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The number of distinct real roots of $x^{4}-4 x+1=0 \text { is :$
Option: 1
$4$

Option: 2
$2$

Option: 3
$1$

Option: 4
$0$

Answers (1)

best_answer

$\mathrm{f(x) =x^{4}-4 x+1} \\$

$\mathrm{f^{\prime}(x) =4 x^{3}-4=0} \\$

$\mathrm{\Rightarrow x=1}$

$\mathrm{f^{\prime \prime}(x)=12 x^{2}, \quad f^{\prime \prime}(1): 12>0}$

$\mathrm{\Rightarrow x= 1}$ is a poit of minima

$\mathrm{f(1): 1-4+1=-2 }$

For $\mathrm{x<1, \quad f(n) }$ is decreasing and for $\mathrm{x>1, \quad f(x)}$ is in increasing

$\mathrm{f(-\infty)=\infty, \quad f(1)=-2, \quad f(\infty)=\infty }$

From intermediate value theorem , $\mathrm{f(x) }$ will have 2 real roots one less than 1 and other greater than 1.

Hence the correct answer is option 2

Posted by

Divya Prakash Singh

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