Question

If $f\left( x \right)=\dfrac{1-\sin x}{{{\left( \pi -2x \right)}^{2}}},$ for $x\ne \dfrac{\pi }{2}$ is continuous at $x=\dfrac{\pi }{2},$ find $f\left( \dfrac{\pi }{2} \right)$

Answer 1

Hint: This question can be solved by using the concept of limits. It is given that $f\left( x \right)=\dfrac{1-\sin x}{{{\left( \pi -2x \right)}^{2}}},$ for $x\ne \dfrac{\pi }{2}$ and is continuous at $x=\dfrac{\pi }{2},$ implying that at $x=\dfrac{\pi }{2},$ the value of the function $f\left( x \right)$ will be the same when $x=\dfrac{\pi }{2}+h$ and $x=\dfrac{\pi }{2}-h.$ Here, h is an infinitesimal value. Therefore, we use the concept of limits to simplify this and solve. We will use L'Hospital's rule because we get a $\dfrac{0}{0}$ form while solving. Then we substitute the value of the limit to obtain the final answer.

Complete step by step solution:
The given function $f\left( x \right)$ is a continuous function at $x=\dfrac{\pi }{2}.$ We use the concepts of limits to solve this since it is continuous at this point.
It is given that $f\left( x \right)=\dfrac{1-\sin x}{{{\left( \pi -2x \right)}^{2}}},$ at $x\ne \dfrac{\pi }{2}$ .
Suppose we apply limits for $x=\dfrac{\pi }{2},$
$\Rightarrow \displaystyle \lim_{x \to \dfrac{\pi }{2}}f\left( x \right)$
$\Rightarrow \displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{1-\sin x}{{{\left( \pi -2x \right)}^{2}}}$
Applying the limit,
$\Rightarrow \dfrac{1-\sin \dfrac{\pi }{2}}{{{\left( \pi -2.\dfrac{\pi }{2} \right)}^{2}}}$
We know that $\sin \dfrac{\pi }{2}$ value is 1 and cancelling the 2s in the numerator and denominator of the denominator term,
$\Rightarrow \dfrac{1-1}{{{\left( \pi -\pi \right)}^{2}}}$
Subtracting,
$\Rightarrow \dfrac{0}{0}$
This is an indeterminate form and can be simplified using the L Hospital’s rule which is nothing but the application of partial differentiation of the numerator and denominator separately.
Differentiating the numerator,
$\Rightarrow \dfrac{d}{dx}\left( 1-\sin x \right)$
We know the differentiation of a constant is 0 and of sin is cos.
$\Rightarrow 0-\cos x$
$\Rightarrow -\cos x$
Differentiating the denominator,
$\Rightarrow \dfrac{d}{dx}{{\left( \pi -2x \right)}^{2}}$
We use the differentiation of the composite function method and differentiate the outside function first and multiply it with the inside function. We know the differentiation of ${{x}^{2}}$ is $2x$ and this is the outside function. Here $x$ is $\left( \pi -2x \right).$ This is multiplied with the differentiation of the inside function which is $\pi -2x$ which is nothing but $-2.$
$\Rightarrow 2.\left( \pi -2x \right).-2$
Taking the product of all the terms,
$\Rightarrow -4\pi +8x$
Now applying limits to this,
$\Rightarrow \displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{-\cos x}{\left( -4\pi +8x \right)}$
Applying the limits,
$\Rightarrow \dfrac{-\cos \dfrac{\pi }{2}}{\left( -4\pi +8\dfrac{\pi }{2} \right)}$
We know the value of $\cos \dfrac{\pi }{2}$ is 0 and cancelling the 8 with 2 in the denominator,
$\Rightarrow \dfrac{0}{\left( -4\pi +4\pi \right)}$
Subtracting the terms in the denominator,
$\Rightarrow \dfrac{0}{0}$
Hence, we again apply the L Hospital’s rule.
Differentiating $-\cos x$ gives $+\sin x$ and differentiating the denominator $-4\pi +8x$ , leaves us with the constant of 8.
Using the limit for this now,
$\Rightarrow \displaystyle \lim_{x \to \dfrac{\pi }{2}}\dfrac{\sin x}{8}$
Applying the limit,
$\Rightarrow \dfrac{\sin \dfrac{\pi }{2}}{8}$
We know the value of $\sin \dfrac{\pi }{2}$ is 1. Substituting this,
$\Rightarrow \dfrac{1}{8}$

Hence, the value of the function $f\left( x \right)$ at $x=\dfrac{\pi }{2},$ that is $f\left( \dfrac{\pi }{2} \right)$ , is $\dfrac{1}{8}.$

Note: Basic derivatives and the different methods of differentiation is an important and necessary concept to solve this question. Students are required to know these formulas to solve them. It is also important for the students to know the basic sin and cos values at standard angle values.