Question:
The function $\mathrm{f}(\mathrm{x})=2 x^{3}-15 x^{2}+36 x+4$ is maximum at $\mathrm{x}=$
(a) 3
(b) 0
(c) 4
(d) 2
Solution:
(d) 2
Given : $f(x)=2 x^{3}-15 x^{2}+36 x+4$
$\Rightarrow f^{\prime}(x)=6 x^{2}-30 x+36$
For a local maxima or a local minima, we must have
$f^{\prime}(x)=0$
$\Rightarrow 6 x^{2}-30 x+36=0$
$\Rightarrow x^{2}-5 x+6=0$
$\Rightarrow(x-2)(x-3)=0$
$\Rightarrow x=2,3$
Now,
$f^{\prime \prime}(x)=12 x-30$
$\Rightarrow f^{\prime \prime}(2)=24-30=-6<0$
So, $x=1$ is a local maxima.
Also,
$f^{\prime \prime}(3)=36-30=6>0$
So, $x=2$ is a local maxima.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.