x = 0$ then the given function is in $\dfrac{0}{0}$ that is in degenerate form.
Therefore, we will use the L'Hospital rule to evaluate the limit.
L’Hospital Rule tells us that if we have an indeterminate $\dfrac{0}{0}$ form all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.
We know differentiation of $\cos 3x$ is $-3\sin3x$ and $x^2$ is $2x$.
Thus, the limit becomes,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\cos 3x - \cos x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 3\sin 3x + \sin x}}{{2x}}$
Now, focus on the right-hand side.
If we directly substitute $x = 0$ then again, the function is in $\dfrac{0}{0}$ form.
Thus, we will again use l'Hospital rule.
Now we know that differentiation of $\sin3x$ is $3\cos 3x$ and $2x$ is $2$.
$\mathop {\lim }\limits_{x \to 0} \dfrac{{ - 3\sin 3x + \sin x}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{ - 9\cos 3x + \cos x}}{2}$
Now we can directly substitute $x = 0$ .
Thus, we get,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{ - 9\cos 3x + \cos x}}{2} = \dfrac{{ - 9(1) + (1)}}{2}$
On simplifying we get,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{ - 9\cos 3x + \cos x}}{2} = - 4$
Therefore, from equation (1), $\lambda = - 4$ .

So, the correct answer is “Option B”.

Note:We started with using the basic definition of continuity and established the relation between the function and its limiting value. We then simplified the limit by using l’Hospital rule. We can alternatively also use basic trigonometric properties and find the limit. But it is always easier to use the L'Hospital rule so we chose it.