Continuity & DifferentiabilityeasyFree

Continuity & Differentiability: Number Values Non Derivable

JEE Maths question with a full step-by-step solution.

Question
Number of values of x∈[0,π]x \in \left[0,\pi\right] where f(x)=[4sin⁡x−7]f(x) = \left[4\sin x-7\right] is non-derivable is ([ ⋅ ][\,\cdot\,] is G.I.F.)
Solution
Answer: 7
Step 1: 77 is an integer, so
f(x)=[4sin⁡x−7]=[4sin⁡x]−7f(x) = \left[4\sin x-7\right] = \left[4\sin x\right]-7
 A constant shift cannot affect derivability, so it is enough to study [4sin⁡x]\left[4\sin x\right]. Step 2: [u]\left[u\right] is locally constant, hence derivable, except where it jumps, and it jumps exactly where uu passes through an integer at which [u]\left[u\right] changes value. As xx runs from 00 to π\pi, sin⁡x\sin x rises from 00 to 11 and falls back to 00, so for u=4sin⁡xu = 4\sin x,
u: 0↗4↘0u:\ 0 \nearrow 4 \searrow 0
 The integer values strictly inside the range that uu crosses are 1,2,31,2,3, and the value 44 is attained at the single point where sin⁡x=1\sin x = 1. Step 3: sin⁡\sin is strictly increasing on [0,π2]\left[0,\tfrac{\pi}{2}\right] and strictly decreasing on [π2,π]\left[\tfrac{\pi}{2},\pi\right], so for 0<c<10<c<1 the equation sin⁡x=c\sin x = c has exactly two roots in (0,π)(0,\pi), one on each side of π2\tfrac{\pi}{2}:
sin⁡x=14: two values in (0,π);sin⁡x=12: x=π6,5π6−two values;\sin x = \tfrac14:\ \text{two values in }(0,\pi);\qquad \sin x = \tfrac12:\ x = \tfrac{\pi}{6},\tfrac{5\pi}{6} - \text{two values};
sin⁡x=34: two values;sin⁡x=1: x=π2−one value\sin x = \tfrac34:\ \text{two values};\qquad \sin x = 1:\ x = \tfrac{\pi}{2} - \text{one value}
 Step 4: At x=0x = 0 and x=πx = \pi, u=0u = 0 and [u]=0\left[u\right] = 0; just inside the interval uu is slightly positive and [u]\left[u\right] is still 00. So there is no jump at either endpoint. Step 5: At x=π2x = \tfrac{\pi}{2}, u=4u = 4 exactly and [u]=4\left[u\right] = 4, while on both sides u<4u<4 gives [u]=3\left[u\right] = 3. The value differs from the surrounding limit, so ff is discontinuous, hence non-derivable, at π2\tfrac{\pi}{2}. Step 6:
2+2+2+1=72+2+2+1 = 7
 Answer: 77
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