Question

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

Answer 1

Hint:
For the function $f(x)$ to be continuous at a point $x = a$ if,
$f(x)$ is defined at a.
1) $\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x)$i.e. left-hand limit is equal to right-hand limit as $x \to a$.
2) $\mathop {\lim }\limits_{x \to a} f(x) = f(a)$i.e. $\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = f(a)$
3) In this question it is given that $f(x)$ is continuous at \[x = \dfrac{\pi }{2}\]. By equating all the above conditions, we can calculate p and q.

Complete step by step solution:
We have given with a function $f(x)$and $f(x)$is continuous at \[x = \dfrac{\pi }{2}\]. We have to find the value of p and q. The function $f(x)$ is given as
\[ \Rightarrow f(x) = \left\{ \begin{gathered}
  \dfrac{{1 - {{\sin }^3}x}}{{3{{\cos }^2}x}},x < \dfrac{\pi }{2} \\
  p,x = \dfrac{\pi }{2} \\
  \dfrac{{q(1 - \sin x)}}{{{{(\pi - 2x)}^2}}},x > \dfrac{\pi }{2} \\
\end{gathered} \right.\]
For the $f(x)$to be continuous, $\mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} f(x) = \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ + }} f(x) = f(\dfrac{\pi }{2})$ ……..(1)
First of all, find the value of left-hand limit i.e. $\mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} f(x)$
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} f(x) = \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} \dfrac{{1 - {{\sin }^3}x}}{{3{{\cos }^2}x}}$
By replacing 1 with cube of 1 as cube of 1 is 1 we get,
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} f(x) = \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} \dfrac{{{1^3} - {{\sin }^3}x}}{{3{{\cos }^2}x}}$
By applying formula ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$ on the numerator of the above equation we get,
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} f(x) = \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} \dfrac{{(1 - \sin x)(1 + \sin x + {{\sin }^2}x)}}{{3{{\cos }^2}x}}$
By applying the trigonometric formula for ${\cos ^2}x$ i.e. ${\cos ^2}x = 1 - {\sin ^2}x$ in the above equation we get,
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} f(x) = \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} \dfrac{{(1 - \sin x)(1 + \sin x + {{\sin }^2}x)}}{{3(1 - {{\sin }^2}x)}}$
By applying formula ${a^2} - {b^2} = (a - b)(a + b)$ on the denominator of the above equation we get,
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} f(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)}}$
By cancelling the common terms from numerator and denominator we get,
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} f(x) = \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} \dfrac{{(1 + \sin x + {{\sin }^2}x)}}{{3(1 + \sin x)}}$
As $x \to \dfrac{\pi }{2}$so open the limit by replacing x with $\dfrac{\pi }{2}$we get,
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} f(x) = \dfrac{{(1 + \sin \dfrac{\pi }{2} + {{\sin }^2}\dfrac{\pi }{2})}}{{3(1 + \sin \dfrac{\pi }{2})}}$
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} f(x) = \dfrac{{(1 + 1 + 1)}}{{3(1 + 1)}}$ as $\sin \dfrac{\pi }{2} = 1$
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} f(x) = \dfrac{3}{{3 \times 2}}$
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} f(x) = \dfrac{1}{2}$ ……..(2)
From equation 1 and 2 we get,
$ \Rightarrow f(\dfrac{\pi }{2}) = \dfrac{1}{2}$ …………(3)
As it is given the $f(x) = p$at $\dfrac{\pi }{2}$. Therefore,
$ \Rightarrow f(\dfrac{\pi }{2}) = p$ …………(4)
From 3 and 4 equation we get,
$ \Rightarrow p = \dfrac{1}{2}$
To find the value of q, solve the right-hand limit i.e.
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ + }} f(x) = \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ + }} \dfrac{{q(1 - \sin x)}}{{{{(\pi - 2x)}^2}}}$
We can replace $\sin x$with $\cos (\dfrac{\pi }{2} - x)$ as $\cos (\dfrac{\pi }{2} - x) = \sin x$ we get,
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ + }} f(x) = \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ + }} \dfrac{{q(1 - \cos (\dfrac{\pi }{2} - x))}}{{{{(\pi - 2x)}^2}}}$
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ + }} f(x) = \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ + }} \dfrac{{2q{{\sin }^2}(\dfrac{\pi }{4} - \dfrac{x}{2})}}{{{{(\pi - 2x)}^2}}}$ (as $1 - \cos x = 2{\sin ^2}\dfrac{x}{2}$)
Taking 4 common from both terms in the denominator and taking it out of bracket while taking 4 out of bracket we will do the square of 4 i.e. 16 as there is power of square in the bracket. We get,
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ + }} f(x) = \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ + }} \dfrac{{2q{{\sin }^2}(\dfrac{\pi }{4} - \dfrac{x}{2})}}{{16{{(\dfrac{\pi }{4} - \dfrac{x}{2})}^2}}}$
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ + }} f(x) = \mathop {\dfrac{{2q}}{{16}}\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ + }} \dfrac{{{{\sin }^2}(\dfrac{\pi }{4} - \dfrac{x}{2})}}{{{{(\dfrac{\pi }{4} - \dfrac{x}{2})}^2}}}$ ……….(5)
We know that $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$, therefore
$ \Rightarrow \mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{{{\sin }^2}(\dfrac{\pi }{4} - \dfrac{x}{2})}}{{{{(\dfrac{\pi }{4} - \dfrac{x}{2})}^2}}} = 1$ ….………(6)
From 5 and 6 we get,
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ + }} f(x) = \dfrac{q}{8}$ ………(7)
From 1, 2 and 7 equation we get,
$
   \Rightarrow \dfrac{q}{8} = \dfrac{1}{2} \\
   \Rightarrow q = 4 \\
$

Hence, $p = \dfrac{1}{2}$ and $q = 4$ is the required answer.

Note:
The student gets confused in taking the left-hand limit and right-hand limit, they start putting the value of x in the function directly. By doing this they will get ‘Not defined’ in the result. So, first simplify it by applying appropriate formulae and cancelling the terms and converting it until we get a denominator not equal to zero. It is done so that when we open the limit, the value will not give ‘Not defined’.
Second, when we take some number out of bracket then the power of number is equal to power of bracket i.e. if the bracket has power 2, the number we take out of bracket should also acquire power 2. E.g.
$
   \Rightarrow {(2x - 4)^2} = {(2(x - 2))^2} \\
   \Rightarrow {(2x - 4)^2} = {2^2}{(x - 2)^2} \\
$