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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Differentiation

NTA Abhyas 2020 · Mathematics · Differentiation

If f (x) = (x – 1) (x – 2)(x – 3)(x – 4)(x – 5) , then the value of...

If f(x)=(x–1)(x–2)(x–3)(x–4)(x–5)f \left(x\right) = \left(x – 1\right) \left(x – 2\right)\left(x – 3\right)\left(x – 4\right)\left(x – 5\right), then the value of f′(5)f^{′} \left(5\right) is equal to

Options

  1. A00
  2. B120120
  3. C2424
  4. D55

Correct answer

C. 2424

Step-by-step solution

f′(x)=(x–2)(x–3)(x–4)(x–5)+(x−1)(x−3)(x−4)(x−5){\text{f}}^{′} \left(\text{x}\right) = \left(\text{x} – 2\right)\left(\text{x} – 3\right)\left(\text{x} – 4\right)\left(\text{x} – 5\right) + \left(\text{x} - 1\right)\left(\text{x} - 3\right)\left(\text{x} - 4\right)\left(\text{x} - 5\right)
+......+(x−1)(x−2)(x−3)(x−4)+ . . . . . . + \left(x - 1\right)\left(x - 2\right)\left(x - 3\right)\left(x - 4\right)
Putting x=5x = 5, we get,
f′(5)=0+0+0+0+(5−1)(5−2)(5−3)(5−4){\text{f}}^{′} \left(5\right) = 0 + 0 + 0 + 0 + \left(5 - 1\right)\left(5 - 2\right)\left(5 - 3\right)\left(5 - 4\right)
f′(5)=24{\text{f}}^{′} \left(5\right) = 24
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