JEE Main 2020 (Online) 5th September Morning ShiftQuestion 1
If $\alpha$ is the positive root of the equation, $p(x)=x^2-x-2=0$, then $\lim_{x\to\alpha^+}\frac{\sqrt{1-\cos(p(x))}}{x+\alpha-4}$ is equal to :
Limits & Continuity
Solution
Concept. $1-\cos p=2\sin^2\frac p2$, and $p(x)>0$ just right of $2$, so the root opens without a sign issue. Formulas used. (1) $1-\cos p=2\sin^2\frac p2$(2) $\lim_{y\to0}\frac{\sin y}{y}=1$Step 1 - Root. $x^2-x-2=(x-2)(x+1)$, positive root $\alpha=2$. Step 2 - Simplify (1). For $x\to2^+$, $p=(x-2)(x+1)\to0^+$, so $\sqrt{1-\cos p}=\sqrt2\,\sin\frac p2$ (positive). Step 3 - Denominator. $x+\alpha-4=x-2$. Step 4 - Limit (2). $$\frac{\sqrt2\sin\frac p2}{x-2}=\sqrt2\cdot\frac{\sin\frac p2}{\frac p2}\cdot\frac{x+1}{2}\to\sqrt2\cdot\frac32=\frac{3}{\sqrt2}$$ Answer: C