JEE Main 2023 — Application of Derivatives Question with Solution

Let, $f\left(x\right)=x^5-20x^3+50x+2$

Now differentiating the function $f\left(x\right)$ with respect to $x$ we get,

 $f^{\text{'}}\left(x\right)=5x^4-60x^2+50$

$\Rightarrow f^{\text{'}}\left(x\right)=5\left(x^4-12x^2+10\right)$

Now equating $f^{\text{'}}\left(x\right)=0$ we get,

$x^4-12x^2+10=0$

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

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

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

$\Rightarrow x\approx \pm 3.31, \pm 0.95$

Now finding the nature of function at integer value near by above value we get,

$f\left(0\right)=2, f\left(1\right)>0, f\left(2\right)<0, f\left(-1\right)<0 \& f\left(-2\right)>0$

Now plotting the graph by observing above values, we get,

Hence, it will cross $x-$axis $5$ times.