JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Application of Derivatives

We have $f^{\prime }(x)=5x^4-5=5(x^4-1)=0$ when $x=\pm 1$. The critical values are $f(-1)=-1+5+a=a+4$ and $f(1)=1-5+a=a-4$. Since $f(x)\to -\infty$ as $x\to -\infty$ and $f(x)\to \infty$ as $x\to \infty$, the function has three real roots when the local maximum $f(-1)=a+4>0$ and the local minimum $f(1)=a-4<0$, i.e., when $-4<a<4$. For $a>4$, both critical values are positive, so $f(x)$ has only one real root. Thus (B) and (D) are correct.