Question 35 of 70intermediate🔧 ApplyLong Answer3 marks

For the function f(x)=x100100+x9999+…+x22+x+1f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \ldots + \frac{x^2}{2} + x + 1 Prove that f′(1)=100f′(0)f'(1) = 100f'(0).

Correct Answer

Given:

f(x)=x100100+x9999+…+x22+x+1f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \ldots + \frac{x^2}{2} + x + 1

Step 1: Find f′(x)f'(x)

By Theorem 6, the derivative of xnx^n is nxn−1nx^{n-1}. Therefore:

f′(x)=ddx(x100100)+ddx(x9999)+…+ddx(x22)+ddx(x)+ddx(1)f'(x) = \frac{d}{dx}\left(\frac{x^{100}}{100}\right) + \frac{d}{dx}\left(\frac{x^{99}}{99}\right) + \ldots + \frac{d}{dx}\left(\frac{x^2}{2}\right) + \frac{d}{dx}(x) + \frac{d}{dx}(1) f′(x)=x99+x98+…+x+1+0f'(x) = x^{99} + x^{98} + \ldots + x + 1 + 0 f′(x)=1+x+x2+…+x99f'(x) = 1 + x + x^2 + \ldots + x^{99}

Step 2: Find f′(0)f'(0)

f′(0)=1+0+0+…+0=1f'(0) = 1 + 0 + 0 + \ldots + 0 = \mathbf{1}

Step 3: Find f′(1)f'(1)

f′(1)=1+1+1+…+1 (100 terms)=100f'(1) = 1 + 1 + 1 + \ldots + 1 \text{ (100 terms)} = \mathbf{100}

Step 4: Verify the relation

f′(1)=100f'(1) = 100 100f′(0)=100×1=100100f'(0) = 100 \times 1 = 100

∴ Therefore, f′(1)=100f′(0)f'(1) = 100f'(0) is proved.

Exercise: EXERCISE 12.2 | Q: 5 | (Page Page 32)
For More Understanding

Explanation

This question applies Theorem 6 (derivative of xnx^n is nxn−1nx^{n-1}) to find the derivative term by term.

The key insight is recognizing that when differentiating each term xnn\frac{x^n}{n}, the result is xn−1x^{n-1}. At x=0x = 0, all terms vanish except the constant 11. At x=1x = 1, all 100 terms equal 11, giving the sum 100100.

The relation follows directly.

Solution Steps

  1. Step 1: Apply Theorem 6 to find f′(x)=1+x+x2+…+x99f'(x) = 1 + x + x^2 + \ldots + x^{99}

  2. Step 2: Evaluate f′(0)=1f'(0) = 1

  3. Step 3: Evaluate f′(1)=100f'(1) = 100

  4. Step 4: Verify f′(1)=100=100×1=100f′(0)f'(1) = 100 = 100 \times 1 = 100f'(0)

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