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.