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:

$$\boxed{-1}$$