JEE MAIN - Mathematics (2023 - 6th April Evening Shift - No. 21)
The number of points, where the curve $$y=x^{5}-20 x^{3}+50 x+2$$ crosses the $$\mathrm{x}$$-axis, is ____________.
Answer
5
Explanation
Given equation of curve
$$ \begin{aligned} & y=x^5-20 x^3+50 x+2 \\\\ & \Rightarrow \frac{d y}{d x}=5 x^4-60 x^2+50 \end{aligned} $$
On putting $\frac{d y}{d x}=0$
$$ \begin{array}{ll} \Rightarrow & 5\left(x^4-12 x^2+10\right)=0 \\\\ \Rightarrow & x^2=\frac{12 \pm \sqrt{144-40}}{2}=6 \pm \sqrt{26} \\\\ \Rightarrow & x^2=6-\sqrt{26}, 6+\sqrt{26} \\\\ \Rightarrow & x^2=6-5.10,6+5.10 \\\\ \Rightarrow & x^2=09,11.1 \\\\ \Rightarrow & x= \pm \sqrt{0.9}, \pm \sqrt{11.1} \\\\ \Rightarrow & x=-0.95,0.95,-3.33,3.33 \end{array} $$
Now,
$$ \begin{aligned} y(0) & =2(+\mathrm{ve}) \Rightarrow y(1)=+\mathrm{ve} \\\\ y(2) & =-\mathrm{ve} \Rightarrow y(3.3)=-\mathrm{ve} \\\\ y(-1) & =-\mathrm{ve} \Rightarrow y(-2)=+\mathrm{ve} \\\\ y(-3.3) & =-\mathrm{ve} \end{aligned} $$
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$$ \because \text { Required number of points }=5 $$
$$ \begin{aligned} & y=x^5-20 x^3+50 x+2 \\\\ & \Rightarrow \frac{d y}{d x}=5 x^4-60 x^2+50 \end{aligned} $$
On putting $\frac{d y}{d x}=0$
$$ \begin{array}{ll} \Rightarrow & 5\left(x^4-12 x^2+10\right)=0 \\\\ \Rightarrow & x^2=\frac{12 \pm \sqrt{144-40}}{2}=6 \pm \sqrt{26} \\\\ \Rightarrow & x^2=6-\sqrt{26}, 6+\sqrt{26} \\\\ \Rightarrow & x^2=6-5.10,6+5.10 \\\\ \Rightarrow & x^2=09,11.1 \\\\ \Rightarrow & x= \pm \sqrt{0.9}, \pm \sqrt{11.1} \\\\ \Rightarrow & x=-0.95,0.95,-3.33,3.33 \end{array} $$
Now,
$$ \begin{aligned} y(0) & =2(+\mathrm{ve}) \Rightarrow y(1)=+\mathrm{ve} \\\\ y(2) & =-\mathrm{ve} \Rightarrow y(3.3)=-\mathrm{ve} \\\\ y(-1) & =-\mathrm{ve} \Rightarrow y(-2)=+\mathrm{ve} \\\\ y(-3.3) & =-\mathrm{ve} \end{aligned} $$
$$ \because \text { Required number of points }=5 $$
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