Question

Let the function $f:\left( 0,\pi \right)\to R$ be a twice differentiable function such that
$\displaystyle \lim_{t \to x},\dfrac{f\left( x \right)\sin t-f\left( t \right)\sin x}{t-x}={{\sin }^{2}}x\text{ }\forall \text{x}\in \left( 0,\pi \right)$.
If $f\left( \dfrac{\pi }{6} \right)=-\dfrac{\pi }{12}$, then which of the following statements is/are true?
Multiple correct answer question
A. $f\left( \dfrac{\pi }{4} \right)=\dfrac{\pi }{4\sqrt{2}}$
B. $f\left( x \right)<\dfrac{{{x}^{4}}}{6}-{{x}^{2}}$ for all $\text{x}\in \left( 0,\pi \right)$
C. There exist $\alpha \in \left( 0,\pi \right)$ such that $f'\left( \alpha \right)=0$
D. $f''\left( \dfrac{\pi }{2} \right)+f\left( \dfrac{\pi }{2} \right)=0$

Answer 1

To solve this question, we should know a rule in limits which is named as L-Hospital rule. The rule states that if the function in the limits $\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}$ which is $\dfrac{f\left( x \right)}{g\left( x \right)}$ when substituted the value of a which is $\dfrac{f\left( a \right)}{g\left( a \right)}$ leads to an undetermined form such as $\dfrac{0}{0},\dfrac{\infty }{\infty }$, we can write the limit as $\displaystyle \lim_{x \to a},\dfrac{f\left( x \right)}{g\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{f'\left( x \right)}{g'\left( x \right)}$. We can use this rule as many times as we can until it leads to a determined form. We can infer that if we substitute the value of t as x in the function, we get $\dfrac{0}{0}$ form which is undetermined form. So, we can apply the L-Hospital rule in the question and we can substitute the value of x to get a differential equation in x. We can solve this differential equation by following the method of exact differential equation. After this step, we get the function of $f\left( x \right)$ and an arbitrary constant will be present. By using the value of $f\left( \dfrac{\pi }{6} \right)=-\dfrac{\pi }{12}$, we get the value of c and the exact value of the function. Option A can be verified by substituting the value of $\dfrac{\pi }{4}$ in the function. Option B can be verified by using the property $\sin x>x-\dfrac{{{x}^{3}}}{6}$. Option C can be using Rolle's theorem for the given range. Option D can be verified by double differentiating and verifying.

Complete step-by-step answer:
We are given a limit $\displaystyle \lim_{t \to x}\dfrac{f\left( x \right)\sin t-f\left( t \right)\sin x}{t-x}={{\sin }^{2}}x\text{ }\forall \text{x}\in \left( 0,\pi \right)$.
We can infer that the substitution of t = x in the function, we will get the form of $\dfrac{0}{0}$ which is undetermined. To solve this, we use the L-Hospital rule.
The rule states that if the function in the limits $\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}$ which is $\dfrac{f\left( x \right)}{g\left( x \right)}$ when substituted the value of a which is $\dfrac{f\left( a \right)}{g\left( a \right)}$ leads to an undetermined form such as $\dfrac{0}{0},\dfrac{\infty }{\infty }$, we can write the limit as $\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{f'\left( x \right)}{g'\left( x \right)}$.
Using this rule by considering x as constant and differentiating with respect to t, we can write the limit as
$\begin{aligned} & \displaystyle \lim_{t \to x}\dfrac{f\left( x \right)\cos t-f'\left( t \right)\sin x}{1}={{\sin }^{2}}x\text{ } \\\ & f\left( x \right)\cos x-f'\left( x \right)\sin x={{\sin }^{2}}x \\\ \end{aligned}$
Let us divide by ${{\sin }^{2}}x$ in L.H.S and R.H.S and take a -1 common in the L.H.S, we get
$-\left( \dfrac{f'\left( x \right)\sin x-f\left( x \right)\cos x}{{{\sin }^{2}}x} \right)=1$
The L.H.S is of the form $d\left( \dfrac{f\left( x \right)}{\sin x} \right)$. Writing this in the above expression, we get
$d\left( \dfrac{f\left( x \right)}{\sin x} \right)=-1$
Integrating with respect to x, we get
$\begin{aligned} & \int{d\left( \dfrac{f\left( x \right)}{\sin x} \right)}=\int{-1} \\\ & \dfrac{f\left( x \right)}{\sin x}=-x+c \\\ \end{aligned}$
It is given in the question that $f\left( \dfrac{\pi }{6} \right)=-\dfrac{\pi }{12}$, substituting $\dfrac{\pi }{6}$ in the above equation, we get

& \dfrac{f\left( \dfrac{\pi }{6} \right)}{\sin \dfrac{\pi }{6}}=-\dfrac{\pi }{6}+c \\\ & \dfrac{-\dfrac{\pi }{12}}{\dfrac{1}{2}}=-\dfrac{\pi }{6}+c \\\ & -\dfrac{\pi }{6}=-\dfrac{\pi }{6}+c \\\ & c=0 \\\ \end{aligned}$$ So, the function $f\left( x \right)=-x\sin x$ Let us consider option-A $f\left( \dfrac{\pi }{4} \right)=-\dfrac{\pi }{4}\sin \dfrac{\pi }{4}=-\dfrac{\pi }{4\sqrt{2}}$ Option-A is wrong Let us consider option-B We know that $\sin x>x-\dfrac{{{x}^{3}}}{6}$ in the given range $\text{x}\in \left( 0,\pi \right)$ Multiplying by –x, we get $\begin{aligned} & -x\sin x<\dfrac{{{x}^{4}}}{6}-x \\\ & f\left( x \right)<\dfrac{{{x}^{4}}}{6}-x \\\ \end{aligned}$ For all $\text{x}\in \left( 0,\pi \right)$. Option B is correct. Let us consider option C. We can write that f(x) is continuous and differentiable in the interval $\text{x}\in \left( 0,\pi \right)$ $\begin{aligned} & f\left( 0 \right)=-0\sin 0=0 \\\ & f\left( \pi \right)=-\pi \sin \pi =0 \\\ \end{aligned}$ From the rolle's theorem which states that if a function f(x) is continuous and differentiable in an interval $\left( a,b \right)$ and $f\left( a \right)=f\left( b \right)$, we can write that there exist a value $c\in \left( a,b \right)$ which satisfies $f'\left( c \right)=0$. Here we can write that there exist an $\alpha $ where $f'\left( \alpha \right)=0$ where $\alpha \in \left( 0,\pi \right)$ Option C is correct Let us consider option D $\begin{aligned} & f\left( x \right)=-x\sin x \\\ & f'\left( x \right)=-\sin x-x\cos x \\\ & f''\left( x \right)=-\cos x-\cos x+x\sin x \\\ \end{aligned}$ Substituting $\dfrac{\pi }{2}$ in the required terms, we get $f''\left( \dfrac{\pi }{2} \right)+f\left( \dfrac{\pi }{2} \right)=\dfrac{\pi }{2}\sin \dfrac{\pi }{2}-2\cos \dfrac{\pi }{2}-\dfrac{\pi }{2}\sin \dfrac{\pi }{2}=0$ So, option D is also correct. **So, the correct answers are “Option B,C and D”.** **Note:** Students can make a mistake while applying the L-Hospital rule. While differentiating the numerator, they will differentiate the terms of x also and end up getting wrong terms. To avoid this, we should be clear that in the question, the dependent variable is t and the independent variable is x and we are differentiating with respect to t and we should keep x as constant.