Question

f\left( x \right) = \left\\{ \begin{array}{l} \dfrac{{1 - {{\sin }^3}x}}{{3{{\cos }^2}x}}\;\;\;\;\;\;\;\;\;\;{\rm{if}}\;x < \dfrac{\pi }{2}\\\ a\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{if }}x = \dfrac{\pi }{2}\\\ \dfrac{{b\left( {1 - \sin x} \right)}}{{{{\left( {\pi - 2x} \right)}^2}}}\;\;\;\;\;\;\;\;{\rm{if }}x > \dfrac{\pi }{2} \end{array} \right. is continuous at $x = \dfrac{\pi }{2}$ find $a$ and $b$ .

Answer 1

Hint : We know the function is continuous at $x = \dfrac{\pi }{2}$ then take limit $\dfrac{{1 - {{\sin }^3}x}}{{3{{\cos }^2}x}}$ at $x < \dfrac{\pi }{2}$ is equal to $a$ . Then since the function is continuous at $x = \dfrac{\pi }{2}$ then take the limit $a$ is equal to limit a $\dfrac{{b\left( {1 - \sin x} \right)}}{{{{\left( {\pi - 2x} \right)}^2}}}$ at $x > \dfrac{\pi }{2}$ .

Complete step-by-step answer :
The given function is f\left( x \right) = \left\\{ \begin{array}{l} \dfrac{{1 - {{\sin }^3}x}}{{3{{\cos }^2}x}}\;\;\;\;\;\;\;\;\;\;{\rm{if}}\;x < \dfrac{\pi }{2}\\\ a\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{if }}x = \dfrac{\pi }{2}\\\ \dfrac{{b\left( {1 - \sin x} \right)}}{{{{\left( {\pi - 2x} \right)}^2}}}\;\;\;\;\;\;\;\;{\rm{if }}x > \dfrac{\pi }{2} \end{array} \right. .
Also, it is given that the function is continuous at $x = \dfrac{\pi }{2}$ .
The definition of continuous function:
We say the function is continuous at some point let us say it as $a$ when limit at $x \to a$ function is equal to function at point $a$ that is,
$\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)$
Since, it is given that the function is continuous at $x = \dfrac{\pi }{2}$ so,

$\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{1 - {{\sin }^3}x}}{{3{{\cos }^2}x}} = a$ ........(1)
Now, let us substitute $\dfrac{\pi }{2}$ in the function we get,
$\dfrac{{1 - {{\sin }^3}\dfrac{\pi }{2}}}{{3{{\cos }^2}\dfrac{\pi }{2}}}$
Since, we know the value of $\sin \dfrac{\pi }{2}$ is $1$ and the value for $\cos \left( {\dfrac{\pi }{2}} \right)$ is $0$ .
So, now let us substitute $\sin \dfrac{\pi }{2}$ value and $\cos \left( {\dfrac{\pi }{2}} \right)$ value in the above equation, we get
$\dfrac{{1 - 1}}{{3 \times 0}} = \dfrac{0}{0}$
Since, the function is $\dfrac{0}{0}$ for now let us use L’hospital rule.
L’ Hospital rule says that if $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{0}{0}$ , then we will take .
So after applying L’Hospital rule in our problem we have,
Now, by chain rule first the derivative for numerator term is,
$\dfrac{d}{{dx}}\left( {1 - {{\sin }^3}x} \right) = - 3{\sin ^2}x\cos x$
Also, second the derivative for denominator term is, we get,
$\dfrac{d}{{dx}}\left( {3{{\cos }^2}x} \right)6\cos x \times \left( { - \sin x} \right)$
On substituting in equations (1) we get,
$\begin{array}{c} \mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{ - 3{{\sin }^2}x\cos x}}{{6\cos x \times \left( { - \sin x} \right)}} = a\\\ \mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{\sin x}}{2} = a\\\ \dfrac{1}{2} = a \end{array}$
The value for $a$ is $\dfrac{1}{2}$ .

Now we need to find the value for $b$ .
Using the continuity property we have, since it is given that the function is continuous at $x = \dfrac{\pi }{2}$ so,
$\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{b\left( {1 - \sin x} \right)}}{{{{\left( {\pi - 2x} \right)}^2}}} = \dfrac{1}{2}$ ...........(2)
Now let us substitute $\dfrac{\pi }{2}$ in the function we get,
$\dfrac{{b\left( {1 - \sin \dfrac{\pi }{2}} \right)}}{{{{\left( {\pi - 2x} \right)}^2}}}$
Since value of $\sin \dfrac{\pi }{2}$ is $1$ , so now let us substitute $\sin \dfrac{\pi }{2}$ value in the above equation, we get,
$\dfrac{{b\left( 0 \right)}}{0} = \dfrac{0}{0}$
Since, the function is $\dfrac{0}{0}$ for now let us use L’Hospital rule.
L’ Hospital rule says that if $\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{0}{0}$ then we take .
So after applying L’Hospital rule in our problem we have,
The derivative for numerator term is,
$\dfrac{d}{{dx}}b\left( {1 - \sin x} \right) = - b\cos x$
The derivative for denominator term is, we get,
$\dfrac{d}{{dx}}{\left( {\pi - 2x} \right)^2} = 2\left( {\pi - 2x} \right)\left( { - 2} \right)$
On substituting the value in (2) we get,
$\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{ - b\cos x}}{{ - 4\left( {\pi - 2x} \right)}} = \dfrac{0}{0}$
Let us use L’Hospital rule again we get,
$\begin{array}{c} \mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} \dfrac{{b\sin x}}{8} = \dfrac{1}{2}\\\ \dfrac{{b\sin \left( {\dfrac{\pi }{2}} \right)}}{8} = \dfrac{1}{2}\\\ \dfrac{b}{8} = \dfrac{1}{2}\\\ b = 4 \end{array}$
The value for $b$ is $4$ .

Note : If the function is continuous at some point we can use $\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} f\left( x \right)$ is equal to $f\left( a \right)$ . Otherwise we cannot use it. For L’Hospital rule if the denominator term is $0$ and numerator term is non zero we cannot use this rule and vice versa.