Question

If $f\left(x\right)=x^5-5x^4+5x^3-10$ has local maxima and minima at $x=p$ and $x=q$, then $p+q$ is ?

• A. $1$
• B. $4$
• C. $3$
• D. None of the above.

Answer 1

The correct option is B $4$

The given equation is:

$f\left(x\right)=x^5-5x^4+5x^3-10$

To find the extremum points we differentiate and equate it to zero

$\Rightarrow f^{\prime }\left(x\right)=5x^4-20x^3+15x^2$

$f^{\prime }\left(x\right)=0$

$\Rightarrow 5x^4-20x^3+15x^2=0$

$\Rightarrow 5x^2(x^2-4x+3)=0$

$\Rightarrow x=0$

$\Rightarrow x^2-4x+3=0$

$\Rightarrow (x-3)(x-1)=0$

$\Rightarrow x=3,1$

Now to find whether at the critical points we find a maxima or minima we use the second derivative test.

$\Rightarrow f^{\prime \prime }\left(x\right)=20x^3-60x^2+30x$

$\Rightarrow f^{\prime \prime }\left(0\right)=0$

$\Rightarrow f^{\prime \prime }\left(3\right)=20\times 3^3-60\times 3^2+30\times 3$

$\Rightarrow f^{\prime \prime }\left(3\right)=1170$

$\Rightarrow f^{\prime \prime }\left(3\right)>0$

$\Rightarrow f^{\prime \prime }\left(1\right)=20\times 1^3-60\times 1^2+30\times 1$

$\Rightarrow f^{\prime \prime }\left(1\right)=-10$

$\Rightarrow f^{\prime \prime }\left(1\right)<0$

Hence we get a minima at x = 3 and a amxima at x =1. Thus,

$p=1$ and $q=3$

$\Rightarrow p+q=4$