QJEEM23Ma4002: JEE Main 2023 Mathematics - Class 12 JEEM23Ma4002 Solution: The number of points, where the curve crosses the -axis, is…

Step-by-Step Verified Solution

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,

$\begin{aligned}y(0) & =2(+\mathrm{ve}) \Rightarrow y(1)=+\mathrm{ve} \\\\y(2) & =-\mathrm{ve} \Rightarrow y(3.3)=-\mathrm{ve} \\\\y(-1) & =-\mathrm{ve} \Rightarrow y(-2)=+\mathrm{ve} \\\\y(-3.3) & =-\mathrm{ve}\end{aligned}$


$\because \text { Required number of points }=5$