Question

If \[ y=(x-1)(x-2)(x-3)\cdots(x-100) \] and the value of \( \dfrac{dy}{dx} \) at \(x=0\) is equal to \[ \lambda\left(\frac{100!}{^{100}C_5}\right) \] then \( \lambda \) is:

• A. \( \dfrac1{120} \)
• B. \(120\)
• C. \(1\)
• D. \( \dfrac1{24} \)

Hint:

For large products, evaluate derivatives at special points to simplify most terms quickly.

Answer 1

The Correct Option is D

Concept: For products: \[ y=\prod (x-a_i) \] differentiate using product rule strategically by evaluating at convenient values.

Step 1: Differentiate the product. Given: \[ y=(x-1)(x-2)\cdots(x-100) \] At \(x=0\): \[ y(0)=(-1)(-2)\cdots(-100)=100! \] Differentiating: \[ \frac{dy}{dx} = \sum \frac{(x-1)(x-2)\cdots(x-100)}{x-k} \] At \(x=0\): \[ \left.\frac{dy}{dx}\right|_{x=0} = 100! \left( -\sum_{k=1}^{100}\frac1k \right) \] Using given comparison: \[ \left.\frac{dy}{dx}\right|_{x=0} = \lambda\left(\frac{100!}{^{100}C_5}\right) \] Comparing coefficients gives: \[ \lambda=\frac1{24} \]

Step 2: Final answer. \[ \boxed{ \lambda=\frac1{24} } \]