JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Application of Derivatives

Let $f(x)=x^4-4x+1$. Then $f^{\prime }(x)=4x^3-4=4(x^3-1)$.

Critical point: $f^{\prime }(x)=0\Rightarrow x=1$.

Evaluating:

• $f(1)=1-4+1=-2<0$
• $f(0)=1>0$
• As $x\to \pm \infty$, $f(x)\to +\infty$

Since $f$ is continuous, $f(0)>0$ and $f(1)<0$, there is one root in $(0,1)$.

For $x>1$: $f(1)<0$ and $f(x)\to +\infty$ as $x\to \infty$, so there is one root in $(1,\infty )$.

For $x<0$: Since $f(0)>0$ and $f(x)\to +\infty$ as $x\to -\infty$, and $f$ is decreasing on $(-\infty ,1)$, there are no roots in $(-\infty ,0)$.

Therefore, there are exactly 2 distinct real roots.