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Domain of 1/sqrt(sin(cos x)) + sin^-1(2x/pi) + 1/{-x} | JEE Advanced

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

Question
If the domain of the function f(x)f(x) which is defined as
f(x)=1sin⁡(cos⁡x)+sin⁡−1(2xπ)+1{−x}+1ln⁡(1−[tan⁡x2]−[−tan⁡x2])f(x) = \frac1{\sqrt{\sin\left(\cos x\right)}}+\sin^{-1}\left(\frac{2x}\pi\right)+\frac1{\left\{-x\right\}} +\frac1{\ln\left(1-\left[\tan\dfrac x2\right]-\left[-\tan\dfrac x2\right]\right)}
 is x∈(a,b)−{c,d,e}x \in \left(a,b\right)-\left\{c,d,e\right\}, then the value of (b+c+d+e)−a\left(b+c+d+e\right)-a is equal to (Where [ . ][\,.\,] represents the greatest integer function and {x}=x−[x]\left\{x\right\} = x-[x])
Solution
Answer: 3.14 (± 0.01)
Step 1: 1sin⁡(cos⁡x)\dfrac1{\sqrt{\sin\left(\cos x\right)}} needs sin⁡(cos⁡x)>0\sin\left(\cos x\right)>0, i.e. cos⁡x∈(0,π)\cos x \in \left(0,\pi\right) modulo 2π2\pi; but cos⁡x∈[−1,1]\cos x \in \left[-1,1\right] and 1<π1<\pi, so this reduces to
cos⁡x>0⟹x∈(−π2,π2)\cos x > 0 \quad\Longrightarrow\quad x \in \left(-\frac\pi2,\frac\pi2\right)
 taking the branch containing the origin, which is what the answer format (a,b)\left(a,b\right) demands. Step 2: sin⁡−1(2xπ)\sin^{-1}\left(\dfrac{2x}\pi\right) needs
∣2xπ∣≤1⟹x∈[−π2,π2]\left|\frac{2x}\pi\right| \le 1 \quad\Longrightarrow\quad x \in \left[-\frac\pi2,\frac\pi2\right]
 which adds nothing beyond Step 1. Next, 1{−x}\dfrac1{\left\{-x\right\}} needs {−x}≠0\left\{-x\right\} \ne 0, and the fractional part {−x}\left\{-x\right\} vanishes exactly when −x-x is an integer, i.e. when x∈Zx \in \mathbb Z. Inside (−π2,π2)≈(−1.5708,1.5708)\left(-\tfrac\pi2,\tfrac\pi2\right) \approx \left(-1.5708,1.5708\right) the integers are
x=−1, 0, 1x = -1,\ 0,\ 1
 all three of which must be removed. Step 3: Put t=tan⁡x2t = \tan\dfrac x2. The standard identity
[t]+[−t]={0,t∈Z−1,t∉Z\left[t\right]+\left[-t\right] = \begin{cases}0, & t \in \mathbb Z\\ -1, & t \notin \mathbb Z\end{cases}
 gives
1−[t]−[−t]={1,t∈Z2,t∉Z.1-\left[t\right]-\left[-t\right] = \begin{cases}1, & t \in \mathbb Z\\ 2, & t \notin \mathbb Z .\end{cases}
 If t∈Zt \in \mathbb Z the logarithm is ln⁡1=0\ln1 = 0 and the term blows up; if t∉Zt \notin \mathbb Z it is ln⁡2≠0\ln2 \ne 0, which is fine. So we need
tan⁡x2∉Z\tan\frac x2 \notin \mathbb Z
 Step 4: For x∈(−π2,π2)x \in \left(-\tfrac\pi2,\tfrac\pi2\right) we have x2∈(−π4,π4)\tfrac x2 \in \left(-\tfrac\pi4,\tfrac\pi4\right) and hence t=tan⁡x2∈(−1,1)t = \tan\tfrac x2 \in \left(-1,1\right). The only integer there is 00, attained at x=0x = 0, already removed in Step 2. So this term excludes nothing new. Step 5:
x∈(−π2, π2)−{−1, 0, 1}x \in \left(-\frac\pi2,\ \frac\pi2\right)-\left\{-1,\ 0,\ 1\right\}
 so a=−π2a = -\dfrac\pi2, b=π2b = \dfrac\pi2, and {c,d,e}={−1,0,1}\left\{c,d,e\right\} = \left\{-1,0,1\right\}. Hence
c+d+e=−1+0+1=0c+d+e = -1+0+1 = 0
(b+c+d+e)−a=π2+0−(−π2)=π≈3.14\left(b+c+d+e\right)-a = \frac\pi2+0-\left(-\frac\pi2\right) = \pi \approx 3.14
 Answer: 3.143.14
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