Question

Find the value of $p$ and $q$ for which

f\left( x \right)=\left\\{\begin{array}{ll} \dfrac{1-{{\sin }^{3}}x}{3{{\cos }^{2}}x} :if\text{ }x<\dfrac{\pi }{2} \\\ p :if\text{ }x=\dfrac{\pi }{2} \\\ \dfrac{q(1-\sin x)}{{{(\pi -2x)}^{2}}} :if\text{ }x>\dfrac{\pi }{2} \\\ \end{array} \right. $$ is continuous at $x=\dfrac{\pi }{2}$.

Answer 1

Hint: We know that $f(x)$ is continuous then $LHL=RHL=f(\dfrac{\pi }{2})$. So take LHL and RHL and simplify it. Then after solving equate it. You will get the values of $p$ and $q$.

“Complete step-by-step answer:”

So first of all, a limit is a value toward which an expression converges as one or more variables approach certain values. Limits are important in calculus and analysis.
Consider the limit of the expression $2x+3$ as $x$ approaches $0$. It is not difficult to see that this limit is $3$, because we can assign the value $0$ to the variable $x$ and perform the calculation directly.
A limit is a value that a function (or sequence) "approaches" as the input (or index) "approaches" some value.
Limits are essential to calculus (and mathematical analysis in general) and are used to define continuity, derivatives, and integrals.
The concept of a limit of a sequence is further generalized to the concept of a limit of a topological net and is closely related to limit and direct limit in category theory.

A function $f(x)$ is said to be continuous at $x=a$ if $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=f(a)$
i.e. L.H.L.$=$ R.H.L$=f(a)$ = value of the function at a i.e.$\underset{x\to a}{\mathop{\lim }}\,f(x)=f(a)$
If $f(x)$ it is not continuous at $x=a$ , we say that$f(x)$ it is discontinuous $x=a$.

But here it is given that $f(x)$ is continuous at $x=\dfrac{\pi }{2}$ .
Now let$f{{(x)}_{1}}=\underset{x\to {{\dfrac{\pi }{2}}^{-}}}{\mathop{\lim }}\,\dfrac{1-{{\sin }^{3}}x}{3{{\cos }^{2}}x}$,$f{{(x)}_{2}}=p$ and $f{{(x)}_{3}}=\underset{x\to {{\dfrac{\pi }{2}}^{+}}}{\mathop{\lim }}\,\dfrac{q(1-\sin x)}{{{(\pi -2x)}^{2}}}$.
We know $x$ is continuous at $\dfrac{\pi }{2}$.
Here $f{{(x)}_{1}}=f{{(x)}_{2}}=f{{(x)}_{3}}$.
So for LHL, here we have taken the limit for $x$ tending to ${{\dfrac{\pi }{2}}^{-}}$.
So now taking function$f{{(x)}_{1}}=\underset{x\to {{\dfrac{\pi }{2}}^{-}}}{\mathop{\lim }}\,\dfrac{1-{{\sin }^{3}}x}{3{{\cos }^{2}}x}$.
$f{{(x)}_{1}}=\underset{x\to {{\dfrac{\pi }{2}}^{-}}}{\mathop{\lim }}\,\dfrac{1-{{\sin }^{3}}x}{3{{\cos }^{2}}x}$
$f{{(x)}_{1}}=\underset{x\to {{\dfrac{\pi }{2}}^{-}}}{\mathop{\lim }}\,\dfrac{{{1}^{3}}-{{\sin }^{3}}x}{3{{\cos }^{2}}x}$
Here we know the identity that ${{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+{{b}^{2}}+ab)$.
Using the above identity we get,
$f{{(x)}_{1}}=\underset{x\to {{\dfrac{\pi }{2}}^{-}}}{\mathop{\lim }}\,\dfrac{(1-\sin x)(1+{{\sin }^{2}}x+\sin x)}{3{{\cos }^{2}}x}$
Also, we know, ${{\cos }^{2}}x=1-{{\sin }^{2}}x=(1+\sin x)(1-\sin x)$.
$f{{(x)}_{1}}=\underset{x\to {{\dfrac{\pi }{2}}^{-}}}{\mathop{\lim }}\,\dfrac{(1-\sin x)(1+{{\sin }^{2}}x+\sin x)}{3(1+\sin x)(1-\sin x)}=\underset{x\to {{\dfrac{\pi }{2}}^{-}}}{\mathop{\lim }}\,\dfrac{(1+{{\sin }^{2}}x+\sin x)}{3(1+\sin x)}$
Put $x=\dfrac{\pi }{2}-h$, when $x=\dfrac{\pi }{2}$ and $h=0$.
So, $x=\dfrac{\pi }{2}$.
$f{{(x)}_{1}}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{(1+{{\sin }^{2}}(\dfrac{\pi }{2}-h)+\sin (\dfrac{\pi }{2}-h))}{3(1+\sin (\dfrac{\pi }{2}-h))}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{(1+{{\cos }^{2}}(h)+\cos (h))}{3(1+\cos (h))}=\dfrac{1+1+1}{3(2)}=\dfrac{1}{2}$
So for RHL, here we have taken the limit for $x$ tending to ${{\dfrac{\pi }{2}}^{+}}$.
$f{{(x)}_{3}}=\underset{x\to {{\dfrac{\pi }{2}}^{+}}}{\mathop{\lim }}\,\dfrac{q(1-\sin x)}{{{(\pi -2x)}^{2}}}$
Put $x=\dfrac{\pi }{2}+h$, so $x=\dfrac{\pi }{2}$ and $h=0$.
$f{{(x)}_{3}}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{q(1-\sin (\dfrac{\pi }{2}+h))}{{{(\pi -2(\dfrac{\pi }{2}+h))}^{2}}}$
$f{{(x)}_{3}}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{q(1-\cosh )}{4{{h}^{2}}}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{q{{\sin }^{2}}\left( \dfrac{h}{2} \right)}{2{{h}^{2}}}$ …… ($1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}$)
Multiplying dividing denominator by 4 and arranging it such that $\underset{h\to 0}{\mathop{\lim }}$ $\dfrac{sinh}{h}$=$1$
Applying $h=0$ we get,
$f{{(x)}_{3}}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{q(1-\cosh )}{4{{h}^{2}}}=\dfrac{q}{8}\underset{h\to 0}{\mathop{\lim }}\,{{\left( \dfrac{\sin \left( \dfrac{h}{2} \right)}{\dfrac{h}{2}} \right)}^{2}}=\dfrac{q}{8}$
Also at $x=\dfrac{\pi }{2}$, $f{{(x)}_{2}}=p$.
So we have been given that $f(x)$ is continuous $x=\dfrac{\pi }{2}$.
Here we get,
$LHL=RHL=f(\dfrac{\pi }{2})$.
So we get $\dfrac{1}{2}=\dfrac{q}{8}=p$
Here, equating we get,
$\dfrac{1}{2}=\dfrac{q}{8}$
$q=4$
and $\dfrac{1}{2}=p$.
So we get the values $p=\dfrac{1}{2}$ and $q=4$.

Note: Students should know what RHL and LHL and are thorough with the concept .If the limit is continuous then $LHL=RHL=f(\dfrac{\pi }{2})$. Students should remember trigonometric identities and formula $1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}$ and important limit values i.e $\underset{h\to 0}{\mathop{\lim }}$ $\dfrac{sinh}{h}$=$1$ for solving these types of questions.