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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

The number of points, where the curve \(y=x^5-20 x^3+50 x+2\) crosses the \(x\)-axis, is \(............\).

  1. A \(4\)
  2. B \(3\)
  3. C \(5\)
  4. D \(1\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(5\)

Step-by-step Solution

Detailed explanation

\(y=x^5-20 x^3+50 x+2\) \(\frac{d y}{d x}=5 x^4-60 x^2+50=5\left(x^4-12 x^2+10\right)\) \(\frac{d y}{d x}=0 \Rightarrow x^4-12 x^2+10=0\) \(\Rightarrow x^2=\frac{12 \pm \sqrt{144-40}}{2}\) \(\Rightarrow x^2=6 \pm \sqrt{26} \Rightarrow x^2 \approx 6 \pm 5.1\)…