The number of distinct real roots of x^4-4 x^ | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$f(x)=x^4-4 x^3+12 x^2+x-1$

Let $f(x)$ has four distinct real roots

$\Rightarrow f^{\prime}(x)=4 x^3-12 x^2+24 x+1$

$f ^{\prime}( x )$ has th

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$f(x)=x^4-4 x^3+12 x^2+x-1$

Let $f(x)$ has four distinct real roots

$\Rightarrow f^{\prime}(x)=4 x^3-12 x^2+24 x+1$

$f ^{\prime}( x )$ has three distinct real roots

$f ^{\prime \prime}( x )=12 x ^2-24 x +24=12\left( x ^2-2 x +2\right) $

$D =-4 < 0$

$f ^{\prime \prime}$ (x) cannot have 2 real solutions.

So, $f ( x )$ cannot have four real distinct roots

It can have 2 or o real roots.

$f(0)=-1, \quad f(1)=9$

$\Rightarrow$ At least one real solution

So, 2 real distinct solutions.

The number of distinct real roots of $x^4-4 x^3+12 x^2+x-1=0$ is

Similar Questions

If $x$ is a solution of the equation, $\sqrt {2x + 1} - \sqrt {2x - 1} = 1, \left( {x \ge \frac{1}{2}} \right)$ , then $\sqrt {4{x^2} - 1} $ is equal to

If the expression $\left( {mx - 1 + \frac{1}{x}} \right)$ is always non-negative, then the minimum value of m must be

The number of real values of $x$ for which the equality $\left| {\,3{x^2} + 12x + 6\,} \right| = 5x + 16$ holds good is

If the equation $\frac{{{x^2} + 5}}{2} = x - 2\cos \left( {ax + b} \right)$ has atleast one solution, then $(b + a)$ can be equal to

The number of real solutions of the equation $|{x^2} + 4x + 3| + 2x + 5 = 0 $are