ISI B.Stat-B.Math Objective Problems Discussion (problem 14)

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)