JEE Main 2023 (Online) 29th January Morning ShiftQuestion 1
Let $x=2$ be a root of the equation $x^{2}+px+q=0$ and $f(x)=\begin{cases}\frac{1-\cos\left(x^{2}-4px+q^{2}+8q+16\right)}{(x-2p)^{4}},&x\neq2p\\0,&x=2p\end{cases}$. Then $\lim_{x\to2p^{+}}[f(x)]$ where $[\cdot]$ denotes greatest integer function, is
Limits & Continuity
Solution
Concept. The cosine argument is a perfect square $(x-2p)^2$; then $f$ approaches $\frac12$ from below. Formulas used. (1) $1-\cos\theta<\frac{\theta^2}2$ for $\theta\ne0$, and $\frac{1-\cos\theta}{\theta^2}\to\frac12$Step 1 - Root. $x=2$: $4+2p+q=0\Rightarrow q+4=-2p$. Step 2 - Argument. $q^2+8q+16=(q+4)^2=4p^2$, so $$x^2-4px+4p^2=(x-2p)^2$$ Step 3 - $f$. With $u=x-2p$, $\theta=u^2$: $f=\frac{1-\cos u^2}{u^4}$. Step 4 - Bounds. By (1), $0<f<\frac12$ for $u\ne0$ small, and $f\to\frac12$. Step 5. So $[f]=0$ near $2p^+$, and the limit is $0$. Answer: C