JEE Advanced 2011 Paper 2Question 1
If $$f(x)=\begin{cases}-x-\frac{\pi}{2}, & x\le -\frac{\pi}{2}\\ -\cos x, & -\frac{\pi}{2}<x\le 0\\ x-1, & 0<x\le 1\\ \ln x, & x>1,\end{cases}$$ then
Limits & Continuity
Solution
Concept. At each join compare values (continuity) and slopes (differentiability). Formulas used. (1) $(-\cos x)'=\sin x$, $(\ln x)'=\frac1x$(2) differentiable $\iff$ continuous and left slope $=$ right slopeStep 1 - (A) $x=-\frac\pi2$. Left: $\frac\pi2-\frac\pi2=0$. Right: $-\cos\big(-\frac\pi2\big)=0$. Continuous: true. Step 2 - (B) $x=0$. Values: $-\cos0=-1$ and $0-1=-1$, continuous. Slopes (1): left $\sin0=0$, right $1$. Unequal, not differentiable: true. Step 3 - (C) $x=1$. Values: $1-1=0$, $\ln1=0$. Slopes: $1$ and $\frac11=1$. Equal, so differentiable (2): true.Step 4 - (D) $x=-\frac32$. $-\frac\pi2\approx-1.571<-1.5$, so $-\frac32$ lies inside $\big(-\frac\pi2,0\big)$, where $f=-\cos x$ is smooth. Differentiable: true. Answer: A,B,C,D