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}
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\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)}
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}
Given:
f(x) = \cos\left([\pi^2]x\right) + \cos\left([-\pi^2]x\right)
Find: f\left(\frac{\pi}{2}\right)
\pi^2 \approx 9.8696 \Rightarrow [\pi^2] = 9,\quad [-\pi^2] = -10
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)
\cos\left(\frac{9\pi}{2}\right) = 0,\quad \cos(-5\pi) = -1
\boxed{-1}
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