Limit Continuity And Differentiability Ques 129
- Let $f, g:[-1,2] \rightarrow R$ be continuous functions which are twice differentiable on the interval $(-1,2)$. Let the values of $f$ and $g$ at the points $-1,0$ and 2 be as given in the following table:
| $x=-1$ | $x=0$ | $x=2$ | |
|---|---|---|---|
| $f(x)$ | 3 | 6 | 0 |
| $g(x)$ | 0 | 1 | -1 |
In each of the intervals $(-1,0)$ and $(0,2)$, the function $(f-3 g)^{\prime \prime}$ does not vanish. Then, the correct statement(s) is/are
(2015 Adv.) (a) $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup(0,2)$
(b) $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$
(c) $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(0,2)$
(d) $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$
Show Answer
Answer:
Correct Answer: 129.$f^{\prime}\left(0^{+}\right)=0, f^{\prime}\left(0^{-}\right)=1$
Solution:
- Given, $y=\left(\log _{\cos x} \sin x\right) \cdot\left(\log _{\sin x} \cos x\right)^{-1}+\sin ^{-1} \frac{2 x}{1+x^{2}}$
$$ \begin{alignedat} & \therefore \quad y={\frac{\log _e(\sin x)}{\log _e(\cos x)}}^{2}+\sin ^{-1} \frac{2 x}{1+x^{2}} \\ & \left(\log _e(\cos x) \cdot \cot x\right. \ & \Rightarrow \quad \frac{d y}{d x}=2 \frac{\log _e(\sin x)}{\log _e(\cos x)} \cdot \frac{\left.+\log _e(\sin x) \cdot \tan x\right)}{{\log _e(\cos x) }^{2}}+\frac{2}{1+x^{2}} \end{aligned} $$
$$ \begin{alignedat} \Rightarrow \frac{d y}{d x} \int _{x=\frac{\pi}{4}}^{x} & =21 \cdot \frac{2 \cdot \log \frac{1}{\sqrt{2}}}{\log \frac{1}{\sqrt{2}}}+\frac{2}{1+\frac{\pi^{2}}{16}} \\ & =-\frac{8}{\log _e 2}+\frac{32}{16+\pi^{2}} \end{aligned} $$
BETA
