Question

Which of the following statements are CORRECT ?
I. If $f$ is differentiable in $(-\infty ,\infty )$ and ${lim}_{x\to -\infty }f\left(x\right)=-\infty$ and ${lim}_{x\to \infty }f\left(x\right)=1,$ then the graph of $y=f\left(x\right)$ must cross x-axis.
II. $\tan ({\sin }^{-1}\left(\frac{-4}{5}\right)-{\cos }^{-1}\left(\frac{-5}{13}\right))$ is equal to $\frac{16}{63}.$
III. Suppose that $f\left(x\right)$ is a rational function of $x$ and that $f\left(0\right)<0$ and $f\left(2\right)>0$ then there must exist a value of $x$ between 0 and 2 where $f\left(x\right)=0$

• A. I and II
• B. II and III
• C. III and I
• D. I , II and III

Answer 1

The correct option is A I and II

I. Given f is differentiable in $(-\infty ,\infty )$

$\to f$ is a continuous curve in $(-\infty ,\infty )$

Since f changes sign from $-\infty$ to $1$ and it is continuous it must pass through the x-axis

II. $\tan ({\sin }^{-1}\left(\frac{-4}{5}\right)-{\cos }^{-1}\left(\frac{-5}{13}\right))$

$=\tan ({\tan }^{-1}\left(\frac{-4}{3}\right)-{\tan }^{-1}\left(\frac{-12}{5}\right))$

$=\tan \left({\tan }^{-1}\left(\frac{-\frac{4}{3}+\frac{12}{5}}{1+\frac{4}{3}\times \frac{12}{5}}\right)\right)$

$=\tan \left({\tan }^{-1}\frac{\frac{36-20}{15}}{\frac{15+48}{15}}\right)$

$=\tan \left({\tan }^{-1}\left(\frac{16}{63}\right)\right)=\frac{16}{63}$

III. $f\left(x\right)$ is a rational function of x

$\Rightarrow f\left(x\right)$ can be discontinuous

Even though f changes sign .

$f\left(x\right)=0$may not exist due to its discontinuity.