Correct Shortcut Method — Find \( f(5) \)

Step 1: Define a helper polynomial:

\[ g(x) = f(x) - (x + 1) \]

Given: \( f(1) = 2, f(2) = 3, f(3) = 4, f(4) = 5 \Rightarrow g(1) = g(2) = g(3) = g(4) = 0 \)

So, \[ g(x) = A(x - 1)(x - 2)(x - 3)(x - 4) \quad \Rightarrow \quad f(x) = A(x - 1)(x - 2)(x - 3)(x - 4) + (x + 1) \]

Step 2: Use \( f(0) = 25 \) to find A:

\[ f(0) = A(-1)(-2)(-3)(-4) + (0 + 1) = 24A + 1 = 25 \Rightarrow A = 1 \]

Step 3: Compute \( f(5) \):

\[ f(5) = (5 - 1)(5 - 2)(5 - 3)(5 - 4) + (5 + 1) = 4 \cdot 3 \cdot 2 \cdot 1 + 6 = 24 + 6 = \boxed{30} \]

✅ Final Answer:   \( \boxed{f(5) = 30} \)

Maximum Value of \( f(x) = (x - 1)^2(x + 1)^3 \)

Step 1: Let’s define the function:

\[ f(x) = (x - 1)^2 (x + 1)^3 \]

Step 2: Take derivative to find critical points

Use product rule:
Let \( u = (x - 1)^2 \), \( v = (x + 1)^3 \)
\[ f'(x) = u'v + uv' = 2(x - 1)(x + 1)^3 + (x - 1)^2 \cdot 3(x + 1)^2 \] \[ f'(x) = (x - 1)(x + 1)^2 [2(x + 1) + 3(x - 1)] \] \[ f'(x) = (x - 1)(x + 1)^2 (5x - 1) \]

Step 3: Find critical points

Set \( f'(x) = 0 \): \[ (x - 1)(x + 1)^2 (5x - 1) = 0 \Rightarrow x = 1,\ -1,\ \frac{1}{5} \]

Step 4: Evaluate \( f(x) \) at these points

• \( f(1) = 0 \)
• \( f(-1) = 0 \)
• \( f\left(\frac{1}{5}\right) = \left(\frac{1}{5} - 1\right)^2 \left(\frac{1}{5} + 1\right)^3 = \left(-\frac{4}{5}\right)^2 \left(\frac{6}{5}\right)^3 \)

\[ f\left(\frac{1}{5}\right) = \frac{16}{25} \cdot \frac{216}{125} = \frac{3456}{3125} \]

Step 5: Compare with given form:

It is given that maximum value is \( \frac{3456}{3125} = 2^p \cdot 3^q / 3125 \)

Factor 3456: \[ 3456 = 2^7 \cdot 3^3 \Rightarrow \text{So } p = 7, \quad q = 3 \]

✅ Final Answer:   \( \boxed{(p, q) = (7,\ 3)} \)

Given: A one-one function from set $\{1,2,3\}$ to set $\{a,b,c,d,e\}$

Step 1: One-one (injective) function means no two elements map to the same output.

We choose 3 different elements from 5 and assign them to 3 inputs in order.

So, total one-one functions = $P(5,3) = 5 \times 4 \times 3 = 60$

✅ Final Answer: $\boxed{60}$

Given:
$$f\left(\frac{1 - x}{1 + x}\right) = x + 2$$

To Find: \( f(1) \)

Let \( \frac{1 - x}{1 + x} = 1 \Rightarrow x = 0 \)

Then, \( f(1) = f\left(\frac{1 - 0}{1 + 0}\right) = 0 + 2 = 2 \)

Answer: $$\boxed{2}$$

? Function with Greatest Integer and Cosine

Given:

\[ f(x) = \cos\left([\pi^2]x\right) + \cos\left([-\pi^2]x\right) \]

Find: \[ f\left(\frac{\pi}{2}\right) \]

Step 1: Estimate Floor Values

\[ \pi^2 \approx 9.8696 \Rightarrow [\pi^2] = 9,\quad [-\pi^2] = -10 \]

Step 2: Plug into the Function

\[ f\left(\frac{\pi}{2}\right) = \cos\left(9 \cdot \frac{\pi}{2}\right) + \cos\left(-10 \cdot \frac{\pi}{2}\right) = \cos\left(\frac{9\pi}{2}\right) + \cos(-5\pi) \]

Step 3: Simplify

\[ \cos\left(\frac{9\pi}{2}\right) = 0,\quad \cos(-5\pi) = -1 \]

✅ Final Answer:

\[ \boxed{-1} \]