Question

Find the value of P, and Q, for which
 \[f(x) = \left\{ \begin{matrix}
  \dfrac{{1 - {{\sin }^3}x}}{{3{{\cos }^2}x}}\,\,\,\,\,\,for\,x < \dfrac{\pi }{2} \\
  P\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,x = \dfrac{\pi }{2} \\
  \dfrac{{q(1 - \sin x)}}{{{{(\pi - 2x)}^2}}}\,\,for\,x > \dfrac{\pi }{2} \\
\end{matrix}\right.\]
is continuous at $x = \dfrac{\pi }{2}$.

Answer 1

Hint: Solve LHL and RHL, and equate them to get values.First, we solve the LHL, using necessary limits and then suitable identities and formulas, we simplify the LHL. Similarly, we solve the RHL as well using the necessary limit and then we get a simplified RHL as well. Since we know that $f(x)$ is continuous at some given value of x. We then equate RHL and LHL at that given point. Thus, we get the value of P and Q.

Complete step by step answer:
Here, we begin by taking the LHL and solve it.
$\mathop {\lim }\limits_{x \to \dfrac{{{\pi ^ - }}}{2}} \dfrac{{\left( {1 - {{\sin }^3}x} \right)}}{{3({1^2} - {{\sin }^2}x)}}$
Now, we convert the numerator in the form of $({a^3} - {b^3})$, such that a formula can be used.

$
   = \mathop {\lim }\limits_{x \to \dfrac{{{\pi ^ - }}}{2}} \dfrac{{\left( {{1^3} - {{\sin }^3}x} \right)}}{{3({1^2} - {{\sin }^2}x)}} \\
  \mathop {\lim }\limits_{x \to \dfrac{{{\pi ^ - }}}{2}} \dfrac{{(1 - \sin x)(1 + \sin x + {{\sin }^2}x)}}{{3(1 - \sin x)(1 + \sin x)}} \\
$
Where $(1 - \sin x) \ne 0$
$ = \mathop {\lim }\limits_{x \to \dfrac{{{\pi ^ - }}}{2}} \dfrac{{1 + \sin x + {{\sin }^2}x}}{{3(1 + \sin x)}}$
Now, we are going to apply the limits the equations and simplify it.
$Put\,x = \dfrac{\pi }{2} - h,as\,x \to \dfrac{\pi }{2},\,\,h \to 0$
Now, we get as
\[
   = \dfrac{{\left( {1 + {{\sin }^2}\dfrac{\pi }{2} + \sin \dfrac{\pi }{2}} \right)}}{{3(1 + \sin \dfrac{\pi }{2})}} \\
   = \dfrac{{1 + 1 + 1}}{{3(1 + 1)}} \\
   = \dfrac{1}{2} \\
\]

Since LHL $ = \dfrac{1}{2}$ and since $f(\dfrac{\pi }{2}) = P$, we can say that \[P = \dfrac{1}{2}\].
Now, we are going to solve RHL, to find the value of Q.
$ = \mathop {\lim }\limits_{x \to \dfrac{{{\pi ^ + }}}{2}} \dfrac{{q(1 - \sin x)}}{{{{(\pi - 2x)}^2}}}$
For, simplifying the calculation, we will convert the $\sin x$ into $\cos x$ such that we can apply an identity, which makes simplification and limit application easy.
$
   = \mathop {\lim }\limits_{x \to \dfrac{{{\pi ^ + }}}{2}} \dfrac{{q(1 - \cos \left( {\dfrac{\pi }{2} - x} \right))}}{{{{(\pi - 2x)}^2}}} \\
   = \mathop {\lim }\limits_{x \to \dfrac{{{\pi ^ + }}}{2}} \dfrac{{q\left( {2{{\sin }^2}\left( {\dfrac{\pi }{2} - \dfrac{x}{2}} \right)} \right)}}{{16{{\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}^2}}} \\
$$
   = \mathop {\lim }\limits_{x \to \dfrac{{{\pi ^ + }}}{2}} \dfrac{{q(1 - \cos \left( {\dfrac{\pi }{2} - x} \right))}}{{{{(\pi - 2x)}^2}}} \\
   = \mathop {\lim }\limits_{x \to \dfrac{{{\pi ^ + }}}{2}} \dfrac{{q\left( {2{{\sin }^2}\left( {\dfrac{\pi }{2} - \dfrac{x}{2}} \right)} \right)}}{{16{{\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}^2}}} \\
$
Now, we apply the limits the equation, and get the value.
$\dfrac{{2q}}{{16}} = \dfrac{q}{8}$$\dfrac{{2q}}{{16}} = \dfrac{q}{8}$
Since, in the question we are given that the functions are continuous at a given point.
So, it means that LHL= RHL at that point.
Which is

$
  \dfrac{q}{8} = \dfrac{1}{2} \\
  q = 4 \\
$
Hence, we have found the value of p and q which is required in the question.

Note: We can also solve the RHL and LHL separately as an individual sums and get the required values. Also, this is the preferred method as we also have been given that the function is continuous at some particular point.