MCQ
The function $\frac{{\sin \,\,(x\, + \,a)}}{{\sin \,\,(x\, + \,b)}}$ has no maxima or minima if
- A$b - a = n \pi , n \in I$
- B$b - a = (2n + 1) \pi , n \in I$
- C$b - a = 2n \pi , n \in I$
- ✓All of these .
$ f '(x) =\,\frac{{\sin (x + b)\, \times \,\cos (x + a)\, - \,\sin (x + a)\,\cos (x + b)}}{{{{\sin }^2}(x + b)}}\, =\,\frac{{\sin (b - a)}}{{{{\sin }^2}(x + b)}}\,$
If $sin(b - a) = 0$ then $f' (x) = 0$ ==> $f (x)$ will be constant
i.e. $b - a = n\pi$ or $n\pi$
or $b - a = (2n + 1)\pi$ or $b - a = 2n\pi$
then $f (x)$ has no minima
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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 $(0,2)$