Question

$f\left(x\right)=x^4-10x^3+35x^2-50x+c$ is a constant. the number of real roots of . f (x) = 0 and
f'' (x) = 0 are respectively

• A. 1 , 0
• B. 3, 2
• C. 1 , 2
• D. 3 , 0

Answer 1

The correct option is A 1 , 0

Given, $f\left(x\right)=x^4-10x^3+35x^2-50x+c$

Then $,f^{\prime }\left(x\right)=\frac{d}{dx}(-10x^3+x^4+35x^2-50x+c)$

$\Rightarrow f^{\prime }\left(x\right)=4x^3-30x^2+70x-50=0$

$\therefore 2x^3-15x^2+35x-25=0$

First, let us factorise $f\left(x\right)=0$

$\Rightarrow x^4-10x^3+35x^2-50x+c=0$

$\Rightarrow (x-1)(x^3-9x^2+26x-c)=0$

$\Rightarrow (x-1)\left(\right(x-2\left)(x^2-7x+\frac{c}{2})\right)=0$

$\Rightarrow f\left(x\right)=(x-1)(x-2)(x-4)(x-3)$

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

$\Rightarrow 2x^3-15x^2+35x-25=0$

$\Rightarrow f^{\prime }\left(x\right)$ has 3 roots which are

$x=\frac{5}{2},\frac{5+\sqrt{5}}{2},\frac{-\sqrt{5}+5}{2}$

Hence $\left(A\right)$ is the correct option.