Let `f(x)=sin x,g(x)=[x+1] and h(x)=gof(x)` where [.] the greatest integer function. Then `h'((pi)/(2))` is

Options

A. 1
B. `-1`
C. non-existent
D. none of these

Explanation

To find \( h'(\frac{\pi}{2}) \) where \( h(x) = g(f(x)) \), we first need to understand the functions involved.

1. **Define the Functions**:
- \( f(x) = \sin x \)
- \( g(x) = [x + 1] \), where \([.]\) denotes the greatest integer function.
- Thus, \( h(x) = g(f(x)) = g(\sin x) = [\sin x + 1] \).

2. **Evaluate \( h(\frac{\pi}{2}) \)**:
- Calculate \( f(\frac{\pi}{2}) \):
\[ f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \]
- Now, substitute into \( g \):
\[ h\left(\frac{\pi}{2}\right) = g(f(\frac{\pi}{2})) = g(1) = [1 + 1] = [2] = 2 \]

3. **Determine the Left-Hand Derivative (LHD)**:
- The left-hand derivative at \( x = \frac{\pi}{2} \) is given by:
\[ \text{LHD} = \lim_{h \to 0^-} \frac{h\left(\frac{\pi}{2} + h\right) - h\left(\frac{\pi}{2}\right)}{h} \]
- For \( x < \frac{\pi}{2} \), \( \sin x < 1 \) so \( h(x) = [\sin x + 1] = 1 \).
- Thus, we have:
\[ \text{LHD} = \lim_{h \to 0^-} \frac{1 - 2}{h} = \lim_{h \to 0^-} \frac{-1}{h} \]
- As \( h \to 0^- \), this approaches \( -\infty \).

4. **Determine the Right-Hand Derivative (RHD)**:
- The right-hand derivative at \( x = \frac{\pi}{2} \) is given by:
\[ \text{RHD} = \lim_{h \to 0^+} \frac{h\left(\frac{\pi}{2} + h\right) - h\left(\frac{\pi}{2}\right)}{h} \]
- For \( x > \frac{\pi}{2} \), \( \sin x > 1 \) so \( h(x) = [\sin x + 1] = 2 \).
- Thus, we have:
\[ \text{RHD} = \lim_{h \to 0^+} \frac{2 - 2}{h} = \lim_{h \to 0^+} \frac{0}{h} = 0 \]

5. **Conclusion**:
- Since LHD \(\neq\) RHD, \( h'(\frac{\pi}{2}) \) does not exist.

Thus, the final answer is that \( h'(\frac{\pi}{2}) \) is **non-existent**.