Number of points, where curve y=x^5-20 x^3+50 x+2 crosses x -axis, is ............ .

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5. Continuity and Differentiation

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

A

$4$

B

$3$

C

$5$

D

$1$

(JEE MAIN-2023)

Solution

$y=x^5-20 x^3+50 x+2$

$\frac{d y}{d x}=5 x^4-60 x^2+50=5\left(x^4-12 x^2+10\right)$

$\frac{d y}{d x}=0 \Rightarrow x^4-12 x^2+10=0$

$\Rightarrow x^2=\frac{12 \pm \sqrt{144-40}}{2}$

$\Rightarrow x^2=6 \pm \sqrt{26} \Rightarrow x^2 \approx 6 \pm 5.1$

$\Rightarrow x^2 \approx 11.1,0.9$

$\Rightarrow x \approx \pm 3.3, \pm 0.95$

$f(0)=2, f(1)=+v e, f(2)=-ve$

$f(-1)=-v e, f(-2)=+v e$

Standard 12

Mathematics

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