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)