Question

Let $f\left(x\right)=\underset{0}{\overset{x}{\int }}\left\{\right(a-1\left)\right(t^2+t+1{)}^2-(a+1)(t^4+t^2+1)\}dt.$ Then the set of values of ${}^{\prime }{a}^{\prime }$ for which $f^{\prime }\left(x\right)=0$ has two distinct real roots is:

• A. $(-\infty ,0)$
• B. $(-\infty ,-1)\cup (1,\infty )$
• C. $(-\infty ,-2)\cup (2,\infty )$
• D. $(-\infty ,-2)\cup (1,\infty )$

Answer 1

The correct option is C $(-\infty ,-2)\cup (2,\infty )$

Differentitating the given equation using Leibnitz theorem, we get
$f^{\prime }\left(x\right)=(a-1)(x^2+x+1{)}^2-(a+1\left)\right(x^2+x+1\left)\right(x^2-x+1).$
Now, $f^{\prime }\left(x\right)=0\Rightarrow (a-1)(x^2+x+1)-(a+1)(x^2-x+1)=0\Rightarrow x^2-ax+1=0$
For distinct real roots
$D>0\Rightarrow a^2-4>0\Rightarrow a^2>4\Rightarrow a\in (-\infty ,-2)\cup (2,\infty )$