  Problem 14

f(x) = tan(sinx) (x > 0)

To understand the graph of a function, easiest and the most proper method is to apply techniques from calculus. We will quickly compute, derivative and second derivative and try to understand,extreme points and convexity of the curve. $f'(x) = cos (x) sec^2 (sin (x))$

Hence when cos(x) is positive (correspondingly negative), derivative is positive (is negative). Therefore from $(0, \frac{\pi}{2} )$ function is increasing, $( \frac{\pi}{2} , \frac{3\pi}{2} )$ the function is decreasing. Also it has critical points on cos (x) is 0 (at $x = \frac{\pi}{2} , \frac{3\pi}{2}$ )

Now we compute the second derivative. $f''(x) = - \sec^2 (\sin(x)) (\sin(x) - 2 \cos^2(x) \tan (\sin(x)))$

At $x = \frac{\pi}{2}$ second derivative is $-\sec^2 1$ [hence we have a maxima] and at $x = \frac{3\pi}{2}$ second derivative is $\sec^2 1$ [hence we have a minima]

Finally we compute $f(\frac{\pi}{2} ) = \tan 1 > 1 , f(\frac{3\pi}{2} ) = - \tan 1 < -1$.

Moreover the function is differentiable at the points of maxima and minima. Hence answer is (B)