Question

Let $f(x) = \dfrac{2x+3}{x-2}, \, x \in \mathbb{R}, \, x \neq 2$ and $h(x) = f(f(x))$. Then $h(h(10))$ is equal to:

• A. $100$
• B. $20$
• C. $10$
• D. $1000$
• E. $1$

Hint:

If repeated function application gives same value (fixed point), further compositions give the same result.

Answer 1

The Correct Option is C

Concept:
• Composition of functions: $h(x) = f(f(x))$
• Evaluate step-by-step instead of expanding fully

Step 1: Find $f(10)$
\[ f(10) = \frac{2(10)+3}{10-2} = \frac{20+3}{8} = \frac{23}{8} \]

Step 2: Find $h(10) = f(f(10)) = f\left(\frac{23}{8}\right)$
\[ f\left(\frac{23}{8}\right) = \frac{2\cdot \frac{23}{8} + 3}{\frac{23}{8} - 2} = \frac{\frac{46}{8} + \frac{24}{8}}{\frac{23}{8} - \frac{16}{8}} = \frac{\frac{70}{8}}{\frac{7}{8}} = 10 \]

Step 3: Find $h(h(10)) = h(10)$ again
\[ h(10) = 10 \Rightarrow h(h(10)) = h(10) = 10 \] Final Conclusion:
\[ h(h(10)) = 10 \]