JEE Main 2023 (Online) 6th April Morning ShiftQuestion 1
Let $a\in\mathbb{Z}$ and $[t]$ be the greatest integer $\le t$. Then the number of points, where the function $f(x)=[a+13\sin x]$, $x\in(0,\pi)$ is not differentiable, is $\_\_\_\_\_\_$ .
Differentiation and its Application
Answer$25$
Solution
Concept. $[a+t]=a+[t]$ for integer $a$. $f$ fails to be differentiable wherever $[13\sin x]$ jumps. Formulas used. (1) $[a+t]=a+[t]$ for $a\in\mathbf Z$(2) A discontinuous function is not differentiableStep 1 - Reduce. By (1), $f=a+[13\sin x]$, and on $(0,\pi)$, $13\sin x\in(0,13]$. Step 2 - Jumps at $k=1,\dots,12$. $13\sin x=k$ has two solutions in $(0,\pi)$, one on each side of $\frac\pi2$. At each, the floor jumps: $12\times2=24$ points. Step 3 - At $x=\frac\pi2$. $13\sin x=13$ only at this point; nearby it is just below $13$, so $f=a+12$ around it but $f\left(\frac\pi2\right)=a+13$. By (2), one more point. Step 4 - Total. $24+1=25$. Answer: 25