MCQ
${d \over {dx}}{\tan ^{ - 1}}(\sec x + \tan x) = $
- A$1$
- ✓$1/2$
- C$\cos x$
- D$\sec x$
$ = \frac{d}{{dx}}{\tan ^{ - 1}}\left( {\frac{{\sin \left( {\frac{x}{2}} \right) + \cos \left( {\frac{x}{2}} \right)}}{{\cos \left( {\frac{x}{2}} \right) - \sin \left( {\frac{x}{2}} \right)}}} \right) = \frac{d}{{dx}}\left( {\frac{\pi }{4} + \frac{x}{2}} \right) = \frac{1}{2}$.
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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 $(f-3 g)^{\prime \prime}$ never vanishes. Then the correct statement(s) is(are)
$(A)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ 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 $(0,2)$