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JEE Main 20236 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
6 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2023 · 6 April · Evening Shift · Mathematics · Application of Derivatives

The number of points, where the curve y = x ^ 5 - 20 x ^ 3 + 50x +...

The number of points, where the curve y=x5−20x3+50x+2y = {x}^{5} - 20{x}^{3} + 50x + 2 crosses the xx-axis, is _____.

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

Correct answer

0

Step-by-step solution

Let, f(x)=x5−20x3+50x+2f ( x ) = {x}^{5} - 20{x}^{3} + 50x + 2

Now differentiating the function ff\left(x\right)withrespecttowith respect toxweget,</p><p>we get,</p><p>{f}^{'} ( x ) = 5{x}^{4} - 60{x}^{2} + 50</p><p></p><p>⇒ {f}^{'} ( x ) = 5\left({x}^{4} - 12{x}^{2} + 10\right)</p><p>Nowequating</p><p>Now equating{f}^{'} ( x ) = 0weget,</p><p>we get,</p><p>{x}^{4} - 12{x}^{2} + 10 = 0</p><p></p><p>⇒ {x}^{2} = \dfrac{12 \pm \sqrt{144 - 40}}{2}</p><p></p><p>⇒ {x}^{2} = 6 \pm \sqrt{26}</p><p></p><p>⇒ {x}^{2} \approx 11 . 1, 0 . 9</p><p></p><p>⇒ x \approx \pm 3 . 31, \pm 0 . 95</p><p>Nowfindingthenatureoffunctionatintegervaluenearbyabovevalueweget,</p><p></p><p>Now finding the nature of function at integer value near by above value we get,</p><p>f\left(0\right) = 2, f\left(1\right) > 0, f\left(2\right) 0</p><p>Nowplottingthegraphbyobservingabovevalues,weget,</p><p><imgsrc="https://cdn−question−pool.getmarks.app/pyq/jeemain/4bbf06d4−950b−4ae8−99cf−74634038d164.png"style="width:299px;height:146px;"/></p><p>Hence,itwillcross</p><p>Now plotting the graph by observing above values, we get,</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/4bbf06d4-950b-4ae8-99cf-74634038d164.png" style="width: 299px; height: 146px;" /></p><p>Hence, it will crossx -axisaxis5 times.

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