Let $h(x) = min[\sin x], [\cos x]]$, for all real numbers $x$. Let S be the set of points in $(0, \frac{\pi}{2})$ where $h(x)$ is not differentiable. Then the cardinality of S is:

The notation $[\sin x]$ and $[\cos x]$ appears to represent the floor function. However, analyzing this in $(0, \frac{\pi}{2})$ would give a constant function. Given the correct answer is 1, the problem likely means:

$h(x) = \min\{\sin x, \cos x\}$

We are following what was given in the actual past-year-paper released by IIM Bangalore.

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In the interval $(0, \frac{\pi}{2})$, both $\sin x$ and $\cos x$ are positive and continuous.

To find where $h(x)$ might not be differentiable, we need to find where the two functions intersect (where the minimum switches from one function to the other).

Setting $\sin x = \cos x$:

$\tan x = 1$

$x = \frac{\pi}{4}$

This is the only point in $(0, \frac{\pi}{2})$ where $\sin x = \cos x$.

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Now let's check the behavior of $h(x)$:

Interval	Comparison	$h(x)$ equals
$0 < x < \frac{\pi}{4}$	$\sin x < \cos x$	$\sin x$
$x = \frac{\pi}{4}$	$\sin x = \cos x$	both (equal)
$\frac{\pi}{4} < x < \frac{\pi}{2}$	$\sin x > \cos x$	$\cos x$

At $x = \frac{\pi}{4}$:

• Left-hand derivative: Since $h(x) = \sin x$ for $x < \frac{\pi}{4}$, the derivative is $\cos x$, giving $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$
• Right-hand derivative: Since $h(x) = \cos x$ for $x > \frac{\pi}{4}$, the derivative is $-\sin x$, giving $-\sin(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$

Since $\frac{1}{\sqrt{2}} \neq -\frac{1}{\sqrt{2}}$, the function is not differentiable at $x = \frac{\pi}{4}$.

This is the only such point in the given interval.

Answer: Option 4 (cardinality = 1)