Limit Continuity and Differentiability 7 Question 27
27. Let $f:(0, \pi) \rightarrow R$ be a twice differentiable function such that $\lim _{t \rightarrow x} \frac{f(x) \sin t-f(t) \sin x}{t-x}=\sin ^{2} x$ for all $x \in(0, \pi)$.
If $f \frac{\pi}{6}=-\frac{\pi}{12}$, then which of the following statement(s) is (are) TRUE?
(2018 Adv.)
(a) $f \frac{\pi}{4}=\frac{\pi}{4 \sqrt{2}}$
(b) $f(x)<\frac{x^{4}}{6}-x^{2}$ for all $x \in(0, \pi)$
(c) There exists $\alpha \in(0, \pi)$ such that $f^{\prime}(\alpha)=0$
(d) $f^{\prime \prime} \frac{\pi}{2}+f \frac{\pi}{2}=0$
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Answer:
Correct Answer: 27. (a, c, d)
Solution:
- Given, $y=f \frac{2 x-1}{x^{2}+1}$
and $f^{\prime}(x)=\sin ^{2} x$
$$ \begin{aligned} \therefore \quad \frac{d y}{d x} & =f^{\prime} \frac{2 x-1}{x^{2}+1} \cdot \frac{d}{d x} \frac{2 x-1}{x^{2}+1} \\ & =\sin ^{2} \frac{2 x-1}{x^{2}+1} \cdot \frac{\left(x^{2}+1\right) \cdot 2-(2 x-1)(2 x)}{\left(x^{2}+1\right)^{2}} \end{aligned} $$
$$ \begin{aligned} & =\sin ^{2} \frac{2 x-1}{x^{2}+1} \cdot \frac{-2 x^{2}+2 x+2}{\left(x^{2}+1\right)^{2}} \\ & =\frac{-2\left(x^{2}-x-1\right)}{\left(x^{2}+1\right)^{2}} \sin ^{2} \frac{2 x-1}{x^{2}+1} \end{aligned} $$
